C H A P T E R 3
Exponential and Logarithmic Functions
Section 3.1 Exponential Functions and Their Graphs . . . . . . . . . 265
Section 3.2 Logarithmic Functions and Their Graphs . . . . . . . . 273
Section 3.3 Properties of Logarithms . . . . . . . . . . . . . . . . . 281
Section 3.4 Exponential and Logarithmic Equations . . . . . . . . . 289
Section 3.5 Exponential and Logarithmic Models . . . . . . . . . . 303
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
265
C H A P T E R 3
Exponential and Logarithmic Functions
Section 3.1 Exponential Functions and Their Graphs
n You should know that a function of the form where is called an exponential
function with base a.
n You should be able to graph exponential functions.
n You should know formulas for compound interest.
(a) For n compoundings per year:
(b) For continuous compoundings: A 5 Pert.
A 5 P11 1r
n2nt
.
a Þ 1,a > 0,f sxd 5 ax,
Vocabulary Check
1. algebraic 2. transcendental 3. natural exponential; natural
4. 5. A 5 PertA 5 P11 1r
n2nt
1. f s5.6d 5 s3.4d5.6 < 946.852 2. f sxd 5 2.3x 5 2.33/2 < 3.488 3. f s2pd 5 52p < 0.006
4. f sxd 5 s23d5x
5 s23d5s0.3d < 0.544 5.
< 1767.767
gsxd 5 5000s2xd 5 5000s221.5d 6.
< 1.274 3 1025
5 200s1.2d12?24
f sxd 5 200s1.2d12x
7.
Increasing
Asymptote:
Intercept:
Matches graph (d).
s0, 1d
y 5 0
f sxd 5 2x 8. rises to the right.
Asymptote:
Intercept:
Matches graph (c).
s0, 2d
y 5 1
f sxd 5 2x 1 1 9.
Decreasing
Asymptote:
Intercept:
Matches graph (a).
s0, 1d
y 5 0
f sxd 5 22x
10. rises to the right.
Asymptote:
Intercept:
Matches graph (b).
s0, 14d
y 5 0
fsxd 5 2x22 11.
Asymptote: y 5 0x
321−1−2−3
5
4
3
2
1
−1
yf sxd 5 s12dx
x 0 1 2
4 2 1 0.250.5f sxd
2122
266 Chapter 3 Exponential and Logarithmic Functions
12.
Asymptote:
x321−1−2−3
5
4
3
2
−1
y
y 5 0
f sxd 5 s12d2x
5 2x
x 0 1 2
1 2 40.50.25f sxd
2122
13.
Asymptote:
x321−1−2−3
5
4
3
1
−1
y
y 5 0
f sxd 5 62x
x 22 21 0 1 2
36 6 1 0.167 0.028f sxd
14.
Asymptote:
x
5
4
3
321−1−2−3
2
1
−1
y
y 5 0
f sxd 5 6x
x 22 21 0 1 2
0.028 0.167 1 6 36f sxd
15.
Asymptote:
x321−1−2−3
5
4
3
2
1
−1
y
y 5 0
f sxd 5 2x21
x 0 1 2
1 20.50.250.125f sxd
2122
16.
Asymptote: y 5 3
x
7
6
5
4
2
1
1−1−2−3 2 3 4 5
yf sxd 5 4x23 1 3
x 0 1 2 3
43.253.0633.0163.004f sxd
21
17.
Because the graph of g can be obtained
by shifting the graph of f four units to the right.
gsxd 5 f sx 2 4d,
gsxd 5 3x24 f sxd 5 3x, 18.
Because the graph of g can be obtained
by shifting the graph of f one unit upward.
gsxd 5 f sxd 1 1,
f sxd 5 4x, gsxd 5 4x 1 1
19.
Because the graph of g can be obtained
by shifting the graph of f five units upward.
gsxd 5 5 1 f sxd,
gsxd 5 5 2 2x f sxd 5 22x, 20.
Because the graph of g can be
obtained by reflecting the graph of f in the y-axis and
shifting f three units to the right. (Note: This is equivalent
to shifting f three units to the left and then reflecting the
graph in the y-axis.)
gsxd 5 f s2x 1 3d,
f sxd 5 10x, gsxd 5 102x13
Section 3.1 Exponential Functions and Their Graphs 267
21.
Because the graph of g can be
obtained by reflecting the graph of f in the x-axis and
y-axis and shifting f six units to the right. (Note: This is
equivalent to shifting f six units to the left and then
reflecting the graph in the x-axis and y-axis.)
gsxd 5 2f s2x 1 6d,
f sxd 5 s72dx
, gsxd 5 2s72d2x16
22.
hence the graph of g can be obtained
by reflecting the graph of f in the x-axis and shifting the
resulting graph five units upward.
gsxd 5 2f sxd 1 5,
gsxd 5 20.3x 1 5 f sxd 5 0.3x,
27. f s34d 5 e23y4 < 0.472 28. f sxd 5 ex 5 e3.2 < 24.533 29. f s10d 5 2e25s10d < 3.857 3 10222
23.
−3
−1
3
3
y 5 22x224.
−3 3
−1
3
y 5 32|x| 25.
−1
0
5
4
fsxd 5 3x22 1 1 26.
−6 3
−3
3
y 5 4x11 2 2
30.
5 1.5e120 < 1.956 3 1052
f sxd 5 1.5es1y2dx 31. f s6d 5 5000e0.06s6d < 7166.647 32.
5 250e0.05s20d < 679.570
f sxd 5 250e0.05x
33.
Asymptote:
x321−1−2−3
5
4
3
2
1
−1
y
y 5 0
f sxd 5 ex
x 0 1 2
1 7.3892.7180.3680.135f sxd
2122
34.
Asymptote:
x
5
4
3
2
1
−1
321−1−2−3
y
y 5 0
f sxd 5 e2x
x 22 21 0 1 2
7.389 2.718 1 0.368 0.135f sxd
35.
Asymptote:
x−1−2−3−4−5−6−7−8 1
8
7
6
5
4
3
2
1
y
y 5 0
f sxd 5 3ex14
x
31.1040.4060.1490.055f sxd
2425262728
36.
Asymptote:
x
6
5
4
3
2
1
−14321−1−2−3
y
y 5 0
f sxd 5 2e20.5x
x 22 21 0 1 2
5.437 3.297 2 1.213 0.736f sxd
268 Chapter 3 Exponential and Logarithmic Functions
37.
Asymptote:
x7654321
9
8
7
6
5
3
2
1
−1−2−3
y
y 5 4
f sxd 5 2ex22 1 4
x 0 1 2
64.7364.2714.1004.037f sxd
2122
38.
Asymptote:
x
8
7
6
5
4
3
1
87654321−1
y
y 5 2
f sxd 5 2 1 ex25
x 0 2 4 5 6
2.007 2.050 2.368 3 4.718f sxd
45.
x 5 2
x 1 1 5 3
3x11 5 33
3x11 5 27 46.
x 5 7
x 2 3 5 4
2x23 5 24
2x23 5 16 47.
x 5 23
x 2 2 5 25
2x22 5 225
2x22 51
32
48.
x 5 24
x 1 1 5 23
s15dx11
5 s15d23
s15dx11
5 53
s15dx11
5 125 49.
x 513
3x 5 1
3x 1 2 5 3
e3x12 5 e3 50.
x 552
2x 5 5
2x 2 1 5 4
e2x21 5 e4
39.
−7
−1
5
7
y 5 1.0825x 40.
−4 8
−2
6
y 5 1.085x 41.
−10
0
23
22
sstd 5 2e0.12t
42.
−16 17
−2
20
sstd 5 3e20.2t 43.
−3
0
3
4
gsxd 5 1 1 e2x 44.
−2 4
0
4
hsxd 5 ex22
51.
x 5 3 or x 5 21
sx 2 3dsx 1 1d 5 0
x2 2 2x 2 3 5 0
x2 2 3 5 2x
ex223 5 e2x 52.
x 5 3 or x 5 2
sx 2 3dsx 2 2d 5 0
x2 2 5x 1 6 5 0
x2 1 6 5 5x
ex216 5 e5x
Section 3.1 Exponential Functions and Their Graphs 269
58. A 5 Pert 5 12,000e0.06t
53.
Compounded times per year:
Compounded continuously: A 5 Pert 5 2500e0.025s10d
A 5 P11 1r
n2nt
5 250011 10.025
n 210n
n
P 5 $2500, r 5 2.5%, t 5 10 years
n 1 2 4 12 365ContinuousCompounding
A $3200.21 $3205.09 $3207.57 $3209.23 $3210.04 $3210.06
54.
Compounded n times per year:
Compounded continuously: A 5 1000e0.04s10d
A 5 100011 10.04
n 210n
P 5 $1000, r 5 4%, t 5 10 years
n 1 2 4 12 365ContinuousCompounding
A $1480.24 $1485.95 $1488.86 $1490.83 $1491.79 $1491.82
55.
Compounded times per year:
Compounded continuously: A 5 Pert 5 2500e0.03s20d
A 5 P11 1r
n2nt
5 250011 10.03
n 220n
n
P 5 $2500, r 5 3%, t 5 20 years
n 1 2 4 12 365ContinuousCompounding
A $4515.28 $4535.05 $4545.11 $4551.89 $4555.18 $4555.30
56.
Compounded n times per year:
Compounded continuously: A 5 1000e0.06s40d
A 5 100011 10.06
n 240n
P 5 $1000, r 5 6%, t 5 40 years
n 1 2 4 12 365ContinuousCompounding
A $10,285.72 $10,640.89 $10,828.46 $10,957.45 $11,021.00 $11,023.18
57. A 5 Pert 5 12,000e0.04t
t 10 20 30 40 50
A $17,901.90 $26,706.49 $39,841.40 $59,436.39 $88,668.67
t 10 20 30 40 50
A $21,865.43 $39,841.40 $72,595.77 $132,278.12 $241,026.44
270 Chapter 3 Exponential and Logarithmic Functions
59. A 5 Pert 5 12,000e0.065t
t 10 20 30 40 50
A $22,986.49 $44,031.56 $84,344.25 $161,564.86 $309,484.08
60. A 5 Pert 5 12,000e0.035t
t 10 20 30 40 50
A $17,028.81 $24,165.03 $34,291.81 $48,662.40 $69,055.23
64.
(a)
(b) When
(c) Since is on the graph in part (a), it
appears that the greatest price that will still yield a
demand of at least 600 units is about $350.
s600, 350.13d
p 5 500011 24
4 1 e20.002s500d2 < $421.12
x 5 500:
0
0
2000
1200
p 5 500011 24
4 1 e20.002x2 65.
(a)
(b)
(c) Vs2d < 1,000,059.63 computers
Vs1.5d < 10,004.472 computers
Vs1d < 10,000.298 computers
Vstd 5 100e4.6052t
61.
< $222,822.57
A 5 25,000es0.0875ds25d 62.
< $212,605.41
A 5 5000es0.075ds50d 63. Cs10d 5 23.95s1.04d10 < $35.45
66. (a)
Since the growth rate is negative,
the population is decreasing.
(b) In 1998, and the population is given by
In 2000, and the population is given by
(c) In 2010, and the population is given by
Ps20d 5 152.26e20.0039s20d 5 140.84 million.
t 5 20
Ps10d 5 152.26e20.0039s10d 5 146.44 million.
t 5 10
Ps8d 5 152.26e20.0039s8d 5 147.58 million.
t 5 8
20.0039 5 20.39%,
P 5 152.26e20.0039t 67.
(a)
(b)
(c)
0 5000
0
30
Qs1000d < 16.21 grams
Qs0d 5 25 grams
Q 5 25s12dty1599
68.
(a)
(b)
< 7.85 grams
When t 5 2000: Q 5 10s12d2000y5715
5 10s1d 5 10 grams
When t 5 0: Q 5 10s12d0y5715
Q 5 10s12d ty5715
(c)
Time (in years)
t80004000
2
4
6
8
10
12
Q
Mas
s of
14C
(in
gra
ms)
Section 3.1 Exponential Functions and Their Graphs 271
69.
(a)
0
0 120
110
y 5100
1 1 7e20.069x
71. True. The line is a horizontal asymptote for the
graph of f sxd 5 10x 2 2.
y 5 22 72. False, e is an irrational number.e Þ271,80199,990 .
73.
Thus, but f sxd 5 hsxd.f sxd Þ gsxd,
5 hsxd
51
9s3xd
5 3x 1 1
322 5 3x322
f sxd 5 3x22 74.
Thus, but gsxd Þ fsxd.gsxd 5 hsxd
5 hsxd
5 64s4xd
5 64s22dx
5 64s22xd
5 22x ? 26
gsxd 5 22x16
75. and
Thus, f sxd 5 gsxd 5 hsxd.
5 gsxd
5 s14dx22
5 hsxd 5 s14d2s22xd
5 16s222xd 5 422x
5 16s22d2x 5 42s42xd
f sxd 5 16s42xd f sxd 5 16s42xd 76.
Thus, none are equal.
hsxd 5 25x23 5 2s5x ? 523d
gsxd 5 532x 5 53 ? 52x
fsxd 5 52x 1 3
(b)x Sample Data Model
0 12 12.5
25 44 44.5
50 81 81.82
75 96 96.19
100 99 99.3
70. (a)
Altitude (in km)
Atm
osp
her
ic p
ress
ure
(in p
asca
ls)
P
h5 10 15 20
40,000
60,000
80,000
25
100,000
120,000
20,000
(b)
5 32,357 pascals
5 107,428e20.150s8d
p 5 107,428e20.150h
(c) When
(d) when
x < 38 masses.
2
3s100d 5
100
1 1 7e20.069x
y 5100
1 1 7e20.069s36d < 63.14%.
x 5 36:
272 Chapter 3 Exponential and Logarithmic Functions
78. (a)
Decreasing:
Increasing:
Relative maximum:
Relative minimum: s0, 0d
s2, 4e22d
s0, 2d
s2`, 0d, s2, `d
−2 7
−1
5
f sxd 5 x2e2x (b)
Decreasing:
Increasing:
Relative maximum: s1.44, 4.25d
s2`, 1.44d
s1.44, `d
−2
−2
10
6
gsxd 5 x232x
81.
y 5 ±!25 2 x2
y2 5 25 2 x2
x2 1 y2 5 25 82.
y 5 x 2 2 and y 5 2sx 2 2d, x ≥ 2
x 2 2 5 |y|x 2 |y| 5 2
79. (Horizontal line)
As
As x → 2`, fsxd → gsxd.
x → `, fsxd → gsxd.
−3 3
0
g
f
4
f sxd 5 11 10.5
x 2x
and gsxd 5 e0.5 80. The functions (c) and (d) are exponential.22x3x
77.
(a)
(b) 4x> 3x when x > 0.
4x< 3x when x < 0.
x21−1−2
−1
1
2
3
y
y = 4x y = 3x
y 5 3x and y 5 4x
x 0 1 2
1 3 9
1 4 1614
1164x
13
193x
2122
83.
Vertical asymptote:
Horizontal asymptote: y 5 0
x 5 29
x3−3−15−18 −6
9
12
6
3
−9
−3
−6
yf sxd 52
9 1 x
x
2 12221f sxd
2728210211
Section 3.2 Logarithmic Functions and Their Graphs 273
Section 3.2 Logarithmic Functions and Their Graphs
n You should know that a function of the form where and is called
a logarithm of x to base a.
n You should be able to convert from logarithmic form to exponential form and vice versa.
n You should know the following properties of logarithms.
(a)
(b)
(c)
(d) Inverse Property
(e) If
n You should know the definition of the natural logarithmic function.
n You should know the properties of the natural logarithmic function.
(a)
(b)
(c)
(d) Inverse Property
(e) If
n You should be able to graph logarithmic functions.
ln x 5 ln y, then x 5 y.
eln x 5 x
ln ex 5 x since ex 5 ex .
ln e 5 1 since e1 5 e.
ln 1 5 0 since e0 5 1.
loge x 5 ln x, x > 0
loga x 5 loga y, then x 5 y.
aloga x 5 x
loga ax 5 x since ax 5 ax .
loga a 5 1 since a1 5 a.
loga 1 5 0 since a0 5 1.
y 5 loga x ⇔ ay 5 x
x > 0,a > 0, a Þ 1,y 5 loga x,
Vocabulary Check
1. logarithmic 2. 10 3. natural;
4. 5. x 5 yaloga x 5 x
e
1. log4 64 5 3 ⇒ 43 5 64 2. log3 81 5 4 ⇒ 34 5 81 3. log7 1
49 5 22 ⇒ 722 51
49
4. log 1
1000 5 23 ⇒ 1023 51
1000 5. log32 4 525 ⇒ 322y5 5 4 6. log16 8 5
34 ⇒ 163y4 5 8
7. log36 6 512 ⇒ 361y2 5 6 8. log8 4 5
23 ⇒ 82y3 5 4 9. 53 5 125 ⇒ log5 125 5 3
10. 82 5 64 ⇒ log8 64 5 2 11. 811y4 5 3 ⇒ log81 3 514 12. 93y2 5 27 ⇒ log9 27 5
32
85. Answers will vary.84.
Domain: s2`, 7g
y
x64 8
−6
−4
−2
2
4
6
2−2−4
f sxd 5 !7 2 x
3 6 7
4 3 2 1 0y
2229x
274 Chapter 3 Exponential and Logarithmic Functions
22.
since
b23 5 b23
gsb23d 5 logb b23 5 23
gsxd 5 logb x 23.
f s45d 5 logs4
5d < 20.097
f sxd 5 log x 24.
f s 1500d 5 log
1500 < 22.699
f sxd 5 log x
25.
f s12.5d < 1.097
f sxd 5 log x 26.
f s75.25d < 1.877
f sxd 5 log x 27. since 34 5 34log3 34 5 4
28.
Since log1.5 1 5 0.1.50 5 1,
log1.5 1 29. since p1 5 p.logp p 5 1 30.
Since 9log9 15 5 15.aloga x 5 x,
9log9 15
31.
Domain: The domain is
x-intercept:
Vertical asymptote:
y 5 log4 x ⇒ 4 y 5 x
x 5 0
s1, 0d
s0, `d.x > 0 ⇒
f sxd 5 log4 x
x 1 4 2
0 11221f sxd
14
−1 1 2 3
−2
−1
1
2
x
y
32.
Domain:
x-intercept:
Vertical asymptote:
y 5 log6 x ⇒ 6y 5 x
x 5 0
s1, 0d
s0, `d
1−1 2 3
−2
−1
1
2
x
ygsxd 5 log6 x
x 1 6
y 0 11221
!616
33.
Domain:
x-intercept:
The x-intercept is
Vertical asymptote:
log3 x 5 2 2 y ⇒ 322y 5 x
y 5 2log3 x 1 2
x 5 0
s9, 0d.
9 5 x
32 5 x
2 5 log3 x
2log3 x 1 2 5 0
s0, `d
y 5 2log3 x 1 2
x 27 9 3 1
y 0 1 2 321
13
2 4 6 8 10 12
−6
−4
−2
2
4
6
x
y 34.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
y 5 log4sx 2 3d ⇒ 4y 1 3 5 x
x 2 3 5 0 ⇒ x 5 3
s4, 0d.
4 5 x
1 5 x 2 3
40 5 x 2 3
log4sx 2 3d 5 0
2 4 6 8 10
−4
−2
2
4
6
x
ys3, `d.
x 2 3 > 0 ⇒ x > 3
hsxd 5 log4sx 2 3d
x 3 4 7 19
y 0 1 221
14
13. 622 5136 ⇒ log6
136 5 22 14. 423 5
164 ⇒ log4
164 5 23 15. 70 5 1 ⇒ log7 1 5 0
16. 1023 5 0.001 ⇒ log10 0.001 5 23 17.
f s16d 5 log2 16 5 4 since 24 5 16
f sxd 5 log2 x 18.
since 161y2 5 4fs4d 5 log16 4 512
fsxd 5 log16 x
19.
f s1d 5 log7 1 5 0 since 70 5 1
f sxd 5 log7 x 20.
since 101 5 10f s10d 5 log 10 5 1
fsxd 5 log x 21.
5 2 by the Inverse Property
gsa2d 5 loga a2
gsxd 5 loga x
Section 3.2 Logarithmic Functions and Their Graphs 275
35.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
62y 2 2 5 x
2y 5 log6sx 1 2d
2 y 5 2log6sx 1 2d
x 1 2 5 0 ⇒ x 5 22
s21, 0d.
21 5 x
1 5 x 1 2
60 5 x 1 2
0 5 log6sx 1 2d
0 5 2log6sx 1 2d
6
−4
−2
2
4
x
ys22, `d.
x 1 2 > 0 ⇒ x > 22
f sxd 5 2log6sx 1 2d
x 4
0 1 221f sxd
21353621
5621
36.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
y 5 log5sx 2 1d 1 4 ⇒ 5y24 1 1 5 x
x 2 1 5 0 ⇒ x 5 1
s626625, 0d.
626625 5 x
1
625 5 x 2 1
524 5 x 2 1
log5sx 2 1d 5 24
log5sx 2 1d 1 4 5 0
2 3 4 5 6
1
2
3
4
5
6
x
ys1, `d.
x 2 1 > 0 ⇒ x > 1
y 5 log5sx 2 1d 1 4
x 1.00032 1.0016 1.008 1.04 1.2
y 21 0 1 2 3
37.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
The vertical asymptote is the y-axis.
x
55 0 ⇒ x 5 0
s5, 0d.
x
55 1 ⇒ x 5 5
x
55 100
log1x
52 5 0
4 6 8
−4
−2
2
4
x
ys0, `d.
x
5> 0 ⇒ x > 0
y 5 log1x
52 x 1 2 3 4 5 6 7
y 0 0.150.0820.1020.2220.4020.70
38.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
y 5 logs2xd ⇒ 210y 5 x
x 5 0
s21, 0d.
21 5 x
100 5 2x
−3 −2 −1 1
−2
−1
1
2
x
ylogs2xd 5 0
s2`, 0d.
2x > 0 ⇒ x < 0
y 5 logs2xdx
y 0 12122
2102121
1021
100
276 Chapter 3 Exponential and Logarithmic Functions
39.
Asymptote:
Point on graph:
Matches graph (c).
The graph of is obtained by shifting the graph of
upward two units.
gsxdf sxd
s1, 2d
x 5 0
f sxd 5 log3 x 1 2 40.
Asymptote:
Point on graph:
Matches graph (f).
reflects in the axis.x-gsxdf sxd
s1, 0d
x 5 0
fsxd 5 2log3 x
41.
Asymptote:
Point on graph:
Matches graph (d).
The graph of is obtained by reflecting the graph of
about the axis and shifting the graph two units to
the left.
x-gsxdf sxd
s21, 0d
x 5 22
f sxd 5 2log3sx 1 2d 42.
Asymptote:
Point on graph:
Matches graph (e).
shifts one unit to the right.gsxdf sxd
s2, 0d
x 5 1
fsxd 5 log3sx 2 1d
43.
Asymptote:
Point on graph:
Matches graph (b).
The graph of is obtained by reflecting the graph of
about the axis and shifting the graph one unit to
the right.
y-gsxdf sxd
s0, 0d
x 5 1
f sxd 5 log3s1 2 xd 5 log3f2sx 2 1dg 44.
Asymptote:
Point on graph:
Matches graph (a).
reflects in the axis then reflects that graph in
the axis.y-
x-gsxdf sxd
s21, 0d
x 5 0
fsxd 5 2log3s2xd
45. ln 12 5 20.693 . . . ⇒ e20.693 . . . 5
12 46. ln
25 5 20.916 . . . ⇒ e20.916 . . . 5
25
47. ln 4 5 1.386 . . . ⇒ e1.386 . . . 5 4 48. ln 10 5 2.302 . . . ⇒ e2.302 . . . 5 10
49. ln 250 5 5.521 . . . ⇒ e5.521 . . . 5 250 50. ln 679 5 6.520 . . . ⇒ e6.520 . . . 5 679
51. ln 1 5 0 ⇒ e0 5 1 52. ln e 5 1 ⇒ e1 5 e
53. e3 5 20.0855 . . . ⇒ ln 20.0855 . . . 5 3 54. e2 5 7.3890 . . . ⇒ ln 7.3890 . . . 5 2
55. e1/251.6487 . . . ⇒ ln 1.6487 . . . 512 56. e1y3 5 1.3956 . . . ⇒ ln 1.3956 . . . 5
13
57. e20.5 5 0.6065 . . . ⇒ ln 0.6065 . . . 5 20.5 58. e24.1 5 0.0165 . . . ⇒ ln 0.0165 . . . 5 24.1
59. ex 5 4 ⇒ ln 4 5 x 60. e2x 5 3 ⇒ ln 3 5 2x
61.
f s18.42d 5 ln 18.42 < 2.913
f sxd 5 ln x 62.
fs0.32d 5 3 ln 0.32 < 23.418
fsxd 5 3 ln x
63.
gs0.75d 5 2 ln 0.75 < 20.575
gsxd 5 2 ln x 64.
gs12d 5 2ln
12 < 0.693
gsxd 5 2ln x
Section 3.2 Logarithmic Functions and Their Graphs 277
65.
by the Inverse Propertygse3d 5 ln e3 5 3
gsxd 5 ln x 66.
gse22d 5 ln e22 5 22
gsxd 5 ln x
67.
by the Inverse Propertygse22y3d 5 ln e22y3 5 223
gsxd 5 ln x 68.
gse25y2d 5 ln e25y2 5 252
gsxd 5 ln x
69.
Domain:
The domain is
intercept:
The intercept is
Vertical asymptote: x 2 1 5 0 ⇒ x 5 1
s2, 0d.x-
2 5 x
e0 5 x 2 1
0 5 lnsx 2 1d
x-
y
x32 4 5
−3
−2
−1
1
2
3
1−1
s1, `d.
x 2 1 > 0 ⇒ x > 1
f sxd 5 lnsx 2 1d
1.5 2 3 4
0 0.69 1.1020.69f sxd
x
70.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
y 5 lnsx 1 1d ⇒ ey 2 1 5 x
x 1 1 5 0 ⇒ x 5 21
s0, 0d.
0 5 x
1 5 x 1 1
e0 5 x 1 1
lnsx 1 1d 5 0
2−2 4 6 8x
6
4
2
ys21, `d.
x 1 1 > 0 ⇒ x > 21
hsxd 5 lnsx 1 1d
x 0 1.72 6.39 19.09
y 0 1 2 3212
20.39
71.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote: 2x 5 0 ⇒ x 5 0
s21, 0d.
21 5 x
e0 5 2x
0 5 lns2xd
s2`, 0d.
−3 −2 −1 1
−2
1
2
x
y2x > 0 ⇒ x < 0
gsxd 5 lns2xd
x
0 1.100.6920.69gsxd
23222120.5
72.
Domain:
The domain is
x-intercept:
The x-intercept is
Vertical asymptote:
y 5 lns3 2 xd ⇒ 3 2 ey 5 x
3 2 x 5 0 ⇒ x 5 3
s2, 0d.
2 5 x
1 5 3 2 x
e0 5 3 2 x
lns3 2 xd 5 0
−2 −1 1 2 4
−3
−2
−1
2
3
x
ys2`, 3d.
3 2 x > 0 ⇒ x < 3
fsxd 5 lns3 2 xd
x 2.95 2.86 2.63 2 0.28
y 23 22 21 0 1
73.
−1
−2
5
2
y1 5 logsx 1 1d 74.
−1 5
−2
2
f sxd 5 logsx 2 1d 75.
0
−3
9
3
y1 5 lnsx 2 1d
278 Chapter 3 Exponential and Logarithmic Functions
76.
−4 5
−3
3
f sxd 5 lnsx 1 2d 77.
0
−1
9
5
y 5 ln x 1 2 78.
−5 10
−6
4
f sxd 5 3 ln x 2 1
79.
x 5 3
x 1 1 5 4
log2sx 1 1d 5 log2 4 80.
x 5 12
x 2 3 5 9
log2sx 2 3d 5 log2 9
81.
x 5 7
2x 1 1 5 15
logs2x 1 1d 5 log 15 82.
x 595
5x 5 9
5x 1 3 5 12
logs5x 1 3d 5 log 12
83.
x 5 4
x 1 2 5 6
lnsx 1 2d 5 ln 6 84.
x 5 6
x 2 4 5 2
lnsx 2 4d 5 ln 2
85.
x 5 ±5
x2 5 25
x2 2 2 5 23
lnsx2 2 2d 5 ln 23 86.
x 5 22 or x 5 3
sx 2 3dsx 1 2d 5 0
x2 2 x 2 6 5 0
x2 2 x 5 6
lnsx2 2 xd 5 ln 6
87.
(a) When
When
(b) Total amounts:
(c) Interest charges:
(d) The vertical asymptote is The closer the pay-
ment is to $1000 per month, the longer the length of the
mortgage will be. Also, the monthly payment must be
greater than $1000.
x 5 1000.
301,123.20 2 150,000 5 $151,123.20
396,234 2 150,000 5 $246,234
s1254.68ds12ds20d 5 $301,123.20
s1100.65ds12ds30d 5 $396,234.00
t 5 12.542 ln1 1254.68
1254.68 2 10002 < 20 years
x 5 $1254.68:
t 5 12.542 ln1 1100.65
1100.65 2 10002 < 30 years
x 5 $1100.65:
t 5 12.542 ln1 x
x 2 10002, x > 1000 88.
(a)
The number of years required to multiply the original
investment by increases with However, the larger
the value of the fewer the years required to increase
the value of the investment by an additional multiple
of the original investment.
(b)
2 4 6 8 10 12
5
10
15
20
25
K
t
K,
K.K
t 5ln K
0.095
K 1 2 4 6 8 10 12
t 0 7.3 14.6 18.9 21.9 24.2 26.2
Section 3.2 Logarithmic Functions and Their Graphs 279
98.
(a)
fsxd 5ln x
x
x 1 5 10
0 0.00001380.000920.0460.2300.322f sxd
106104102
90.
(a)
(b)
(c) No, the difference is due to the logarithmic
relationship between intensity and number of
decibels.
b 5 10 log1 1022
102122 5 10 logs1010d 5 100 decibels
b 5 10 log1 1
102122 5 10 logs1012d 5 120 decibels
b 5 10 log1 I
10212289.
(a)
(b)
(c)
(d) f s10d 5 80 2 17 log 11 < 62.3
f s4d 5 80 2 17 log 5 < 68.1
f s0d 5 80 2 17 log 1 5 80.0
0
0 12
100
f std 5 80 2 17 logst 1 1d, 0 ≤ t ≤ 12
91. False. Reflecting about the line will determine
the graph of f sxd.y 5 xgsxd 92. True, log3 27 5 3 ⇒ 33 5 27.
93.
f and g are inverses. Their graphs
are reflected about the line y 5 x.
−2 −1 1 2
−2
−1
1
2
x
g
f
y
f sxd 5 3x, gsxd 5 log3 x 94.
and are inverses. Their graphs
are reflected about the line y 5 x.
gf
−2 −1 1 2
−2
−1
1
2
x
g
f
y
f sxd 5 5x, gsxd 5 log5 x 95 .
f and g are inverses. Their graphs
are reflected about the line y 5 x.
−2 −1 1 2
−2
−1
1
2
x
g
f
y
fsxd 5 ex, gsxd 5 ln x
96.
and are inverses. Their graphs are reflected
about the line y 5 x.
gf
−2 −1 1 2
−2
−1
1
2
x
g
f
y
f sxd 5 10x, gsxd 5 log10 x 97. (a)
The natural log function
grows at a slower rate
than the square root
function.
(b)
The natural log function
grows at a slower rate than
the fourth root function.0
0
20,000
g
f
15gsxd 5 4!xfsxd 5 ln x,
0
0
1000
f
g
40gsxd 5 !xfsxd 5 ln x,
(b) As
(c)
0
0
100
0.5
x → `, fsxd → 0 .
280 Chapter 3 Exponential and Logarithmic Functions
99. (a) False. If were an exponential function of then
but not 0. Because one point is
is not an exponential function of
(c) True.
For
y 5 3, 23 5 8
y 5 1, 21 5 2
y 5 0, 20 5 1
a 5 2, x 5 2y.
x 5 ay
x.ys1, 0d,a1 5 a,y 5 ax,
x,y (b) True.
For
(d) False. If were a linear function of the slope
between and and the slope between
and would be the same. However,
and
Therefore, is not a linear function of x.y
m2 53 2 1
8 2 25
2
65
1
3.m1 5
1 2 0
2 2 15 1
s8, 3ds2, 1ds2, 1ds1, 0d
x,y
x 5 8, log2 8 5 3
x 5 2, log2 2 5 1
x 5 1, log2 1 5 0
a 5 2, y 5 log2 x.
y 5 loga x
100. so, for example, if there is no value of y for which If then every power
of a is equal to 1, so x could only be 1. So, is defined only for and a > 1.0 < a < 1loga x
a 5 1,s22dy 5 24.a 5 22,y 5 loga x ⇒ ay 5 x,
101.
(a) (b) Increasing on
Decreasing on
(c) Relative minimum:
s1, 0d
s0, 1ds1, `d
−1
−2
8
4
fsxd 5 |ln x| 102. (a)
−9
−4
9
8
hsxd 5 lnsx2 1 1d
For Exercises 103–108, use and gxxc 5 x32 1.f xxc 5 3x 1 2
103.
5 15
5 8 1 7
5 f3s2d 1 2g 1 fs2d3 2 1g
s f 1 gds2d 5 f s2d 1 gs2d 104.
Therefore,
5 1.
5 23 1 1 1 3
s f 2 gds21d 5 3s21d 2 s21d3 1 3
5 3x 2 x3 1 3
5 3x 1 2 2 x3 1 1
f sxd 2 gsxd 5 3x 1 2 2 sx3 2 1d
105.
5 4300
5 s20ds215d
5 f3s6d 1 2gfs6d3 2 1g
s fgds6d 5 f s6dgs6d 106.
Therefore, 1 f
g 2s0d 53 ? 0 1 2
03 2 15 22.
f sxdgsxd 5
3x 1 2
x3 2 1
107.
5 1028
5 3s342d 1 2
5 f s342d
5 f ss7d3 2 1d
s f 8 gds7d 5 f sgs7dd 108.
Therefore,
5 273 2 1 5 2344.
sg 8 f ds23d 5 s3 ? s23d 1 2d3 2 1
sg 8 f d(xd 5 gs f sxdd 5 gs3x 1 2d 5 s3x 1 2d3 2 1
(b) Increasing on
Decreasing on
(c) Relative minimum:
s0, 0d
s2`, 0ds0, `d
Section 3.3 Properties of Logarithms 281
12. log1y4 5 5log 5
logs1y4d 5ln 5
lns1y4d< 21.161 13. log9s0.4d 5
log 0.4
log 95
ln 0.4
ln 9< 20.417
14. log20 0.125 5log 0.125
log 205
ln 0.125
ln 20< 20.694 15. log15 1250 5
log 1250
log 155
ln 1250
ln 15< 2.633
16. log3 0.015 5log 0.015
log 35
ln 0.015
ln 3< 23.823 17. log4 8 5
log2 8
log2 45
log2 23
log2 22
53
2
Section 3.3 Properties of Logarithms
n You should know the following properties of logarithms.
(a)
(b)
(c)
(d)
n You should be able to rewrite logarithmic expressions using these properties.
ln un 5 n ln uloga un 5 n loga u
ln1u
v2 5 ln u 2 ln vloga1u
v2 5 loga u 2 loga v
lnsuvd 5 ln u 1 ln vlogasuvd 5 loga u 1 loga v
loga x 5ln x
ln aloga x 5
log10 x
log10 aloga x 5
logb x
logb a
Vocabulary Check
1. change-of-base 2.
3. 4.
This is the Product Property. Matches (c). This is the Power Property. Matches (a).
5.
This is the Quotient Property. Matches (b).
loga u
v5 loga u 2 loga v
ln un 5 n ln ulogasuvd 5 loga u 1 loga v
log x
log a5
ln x
ln a
1. (a)
(b) log5 x 5ln x
ln 5
log5 x 5log x
log 52. (a)
(b) log3 x 5ln x
ln 3
log3 x 5log x
log 33. (a)
(b) log1y5 x 5ln x
lns1y5d
log1y5 x 5log x
logs1y5d
4. (a)
(b) log1y3 x 5ln x
lns1y3d
log1y3 x 5log x
logs1y3d5. (a)
(b) logx 3
105
lns3y10dln x
logx 3
105
logs3y10dlog x
6. (a)
(b) logx 3
45
lns3y4dln x
logx 3
45
logs3y4dlog x
7. (a)
(b) log2.6 x 5ln x
ln 2.6
log2.6 x 5log x
log 2.68. (a)
(b) log7.1 x 5ln x
ln 7.1
log7.1 x 5log x
log 7.19. log3 7 5
log 7
log 35
ln 7
ln 3< 1.771
10. log7 4 5log 4
log 75
ln 4
ln 7< 0.712 11. log1y2 4 5
log 4
logs1y2d5
ln 4
lns1y2d5 22.000
282 Chapter 3 Exponential and Logarithmic Functions
18.
5 4 1 4 log2 3
5 4 log2 2 1 4 log2 3
5 2 log2 22 1 4 log2 3
5 2 log2 4 1 4 log2 3
log2s42 ? 34d 5 log2 42 1 log2 3
4 19.
5 23 2 log5 2
5 log5 523 1 log5 2
21
5 log5 1
125 1 log5 12
log5 1
250 5 log5s 1125 ? 1
2d 20.
5 log 3 2 2
5 log 3 2 2 log 10
5 log 3 2 log 102
5 log 3 2 log 100
log 9
300 5 log 3
100
21.
5 6 1 ln 5
5 ln 5 1 6
lns5e6d 5 ln 5 1 ln e6 22.
5 ln 6 2 2
5 ln 6 2 2 ln e
ln 6
e25 ln 6 2 ln e2 23. log3 9 5 2 log3 3 5 2
24. log5 1
125 5 log5 523 5 23 log5 5 5 23s1d 5 23 25. log2
4!8 514 log2 2
3 534 log2 2 5
34s1d 5
34
26. log6 3!6 5 log6 61y3 5
13 log6 6 5
13s1d 5
13 27. log4 161.2 5 1.2slog4 16d 5 1.2 log4 4
2 5 1.2s2d 5 2.4
28.
5 20.2s4d 5 20.8
5 20.2 log3 34
log3 8120.2 5 20.2 log3 81 29. is undefined. is not in the domain of log3 x.29log3s29d
30. is undefined because
is not in the domain of
log2 x.
216
log2s216d 31. ln e4.5 5 4.5 32.
5 12
5 12s1d
3 ln e4 5 s3ds4d ln e
33.
5 21
2
5 0 21
2s1d
5 0 21
2 ln e
ln 1
!e5 ln 1 2 ln!e 34.
53
4
53
4s1d
53
4 ln e
ln 4!e3 5 ln e3y4 35. ln e2 1 ln e5 5 2 1 5 5 7
36.
5 7
5 ln e7
5 ln e12
e5
2 ln e6 2 ln e5 5 ln e12 2 ln e5
37.
5 2
5 2 log5 5
5 log5 52
5 log5 25
log5 75 2 log5 3 5 log5 75
3
38.
5 3
512s1d 1
52s1d
512 log4 4 1
52 log4 4
log4 2 1 log4 32 5 log4 41y2 1 log4 4
5y2 39. log4 5x 5 log4 5 1 log4 x
Section 3.3 Properties of Logarithms 283
55.
5 4 ln x 11
2 ln y 2 5 ln z
5 ln x4 1 ln !y 2 ln z5
ln1x4!y
z5 2 5 ln x4!y 2 ln z5 56.
51
2 log2 x 1 4 log2 y 2 4 log2 z
5 log2 !x 1 log2 y4 2 log2 z
4
log2 !x y4
z45 log2 !x y4 2 log2 z
4
53.
51
3 ln x 2
1
3 ln y
51
3fln x 2 ln yg
ln 3!x
y5
1
3 ln
x
y54.
5 ln x 23
2 ln y
51
2s2 ln x 2 3 ln yd
51
2sln x2 2 ln y3d
ln!x2
y35 ln1x2
y321y2
51
2 ln1x2
y32
57.
5 2 log5 x 2 2 log5 y 2 3 log5 z
5 log5 x2 2 slog5 y
2 1 log5 z3d
log5 1 x2
y2z32 5 log5 x2 2 log5 y
2z3 58.
5 log x 1 4 log y 2 5 log z
5 log x 1 log y4 2 log z5
log xy4
z55 log xy4 2 log z5
59.
534 ln x 1
14
lnsx2 1 3d
514f3 ln x 1 lnsx2 1 3dg
514fln x3 1 lnsx2 1 3dg
ln 4!x3sx2 1 3d 514 ln x3sx2 1 3d 60.
5 ln x 112 lnsx 1 2d
5 ln x 1 lnsx 1 2d1y2
5 lnfxsx 1 2d1y2g
ln!x2sx 1 2d 5 lnfx2sx 1 2dg1y2
40. log3 10z 5 log3 10 1 log3 z 41. log8 x4 5 4 log8 x 42. log
y
25 log y 2 log 2
43.
5 1 2 log5 x
log5 5
x5 log5 5 2 log5 x 44. log6 z
23 5 23 log6 z 45. ln!z 5 ln z1y2 51
2 ln z
49.
5 ln z 1 2 lnsz 2 1d, z > 1
ln zsz 2 1d2 5 ln z 1 lnsz 2 1d2 50.
5 lnsx 1 1d 1 lnsx 2 1d 2 3 ln x
5 lnfsx 1 1dsx 2 1dg 2 ln x3
ln1x2 2 1
x3 2 5 lnsx2 2 1d 2 ln x3
46. ln 3!t 5 ln t1y3 513 ln t 47.
5 ln x 1 ln y 1 2 ln z
ln xyz2 5 ln x 1 ln y 1 ln z2 48.
5 log 4 1 2 log x 1 log y
log 4x2y 5 log 4 1 log x2 1 log y
51.
51
2 log2sa 2 1d 2 2 log2 3, a > 1
51
2 log2sa 2 1d 2 log2 3
2
log2 !a 2 1
95 log2
!a 2 1 2 log2 9 52.
5 ln 6 21
2 lnsx2 1 1d
5 ln 6 2 lnsx2 1 1d1y2
ln 6
!x2 1 15 ln 6 2 ln!x2 1 1
284 Chapter 3 Exponential and Logarithmic Functions
61. ln x 1 ln 3 5 ln 3x 62. ln y 1 ln t 5 ln yt 5 ln ty 63. log4 z 2 log4 y 5 log4 z
y
64. log5 8 2 log5 t 5 log5 8
t65. 2 log2sx 1 4d 5 log2sx 1 4d2 66.
2
3 log7sz 2 2d 5 log7sz 2 2d2y3
67. 1
4 log3 5x 5 log3s5xd1y4 5 log3
4!5x 68. 24 log6 2x 5 log6s2xd24 5 log6 1
16x4
69.
5 ln x
sx 1 1d3
ln x 2 3 lnsx 1 1d 5 ln x 2 lnsx 1 1d3 70.
5 ln 64sz 2 4d5
5 ln 64 1 lnsz 2 4d5
2 ln 8 1 5 lnsz 2 4d 5 ln 82 1 lnsz 2 4d5
71.
5 log x
y21 log z
3 5 log xz3
y2
log x 2 2 log y 1 3 log z 5 log x 2 log y2 1 log z3 72.
5 log3 x3y4
z4
5 log3 x3y4 2 log3 z
4
3 log3 x 1 4 log3 y 2 4 log3 z 5 log3 x3 1 log3 y
4 2 log3 z4
73.
5 ln x
sx2 2 4d4
5 ln x 2 lnsx2 2 4d4
5 ln x 2 4 lnsx2 2 4d
ln x 2 4flnsx 1 2d 1 lnsx 2 2dg 5 ln x 2 4 lnsx 1 2dsx 2 2d
74.
5 ln z4sz 1 5d4
sz 2 5d2
5 lnfzsz 1 5dg4 2 lnsz 2 5d2
4fln z 1 lnsz 1 5dg 2 2 lnsz 2 5d 5 4fln zsz 1 5dg 2 lnsz 2 5d2
75.
5 ln 3!xsx 1 3d2
x2 2 1
51
3 ln
xsx 1 3d2
x2 2 1
51
3fln xsx 1 3d2 2 lnsx2 2 1dg
1
3 f2 lnsx 1 3d 1 ln x 2 lnsx2 2 1dg 5
1
3 flnsx 1 3d2 1 ln x 2 lnsx2 2 1dg
76.
5 ln1 x3
x2 2 122
5 2 ln x3
x2 2 1
5 2fln x3 2 flnsx 1 1dsx 2 1dgg
5 2fln x3 2 flnsx 1 1d 1 lnsx 2 1dgg
2f3 ln x 2 lnsx 1 1d 2 lnsx 2 1dg 5 2fln x3 2 lnsx 1 1d 2 lnsx 2 1dg
Section 3.3 Properties of Logarithms 285
80.
51
21 log7
!10 by Property 1 and Property 3
51
21
1
2 log7 10
51
2f1 1 log7 10g
log7!70 5
1
2 log7 70 5
1
2flog7 7 1 log7 10g 81.
When
5 60 decibels
5 120 1 10s26d
b 5 120 1 10 log 1026
I 5 1026:
5 120 1 10 log I
5 10flog I 1 12g
5 10flog I 2 log 10212g
b 5 10 log1 I
102122
82.
5 24 dB
5 10slogs250.79dd
5 10slogs2.5079 3 102dd
5 101log13.16 3 107
1.26 3 10522 5 10slogs3.16 3 107d 2 logs1.26 3 105dd
Difference 5 10 log13.16 3 1025
10212 2 2 10 log11.26 3 1027
10212 2
b 5 10 log1 I
102122 83.
With both stereos playing, the music is
decibels louder.
10 log 2 < 3
5 s120 1 10 log I d 1 10 log 2
5 120 1 10slog 2 1 log I d
b 5 120 1 10 logs2I d
77.
5 log 81 3!y s y 1 4d2
y 2 1 2 5 log8
3!y s y 1 4d2 2 log8s y 2 1d
51
3 log8 ys y 1 4d2 2 log8s y 2 1d
1
3 flog8 y 1 2 log8s y 1 4dg 2 log8s y 2 1d 5
1
3 flog8 y 1 log8s y 1 4d2g 2 log8s y 2 1d
78.
5 log4fx6sx 2 1d!x 1 1g 5 log4f!x 1 1sx 2 1dg 1 log4 x
6
512flog4sx 1 1dsx 2 1d2g 1 log4 x
6
12flog4sx 1 1d 1 2 log4sx 2 1dg 1 6 log4 x 5
12flog4sx 1 1d 1 log4sx 2 1d2g 1 log4 x
6
79.
The second and third expressions are equal by Property 2.
log2 32
45 log2 32 2 log2 4 Þ
log2 32
log2 4
286 Chapter 3 Exponential and Logarithmic Functions
84.
(a)
(b)
(c)
(d)
(e)
120
95
70
f s12d 5 90 2 15 ? logs12 1 1d 5 73.3
f s4d 5 90 2 15 ? logs4 1 1d 5 79.5
f s0d 5 90
f std 5 90 2 logst 1 1d15
f std 5 90 2 15 logst 1 1d, 0 ≤ t ≤ 12
85. By using the regression feature on a graphing calculator we obtain y < 256.24 2 20.8 ln x.
(f ) The average score will be 75 when months. See
graph in (e).
(g)
t 5 9 months
101 5 t 1 1
1 5 logst 1 1d
215 5 215 logst 1 1d
75 5 90 2 15 logst 1 1d
t 5 9
(c)
This graph is identical to in (b).T
T 5 e20.037t14 1 21
lnsT 2 21d 5 20.037t 1 4
0
0
30
5
86. (a)
(b)
See graph in (a).
(d)
(e) Since the scatter plot of the original data is so
nicely exponential, there is no need to do the
transformations unless one desires to deal with
smaller numbers. The transformations did not
make the problem simpler.
Taking logs of temperatures led to a linear scatter
plot because the log function increases very slowly
as the values increase. Taking the reciprocals of
the temperatures led to a linear scatter plot because
of the asymptotic nature of the reciprocal function.
x-
0
0
30
80
0
0
30
0.07
T 51
0.0012t 1 0.0161 21
1
T 2 215 0.0012t 1 0.016
T 5 54.4s0.964d t 1 21
T 2 21 5 54.4s0.964dt
0
0
30
80
(in minutes)
0 78 57 4.043 0.0175
5 66 45 3.807 0.0222
10 57.5 36.5 3.597 0.0274
15 51.2 30.2 3.408 0.0331
20 46.3 25.3 3.231 0.0395
25 42.5 21.5 3.068 0.0465
30 39.6 18.6 2.923 0.0538
1ysT 2 21dlnsT 2 21dT 2 21 s8CdT s8Cdt
87.
False, since 0 is not in the domain of
f s1d 5 ln 1 5 0
f sxd.f s0d Þ 0
f sxd 5 ln x 88.
True, because fsaxd 5 ln ax 5 ln a 1 ln x 5 f sad 1 f sxd.
x > 0a > 0,fsaxd 5 fsad 1 fsxd,
Section 3.3 Properties of Logarithms 287
94. Let then and
logb un 5 logb b
nx 5 nx 5 n logb u
un 5 bnx.u 5 bxx 5 logb u,
95.
−3
−3
6
3
f sxd 5 log2 x 5log x
log 25
ln x
ln 296.
−1 5
−2
2
f sxd 5 log4 x 5log x
log 45
ln x
ln 497.
−3
−3
6
3
5log x
logs1y2d 5ln x
lns1y2d
f sxd 5 log1y2 x
98.
−1 5
−2
2
5log x
logs1y4d 5ln x
lns1y4d
f sxd 5 log1y4 x 99.
−1
−2
5
2
5log x
log 11.85
ln x
ln 11.8
f sxd 5 log11.8 x 100.
−1 5
−2
2
5log x
log 12.45
ln x
ln 12.4
f sxd 5 log12.4 x
101.
by Property 2f sxd 5 hsxd
1 2 3 4
−2
−1
1
2
x
f = hg
y
f sxd 5 ln x
2, gsxd 5
ln x
ln 2, hsxd 5 ln x 2 ln 2
93. Let and then
Then logbsuyvd 5 logbsbx2yd 5 x 2 y 5 logb u 2 logb v.
u
v5
bx
by5 bx2y
bx 5 u and by 5 v.y 5 logb v,x 5 logb u
89. False. fsxd 2 fs2d 5 ln x 2 ln 2 5 ln x
2Þ lnsx 2 2d 90. false
can’t be simplified further.
f s!xd 5 ln!x 5 ln x1y2 51
2 ln x 5
1
2 fsxd
!f sxd 5 !ln x
!f sxd 51
2 fsxd;
91. False.
fsud 5 2fsvd ⇒ ln u 5 2 ln v ⇒ ln u 5 ln v2 ⇒ u 5 v2
92. If then
True
0 < x < 1.f sxd < 0,
288 Chapter 3 Exponential and Logarithmic Functions
102.
ln 20 5 lns5 ? 22d 5 ln 5 1 ln 22 5 ln 5 1 2 ln 2 < 1.6094 1 2s0.6931d 5 2.9956
ln 18 5 lns32 ? 2d 5 ln 32 1 ln 2 5 2 ln 3 1 ln 2 < 2s1.0986d 1 0.6931 5 2.8903
ln 16 5 ln 24 5 4 ln 2 < 4s0.6931d 5 2.7724
ln 15 5 lns5 ? 3d 5 ln 5 1 ln 3 < 1.6094 1 1.0986 5 2.7080
ln 12 5 lns22 ? 3d 5 ln 22 1 ln 3 5 2 ln 2 1 ln 3 < 2s0.6931d 1 1.0986 5 2.4848
ln 10 5 lns5 ? 2d 5 ln 5 1 ln 2 < 1.6094 1 0.6931 5 2.3025
ln 9 5 ln 32 5 2 ln 3 < 2s1.0986d 5 2.1972
ln 8 5 ln 23 5 3 ln 2 < 3s0.6931d 5 2.0793
ln 6 5 lns2 ? 3d 5 ln 2 1 ln 3 < 0.6931 1 1.0986 5 1.7917
ln 5 < 1.6094
ln 4 5 lns2 ? 2d 5 ln 2 1 ln 2 < 0.6931 1 0.6931 5 1.3862
ln 3 < 1.0986
ln 2 < 0.6931
ln 2 < 0.6931, ln 3 < 1.0986, ln 5 < 1.6094
103.24xy22
16x23y5
24xx3
16yy25
3x4
2y3, x Þ 0 104. 12x2
3y 223
5 1 3y
2x223
5s3yd3
s2x2d35
27y3
8x6
105. s18x3y4d23s18x3y4d3 5s18x3y4d3
s18x3y4d35 1 if x Þ 0, y Þ 0. 106.
5xy
s y 1 xdyxy5
sxyd2
x 1 y
5xy
s1yxd 1 s1yyd
xysx21 1 y21d21 5xy
x21 1 y21
107.
x 1 1 5 0 ⇒ x 5 21
3x 2 1 5 0 ⇒ x 513
s3x 2 1dsx 1 1d 5 0
3x2 1 2x 2 1 5 0 108.
The zeros are x 514, 1.
x 2 1 5 0 ⇒ x 5 1
4x 2 1 5 0 ⇒ x 514
s4x 2 1dsx 2 1d 5 0
4x2 2 5x 1 1 5 0
109.
521 ± !97
6
x 521 ± !12 2 4s3ds28d
2s3d
3x2 1 x 2 8 5 0
s3x 1 1dsxd 5 s2ds4d
2
3x 1 15
x
4 110.
The zeros are 1 ±!31
2.
1 ±!31
25 x
2 ±!124
45 x
2s22d ±!s22d2 2 4s2ds215d
2s2d 5 x
0 5 2x2 2 2x 2 15
15 5 2x2 2 2x
5s3d 5 2xsx 2 1d
5
x 2 15
2x
3
Section 3.4 Exponential and Logarithmic Equations 289
Section 3.4 Exponential and Logarithmic Equations
Vocabulary Check
1. solve 2. (a) (b) 3. extraneous
(c) (d) xx
x 5 yx 5 y
1.
(a)
Yes, is a solution.
(b)
No, is not a solution.x 5 2
42s2d27 5 423 5164 Þ 64
x 5 2
x 5 5
42s5d27 5 43 5 64
x 5 5
42x27 5 64 2.
(a)
No, is not a solution.
(b)
No, is not a solution.x 5 2
23s2d11 5 27 5 128
x 5 2
x 5 21
23s21d11 5 222 514
x 5 21
23x11 5 32
3.
(a)
No, is not a solution.
(b)
Yes, is a solution.
(c)
Yes, is a solution.x < 1.219
3e1.21912 5 3e3.219 < 75
x < 1.219
x 5 22 1 ln 25
3es221 ln 25d12 5 3eln 25 5 3s25d 5 75
x 5 22 1 ln 25
x 5 22 1 e25
3es221e25d12 5 3ee25Þ 75
x 5 22 1 e25
3ex12 5 75 4.
(a)
Yes, is a solution.
(b)
No, is not a solution.
(c)
Yes, is an approximate solution.x 5 20.0416
2e5s20.0416d12 5 2e1.792 < 2s6.00144d < 12
x 5 20.0416
x 5ln 6
5 ln 2
< 2 ? 97.9995 5 195.999
< 2e2.58512
2e5[ln 6ys5 ln 2dg12 5 2esln 6yln 2d12
x 5ln 6
5 ln 2
x 51
5s22 1 ln 6d
5 2eln 6 5 2 ? 6 5 12
2e5fs1y5ds221 ln 6dg12 5 2e221 ln 612
x 51
5s22 1 ln 6d
2e5x12 5 12
n To solve an exponential equation, isolate the exponential expression, then take the logarithm of both sides.
Then solve for the variable.
1. 2.
n To solve a logarithmic equation, rewrite it in exponential form. Then solve for the variable.
1. 2.
n If and we have the following:
1.
2.
n Check for extraneous solutions.
ax 5 ay ⇔ x 5 y
loga x 5 loga y ⇔ x 5 y
a Þ 1a > 0
eln x 5 xaloga x 5 x
ln ex 5 xloga ax 5 x
290 Chapter 3 Exponential and Logarithmic Functions
5.
(a)
Yes, 21.333 is an approximate solution.
(b)
No, is not a solution.
(c)
Yes, is a solution.x 5643
3s643 d 5 64
x 5643
x 5 24
3s24d 5 212 Þ 64
x 5 24
3s21.333d < 64
x < 21.333
log4s3xd 5 3 ⇒ 3x 5 43 ⇒ 3x 5 64 6.
(a)
Since is a solution.
(b)
Since is not a solution.
(c)
Since is not a solution.102 2 3210 Þ 100,
log2s97 1 3d 5 log2s100d
x 5 102 2 3 5 97
x 5 17210 Þ 20,
log2s17 1 3d 5 log2s20d
x 5 17
x 5 1021210 5 1024,
log2s1021 1 3d 5 log2s1024d
x 5 1021
log2sx 1 3d 5 10
7.
(a)
No, is not a solution.
(b)
Yes, is a solution.
(c)
Yes, is an approximate solution.x < 163.650
lnf2s163.650d 1 3g 5 ln 330.3 < 5.8
x < 163.650
x 512s23 1 e5.8d
lnf2s12ds23 1 e5.8d 1 3g 5 lnse5.8d 5 5.8
x 512s23 1 e5.8d
x 512s23 1 ln 5.8d
lnf2s12ds23 1 ln 5.8d 1 3g 5 lnsln 5.8d Þ 5.8
x 512s23 1 ln 5.8d
lns2x 1 3d 5 5.8 8.
(a)
Yes, is a solution.
(b)
Yes, is an approximate solution.
(c)
No, is not a solution.x 5 1 1 ln 3.8
lns1 1 ln 3.8 2 1d 5 lnsln 3.8d < 0.289
x 5 1 1 ln 3.8
x < 45.701
lns45.701 2 1d 5 lns44.701d < 3.8
x < 45.701
x 5 1 1 e3.8
lns1 1 e3.8 2 1d 5 ln e3.8 5 3.8
x 5 1 1 e3.8
lnsx 2 1d 5 3.8
9.
x 5 2
4x 5 42
4x 5 16 10.
x 5 5
3x 5 35
3x 5 243 11.
x 5 25
2x 5 5
22x 5 25
s12dx
5 32 12.
x 5 23
2x 5 3
42x 5 43
s14dx
5 64
13.
x 5 2
ln x 5 ln 2
ln x 2 ln 2 5 0 14.
x 5 5
ln x 5 ln 5
ln x 2 ln 5 5 0 15.
x < 0.693
x 5 ln 2
ln ex 5 ln 2
ex 5 2 16.
x < 1.386
x 5 ln 4
ln ex 5 ln 4
ex 5 4
17.
x < 0.368
x 5 e21
eln x 5 e21
ln x 5 21 18.
x < 0.000912
x 5 e27
eln x 5 e27
ln x 5 27 19.
x 5 64
x 5 43
4log4 x 5 43
log4 x 5 3 20.
or 0.008 x 51
125
x 5 523
log5 x 5 23
Section 3.4 Exponential and Logarithmic Equations 291
28.
x 5 0, x 5 1
2xsx 2 1d 5 0
2x2 2 2x 5 0
2x2 5 x2 2 2x
e2x25 ex222x 29.
x < 1.465
x 5 log3 5 5log 5
log 3 or
ln 5
ln 3
log3 3x 5 log3 5
3x 5 5
4s3xd 5 20 30.
x < 1.723
x 5ln 16
ln 5
x 5 log516
5x 5 16
2s5xd 5 32
31.
x 5 ln 5 < 1.609
ln ex 5 ln 5
ex 5 5
2ex 5 10 32.
x 5 ln 914 < 3.125
ln ex 5 ln 914
ex 5914
4ex 5 91 33.
x 5 ln 28 < 3.332
ln ex 5 ln 28
ex 5 28
ex 2 9 5 19
34.
x < 2.015
x 5ln 37
ln 6
x 5 log6 37
6x 5 37
6x 1 10 5 47 35.
x 5ln 80
2 ln 3< 1.994
2x ln 3 5 ln 80
ln 32x 5 ln 80
32x 5 80 36.
x 5ln 3000
5 ln 6< 0.894
5x 5ln 3000
ln 6
s5xd ln 6 5 ln 3000
ln 65x 5 ln 3000
65x 5 3000
37.
t 5 2
2t
25 21
52ty2 5 521
52ty2 51
5
52ty2 5 0.20 38.
t 5 2ln 0.10
3 ln 4< 0.554
23t 5ln 0.10
ln 4
s23td ln 4 5 ln 0.10
ln 423t 5 ln 0.10
423t 5 0.10 39.
x 5 4
x 2 1 5 3
3x21 5 33
3x21 5 27
21.
Point of intersection:
s3, 8d
x 5 3
2x 5 23
2x 5 8
f sxd 5 gsxd 22.
Point of intersection:
s23, 9d
x 523
27x 5 272y3
27x 5 9
fsxd 5 gsxd 23.
Point of intersection:
s9, 2d
x 5 9
x 5 32
log3 x 5 2
f sxd 5 gsxd 24.
Point of intersection: s5, 0d
x 5 5
x 2 4 5 1
elnsx24d5e0
lnsx 2 4d 5 0
fsxd 5 gsxd
25.
x 5 21 or x 5 2
0 5 sx 1 1dsx 2 2d
0 5 x2 2 x 2 2
x 5 x2 2 2
e x 5 ex222 26.
x 5 22, x 5 4
sx 2 4dsx 1 2d 5 0
x2 2 2x 2 8 5 0
2x 5 x2 2 8
e2 x 5 ex228 27.
By the Quadratic Formula
or x < 20.618.x < 1.618
x2 2 x 2 1 5 0
x2 2 3 5 x 2 2
ex223 5 ex22
40.
x 5 8
x 2 3 5 5
x 2 3 5 log2 32
2x23 5 32
292 Chapter 3 Exponential and Logarithmic Functions
41.
5 3 2ln 565
ln 2< 26.142
x 53 ln 2 2 ln 565
ln 2
x ln 2 5 3 ln 2 2 ln 565
2x ln 2 5 ln 565 2 3 ln 2
3 ln 2 2 x ln 2 5 ln 565
s3 2 xd ln 2 5 ln 565
ln 232x 5 ln 565
232x 5 565 42.
x 52ln 431 2 ln 64
ln 8< 24.917
x ln 8 5 2ln 431 2 ln 64
2x ln 8 5 ln 431 1 ln 82
22 ln 8 2 x ln 8 5 ln 431
s22 2 xd ln 8 5 ln 431
ln 8222x 5 ln 431
8222x 5 431
43.
< 0.059
x 5 1
3 log13
22
3x 5 log13
22
log 103x 5 log13
22
103x 512
8
8s103xd 5 12 44.
< 6.146
x 5 6 1 log 7
5
x 2 6 5 log 7
5
log 10x26 5 log 7
5
10x26 57
5
5s10x26d 5 7 45.
x 5 1 1ln 7
ln 5< 2.209
x 2 1 5ln 7
ln 5
sx 2 1d ln 5 5 ln 7
ln 5x21 5 ln 7
5x21 5 7
3s5x21d 5 21
46.
x 5 6 2ln 5
ln 3< 4.535
2x 5ln 5
ln 32 6
6 2 x 5ln 5
ln 3
s6 2 xd ln 3 5 ln 5
ln 362x 5 ln 5
362x 5 5
8s362xd 5 40 47.
x 5ln 12
3< 0.828
3x 5 ln 12
e3x 5 12 48.
x 5ln 50
2< 1.956
2x 5 ln 50
ln e2x 5 ln 50
e2x 5 50
49.
5 ln 53 < 0.511
x 5 2ln 35
2x 5 ln 35
e2x 535
500e2x 5 300 50.
< 0.648
x 5 214 ln
340
24x 5 ln 340
ln e24x 5 ln 340
e24x 5340
1000e24x 5 75 51.
x 5 ln 1 5 0
ex 5 1
22ex 5 22
7 2 2ex 5 5 52.
< 2.120
x 5 ln 253
ln ex 5 ln 253
ex 5253
3ex 5 25
214 1 3ex 5 11
Section 3.4 Exponential and Logarithmic Equations 293
54.
x 5 3 2ln 3.5
2 ln 4< 2.548
22x 5 26 1ln 3.5
ln 4
6 2 2x 5ln 3.5
ln 4
6 2 2x 5 log4 3.5
4622x 5 3.5
8s4622xd 5 28
8s4622xd 1 13 5 41
55.
or
(No solution) x 5 ln 5 < 1.609
ex 5 5ex 5 21
sex 1 1dsex 2 5d 5 0
e2x 2 4ex 2 5 5 0 56.
x 5 ln 2 < 0.693 or x 5 ln 3 < 1.099
ex 5 2 or ex 5 3
sex 2 2dsex 2 3d 5 0
e2x 2 5ex 1 6 5 0
57.
Not possible since for all x.
ex 2 450 ⇒ ex 5 4 ⇒ x 5 ln 4 < 1.386
ex> 0
ex 1 150 ⇒ ex 5 21
sex 1 1dsex 2 4d 5 0
e2x 2 3ex 2 4 5 0 58.
Because the discriminant is there
is no solution.
92 2 4s1ds36d 5 263,
sexd219ex 1 36 5 0
e2x 1 9ex 1 36 5 0
53.
x 51
3 3logs8y3d
log 21 14 < 0.805
3x 2 15 log218
32 5logs8y3d
log 2 or
lns8y3dln 2
log2 23x21 5 log218
32
23x21 58
3
6s23x21d 5 16
6s23x21d 2 7 5 9
59.
x 5 2 ln 75 < 8.635
x
25 ln 75
exy2 5 75
25 5 100 2 exy2
500 5 20s100 2 exy2d
500
100 2 e xy25 20 60.
x 5 ln 7 < 1.946
2x 5 2ln 7
2x 5 ln 721
2x 5 ln 1
7
ln 1
75 ln e2x
1
75 e2x
8
72 1 5 e2x
8
75 1 1 e2x
400 5 350s1 1 e2xd
400
1 1 e2x5 350 61.
x 5ln 1498
2< 3.656
ln 1498 5 2x
1498 5 e2x
1500 5 2 1 e2x
3000 5 2s2 1 e2xd
3000
2 1 e2x5 2
294 Chapter 3 Exponential and Logarithmic Functions
63.
t 5ln 4
365 lns1 10.065365 d < 21.330
365t ln11 10.065
365 2 5 ln 4
ln11 10.065
365 2365t
5 ln 4
11 10.065
365 2365t
5 4
64.
t 5ln 21
9 ln 3.938225< 0.247
9t ln 3.938225 5 ln 21
ln 3.9382259t 5 ln 21
3.9382259t 5 21
14 22.471
40 29t
5 21 65.
t 5ln 2
12 lns1 10.1012 d < 6.960
12t ln11 10.10
12 2 5 ln 2
ln11 10.10
12 212t
5 ln 2
11 10.10
12 212t
5 2
66.
t 5ln 30
3 lns16 20.878
26 d < 0.409
3t ln116 20.878
26 2 5 ln 30
ln116 20.878
26 23t
5 ln 30
116 20.878
26 23t
5 30 67.
Algebraically:
The zero is x < 20.427.
x < 20.427
x 5 1 2 ln125
6 2
1 2 x 5 ln125
6 2
e12x 525
6
6e12x 5 25
−6 15
−30
6gsxd 5 6e12x 2 25
68.
The zero is 22.322.
22.322 < x
21 2 ln 3.75 5 x
1 1 ln 3.75 5 2x
ln 3.75 5 2x 2 1
3.75 5 e2x21
215 5 24e2x21
0 5 24e2x21 1 15−5 5
−20
20f sxd 5 24e2x21 1 15
62.
x 5ln 31
6< 0.572
ln 31 5 6x
ln 31 5 ln e6x
31 5 e6x
17 5 e6x 2 14
119 5 7se6x 2 14d
119
e6x 2 145 7
69.
Algebraically:
The zero is x < 3.847.
x < 3.847
x 52
3 ln1962
3 2
3x
25 ln1962
3 2
e3xy2 5962
3
3e3xy2 5 962
−6 9
−1200
300f sxd 5 3e3xy2 2 962
Section 3.4 Exponential and Logarithmic Equations 295
72.
The zero is 1.081.
−5 5
−7
13
x < 1.081
x 5ln 7
1.8
1.8x 5 ln 7
e1.8x 5 7
2e1.8x 5 27
2e1.8x 1 7 5 0
f sxd 5 2e1.8x 1 7 73.
Algebraically:
The zero is
−40 40
−10
2
t < 16.636.
t < 16.636
t 5 ln 8
0.125
0.125t 5 ln 8
e0.125t 5 8
e0.125t 2 850
hstd 5 e0.125t 2 8 74.
The zero is 1.236.
−5 5
−35
10
x < 1.236
x 5ln 29
2.724
2.724x 5 ln 29
e2.724x 5 29
f sxd 5 e2.724x 2 29
70.
The zero is 20.478.
x < 20.478
x 5 21.5 ln 1.375
22x
35 ln 1.375
e22xy3 5 1.375
8e22xy3 5 11 −3 7
−15
5 gsxd 5 8e22xy3 2 11 71.
Algebraically:
The zero is t < 12.207.
t < 12.207
t 5 ln 3
0.09
0.09t 5 ln 3
e0.09t 5 3 −20 40
−4
8gstd 5 e0.09t 2 3
75.
x 5 e23 < 0.050
ln x 5 23 76.
x 5 e2 < 7.389
eln x 5 e2
ln x 5 2 77.
x 5e2.4
2< 5.512
2x 5 e2.4
ln 2x 5 2.4 78.
x 5e
4< 0.680
4x 5 e
eln 4x 5 e1
ln 4x 5 1
79.
5 1,000,000.000
x 5 106
log x 5 6 80.
z 5100
3< 33.333
3z 5 100
10log 3z 5 102
log 3z 5 2 81.
x 5e10y3
5< 5.606
5x 5 e10y3
ln 5x 510
3
3 ln 5x 5 10 82.
x 5 e7y2 < 33.115
eln x 5 e7y2
ln x 57
2
2 ln x 5 7
83.
< 5.389
x 5 e2 2 2
x 1 2 5 e2
!x 1 2 5 e1
ln !x 1 2 5 1 84.
< 22,034.466
x 5 e10 1 8
x 2 8 5 e10
!x 2 8 5 e5
eln!x28 5 e5
ln!x 2 8 5 5 85.
< 0.513
x 5 e22y3
ln x 5 223
3 ln x 5 22
7 1 3 ln x 5 5 86.
< 0.264
x 5 e24y3
eln x 5 e24y3
ln x 5 243
26 ln x 5 8
2 2 6 ln x 5 10
296 Chapter 3 Exponential and Logarithmic Functions
87.
x 5 2s311y6d < 14.988
0.5x 5 311y6
3log3s0.5xd 5 311y6
log3s0.5xd 5116
6 log3s0.5xd 5 11 88.
x 5 1011y5 1 2 < 160.489
x 2 2 5 1011y5
10log10sx22d 5 1011y5
log10sx 2 2d 5115
5 log10sx 2 2d 5 11
89.
This negative value is extraneous. The equation has
no solution.
x 5e2
1 2 e2< 21.157
xs1 2 e2d 5 e2
x 2 e2x 5 e2
x 5 e2x 1 e2
x 5 e2sx 1 1d
x
x 1 15 e2
ln1 x
x 1 12 5 2
ln x 2 lnsx 1 1d 5 2 90.
The only solution is x 521 1 !1 1 4e
2< 1.223.
x 521 ± !1 1 4e
2
x2 1 x 2 e 5 0
xsx 1 1d 5 e1
elnfxsx11dg 5 e1
lnfxsx 1 1dg 5 1
ln x 1 lnsx 1 1d 5 1
91.
The negative value is extraneous. The only solution is
x 5 1 1 !1 1 e < 2.928.
5 1 ± !1 1 e 52 ± 2!1 1 e
2
x 52 ± !4 1 4e
2
x2 2 2x 2 e 5 0
xsx 2 2d 5 e1
lnfxsx 2 2dg 5 1
ln x 1 lnsx 2 2d 5 1 92.
The only solution is .x 523 1 !9 1 4e
2< 0.729
x 523 ± !9 1 4e
2
x2 1 3x 2 e 5 0
xsx 1 3d 5 e1
elnfxsx13dg 5 e1
lnfxsx 1 3dg 5 1
ln x 1 lnsx 1 3d 5 1
93.
Both of these solutions are extraneous, so the equation
has no solution.
x 5 22 or x 5 23
sx 1 2dsx 1 3d 5 0
x2 1 5x 1 6 5 0
x2 1 6x 1 5 5 x 2 1
sx 1 5dsx 1 1d 5 x 2 1
x 1 5 5x 2 1
x 1 1
lnsx 1 5d 5 ln1x 2 1
x 1 12ln sx 1 5d 5 lnsx 2 1d 2 lnsx 1 1d 94.
(The negative apparent solution is extraneous.)
3.303 < x
3 ±!13
25 x
2s23d ±!s23d2 2 4s1ds21d
2s1d 5 x
0 5 x2 2 3x 2 1
x 1 1 5 x2 2 2x
x 1 1
x 2 25 x
ln1x 1 1
x 2 22 5 ln x
lnsx 1 1d 2 lnsx 2 2d 5 ln x
95.
x 5 7
2x 2 3 5 x 1 4
log2s2x 2 3d 5 log2sx 1 4d
Section 3.4 Exponential and Logarithmic Equations 297
100.
The value is extraneous. The only solution is
x 5 9.
x 5 21
x 5 9 or x 5 21
sx 2 9dsx 1 1d 5 0
x2 2 8x 2 9 5 0
x2 2 8x 5 9
3log3sx228xd 5 32
log3fxsx 2 8dg 5 2
log3 x 1 log3sx 2 8d 5 2
97.
Quadratic Formula
Choosing the positive value of x (the negative value is
extraneous), we have
x 521 1 !17
2< 1.562.
x 521 ± !17
2
0 5 x2 1 x 2 4
x 1 4 5 x2 1 2x
x 1 4
x5 x 1 2
log1x 1 4
x 2 5 logsx 1 2d
logsx 1 4d 2 log x 5 logsx 1 2d 98.
The value is extraneous. The only solution is
x 5 2.
x 5 23
x 5 23 or x 5 2
sx 1 3dsx 2 2d 5 0
x2 1 x 2 6 5 0
xsx 1 2d 5 x 1 6
log2fxsx 1 2dg 5 log2sx 1 6d
log2 x 1 log2sx 1 2d 5 log2sx 1 6d
99.
x 5 2
2x 5 22
x 5 2x 2 2
x 5 2sx 2 1d
x
x 2 15 41y2
4log4fxysx21dg 5 41y2
log41 x
x 2 12 51
2
log4 x 2 log4sx 2 1d 51
2
96.
The apparent solution is extraneous, because the domain of the logarithm function is positive numbers,
and and are negative. There is no solution.2s27d 1 127 2 6
x 5 27
27 5 x
x 2 6 5 2x 1 1
logsx 2 6d 5 logs2x 1 1d
101.
The only solution is x 525s29 1 5!33d
8< 180.384.
x < 0.866 (extraneous) or x < 180.384
525s29 ± 5!33d
85
725 ± !515,625
8 x 5
725 ± !7252 2 4s4ds625d2s4d
4x2 2 725x 1 625 5 0
4x2 2 100x 1 625 5 625x
s2x 2 25d2 5 s25!xd2
2x 2 25 5 25!x
2x 5 25s1 1 !xd 5 25 1 25!x
8x 5 100s1 1 !xd
8x
1 1 !x5 102
log 8x
1 1 !x5 2
log 8x 2 logs1 1 !xd 5 2
298 Chapter 3 Exponential and Logarithmic Functions
102.
The only solution is x 51225 1 125!73
2< 1146.500.
x < 78.500 sextraneousd or x < 1146.500
x 51225 ± 125!73
2
x 51225 ± !1,140,625
2
x 51225 ± !s21225d2 2 4s1ds90,000d
2
x2 2 1225x 1 90,000 5 0
x2 2 600x 1 90,000 5 625x
sx 2 300d2 5 s25!x d2
x 2 300 5 25!x
4x 2 1200 5 100!x
4x 5 1200 1 100!x
4x 5 100s12 1 !x d
4x
12 1 !x5 100
10logs4xy(121!x dd5 102
log1 4x
12 1 !x2 5 2
log 4x 2 logs12 1 !x d 5 2
103.
From the graph we have
when
Algebraically:
x 5ln 7
ln 2< 2.807
x ln 2 5 ln 7
ln 2x 5 ln 7
2x 5 7
y 5 7.x < 2.807
y2 5 2x
−8 10
−2
10y1 5 7 104.
The solution is x < 2.197.
2.197 < x
22 ln 1
35 x
ln 1
35 2
x
2
1
35 e2xy2
−2
−200
10
800 500 5 1500e2xy2
105.
From the graph we have
when
Algebraically:
x 5 e3 < 20.086
ln x 5 3
3 2 ln x 5 0
y 5 3.x < 20.086
y2 5 ln x
−5
−1
30
5y1 5 3 106.
The solution is x < 14.182.
x < 14.182
x 5 e2.5 1 2
x 2 2 5 e2.5
elnsx22d 5 e2.5
lnsx 2 2d 5 2.5
24 lnsx 2 2d 5 210
−5
−3
30
18 10 2 4 lnsx 2 2d 5 0
Section 3.4 Exponential and Logarithmic Equations 299
107. (a)
t < 8.2 years
ln 2
0.0855 t
ln 2 5 0.085t
2 5 e0.085t
5000 5 2500e0.085t
A 5 Pert (b)
t < 12.9 years
ln 3
0.0855 t
ln 3 5 0.085t
3 5 e0.085t
7500 5 2500e0.085t
A 5 Pert 108. (a)
t < 5.8 years
ln 2
0.125 t
ln 2 5 0.12t
ln 2 5 ln e0.12t
2 5 e0.12t
5000 5 2500e0.12t
A 5 Pert
r 5 0.12 (b)
t 5 9.2 years
ln 3
0.125 t
ln 3 5 0.12t
ln 3 5 ln e0.12t
3 5 e0.12t
7500 5 2500e0.12t
A 5 Pert
r 5 0.12
109.
(a)
x < 1426 units
0.004x 5 ln 300
300 5 e0.004x
350 5 500 2 0.5se0.004xd
p 5 350
p 5 500 2 0.5se0.004xd
(b)
x < 1498 units
0.004x 5 ln 400
400 5 e0.004x
300 5 500 2 0.5se0.004xd
p 5 300
110.
(a) When
x 5 2lns6y11d
0.002< 303 units
ln 6
115 20.002x
ln 6
115 ln e20.002x
6
115 e20.002x
0.48 5 0.88e20.002x
4 5 3.52 1 0.88e20.002x
4
4 1 e20.002x5 0.88
0.12 5 1 24
4 1 e20.002x
600 5 500011 24
4 1 e20.002x2p 5 $600:
p 5 500011 24
4 1 e20.002x2(b) When
x 5 2lns8y23d
0.002< 528 units
ln 8
235 20.002x
ln 8
235 ln e20.002x
8
235 e20.002x
0.32 5 0.92e20.002x
4 5 3.68 1 0.92e20.002x
4
4 1 e20.002x5 0.92
0.08 5 1 24
4 1 e20.002x
400 5 500011 24
4 1 e20.002x2p 5 $400:
111.
(a)
0
0
1500
10
V 5 6.7e248.1yt , t ≥ 0
(b) As
Horizontal asymptote:
The yield will approach
6.7 million cubic feet per acre.
V 5 6.7
t → `, V → 6.7. (c)
t 5248.1
lns13y67d< 29.3 years
ln113
672 5248.1
t
1.3
6.75 e248.1yt
1.3 5 6.7e248.1yt
300 Chapter 3 Exponential and Logarithmic Functions
112.
When
x 5 2log10s21y68d
0.04< 12.76 inches
log10 21
685 20.04x
21
685 1020.04x
21 5 68s1020.04xd
N 5 21:
N 5 68s1020.04xd 113.
corresponds to the year 2001.t < 11
t 5 e2.4 < 11
ln t 5 2.4
2630.0 ln t 5 21512
7312 2 630.0 ln t 5 5800
y 5 7312 2 630.0 ln t, 5 ≤ t ≤ 12
114.
Since represents 1995, indicates that the number of daily fee golf facilities in the U.S.
reached 9000 in 2001.
t 5 11.6t 5 5
t 5 e2.45222 5 11.6
ln t 54619
1883.65 2.45222
4619 5 1883.6 ln t
9000 5 4381 1 1883.6 ln t
y 5 4381 1 1883.6 ln t, 5 ≤ t ≤ 13
115. (a) From the graph shown in the textbook, we see horizontal asymptotes at and
These represent the lower and upper percent bounds; the range falls between 0% and 100%.
y 5 100.y 5 0
(b) Males
x 5 69.71 inches
20.6114sx 2 69.71d 5 0
20.6114sx 2 69.71d 5 ln 1
e20.6114sx269.71d 5 1
1 1 e20.6114sx269.71d 5 2
50 5100
1 1 e20.6114sx269.71d
Females
x 5 64.51 inches
20.66607sx 2 64.51d 5 0
20.66607sx 2 64.51d 5 ln 1
e20.6667sx264.51d 5 1
1 1 e20.66607sx264.51d 5 2
50 5100
1 1 e20.66607sx264.51d
(c) When
n 5 2
ln10.83
0.602 12
0.2< 5 trials
20.2n 5 ln10.83
0.602 12
ln e20.2n 5 ln10.83
0.602 12
e20.2n 50.83
0.602 1
1 1 e20.2n 50.83
0.60
0.60 50.83
1 1 e20.2n
P 5 60% or P 5 0.60:
116.
(a)
(b) Horizontal asymptotes:
The upper asymptote, indicates that the
proportion of correct responses will approach 0.83
as the number of trials increases.
P 5 0.83,
P 5 0, P 5 0.83
0
0 40
1.0
P 50.83
1 1 e20.2n
Section 3.4 Exponential and Logarithmic Equations 301
117.
(a)
(b)
The model seems to fit the data well.
(c) When
Add the graph of to the graph in part (a) and
estimate the point of intersection of the two graphs.
We find that meters.
(d) No, it is probably not practical to lower the number
of gs experienced during impact to less than 23
because the required distance traveled at is
meters. It is probably not practical to
design a car allowing a passenger to move forward
2.27 meters (or 7.45 feet) during an impact.
x < 2.27
y 5 23
x < 1.20
y 5 30
30 5 23.00 1 11.88 ln x 136.94
x
y 5 30:
0
0
1.2
200
y 5 23.00 1 11.88 ln x 136.94
x
x 0.2 0.4 0.6 0.8 1.0
y 162.6 78.5 52.5 40.5 33.9
119.
True by Property 1 in Section 3.3.
logasuvd 5 loga u 1 loga v 120.
False.
2.04 < log10s10 1 100d Þ slog10 10dslog10 100d 5 2
logasu 1 vd 5 sloga udsloga vd
121.
False.
Þ log 100 2 log 10 5 1
1.95 < logs100 2 10d
logasu 2 vd 5 loga u 2 loga v 122.
True by Property 2 in Section 3.3.
loga1u
v2 5 loga u 2 loga v 123. Yes, a logarithmic equation can
have more than one extraneous
solution. See Exercise 93.
124.
(a) This doubles your money.
(b)
(c)
Doubling the interest rate yields the same result as
doubling the number of years.
If (i.e., ), then doubling your
investment would yield the most money. If
then doubling either the interest rate
or the number of years would yield more money.
rt > ln 2,
rt < ln 22 > ert
A 5 Pers2td 5 Pertert 5 ertsPertd
A 5 Pes2rdt 5 Pertert 5 ertsPertd
A 5 s2Pdert 5 2sPertd
A 5 Pert 125. Yes.
Time to Double Time to Quadruple
Thus, the time to quadruple is twice as long as the
time to double.
2 ln 2
r5 t
ln 2
r5 t
ln 4 5 rt ln 2 5 rt
4 5 ert 2 5 ert
4P 5 Pert 2P 5 Pert
118.
(a) From the graph in the textbook we see a horizontal
asymptote at This represents the room
temperature.
(b)
h < 0.81 hour
lns4y7d2ln 2
5 h
ln14
72 5 2h ln 2
ln14
72 5 ln 22h
4
75 22h
4 5 7s22hd
5 5 1 1 7s22hd
100 5 20f1 1 7s22hdg
T 5 20.
T 5 20f1 1 7s22hdg
302 Chapter 3 Exponential and Logarithmic Functions
126. (a) When solving an exponential equation, rewrite the
original equation in a form that allows you to use the
One-to-One Property if and only if or
rewrite the original equation in logarithmic form and
use the Inverse Property loga ax 5 x.
x 5 yax 5 ay
(b) When solving a logarithmic equation, rewrite the
original equation in a form that allows you to use the
One-to-One Property if and only if
or rewrite the original equation in exponential
form and use the Inverse Property aloga x 5 x.
x 5 y
loga x 5 loga y
127.
5 4|x|y2!3y
!48x2y5 5 !16x2y43y 128.
5 4!2 2 10
!32 2 2!25 5 !16 ? 2 2 2s5d 129.
5 3!125 ? 3 5 5 3!3
3!25 3!15 5 3!375
130.
51
2!10 1 1
5!10 1 2
2
53s!10 1 2d
6
53s!10 1 2d
10 2 4
3
!10 2 25
3
!10 2 2?!10 1 2
!10 1 2131.
Domain: all real numbers
intercept:
axis symmetryy-
s0, 9dy-
x
y
x−2−4−6−8 2 4 6 8
−2
2
4
6
8
12
14
f sxd 5 |x| 1 9
0
9 10 11 12y
±3±2±1x
132. y
x−2−4−6 2 4 6 8
−2
−4
−6
2
4
6
8
133.
Domain: all real numbers
intercept:
intercept: s0, 4dy-
s2, 0dx-
x
y
x−1−2−3−4 1 3 4
−3
1
2
3
4
5
gsxd 5 52x,
2x2 1 4,
x < 0
x ≥ 0
0 1 2 3
4 3 2 2521222426y
20.5212223x
134. y
x64
−6
−2
1
4
6
2−2−4−6
135. log6 9 5log10 9
log10 65
ln 9
ln 6< 1.226 136. log3 4 5
log10 4
log10 35
ln 4
ln 3< 1.262
137. log3y4 5 5log10 5
log10s3y4d 5ln 5
lns3y4d < 25.595 138. log8 22 5log10 22
log10 85
ln 22
ln 8< 1.486
Section 3.5 Exponential and Logarithmic Models
Section 3.5 Exponential and Logarithmic Models 303
n You should be able to solve growth and decay problems.
(a) Exponential growth if
(b) Exponential decay if
n You should be able to use the Gaussian model
n You should be able to use the logistic growth model
n You should be able to use the logarithmic models
y 5 a 1 b log x.y 5 a 1 b ln x,
y 5a
1 1 be2rx.
y 5 ae2sx2bd2yc.
b > 0 and y 5 ae2bx.
b > 0 and y 5 aebx.
Vocabulary Check
1. 2. 3. normally distributed
4. bell; average value 5. sigmoidal
y 5 a 1 b ln x; y 5 a 1 b log xy 5 aebx; y 5 ae2bx
1.
This is an exponential growth
model. Matches graph (c).
y 5 2exy4 2.
This is an exponential decay
model. Matches graph (e).
y 5 6e2xy4 3.
This is a logarithmic function
shifted up six units and left two
units. Matches graph (b).
y 5 6 1 logsx 1 2d
4.
This is a Gaussian model.
Matches graph (a).
y 5 3e2sx22d2y5 5.
This is a logarithmic model shifted
left one unit. Matches graph (d).
y 5 lnsx 1 1d 6.
This is a logistic growth model.
Matches graph (f).
y 54
1 1 e22x
7. Since the time to double is given by
and we have
Amount after 10 years: A 5 1000e0.35 < $1419.07
t 5ln 2
0.035< 19.8 years.
ln 2 5 0.035t
ln 2 5 ln e0.035t
2 5 e0.035t
2000 5 1000e0.035t
A 5 1000e0.035t, 8. Since the time to double is given by
and we have
Amount after 10 years: A 5 750e0.105s10d < $2143.24
t 5ln 2
0.105< 6.60 years.
ln 2 5 0.105t
ln 2 5 ln e0.105t
2 5 e0.105t
1500 5 750e0.105t
1500 5 750e0.105t,
A 5 750e0.105t,
304 Chapter 3 Exponential and Logarithmic Functions
9. Since when we have
the following.
Amount after 10 years: A 5 750e0.089438s10d < $1834.37
r 5ln 2
7.75< 0.089438 5 8.9438%
ln 2 5 7.75r
ln 2 5 ln e7.75r
2 5 e7.75r
1500 5 750e7.75r
t 5 7.75,A 5 750ert and A 5 1500 10. Since and when
we have
Amount after 10 years:
A 5 10,000e0.057762s10d < $17,817.97
r 5ln 2
12< 0.057762 5 5.7762%.
ln 2 5 12r
ln 2 5 ln e12r
2 5 e12r
20,000 5 10,000e12r
t 5 12,A 5 20,000A 5 10,000ert
11. Since when
we have the following.
The time to double is given by
t 5ln 2
0.110< 6.3 years.
1000 5 500e0.110t
r 5lns1505.00y500d
10< 0.110 5 11.0%
1505.00 5 500e10r
t 5 10,
A 5 500ert and A 5 $1505.00 12. Since and when we have
The time to double is given by
t 5ln 2
0.3466< 2 years.
1200 5 600e0.3466t
r 5lns19,205y600d
10< 0.3466 or 34.66%.
ln119,205
600 2 5 10r
ln119,205
600 2 5 ln e10r
19,205
6005 e10r
19,205 5 600e10r
t 5 10,A 5 19,205A 5 600ert
13. Since and when
we have the following.
The time to double is given by t 5ln 2
0.045< 15.40 years.
10,000.00
e0.045s10d 5 P < $6376.28
10,000.00 5 Pe0.045s10d
t 5 10,A 5 10,000.00A 5 Pe0.045t 14. Since and when we have
The time to double is given by years.t 5ln 2
0.025 34.7
P 52000
e0.02s10d 5 $1637.46.
2000 5 Pe0.02s10d
t 5 10,A 5 2000A 5 Pe0.02t
15.
5500,000
1.00625240< $112,087.09
P 5500,000
11 10.075
12 212s20d
500,000 5 P11 10.075
12 212s20d16.
P 5 $4214.16
500,000 5 P11 10.12
12 212(40)
A 5 P11 1r
n2nt
Section 3.5 Exponential and Logarithmic Models 305
17.
(a)
(c)
t 5ln 2
365 lns1 10.11365 d < 6.302 years
365t ln11 10.11
365 2 5 ln 2
11 10.11
365 2365t
5 2
n 5 365
t 5ln 2
ln 1.11< 6.642 years
t ln 1.11 5 ln 2
s1 1 0.11dt 5 2
n 5 1
P 5 1000, r 5 11%
(b)
(d) Compounded continuously
t 5ln 2
0.11< 6.301 years
0.11t 5 ln 2
e0.11t 5 2
t 5ln 2
12 lns1 10.1112 d < 6.330 years
12t ln11 10.11
12 2 5 ln 2
11 10.11
12 212t
5 2
n 5 12
18.
(a)
(c)
t 5ln 2
365 lns1 10.105365 d < 6.602 years
n 5 365
t 5ln 2
lns1 1 0.105d< 6.94 years
n 5 1
r 5 10.5% 5 0.105P 5 1000,
(b)
(d) Compounded continuously
t 5ln 2
0.105< 6.601 years
t 5ln 2
12 lns1 10.105
12 d < 6.63 years
n 5 12
19.
ln 3
r5 t
ln 3 5 rt
3 5 ert
3P 5 Pert
r 2% 4% 6% 8% 10% 12%
(years) 9.1610.9913.7318.3127.4754.93t 5ln 3
r
20.
Using the power regression feature of a graphing utility, t 5 1.099r21.
0
0
0.16
60
21.
ln 3
lns1 1 rd5 t
ln 3 5 t lns1 1 rd
ln 3 5 lns1 1 rdt
3 5 s1 1 rdt
3P 5 Ps1 1 rdt
r 2% 4% 6% 8% 10% 12%
(years) 55.48 28.01 18.85 14.27 11.53 9.69t 5ln 3
lns1 1 rd
306 Chapter 3 Exponential and Logarithmic Functions
22.
Using the power regression feature of a graphing utility,
t 5 1.222r21.
0
0
0.16
60 23. Continuous compounding results in faster growth.
Time (in years)
Am
ount
(in d
oll
ars)
t
A t= 1 + 0.075 [[ [[
A e= 0.07t
A
1.00
1.25
1.50
1.75
2.00
2 4 6 8 10
A 5 1 1 0.075v t b and A 5 e0.07t
24.
From the graph, compounded
daily grows faster than 6% simple
interest.
512%
0
2
A t= 1 + 0.06[[ [[
A = 1 +0.055365( ) 365t[[ [[
0 10
25.
Given grams after
1000 years, we have
< 6.48 grams.
y 5 10efsln 0.5dy1599gs1000d
C 5 10
k 5ln 0.5
1599
ln 0.5 5 ks1599d
ln 0.5 5 ln eks1599d
0.5 5 eks1599d
1
2C 5 Ceks1599d 26.
Given grams after 1000
years, we have
C < 2.31 grams.
1.5 5 Ceflns1y2dy1599gs1000d
y 5 1.5
k 5lns1y2d1599
ln 1
25 ks1599d
ln 1
25 ln eks1599d
1
25 eks1599d
1
2C 5 Ceks1599d
27.
Given grams after 1000
years, we have
C < 2.26 grams.
2 5 Cefsln 0.5dy5715gs1000d
y 5 2
k 5ln 0.5
5715
ln 0.5 5 ks5715d
ln 0.5 5 ln eks5715d
0.5 5 eks5715d
1
2C 5 Ceks5715d 28.
Given grams, after 1000
years we have
y < 2.66 grams.
y 5 3eflns1y2dy5715gs1000d
C 5 3
k 5lns1y2d
5715
ln 1
25 ks5715d
ln 1
25 ln eks5715d
1
25 eks5715d
1
2C 5 Ceks5715d 29.
Given grams after 1000
years, we have
C < 2.16 grams.
2.1 5 Cefsln 0.5dy24,100gs1000d
y 5 2.1
k 5ln 0.5
24,100
ln 0.5 5 ks24,100d
ln 0.5 5 ln eks24,100d
0.5 5 eks24,100d
1
2C 5 Ceks24,100d
Section 3.5 Exponential and Logarithmic Models 307
36.
—CONTINUED—
30.
Given grams after 1000
years, we have
C < 0.41 grams.
0.4 5 Ceflns1y2dy24,100gs1000d
y 5 0.4
k 5lns1y2d 24,100
ln 1
25 ks24,100d
ln 1
25 ln eks24,100d
1
25 eks24,100d
1
2C 5 Ceks24,100d 31.
Thus, y 5 e0.7675x .
ln 10
35 b ⇒ b < 0.7675
ln 10 5 3b
10 5 ebs3d
1 5 aebs0d ⇒ 1 5 a
y 5 aebx 32.
Thus, y 512e0.5756x.
ln 10
45 b ⇒ b < 0.5756
ln 10 5 4b
ln 10 5 ln e4b
10 5 e4b
5 51
2ebs4d
1
25 aebs0d ⇒ a 5
1
2
y 5 aebx
33.
Thus, y 5 5e20.4024x.
lns1y5d
45 b ⇒ b < 20.4024
ln11
52 5 4b
1
55 e4b
1 5 5ebs4d
5 5 aebs0d ⇒ 5 5 a
y 5 aebx 34.
Thus, y 5 e20.4621x .
lns1y4d
35 b ⇒ b < 20.4621
ln11
42 5 3b
ln11
42 5 ln e3b
1
45 ebs3d
1 5 aebs0d ⇒ 1 5 a
y 5 aebx
35.
(a) Since the exponent is negative, this is an exponential
decay model. The population is decreasing.
(b) For 2000, let thousand people
For 2003, let thousand peopleP < 2408.95t 5 3:
P 5 2430t 5 0:
P 5 2430e20.0029t
(c)
The population will reach 2.3 million (according to
the model) during the later part of the year 2018.
t 5lns2300y2430d
20.0029< 18.96
ln12300
24302 5 20.0029t
2300
24305 e20.0029t
2300 5 2430e20.0029t
2.3 million 5 2300 thousand
Country 2000 2010
Bulgaria 7.8 7.1
Canada 31.3 34.3
China 1268.9 1347.6
United Kingdom 59.5 61.2
United States 282.3 309.2
308 Chapter 3 Exponential and Logarithmic Functions
36. —CONTINUED—
(a) Bulgaria:
For 2030, use
million
China:
For 2030, use
million
United Kingdom:
For 2030, use
milliony 5 59.5e0.00282s30d < 64.7
t 5 30.
ln 61.2
59.55 10b ⇒ b < 0.00282
61.2 5 59.5ebs10d
a 5 59.5
y 5 1268.9e0.00602s30d < 1520.06
t 5 30.
ln 1347.6
1268.95 10b ⇒ b < 0.00602
1347.6 5 1268.9ebs10d
a 5 1268.9
y 5 7.8e20.0094s30d 5 5.88
t 5 30.
ln 7.1
7.85 10b ⇒ b 5 20.0094
7.1 5 7.8ebs10d
a 5 7.8
Canada:
For 2030, use
million
United States:
For 2030, use
milliony 5 282.3e0.0091s30d < 370.9
t 5 30.
ln 309.2
282.35 10b ⇒ b < 0.0091
309.2 5 282.3ebs10d
a 5 282.3
y 5 31.3e0.00915s30d < 41.2
t 5 30.
ln 34.3
31.35 10b ⇒ b < 0.00915
34.3 5 31.3ebs10d
a 5 31.3
(b) The constant b determines the growth rates. The greater the rate of growth, the greater the value of b.
(c) The constant b determines whether the population is increasing or decreasing .sb < 0dsb > 0d
37.
When
When hitsy 5 4080e0.2988s24d < 5,309,734t 5 24:
k 5lns10,000y4080d
3< 0.2988
ln110,000
4080 2 5 3k
10,000
40805 e3k
10,000 5 4080eks3d
y 5 10,000:t 5 3,
y 5 4080ekt 38.
For 2010,
milliony 5 10es0.1337ds20d 5 $144.98
t 5 20:
ln165
102 5 14k ⇒ k 5 0.1337
65 5 10eks14d
y 5 10ekt
Section 3.5 Exponential and Logarithmic Models 309
39.
t 5ln 2
0.2197< 3.15 hours
200 5 100e0.2197t
N 5 100e0.2197t
k 5ln 3
5< 0.2197
ln 3 5 5k
ln 3 5 ln e5k
3 5 e5k
300 5 100e5k
N 5 100ekt 40.
hours t 5ln 2
sln 1.12dy10< 61.16
ln 2 5 1ln 1.12
10 2t
2 5 efsln 1.12dy10gt
500 5 250efsln 1.12dy10gt
N 5 250efsln 1.12dy10gt
k 5ln 1.12
10
1.12 5 e10k
280 5 250eks10d
N 5 250ekt
41.
(a)
(b)
t 5 28223 ln11012
13112 < 4797 years old
2t
82235 ln11012
13112
e2ty8223 51012
1311
1
1012e2ty8223 5
1
1311
t 5 28223 ln11012
814 2 < 12,180 years old
2t
82235 ln11012
814 2
e2ty8223 51012
814
1
1012e2ty8223 5
1
814
R 51
814
R 51
1012e2ty8223 42.
The ancient charcoal has only 15% as much radioactive
carbon.
years t 55715 ln 0.15
ln 0.5< 15,642
ln 0.15 5ln 0.5
5715t
0.15C 5 Cefsln 0.5dy5715gt
k 5lns1y2d5715
ln 1
25 5715k
1
2C 5 Ce5715k
y 5 Cekt
43.
(a)
Linear model:
—CONTINUED—
V 5 26394t 1 30,788
b 5 30,788
m 518,000 2 30,788
2 2 05 26394
s0, 30,788d, s2, 18,000d
(b)
Exponential model: V 5 30,788e20.268t
k 51
2 ln14500
76972 < 20.268
ln14500
76972 5 2k
4500
76975 e2k
18,000 5 30,788eks2d
a 5 30,788 (c)
The exponential model
depreciates faster in the
first two years.
0 4
0
32,000
310 Chapter 3 Exponential and Logarithmic Functions
43. —CONTINUED—
(d) (e) The linear model gives a higher value for the car for the
first two years, then the exponential model yields a higher
value. If the car is less than two years old, the seller would
most likely want to use the linear model and the buyer the
exponential model. If it is more than two years old, the
opposite is true.
1 3
$24,394 $11,606
$23,550 $13,779V 5 30,788e20.268t
V 5 26394t 1 30,788
t
44.
(a)
(c)
The exponential model depreciates faster in the
first two years.
(e) The slope of the linear model means that the comput-
er depreciates $300 per year, then loses all value in
the third year. The exponential model depreciates
faster in the first two years but maintains value longer.
0 4
0
1200
V 5 2300t 1 1150
m 5550 2 1150
2 2 05 2300
s0, 1150d, s2, 550d
(b)
(d)
V 5 1150e20.369t
ln1 550
11502 5 2k ⇒ k < 20.369
550 5 1150eks2d
t 1 3
$850 $250
$795 $380V 5 1150e20.369t
V 5 2300t 1 1100
45.
(a)
Sstd 5 100s1 2 e20.1625td
k < 20.1625
k 5 ln 0.85
ln 0.85 5 ln ek
0.85 5 ek
85100 5 ek
285 5 2100ek
15 5 100s1 2 eks1dd
Sstd 5 100s1 2 ektd
(b)
(c) Ss5d 5 100s1 2 e20.1625s5dd < 55.625 5 55,625 units
Time (in years)
Sal
es
(in t
housa
nds
of
unit
s)
S
t5 3025201510
30
60
90
120
46.
(a)
So, N 5 30s1 2 e20.050d.
k 5 20.050
20k 5 ln111
302
ln e20k 5 ln111
302
e20k 511
30
30e20k 5 11
19 5 30s1 2 e20kd
N 5 19, t 5 20
N 5 30s1 2 ektd
(b)
t 5lns5y30d20.050
5 36 days
ln1 5
302 5 20.050t
ln1 5
302 5 ln e20.050t
5
305 e20.050t
25 5 30s1 2 e20.050td
N 5 25
Section 3.5 Exponential and Logarithmic Models 311
47.
(a)
(b) The average IQ score of an adult student is 100.
70 115
0
0.04
y 5 0.0266e2sx2100d2y450, 70 ≤ x ≤ 116 48. (a)
(b) The average number of hours per week a student uses
the tutor center is 5.4.
4 7
0
0.9
49.
(a) animals
(b)
months
(c)
The horizontal asymptotes are and
The asymptote with the larger -value,
indicates that the population size will approach 1000
as time increases.
p 5 1000,p
p 5 1000.p 5 0
0
0
40
1200
t 5 2lns1y9d0.1656
< 13
e20.1656t 51
9
9e20.1656t 5 1
1 1 9e20.1656t 5 2
500 51000
1 1 9e20.1656t
ps5d 51000
1 1 9e20.1656s5d < 203
pstd 51000
1 1 9e20.1656t50.
(a)
So,
(b) When
S 5500,000
1 1 0.6e0.0263s8d < 287,273 units sold.
t 5 8:
S 5500,000
1 1 0.6e0.0263t.
k 51
4 ln110
9 2 < 0.0263
4k 5 ln110
9 2
e4k 510
9
0.6e4k 52
3
1 1 0.6e4k 55
3
300,000 5500,000
1 1 0.6e4k
S 5500,000
1 1 0.6ekt
51. since
(a)
(b)
(c) 4.2 5 log I ⇒ I 5 104.2 < 15,849
8.3 5 log I ⇒ I 5 108.3 < 199,526,231
7.9 5 log I ⇒ I 5 107.9 < 79,432,823
I0 5 1.R 5 log I
I0
5 log I 52. since
(a)
(b)
(c) R 5 log 251,200 5 5.40
R 5 log 48,275,000 5 7.68
R 5 log 80,500,000 5 7.91
I0 5 1.R 5 log I
I0
5 log I
53.
(a)
(c) b 5 10 log 1028
102125 10 log 104 5 40 decibels
b 5 10 log 10210
102125 10 log 102 5 20 decibels
b 5 10 log I
I0
where I0 5 10212 wattym2.
(b)
(d) b 5 10 log 1
102125 10 log 1012 5 120 decibels
b 5 10 log 1025
102125 10 log 107 5 70 decibels
312 Chapter 3 Exponential and Logarithmic Functions
54. where
(a)
(c) bs1024d 5 10 log 1024
102125 10 log 108 5 80 decibels
bs10211d 5 10 log 10211
102125 10 log 101 5 10 decibels
I0 5 10212 wattym2bsId 5 10 log I
I0
(b)
(d) bs1022d 5 10 log 1022
102125 10 log 1010 5 100 decibels
bs102d 5 10 log 102
102125 10 log 1014 5 140 decibels
55.
% decrease 5I0109.3 2 I0108.0
I0109.33 100 < 95%
I 5 I010by10
10by10 5I
I0
10by10 5 10log IyI0
b
105 log
I
I0
b 5 10 log I
I0
56.
% decrease 5I0108.8 2 I0107.2
I0108.83 100 < 97%
I 5 I010by10
10by10 5I
I0
b 5 10 log10 I
I0
57.
2logs2.3 3 1025d < 4.64
pH 5 2logfH1g 58.
2logf11.3 3 1026g < 4.95
pH 5 2logfH1g
59.
fH1g < 1.58 3 1026 mole per liter
1025.8 5 fH1g
1025.8 5 10logfH1g
25.8 5 logfH1g
5.8 5 2logfH1g 60.
mole per liter fH1g < 6.3 3 1024
1023.2 5 fH1g
3.2 5 2logfH1g
61.
times the hydrogen ion
concentration of drinking water
1022.9
10285 105.1
fH1g 5 1028 for the drinking water
28.0 5 logfH1g
8.0 5 2logfH1g
fH1g 5 1022.9 for the apple juice
22.9 5 logfH1g
2.9 5 2logfH1g 62.
The hydrogen ion concentration is increased by
a factor of 10.
102pH ? 10 5 fH1g
102pH11 5 fH1g
102spH21d 5 fH1g
2spH 2 1d 5 logfH1g
pH 2 1 5 2logfH1g
63.
At 9:00 A.M. we have:
hours
From this you can conclude that the person died at 3:00 A.M.
t 5 210 ln 85.7 2 70
98.6 2 70< 6
t 5 210 ln T 2 70
98.6 2 70
Section 3.5 Exponential and Logarithmic Models 313
64. Interest:
Principal:
(a)
(b) In the early years of the mortgage, the majority of the
monthly payment goes toward interest. The principal
and interest are nearly equal when t < 26 years.
0
0
35
v
u
800
P 5 120,000, t 5 35, r 5 0.075, M 5 809.39
v 5 1M 2Pr
12211 1r
12212t
u 5 M 2 1M 2Pr
12211 1r
12212t
(c)
The interest is still the majority of the monthly payment
in the early years. Now the principal and interest are
nearly equal when t < 10.729 < 11 years.
u
v
0
0
20
800
P 5 120,000, t 5 20, r 5 0.075, M 5 966.71
65.
(a)
0
240
150,000
u 5 120,0003 0.075t
1 2 1 1
1 1 0.075y12212t
2 14(b) From the graph, when years. It
would take approximately 37.6 years to pay $240,000 in
interest. Yes, it is possible to pay twice as much in interest
charges as the size of the mortgage. It is especially likely
when the interest rates are higher.
t < 21u 5 $120,000
66.
(a) Linear model:
Exponential model: or
(b)
20
0
100
t3
t2
t1
t4
25
t4 5 1.5385s1.0296ds
t4 5 1.5385e0.02913s
t3 5 0.2729s 2 6.0143
t2 5 1.2259 1 0.0023s2
t1 5 40.757 1 0.556s 2 15.817 ln s
(c)
Note: Table values will vary slightly depending on the
model used for t4.
s 30 40 50 60 70 80 90
3.6 4.6 6.7 9.4 12.5 15.9 19.6
3.3 4.9 7.0 9.5 12.5 15.9 19.9
2.2 4.9 7.6 10.4 13.1 15.8 18.5
3.7 4.9 6.6 8.8 11.8 15.8 21.2t4
t3
t2
t1
(d) Model :
Model :
Model :
Model :
The quadratic model, best fits the data.t2,
|15.8 2 15.9| 1 |20 2 21.2| 5 2.6
S4 5 |3.4 2 3.7| 1 |5 2 4.9| 1 |7 2 6.6| 1 |9.3 2 8.9| 1 |12 2 11.9| 1t4
|15.8 2 15.8| 1 |20 2 18.5| 5 5.6
S3 5 |3.4 2 2.2| 1 |5 2 4.9| 1 |7 2 7.6| 1 |9.3 2 10.4| 1 |12 2 13.1| 1t3
|15.8 2 15.9| 1 |20 2 19.9| 5 1.1
S2 5 |3.4 2 3.3| 1 |5 2 4.9| 1 |7 2 7| 1 |9.3 2 9.5| 1 |12 2 12.5| 1t2
|15.8 2 15.9| 1 |20 2 19.6| 5 2.0
S1 5 |3.4 2 3.6| 1 |5 2 4.6| 1 |7 2 6.7| 1 |9.3 2 9.4| 1 |12 2 12.5| 1t1
314 Chapter 3 Exponential and Logarithmic Functions
69. False. The graph of is the graph of shifted
upward five units.
gsxdf sxd 70. True. Powers of e are always positive, so if a
Gaussian model will always be greater than 0, and if
a Gaussian model will always be less than 0.a < 0,
a > 0,
67. False. The domain can be the set of real numbers for a
logistic growth function.
68. False. A logistic growth function never has an x-intercept.
71. (a) Logarithmic
(b) Logistic
(c) Exponential (decay)
(d) Linear
(e) None of the above (appears to be a combination
of a linear and a quadratic)
(f) Exponential (growth)
72. Answers will vary.
73.
(a)
(b)
(c) Midpoint:
(d) m 55 2 2
0 2 s21d 53
15 3
121 1 0
2,
2 1 5
2 2 5 121
2,
7
22d 5 !s0 2 s21dd2 1 s5 2 2d2 5 !12 1 32 5 !10
y
x−1−2−3 1 2 3
−1
1
2
3
5 (0, 5)
(−1, 2)
s21, 2d, s0, 5d 74.
(a)
(b)
(c) Midpoint:
(d) m 523 2 1
4 2 s26d 524
105 2
2
5
126 1 4
2,
23 1 1
2 2 5 s21, 21d
5 !100 1 16 5 !116 5 2!29
d 5 !s26 2 4d2 1 s1 2 s23dd2
y
x
(−6, 1)
(4, −3)
−4−6 2 4 6
−2
−4
−6
2
4
6
s4, 23d, s26, 1d
75.
(a)
(b)
(c) Midpoint:
(d) m 522 2 3
14 2 35 2
5
11
13 1 14
2,
3 1 s22d2 2 5 117
2,
1
22 5 !112 1 s25d2 5 !146
d 5 !s14 2 3d2 1 s22 2 3d2
y
x−2 2 4 6 8 10 14
−2
−4
−6
−8
2
4
6
8
(3, 3)
(14, −2)
s3, 3d, s14, 22d 76.
(a)
(b)
(c) Midpoint:
(d) m 54 2 0
10 2 75
4
3
17 1 10
2,
0 1 4
2 2 5 117
2, 22
5 !9 1 16 5 !25 5 5
d 5 !s10 2 7d2 1 s4 2 0d2
y
x64 8 10
−6
−4
−2
2
4
6
2−2
(7, 0)
(10, 4)
s10, 4d, s7, 0d
Section 3.5 Exponential and Logarithmic Models 315
77.
(a)
(b)
(c) Midpoint:
(d) m 50 2 s21y4d
s3y4d 2 s1y2d 51y4
1y45 1
1s1y2d 1 s3y4d2
, s21y4d 1 0
2 2 5 15
8, 2
1
82
5!11
422
1 11
422
5!1
8
d 5!13
42
1
222
1 10 2 121
4222
y
x1
1
1
2
1
4( (, −
3
4( (, 0
1
2−
1
2
1
2
−
11
2, 2
1
42, 13
4, 02 78.
(a)
(b)
(c) Midpoint:
(d) m 5s21y3d 2 s1y6ds22y3d 2 s7y3d 5
21y2
235
1
6
1s22y3d 1 s7y3d2
, s21y3d 1 s1y6d
2 2 5 15
6, 2
1
122
5!s23d2 1 11
222
5 !9.25
d 5!122
32
7
322
1 121
32
1
622
y
x
2
3
1
3( (, −−
−1 1 2 3
−2
1
2
7
3
1
6( (,
17
3,
1
62, 122
3, 2
1
32
79.
Line
Slope:
y-intercept: s0, 10d
m 5 23
x
8
8
6
4
2
62−2−2
10 12
10
yy 5 10 2 3x 80.
Line
Slope:
y-intercept: s0, 21d
m 5 24
x
3
−2
−3
1−1−2−3
2
−1
32
yy 5 24x 2 1
81.
Parabola
Vertex: s0, 23d
y 5 22sx 2 0d2 2 3x
642
2
−2
−2−4−6
yy 5 22x2 2 3 82.
Parabola
Vertex:
x-intercepts: s252, 0d, s6, 0d
s74, 2
2898 d
5 2sx 274d2
2289
8
5 s2x 1 5dsx 2 6d x
−5
−30
−35
2 4−4 8
y y 5 2x2 2 7x 2 30
83.
Parabola
Vertex:
Focus:
Directrix: y 5 213
s0, 13d
s0, 0d
x2 543 y
3x2 5 4y
x
4321−1−2−3−4
7
6
5
4
3
2
1
y 3x2 2 4y 5 0 84.
Parabola
Vertex:
Focus:
Directrix: y 5 2
s0, 22d
s0, 0d
x2 5 28yx
−2
−4
−6
−10
64−4−6
2
−8
y 2x2 2 8y 5 0
316 Chapter 3 Exponential and Logarithmic Functions
85.
Vertical asymptote:
Horizontal asymptote:
x
21−1
−1
−2
−3
−2−3
3
1
y
y 5 0
x 51
3
y 54
1 2 3x86.
Vertical asymptote:
Slant asymptote:
x
−4−6−8 4
2
4
6
8
10
y
y 5 2x 1 2
x 5 22
y 5x2
2x 2 25 2x 1 2 1
4
2x 2 2
87.
Circle
Center:
Radius: 5
s0, 8d
x
8642
14
12
10
8
6
4
2
−2−4−6−8
y x2 1 sy 2 8d2 5 25 88.
Parabola
Vertex:
Focus:
Directrix: y 5 22.75
s4, 23.25d
P 5 214
s4, 23d x
−4
4
−6
−8
2−2 6 8
−2
−10
y
sx 2 4d2 5 2s y 1 3d
sx 2 4d2 5 2y 2 7 1 4
sx 2 4d2 1 s y 1 7d 5 4
89.
Horizontal asymptote:
x
8642−2−4−6 10
14
10
8
4
6
2
12
y
y 5 5
f sxd 5 2x21 1 5
x 25 23 21 0 1 3 5
5.02 5.06 5.3 5.5 6 9 21f sxd
90.
Horizontal asymptote:
x
−4
−6
−10
2
−8
−2
y
y 5 21
f sxd 5 222x21 2 1
x 0 1 2
2982
542
322223f sxd
2122
91.
Horizontal asymptote: y 5 24
x
4
5
4
3
2
1
32−1−2
−2
−3
−5
−3−4−5−6
yf sxd 5 3x 2 4
x 24 22 21 0 1 2
23.99 23.89 23.67 23 21 5f sxd
Review Exercises for Chapter 3 317
92.
Horizontal asymptote: y 5 4
x
2
1
5
1 2 3 4 5−1−2
−2
−3
−4
−5
−3−4−5
yf sxd 5 23x1 4
x 0 1 2
3 1 253233
89f sxd
2122
Review Exercises for Chapter 3
1.
f s2.4d 5 6.12.4 < 76.699
f sxd 5 6.1x 2.
f s!3 d 5 30!35 361.784
f sxd 5 30x 3.
f spd 5 220.5spd < 0.337
f sxd 5 220.5x
4.
f s1d 5 12781y55 4.181
f sxd 5 1278xy5 5.
< 1456.529
f s2!11 d 5 7s0.22!11d f sxd 5 7s0.2xd 6.
f s20.8d 5 214s520.8d 5 23.863
f sxd 5 214s5xd
7.
Intercept:
Horizontal asymptote: x-axis
Increasing on:
Matches graph (c).
s2`, `d
s0, 1d
f sxd 5 4x 8.
Intercept:
Horizontal asymptote:
Decreasing on:
Matches graph (d).
s2`, `d
y 5 0
s0,1d
fsxd 5 42x 9.
Intercept:
Horizontal asymptote: x-axis
Decreasing on:
Matches graph (a).
s2`, `d
s0, 21d
f sxd 5 24x
10.
Intercept:
Horizontal asymptote:
Increasing on:
Matches graph (b).
s2`, `d
y 5 1
s0, 2d
fsxd 5 4x1 1 11.
Since the graph
of g can be obtained by shifting
the graph of f one unit to the right.
gsxd 5 f sx 2 1d,
gsxd 5 5x21
f sxd 5 5x 12.
Because the
graph of g can be obtained by
shifting the graph of f three units
downward.
gsxd 5 f sxd 2 3,
f sxd 5 4x, gsxd 5 4x2 3
13.
Since the graph of g can be obtained
by reflecting the graph of f about the x-axis and shifting
two units to the left.2f
gsxd 5 2f sx 1 2d,
gsxd 5 2s12dx12
f sxd 5 s12dx
14.
Because the graph of g can be
obtained by reflecting the graph of f in the x-axis and
shifting the graph of f eight units upward.
gsxd 5 2f sxd 1 8,
f sxd 5 s23dx
, gsxd 5 8 2 s23dx
15.
Horizontal asymptote: y 5 4
x
2
2
8
4−2−4
yf sxd 5 42x1 4
x 0 1 2 3
8 5 4.0164.0634.25f sxd
21
93. Answers will vary.
318 Chapter 3 Exponential and Logarithmic Functions
16.
Horizontal asymptote:
x
321−1−2−3−4−5−6
1
−2
−3
−4
−5
−6
−7
−8
y
y 5 23
f sxd 5 24x2 3
x 22 21 0 1 2
23.063 23.25 24 27 219f sxd
17.
Horizontal asymptote:
x
−3
−6
−9
−12
−15
63−3−6 9
y
y 5 0
f sxd 5 22.65x11
x 0 1 2
218.6127.02322.652120.377f sxd
2122
18.
Horizontal asymptote:
x321−1−2−3
5
4
3
2
1
−1
y
y 5 0
f sxd 5 2.65x21
x 23 21 0 1 3
0.020 0.142 0.377 1 7.023f sxd
19.
Horizontal asymptote:
x
6
2
2
8
4−2−4
y
y 5 4
f sxd 5 5x221 4
x 0 1 2 3
5 94.24.044.008f sxd
21
20.
Horizontal asymptote:
x
6
4
2
−2
−4
−6
642−2 10
y
y 5 25
f sxd 5 2x262 5
x 0 5 6 7 8 9
24.984 24.5 24 23 21 3f sxd
21.
Horizontal asymptote:
x
2
2
6
8
4−2−4
y
y 5 3
f sxd 5 s12d2x
1 3 5 2x1 3
x 0 1 2
4 5 73.53.25f sxd
2122
Review Exercises for Chapter 3 319
22.
Horizontal asymptote: y 5 25
x
4
2
−2
−4
−6
2−4
yf sxd 5 s18dx12
2 5
x 23 22 21 0 2
3 24 24.875 24.984 25f sxd
23.
x 5 24
x 1 2 5 22
3x125 322
3x125
19 24.
x 5 22
x 2 2 5 24
s13dx22
5 s13d24
s13dx22
5 34
s13dx22
5 81 25.
x 5225
5x 5 22
5x 2 7 5 15
e5x275 e15 26.
x 5112
22x 5 211
8 2 2x 5 23
e822x5 e23
27. e8 < 2980.958 28. e5y8 < 1.868 29. e21.7 < 0.183 30. e0.2785 1.320
31.
−4 −3 −2 −1 1 2 3 4
2
3
4
5
6
7
x
y
hsxd 5 e2xy2
x 0 1 2
2.72 1.65 1 0.61 0.37hsxd
2122
32.
−4 −3 −1 1 2 3 4
−5
−4
−3
−2
3
x
y
hsxd 5 2 2 e2xy2
x 22 21 0 1 2
y 20.72 0.35 1 1.39 1.63
33.
−6 −5 −4 −3 −2 −1 1 2
1
2
6
7
x
y
f sxd 5 ex12
x 0 1
1 20.097.392.720.37f sxd
212223
34.
1 2 3 4 5
1
2
3
4
5
t
y
sstd 5 4e22yt, t > 0
t 1 2 3 4
y 0.07 0.54 1.47 2.05 2.43
12
320 Chapter 3 Exponential and Logarithmic Functions
35. A 5 350011 10.065
n 210n
or A 5 3500es0.065ds10d
n 1 2 4 12 365ContinuousCompounding
A $6704.39$6704.00$6692.64$6669.46$6635.43$6569.98
36. A 5 200011 10.05
n 230n
or A 5 2000es0.05ds30d
n 1 2 4 12 365 Continuous
A $8643.88 $8799.58 $8880.43 $8935.49 $8962.46 $8963.38
37.
(a) (b) (c) Fs5d < 0.811Fs2d < 0.487Fs 12d < 0.154
Fst) 5 1 2 e2ty3
38.
(a)
(b)
(c) According to the model, the car depreciates most
rapidly at the beginning. Yes, this is realistic.
Vs2d 5 14,000s34d2
5 $7875
0
0
10
15,000
Vstd 5 14,000s34d
t39. (a)
(b) The doubling time is ln 2
0.0875< 7.9 years.
A 5 50,000es0.0875ds35d < $1,069,047.14
40.
(a) For grams (b) For grams
(c)
Time (in years)
Mas
s of
24
1P
u (
in g
ram
s) 100
80
60
40
20
20 40 60 80 100
Q
t
t 5 10: Q 5 100s12d10y14.4 < 61.79t 5 0: Q 5 100s1
2d0y14.45 100
Q 5 100s12dty14.4
41.
log4 64 5 3
435 64 42.
log25 125 532
253y25 125 43.
ln 2.2255 . . . 5 0.8
e0.85 2.2255 . . .
44.
ln 1 5 0
e05 1 45.
5 log 1035 3
f s1000d 5 log 1000
f sxd 5 log x 46. log9 3 5 log9 91y2
512
Review Exercises for Chapter 3 321
47.
gs18d 5 log2s1
8d 5 log2 223
5 23
gsxd 5 log2 x 48.
f s14d 5 log4
14 5 21
f sxd 5 log4 x 49.
x 5 7
x 1 7 5 14
log4sx 1 7d 5 log4 14
50.
x 5 5
3x 5 15
3x 2 10 5 5
log8s3x 2 10d 5 log8 5 51.
x 5 25
x 1 9 5 4
lnsx 1 9d 5 ln 4 52.
x 5 6
2x 5 12
2x 2 1 5 11
lns2x 2 1d 5 lns11d
53.
Domain:
x-intercept:
Vertical asymptote: x 5 0
s1, 0d
s0, `d
x4321−1−2
4
3
2
1
−1
−2
ygsxd 5 log7 x ⇒ x 5 7y
x 1 7 49
0 1 221gsxd
17
54.
Domain:
x-intercept:
Vertical asymptote: x 5 0
s1, 0d
x 5 1
x 5 50
log5 x 5 0
s0, `d
−1 1 2 3 4 5
−3
−2
−1
1
2
3
x
ygsxd 5 log5 x ⇒ 5y5 x
x 1 5 25
22 21 0 1 2gsxd
15
125
55.
Domain:
x-intercept:
Vertical asymptote:
x 5 0
s3, 0d
x
5432−1
3
2
1
−1
−2
−3
ys0, `d
f sxd 5 log1x
32 ⇒ x
35 10y ⇒ x 5 3s10yd
x 3 30
0 12122f sxd
0.30.03
56.
Domain:
x-intercept:
Vertical asymptote: x 5 0
s0.000001, 0d
x 5 0.000001
x 5 1026
log x 5 26
6 1 log x 5 0
s0, `d
x8642−2
−2
10
8
10
6
4
2
yf sxd 5 6 1 log x
x 1 2 4 6 8 10
6 6.3 6.6 6.8 6.9 7f sxd
57.
Domain:
x-intercept:
Vertical asymptote: x 5 25
x 5 1042 5 5 9995.
x 1 5 5 104
Since 4 2 logsx 1 5d 5 0 ⇒ logsx 1 5d 5 4
s9995, 0d
s25, `d
x21−1−2−3−4−6
7
6
5
4
3
2
1
yf sxd 5 4 2 logsx 1 5dx 0 1
4 3.223.303.403.523.70f sxd
21222324
322 Chapter 3 Exponential and Logarithmic Functions
58.
Domain:
x-intercept:
Vertical asymptote: x 5 3
s3.1, 0d
x 5 3.1
x 2 3 5 1021
logsx 2 3d 5 21
logsx 2 3d 1 1 5 0
s3, `d
x98765
5
4
3
2
1
−2
−3
−4
−5
421−1
yf sxd 5 logsx 2 3d 1 1
x 4 5 6 7 8
1 1.3 1.5 1.6 1.7f sxd
59. ln 22.6 < 3.118 60. ln 0.98 < 20.020 61. ln e2125 212
62. ln e75 7 63. lns!7 1 5d < 2.034 64. ln1!3
8 2 < 21.530
65.
Domain:
x-intercept:
Vertical asymptote: x 5 0
se23, 0d
x 5 e23
ln x 5 23
ln x 1 3 5 0
s0, `d
1−1 2 3 4 5
1
2
3
4
5
6
x
yf sxd 5 ln x 1 3
x 1 2 3
3 1.612.314.103.69f sxd
14
12
66.
Domain:
x-intercept:
Vertical asymptote: x 5 3
s4, 0d
x 5 4
x 2 3 5 e0
lnsx 2 3d 5 0
s3, `d
2 4 6 8
−4
−2
2
4
x
yfsxd 5 lnsx 2 3d
x 3.5 4 4.5 5 5.5
y 20.69 0 0.41 0.69 0.92
67.
Domain:
x-intercepts:
Vertical asymptote: x 5 0
s±1, 0d
s2`, 0d < s0,`d
x4321−1−2−3−4
4
3
2
1
−3
−4
yhsxd 5 lnsx2d 5 2 ln|x|
x
y 0 2.772.201.3921.39
±4±3±2±1±0.5
68.
Domain:
x-intercept:
Vertical asymptote: x 5 0
s1, 0d
x 5 1
x 5 e0
ln x 5 0
14 ln x 5 0
s0, `d
1 2 3 4 5 6
−3
−2
−1
1
2
3
x
yfsxd 5
14 ln x
x 1 2 3
y 20.17 0 0.10 0.17 0.23 0.27
52
32
12
69.
< 53.4 inches
hs55d 5 116 logs55 1 40d 2 176
h 5 116 logsa 1 40d 2 176 70.
< 27.16 miles
s 5 25 213 lns10y12d
ln 371.
log4 9 5ln 9
ln 4< 1.585
log4 9 5log 9
log 4< 1.585
Review Exercises for Chapter 3 323
72.
log12 200 5ln 200
ln 12 < 2.132
log12 200 5log 200
log 12 < 2.132 73.
log1y2 5 5ln 5
lns1y2d< 22.322
log1y2 5 5log 5
logs1y2d< 22.322 74.
log3 0.28 5ln 0.28
ln 3< 21.159
log3 0.28 5log 0.28
log 3< 21.159
75.
< 1.255
5 log 2 1 2 log 3
log 18 5 logs2 ? 32d 76.
< 23.585
5 22 log2 22
2 log2 3 5 22 2log 3
log 2
log2 1
125 log2 1 2 log2 12 5 0 2 logs22
? 3d
77.
5 2 ln 2 1 ln 5 < 2.996
ln 20 5 lns22? 5d 78.
< 22.90
5 ln 3 2 4
ln 3e245 ln 3 1 ln e24 79.
5 1 1 2 log5 x
log5 5x25 log5 5 1 log5 x
2
80.
5 log 7 1 4 log x
log10 7x45 log 7 1 log x
4 81.
5 1 1 log3 2 21
3 log3 x
5 log3 3 1 log3 2 21
3 log3 x
5 log3s3 ? 2d 2 log3 x1y3
log3 6
3!x5 log3 6 2 log3
3!x 82.
51
2 log7 x 2 log7 4
5 log7 x1y2
2 log7 4
log7 !x
45 log7 !x 2 log7 4
83.
5 2 ln x 1 2 ln y 1 ln z
ln x2y2z 5 ln x21 ln y2
1 ln z 84.
5 ln 3 1 ln x 1 2 ln y
ln 3xy25 ln 3 1 ln x 1 ln y2
85.
5 lnsx 1 3d 2 ln x 2 ln y
5 lnsx 1 3d 2 fln x 1 ln yg
ln1x 1 3
xy 2 5 lnsx 1 3d 2 ln xy 86.
5 2 lnsy 2 1d 2 ln 16, y > 1
5 2 lnsy 2 1d 2 2 ln 4
ln1y 2 1
4 22
5 2 ln1y 2 1
4 2
87. log2 5 1 log2 x 5 log2 5x 88.
5 log6 y
z2
log6 y 2 2 log6 z 5 log6 y 2 log6 z2
89. ln x 21
4 ln y 5 ln x 2 ln 4!y 5 ln1 x
4!y2 90.
5 ln x3sx 1 1d2
3 ln x 1 2 lnsx 1 1d 5 ln x31 lnsx 1 1d2
91.
5 log8sy7 3!x 1 4d
1
3 log8sx 1 4d 1 7 log8 y 5 log8
3!x 1 4 1 log8 y7 92.
5 log 1
x2sx 1 6d5
5 log x22
sx 1 6d5
22 log x 2 5 logsx 1 6d 5 log x222 logsx 1 6d5
324 Chapter 3 Exponential and Logarithmic Functions
94.
5 ln sx 2 2d5
x3sx 1 2d
5 lnsx 2 2d52 ln x3sx 1 2d
5 lnsx 2 2d52 flnsx 1 2d 1 ln x3g
5 lnsx 2 2d 2 lnsx 1 2d 2 3 ln x 5 lnsx 2 2d52 lnsx 1 2d 2 ln x3
95.
(a) Domain:
(b)
Vertical asymptote: h 5 18,000
0 20,000
0
100
0 ≤ h < 18,000
t 5 50 log 18,000
18,000 2 h
(c) As the plane approaches its absolute ceiling, it climbs at
a slower rate, so the time required increases.
(d) minutes50 log 18,000
18,000 2 4000< 5.46
96. Using a calculator gives
s 5 84.66 1 s211 ln td.
97.
x 5 3
8x5 83
8x5 512 98.
x 5 23
6x5 623
6x5
1216 99.
x 5 ln 3
ex5 3
100.
x 5 ln 6 < 1.792
ln ex5 ln 6
ex5 6 101.
x 5 425 16
log4 x 5 2 102.
x 516
6log6 x 5 621
log6 x 5 21 103.
x 5 e4
ln x 5 4
104.
x 5 e23 < 0.0498
ln x 5 23 105.
x 5 ln 12 < 2.485
ln ex5 ln12
ex5 12 106.
x 5ln 25
3< 1.073
3x 5 ln 25
ln e3x5 ln 25
e3x5 25 107.
x 5 1 or x 5 3
0 5 sx 2 1dsx 2 3d
0 5 x22 4x 1 3
4x 5 x21 3
e4x5 ex2
13
108.
x 5sln 40d 2 2
3< 0.563
3x 1 2 5 ln 40
ln e3x125 ln 40
e3x125 40
14e3x125 560 109.
x < 4.459
5log 22
log 2 or
ln 22
ln 2
x 5 log2 22
2x5 22
2x1 13 5 35 110.
x 5ln 20
ln 6< 1.672
x 5 log6 20
log6 6x
5 log6 20
6x5 20
6x2 28 5 28
93.
5 ln !2x 2 1
sx 1 1d2
1
2 lns2x 2 1d 2 2 lnsx 1 1d 5 ln!2x 2 1 2 lnsx 1 1d2
Review Exercises for Chapter 3 325
113.
x 5 ln 2 < 0.693 x 5 ln 5 < 1.609
ln ex5 ln 2 ln e x
5 ln 5
ex5 2 or e x
5 5
sex2 2dsex
2 5d 5 0
e2x2 7ex
1 10 5 0 114.
x < 0.693 x < 1.386
x 5 ln 2 x 5 ln 4
ex5 2 or ex
5 4
sex2 2dsex
2 4d 5 0
e2x2 6ex
1 8 5 0
115.
Graph .
The x-intercepts are at
x < 0.392 and at x < 7.480.
y1 5 20.6x2 3x
−10
−10
10
1020.6x2 3x 5 0 116.
Graph .
The x-intercepts are at
x < 21.527.
x < 27.038 and at
y1 5 420.2x1 x
−3
−12 6
9420.2x1 x 5 0
117.
Graph and
The graphs intersect at
x < 2.447.
y2 5 12.
y1 5 25e20.3x
−6
16
−2
18
25e20.3x5 12 118.
Graph and
The graphs intersect at
x < 0.676.
y2 5 9.y1 5 4e1.2x
6−6
12
−2
4e1.2x5 9
119.
x 5e8.2
3< 1213.650
3x 5 e8.2
eln 3x5 e8.2
ln 3x 5 8.2 120.
x 5e7.2
5< 267.886
5x 5 e7.2
ln 5x 5 7.2 121.
x 51
4e7.5 < 452.011
4x 5 e7.5
eln 4x5 e7.5
ln 4x 515
2
2 ln 4x 5 15
122.
x 5e15y4
3< 14.174
3x 5 e15y4
ln 3x 515
4
4 ln 3x 5 15 123.
x 5 3e2 < 22.167
x
35 e2
elnsxy3d5 e2
ln x
35 2
ln x 2 ln 3 5 2 124.
x 5 e62 8 < 395.429
x 1 8 5 e6
lnsx 1 8d 5 6
1
2 lnsx 1 8d 5 3
ln!x 1 8 5 3
111.
x 5ln 17
ln 5< 1.760
x ln 5 5 ln 17
ln 5x5 ln 17
5x5 17
24s5xd 5 268 112.
x 5ln 95
ln 12< 1.833
x ln 12 5 ln 95
ln 12x5 ln 95
12x5 95
2s12xd 5 190
326 Chapter 3 Exponential and Logarithmic Functions
127.
Since is not in the domain of or of
it is an extraneous solution. The equation
has no solution.
log8sx 2 2d,log8sx 2 1dx 5 0
x 5 0
x25 0
x21 x 2 2 5 x 2 2
sx 2 1dsx 1 2d 5 x 2 2
x 2 1 5x 2 2
x 1 2
log8sx 2 1d 5 log81x 2 2
x 1 22log8sx 2 1d 5 log8sx 2 2d 2 log8sx 1 2d 128.
Quadratic Formula
Only is a valid solution.x 5 22 1 !6 < 0.449
x 5 22 ± !6,
0 5 x21 4x 2 2
x 1 2 5 x21 5x
x 1 2
x5 x 1 5
log61x 1 2
x 2 5 log6sx 1 5d
log6sx 1 2d 2 log6 x 5 log6sx 1 5d
129.
x 5 0.900
1 2110 5 x
1 2 x 5 1021
logs1 2 xd 5 21 130.
x 5 2104
2x 5 100 1 4
2x 2 4 5 102
logs2x 2 4d 5 2
131.
Graph
The graphs intersect at approximately
The solution of the equation is x < 1.643.
s1.643, 8d.
−9 9
−2
10
(1.64, 8)
y1 5 2 lnsx 1 3d 1 3x and y2 5 8.
2 lnsx 1 3d 1 3x 5 8 132.
Graph
The x-intercepts are at x 5 0, x < 0.416, and x < 13.627.
−4
−8 16
12
y1 5 6 logsx21 1d 2 x.
6 logsx21 1d 2 x 5 0
133.
Graph
The graphs do not intersect. The equation has no solution.
−6
11
−1
12
y1 5 4 lnsx 1 5d 2 x and y2 5 10.
4 lnsx 1 5d 2 x 5 10
125.
x 5 e42 1 < 53.598
x 1 1 5 e4
elnsx11d5 e4
lnsx 1 1d 5 4
1
2 lnsx 1 1d 5 2
ln!x 1 1 5 2 126.
x 5 5e4 < 272.991
x
55 e4
ln x
55 4
ln x 2 ln 5 5 4
Review Exercises for Chapter 3 327
134.
Let
The x-intercepts are at x < 23.990 and x < 1.477.
−4
−8 16
12
y1 5 x 2 2 logsx 1 4d.
x 2 2 logsx 1 4d 5 0 135.
t 5ln 3
0.0725< 15.2 years
ln 3 5 0.0725t
ln 3 5 ln e0.0725t
3 5 e0.0725t
3s7550d 5 7550e0.0725t
136.
d 5 10s218y93d < 220.8 miles
logsdd 5218
93
218 5 93 logsdd
283 5 93 logsdd 1 65
S 5 93 logsdd 1 65 137.
Exponential decay model
Matches graph (e).
y 5 3e22xy3
138.
Exponential growth model
Matches graph (b).
y 5 4e2xy3 139.
Logarithmic model
Vertical asymptote:
Graph includes
Matches graph (f).
s22, 0d
x 5 23
y 5 lnsx 1 3d
140.
Logarithmic model
Vertical asymptote:
Matches graph (d).
x 5 23
y 5 7 2 logsx 1 3d 141.
Gaussian model
Matches graph (a).
y 5 2e2sx14d2y3 142.
Logistics growth model
Matches graph (c).
y 56
1 1 2e22x
143.
Thus, y < 2e0.1014x.
ln 1.5 5 4b ⇒ b < 0.1014
1.5 5 e4b
3 5 2ebs4d
2 5 aebs0d ⇒ a 5 2
y 5 aebx 144.
y 51
2e0.4605x
b < 0.4605
ln 10
55 b
ln 10 5 5b
10 5 e5b
5 51
2ebs5d
1
25 aebs0d
⇒ a 51
2
y 5 aebx
328 Chapter 3 Exponential and Logarithmic Functions
145.
According to this model, the population of South
Carolina will reach 4.5 million during the year 2008.
t 5lns4500y3499d
0.0135< 18.6 years
ln14500
34992 5 0.0135t
4500
34995 e0.0135t
4500 5 3499e0.0135t
4.5 million 5 4500 thousand
P 5 3499e0.0135t 146.
When we have
After 5000 years, approximately 98.6% of the
radioactive uranium II will remain.
y 5 Ceflns1y2dy250,000gs5000d < 0.986C 5 98.6%C.
t 5 5000,
k 5lns1y2d250,000
ln 1
25 250,000k
ln 1
25 ln es250,000dk
1
2C 5 Ces250,000dk
y 5 Cekt
150.
(a) When
weeks t 5lns107y270d
20.12< 7.7
20.12t 5 ln 107
270
e20.12t5
107
270
5.4e20.12t5
107
50
1 1 5.4e20.12t5
157
50
50 5157
1 1 5.4e20.12t
N 5 50:
N 5157
1 1 5.4e20.12t
(b) When
weeks t 5lns82y405d
20.12< 13.3
20.12t 5 ln 82
405
e20.12t5
82
405
5.4e20.12t5
82
75
1 1 5.4e20.12t5
157
75
75 5157
1 1 5.4e20.12t
N 5 75:
147. (a)
(b)
< $11,486.98
A 5 10,000e0.138629
5 13.8629%
r < 0.138629
ln 2
55 r
ln 2 5 5r
2 5 e5r
20,000 5 10,000ers5d 148. and so
and:
The population one year ago:
5 1243 bats
Ns4d 5 2000e20.11889s4d
k 5lns7y10d
35 20.11889
3k 5 ln1 7
102
7
105 e3k
1400 5 2000e3k
N 5 2000ekt
N3 5 1400N0 5 2000 149.
(a) Graph
(b) The average test score is 71.
40 100
0
0.05
y1 5 0.0499e2sx271d2y128.
40 ≤ x ≤ 100
y 5 0.0499e2sx271d2y128,
Problem Solving for Chapter 3 329
151.
I 5 1023.5 wattycm2
1012.5 5I
10216
12.5 5 log1 I
102162
125 5 10 log1 I
102162
b 5 10 log1 I
102162 152.
(a)
(b)
(c)
I 5 109.1 < 1,258,925,412
log I 5 9.1
I 5 106.85 < 7,079,458
log I 5 6.85
I 5 108.4 < 251,188,643
log I 5 8.4
R 5 log I since I0 5 1.
153. True. By the inverse properties, logb b2x 5 2x. 154. False. ln x 1 ln y 5 lnsxyd Þ lnsx 1 yd
155. Since graphs (b) and (d) represent exponential decay, b and d are negative.
Since graph (a) and (c) represent exponential growth, a and c are positive.
Problem Solving for Chapter 3
1.
The curves and cross the line
From checking the graphs it appears that will cross
for 0 ≤ a ≤ 1.44.y 5 ax
y 5 x
y 5 x.y 5 1.2xy 5 0.5x
y4 5 x
y3 5 2.0x
y2 5 1.2x
y1 5 0.5x
y
x
−4 −3 −2 −1 1 2
7
6
5
4
3
2
3 4
y1
y3
y4
y2
y 5 ax 2.
The function that increases at the fastest rate for “large”
values of x is (Note: One of the intersection
points of and is approximately
and past this point This is not shown on the
graph above.)
ex> x3.
s4.536, 93dy 5 x3y 5 ex
y1 5 ex.
y5 5 |x| y4 5 !x
y3 5 x3
y2 5 x2
0
0 6
24
y1
y5
y3 y2
y4
y1 5 ex
3. The exponential function, increases at a faster rate
than the polynomial function y 5 xn.
y 5 ex, 4. It usually implies rapid growth.
5. (a)
(b)
5 f f sxdg2
5 saxd2
f s2xd 5 a2x
5 f sud ? f svd
5 au ? av
f su 1 vd 5 au1v 6.
5 1
54
4
5 1e2x 1 2 1 e22x
4 2 2 1e2x 2 2 1 e22x
4 2
f f sxdg2 2 fg sxdg2 5 1ex 1 e2x
2 22
2 1ex 2 e2x
2 22
7. (a)
−2
−6 6
6
y = ex y1
(b)y = ex
y2
−2
−6 6
6 (c)y = ex
y3
−2
−6 6
6
330 Chapter 3 Exponential and Logarithmic Functions
8.
As more terms are added, the polynomial approaches
ex 5 1 1x
1!1
x2
2!1
x3
3!1
x4
4!1
x5
5!1 . . .
ex.
−2
−6 6
6
y = exy4
y4 5 1 1x
1!1
x2
2!1
x3
3!1
x4
4!9.
Quadratic Formula
Choosing the positive quantity for we have
Thus, f 21sxd 5 ln1x 1!x2 1 4
2 2.y 5 ln1x 1 !x2 1 4
2 2.
ey
ey 5x ± !x2 1 4
2
e2y 2 xey 2 150
xey 5 e2y 2 1
x 5e2y 2 1
ey
x 5 ey 2 e2y
y 5 ex 2 e2x
f sxd 5 ex 2 e2x
10.
y 5 loga1x 1 1
x 2 12 5
ln1x 1 1
x 2 12ln a
5 f 21sxd
ay 5x 1 1
x 2 1
aysx 2 1d 5 x 1 1
xay 2 ay 5 x 1 1
xsay 2 1d 5 ay 1 1
x 5ay 1 1
ay 2 1
f sxd 5ax 1 1
ax 2 1, a > 0, a Þ 1 11. Answer (c).
The graph passes through and neither (a) nor (b) pass
through the origin. Also, the graph has y-axis symmetry and
a horizontal asymptote at y 5 6.
s0, 0d
y 5 6s1 2 e2x2y2d
12. (a) The steeper curve represents the investment earning compound interest,
because compound interest earns more than simple interest. With simple
interest there is no compounding so the growth is linear.
(b) Compound interest formula:
Simple interest formula:
(c) One should choose compound interest since the earnings would be higher.
A 5 Prt 1 P 5 500s0.07dt 1 500
A 5 500s1 10.07
1 ds1dt5 500s1.07dt
Time (in years)
Gro
wth
of
inves
tmen
t(i
n d
oll
ars)
A
t5 10 15 20 25
1000
2000
3000
4000
Compounded Interest
Simple Interest
30
13. and
t 5ln c1 2 ln c2
fs1yk2d 2 s1yk1dg lns1y2d
ln c1 2 ln c2 5 t1 1
k2
21
k12 ln11
22
ln1c1
c22 5 1 t
k2
2t
k12 ln11
22
c1
c2
5 11
22styk22tyk1d
c111
22tyk1
5 c211
22tyk2
y2 5 c211
22tyk2
y1 5 c111
22tyk1
14. through and
B 5 500afs1y2d logas2y5dgt 5 500 faloga s2y5dgty2 5 50012
52ty2
1
2 loga12
52 5 k
loga12
52 5 2k
2
55 a2k
200 5 500aks2d
B0 5 500
s2, 200ds0, 500dB 5 B0akt
y
x−4 −3 −2 −1 1 2
4
3
2
1
−4
3 4
Problem Solving for Chapter 3 331
15. (a)
(b)
(c)
(d) Both models appear to be “good fits” for the data, but
neither would be reliable to predict the population of
the United States in 2010. The exponential model
approaches infinity rapidly.
200,000
0 85
2,900,000
y2
y1
y < 400.88t 2 2 1464.6t 1 291,782
y < 252.606s1.0310dt16. Let and Then and
1 1 loga 1
b5
loga x
logayb x
1 1 loga 1
b5
m
n
loga 1
b5
m
n2 1
amyn21 51
b
amyn 5a
b
am 5 1a
b2n
x 5 saybdn.
x 5 amlogayb x 5 n.loga x 5 m
17.
x 5 1 or x 5 e2
ln x 5 0 or ln x 5 2
ln xsln x 2 2d 5 0
sln xd2 2 2 ln x 5 0
sln xd2 5 ln x2
18.
(a) (b) (c)y = ln xy3
−4
−3 9
4
y = ln x
y2
−4
−3 9
4
−4
−3 9
4
y = ln xy1
y3 5 sx 2 1d 212sx 2 1d2 1
13sx 2 1d3
y2 5 sx 2 1d 212sx 2 1d2
y1 5 x 2 1
y 5 ln x
19.
The pattern implies that
ln x 5 sx 2 1d 212sx 2 1d2 1
13sx 2 1d3 2
14sx 2 1d4 1 . . . .
−4
−3 9
4
y = ln x
y4
y4 5 sx 2 1d 212sx 2 1d2 1
13sx 2 1d3 2
14sx 2 1d4
20.
Slope:
y-intercept: s0, ln ad
m 5 ln b
ln y 5 sln bdx 1 ln a
ln y 5 ln a 1 x ln b
ln y 5 ln a 1 ln bx
ln y 5 lnsabxd
y 5 abx 21.
ys300d 5 80.4 2 11 ln 300 < 17.7 ft3/min
0
100 1500
30
y 5 80.4 2 11 ln x
Slope:
y-intercept: s0, ln ad
m 5 b
ln y 5 b ln x 1 ln a
ln y 5 ln a 1 b ln x
ln y 5 ln a 1 ln x b
ln y 5 lnsaxbd
y 5 axb
332 Chapter 3 Exponential and Logarithmic Functions
22. (a) cubic feet per minute
(b)
x < 382 cubic feet of air space per child.
x 5 e65.4y11
ln x 565.4
11
11 ln x 5 65.4
15 5 80.4 2 11 ln x
450
305 15 (c) Total air space required: cubic feet
Let floor space in square feet and feet.
If the ceiling height is 30 feet, the minimum number of
square feet of floor space required is 382 square feet.
x 5 382
11,460 5 xs30d
V 5 xh
h 5 30x 5
382s30d 5 11,460
23. (a)
(b) The data could best be modeled by a logarithmic
model.
(c) The shape of the curve looks much more logarithmic
than linear or exponential.
(d)
(e) The model is a good fit to the actual data.
0
0 9
9
y < 2.1518 1 2.7044 ln x
0 9
0
9 24. (a)
(b) The data could best be modeled by an exponential
model.
(c) The data scatter plot looks exponential.
(d)
(e) The model graph hits every point of the scatter plot.
0
0 9
36
y < 3.114s1.341dx
0 9
0
36
25. (a)
(b) The data could best be modeled by a linear model.
(c) The shape of the curve looks much more linear than
exponential or logarithmic.
(d)
(e) The model is a good fit to the actual data.
0
0 9
9
y < 20.7884x 1 8.2566
0 9
0
9 26. (a)
(b) The data could best be modeled by a logarithmic
model.
(c) The data scatter plot looks logarithmic.
(d)
(e) The model graph hits every point of the scatter plot.
0
0 9
10
y < 5.099 1 1.92 lnsxd
0 9
0
10
Practice Test for Chapter 3 333
Chapter 3 Practice Test
1. Solve for x: x3@55 8.
2. Solve for x: 3x215
181.
3. Graph fsxd 5 22x.
4. Graph gsxd 5 ex1 1.
5. If $5000 is invested at 9% interest, find the amount after three years if the interest is compounded
(a) monthly. (b) quarterly. (c) continuously.
6. Write the equation in logarithmic form: 7225
149.
7. Solve for x: x 2 4 5 log2 164.
8. Given evaluate logb 4!8y25.logb 2 5 0.3562 and logb 5 5 0.8271,
9. Write as a single logarithm.5 ln x 212 ln y 1 6 ln z
10. Using your calculator and the change of base formula, evaluate log9 28.
11. Use your calculator to solve for N: log10 N 5 0.6646
12. Graph y 5 log4 x.
13. Determine the domain of f sxd 5 log3sx22 9d.
14. Graph y 5 lnsx 2 2d.
15. True or false:ln x
ln y5 lnsx 2 yd
16. Solve for x: 5x5 41
17. Solve for x: x 2 x25 log5
125
18. Solve for x: log2 x 1 log2sx 2 3d 5 2
19. Solve for x:ex
1 e2x
35 4
20. Six thousand dollars is deposited into a fund at an annual interest rate of 13%. Find the time required
for the investment to double if the interest is compounded continuously.