59 EXPONENTIAL AND LOGARITHMIC FUNCTIONS - LESSON 6 LESSON 6 EXPONENTIAL AND LOGARITHMIC FU TIONS Exponential functions are of the form y = a x where a is a constant greater than zero and not equal to one and x is a variable. Both y = 2 x and y = e x are exponen- tial functions. e function, e x , is extensively used in calculus. You should memo- rize its approximate value when x = 1. (e 1 ≈ 2.718) You should also be able to quickly graph y = e x without the aid of a calculator. A simple graph of y = e x is shown below. Figure 1 1 –1 2 3 12 3 y = e x LOGARITHMIC FUNCTIONS e equation y = log a x is the same as a y = x. e inverse of the exponential function is y = a x . In this course we will restrict our study of logarithms to log base e which will be written as ln(x). e equation y = ln(x) is the inverse function of y = e x . Notice that the graph of ln(x) is a reflection of graph of e x around the line y = x. You should be able to quickly sketch from memory y = ln(x). It will also be important to remember the basic logarithm rules listed on the next page. LESSON 6 Exponential and Logarithmic Functions
Transcript
59eXponenTIAL AnD LoGARITHMIc FUncTIonS - LeSSon 6
LeSSon 6
eXponenTIAL AnD LoGARITHMIc FUnc-TIonS
Exponential functions are of the form y = ax where a is a constant greater than zero and not equal to one and x is a variable. Both y = 2x and y = ex are exponen-tial functions. The function, ex, is extensively used in calculus. You should memo-rize its approximate value when x = 1. (e1 ≈ 2.718) You should also be able to quickly graph y = ex without the aid of a calculator. A simple graph of y = ex is shown below.
Figure 1
1
–1
2
3
1 2 3
y = ex
LoGARITHMIc FUncTIonS The equation y = logax is the same as ay = x. The inverse of the exponential function is y = ax. In this course we will restrict our study of logarithms to log base e which will be written as ln(x). The equation y = ln(x) is the inverse function of y = ex. Notice that the graph of ln(x) is a reflection of graph of ex around the line y = x. You should be able to quickly sketch from memory y = ln(x). It will also be important to remember the basic logarithm rules listed on the next page.
LeSSon 6 Exponential and Logarithmic Functions
60 LeSSon 6 - eXponenTIAL AnD LoGARITHMIc FUncTIonS cALcULUS
Figure 2
y = ln(x)
I. ln(1) = 0
II. ln(e) =1
III. ln(ex) = x
IV. eln(x) = x
V. Product: ln(xy) = ln(x) + ln(y)
VI. Quotient: ln(x/y) = ln(x) – ln(y)
VII. Power: ln(xa) = a ln(x)
Remember that the natural log of a negative number is undefined. Some books specify ln(x) as ln |x|. We will use ln(x) for this book. Be careful to use only posi-tive, non-zero values for x when employing the natural log function.
61eXponenTIAL AnD LoGARITHMIc FUncTIonS - LeSSon 6cALcULUS
The natural log function can be used to free variable exponents from their exponential functions. Conversely, the exponential function can do the same for the natural log functions.
Example 1
Solve for x.
e2x = 1
Taking ln of both sides:
ln lnexx
x2 12 0
0
( ) = ( )==
checking e2(0) = e0 = 1
Example 2
Solve for x.
ln(x + 5) = 0
Use each side of the equation as the exponent for e.
eln(x + 5) = e0
x + 5 = 1; so x = –4
Sometimes the equations are complex and we need to use substitution to solve them. See example 3 on the next page.
62 LeSSon 6 - eXponenTIAL AnD LoGARITHMIc FUncTIonS cALcULUS
Example 3
Solve for x.
e2x – 4ex + 3 = 0
Substituting u = ex, u2 – 4u + 3 = 0.
Factoring, we get (u – 3)(u – 1) = 0.
Replacing u with ex, we get (ex – 3)(ex –1) = 0.
Solving each factor, we get: ex = 3; ex = 1.
Taking the ln of both sides:
ln ln
ln
ln lne
x
ex
x x( ) = ( )= ( )
( ) = ( )=
3
3
10
Example 4
Draw the graph of y = 2ex and its inverse.
y 2e
x 2e variables
x
y
=
=
=
= ( )
switch
x e
x e
y
y
12
12
ln ln
l
( )nn
ln
( )
( )
12
12
1
x y
x x
=
( ) =−f
y = 2ex
y = x
f −1 x( ) = ln( 1
2x)
CALCULUS LESSoN PrACTICE 6A 45
lesson practice 6AAnswer the question.
Draw the graph of y = 1. ex
3. Find the inverse function. Graph it.
Draw the graph of y = 2e2. x. Find the inverse function. Graph it.
Solve for x.3.
A. e2x + 1 = 1
B. 2e3x = e0
C. 0 = ln(2x +5)
LESSoN PrACTICE 6A
46 CALCULUS
D. ln(x) + ln(5) = 6
Solve for x. (4. Hint: Substitute and factor.)
A. e2x – 5ex = –6
B. 2e2x + 7ex = 4
CALCULUS LESSoN PrACTICE 6B 47
lesson practice 6BAnswer the question.
Draw the graph of y = e1. x + 1. Find the inverse function. Graph it.
Draw the graph of y = e2.
x2
. Find the inverse function. Graph it.
Solve for x.3.
A. ex + ln(3) = 2
B. ex + 1 = e2x – 2
LESSoN PrACTICE 6B
48 CALCULUS
C. ln(x2 + 3x + 5) = ln(1 – x)
D. ln( )x2
= 3
Solve for x.4.
A. 2ln2(x) + 3 = 7ln(x)
B. e2x = 2ex
CALCULUS LESSoN PrACTICE 6C 49
lesson practice 6CAnswer the question.
Draw the graph of y = 2x1. 2. Find the inverse function. Graph it.
Solve for x.
e2. 4x = e
ln(3x – 1) = 13.
LESSoN PrACTICE 6C
50 CALCULUS
e4. 2x – 7ex + 10 = 0
ln5. 2(x) = 2 ln(x)
e6. 2x – 3ex + 2 = 0
CALCULUS LESSoN PrACTICE 6D 51
lesson practice 6DSolve for x.
e1. 2x + 2 = 5
2e2. 2x + 5ex = 3
ln(x + 2) = 23.
LESSoN PrACTICE 6D
52 CALCULUS
ln(x + 1) + ln(4) = 34.
Solve for x: ln(2x – 4) = 2.5.
Draw the graph of y = e6. 3x. Find the inverse function. Graph it.
cAlcUlUS TeST 6 17
test 6Circle your answer.
Simplify 1. ln 92( ) =
A. ln(4.5)
B. 12
ln(4.5)
c. ln(3)
D. cannot be simplified
Solve for x: ln(x) – ln(4) = 2.2.
A. 14
e2
B. e4
c. e2
D. 4e2
ln3. ( )63
is the same as:
A. 2
B. ln(18)
c. ln(3)
D. ln(6) – ln(3)
Find the inverse function: 4. f (x) = ln(x – 2).
A. f –1(x) = ln(x + 2)
B. f –1(x) = ex + 2
c. f –1(x) = 2ex – 2
D. f –1(x) = 2ex
Simplify ln5. 2( ) + ln 10( ).
A. ln 12( )
B. 20
c. ln 2 5 ( )
D. cannot be simplified
TeST 6
18 cAlcUlUS
Solve for x: ln6. 2(x) – 5 ln(x) = –4
A. x = e, e4
B. x = ln(4), ln(5)
c. x = e5, e
D. x = ln(4), e
7. eX and ln(x) are inverse functions. The graph of y = ln(x) is the reflection of the graph
of y = ex around the:
A. x-axis B. y-axis c. origin D. line y = x
8. Solve for x: e2x = 3ex
A. x = ex – 3
B. x = ln(3)
c. x = ln(3) and x = 0
D. x = 0
9. Solve for x: ln(2) + ln(x) = 7
A. x = e2
7
B. x = 7e2
c. x = 2e7
D. x = e7
2
10. Solve for x: e3x – 1 = 1
A. 13
B. – 13
c. 3
D. e3
LeSSon pRAcTIce 5D - LeSSon pRAcTIce 6A
SoLUTIonScALcULUS 269
Lesson Practice 5D1. f x x( ) = + +( )2 4 2 π
amplitud
sin
ee
vertical shift
period π π
phase shift π
=
=
= =
= −
4
2
22
2.. f x x( ) = − −( )
=
1 3 12
3
π
amplitude
vertical shift
sin
==
=
= =
( ) = +
1
212
4
3 2 4
phase shift π
period π π
3. f x cos (xx −
=
=
=
π
amplitude
vertical shift
phase shift π
22
3
2
)
pperiod π π
π
amplitude
ver
= =
( ) = − +( )=
24 2
2 4 2
4
4. f x xcos
ttical shift
phase shift π
period π π
=
= −
= =
2
22
25. csc( ππ
π
3120 2
3
2 33
34
135 1
)( )
= = =
= ( ) =
csc( º )
tan tan º –
c
6.
7. oos
cos( )
2 0
02
x( ) =
= =
for x in [0, π]
when πθ θ ,, ,32
52
22
2 32
2 52
π π etc
π π π
π
x x x
x
= = =
=44
34
54
434
12
π π
π π
x x
x and
x
= =
=
8. tan( )==( ) = =
02
0 0
0, π
when π, 2π, 3π etc.
[ ]tan ,θ θ
π
x π
x is the o
12
0 12
0 2
0
x x
x
= =
= =
= nnly answer in 0, π[ ]2
Lesson Practice 6ALesson Practice 6A
1. y e
x e
x
y
=
=
3
3Switch variabless
( )
=
( ) = ( )( ) =( ) = ( )−
3
3
3
31
x e
x e
x y
x x
y
yln ln
ln
lnf
2.
reverse variables
y e
x e
x e
x
y
y
=
= ( )( ) =
2
2
2ln ln(( )( ) = +
( ) − ( ) =( ) = ( ) − ( )−
ln ln
ln ln
ln ln
x y
x y
x x
no
2
2
21f
tte: This problem is the same as
example 4 in thee instruction
manual but solved differently.
Booth solutions are correct.
ln ln lnx x( ) − ( ) =22( )
3.. A e
e
xx
x
B e
x
x
.
ln ln
.
2 1
2 1
3
1
1
2 1 02 1
12
2
+
+
=
( ) = ( )+ =
= −
= −
xx
x
e
e e
x
x
x
=
( ) = ( )( ) + =
= − ( )
= −( )
0
3 02
2 3 0
3 2
2
ln ln
ln
ln
ln33
0 2 5
1 2 54 2
42
2
0 2 5
c x
e exx
x
D
x
. ln
.
ln
= +( )
== +
− =
= − = −
+( )
lln ln
lnln
x
x
e e
x e
x e or
x
( ) + ( ) =( ) =
=
=
=
( )
5 6
5 6
5
51
5 6
6
6
55
5 6
5 6 0
2 3 0
6
2
2
e
A e e
y y
y y
Letx x4. . − = −
− + =
−( ) −( ) =
.
ln ln
ln
y e
y
y
e
e
x
x
x
x
=
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
ln ln
ln
.
y
y
e
e
x
B e e
x
x
x x
− =
=
=
( ) = ( )= ( )
+ =
3 0
3
3
3
3
2 7 42 Let y e
y y
y y
y
y
x=
+ − =
−( ) +( ) =− =
=
.
2 7 4 0
2 1 4 0
2 1 0
2 1
2
yy
e
e
x
x
x
x
x
=
=
( ) == ( ) − ( )= − ( )=
1212
12
1 2
0 2
ln ln
ln ln
ln
( )
−− ( )
+ =
= −
= −
( ) = ( )
ln
ln ln –
2
4 0
4
4
4
y
y
e
e
x
x
error:
no soluution
y = x
f−1 (x) = ln(3x)
f (x) = ex
3
Lesson Practice 6A
1. y e
x e
x
y
=
=
3
3Switch variabless
( )
=
( ) = ( )( ) =( ) = ( )−
3
3
3
31
x e
x e
x y
x x
y
yln ln
ln
lnf
2.
reverse variables
y e
x e
x e
x
y
y
=
= ( )( ) =
2
2
2ln ln(( )( ) = +
( ) − ( ) =( ) = ( ) − ( )−
ln ln
ln ln
ln ln
x y
x y
x x
no
2
2
21f
tte: This problem is the same as
example 4 in thee instruction
manual but solved differently.
Booth solutions are correct.
ln ln lnx x( ) − ( ) =22( )
3.. A e
e
xx
x
B e
x
x
.
ln ln
.
2 1
2 1
3
1
1
2 1 02 1
12
2
+
+
=
( ) = ( )+ =
= −
= −
xx
x
e
e e
x
x
x
=
( ) = ( )( ) + =
= − ( )
= −( )
0
3 02
2 3 0
3 2
2
ln ln
ln
ln
ln33
0 2 5
1 2 54 2
42
2
0 2 5
c x
e exx
x
D
x
. ln
.
ln
= +( )
== +
− =
= − = −
+( )
lln ln
lnln
x
x
e e
x e
x e or
x
( ) + ( ) =( ) =
=
=
=
( )
5 6
5 6
5
51
5 6
6
6
55
5 6
5 6 0
2 3 0
6
2
2
e
A e e
y y
y y
Letx x4. . − = −
− + =
−( ) −( ) =
.
ln ln
ln
y e
y
y
e
e
x
x
x
x
=
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
ln ln
ln
.
y
y
e
e
x
B e e
x
x
x x
− =
=
=
( ) = ( )= ( )
+ =
3 0
3
3
3
3
2 7 42 Let y e
y y
y y
y
y
x=
+ − =
−( ) +( ) =− =
=
.
2 7 4 0
2 1 4 0
2 1 0
2 1
2
yy
e
e
x
x
x
x
x
=
=
( ) == ( ) − ( )= − ( )=
1212
12
1 2
0 2
ln ln
ln ln
ln
( )
−− ( )
+ =
= −
= −
( ) = ( )
ln
ln ln –
2
4 0
4
4
4
y
y
e
e
x
x
error:
no soluution
y = x
y = 2ex
(x)= ln( )x – ln(2)f –1
Lesson Practice 6A
1. y e
x e
x
y
=
=
3
3Switch variabless
( )
=
( ) = ( )( ) =( ) = ( )−
3
3
3
31
x e
x e
x y
x x
y
yln ln
ln
lnf
2.
reverse variables
y e
x e
x e
x
y
y
=
= ( )( ) =
2
2
2ln ln(( )( ) = +
( ) − ( ) =( ) = ( ) − ( )−
ln ln
ln ln
ln ln
x y
x y
x x
no
2
2
21f
tte: This problem is the same as
example 4 in thee instruction
manual but solved differently.
Booth solutions are correct.
ln ln lnx x( ) − ( ) =22( )
3.. A e
e
xx
x
B e
x
x
.
ln ln
.
2 1
2 1
3
1
1
2 1 02 1
12
2
+
+
=
( ) = ( )+ =
= −
= −
xx
x
e
e e
x
x
x
=
( ) = ( )( ) + =
= − ( )
= −( )
0
3 02
2 3 0
3 2
2
ln ln
ln
ln
ln33
0 2 5
1 2 54 2
42
2
0 2 5
c x
e exx
x
D
x
. ln
.
ln
= +( )
== +
− =
= − = −
+( )
lln ln
lnln
x
x
e e
x e
x e or
x
( ) + ( ) =( ) =
=
=
=
( )
5 6
5 6
5
51
5 6
6
6
55
5 6
5 6 0
2 3 0
6
2
2
e
A e e
y y
y y
Letx x4. . − = −
− + =
−( ) −( ) =
.
ln ln
ln
y e
y
y
e
e
x
x
x
x
=
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
ln ln
ln
.
y
y
e
e
x
B e e
x
x
x x
− =
=
=
( ) = ( )= ( )
+ =
3 0
3
3
3
3
2 7 42 Let y e
y y
y y
y
y
x=
+ − =
−( ) +( ) =− =
=
.
2 7 4 0
2 1 4 0
2 1 0
2 1
2
yy
e
e
x
x
x
x
x
=
=
( ) == ( ) − ( )= − ( )=
1212
12
1 2
0 2
ln ln
ln ln
ln
( )
−− ( )
+ =
= −
= −
( ) = ( )
ln
ln ln –
2
4 0
4
4
4
y
y
e
e
x
x
error:
no soluution
LeSSon pRAcTIce 6A - LeSSon pRAcTIce 6B
SoLUTIonS cALcULUS270
Lesson Practice 6A
1. y e
x e
x
y
=
=
3
3Switch variabless
( )
=
( ) = ( )( ) =( ) = ( )−
3
3
3
31
x e
x e
x y
x x
y
yln ln
ln
lnf
2.
reverse variables
y e
x e
x e
x
y
y
=
= ( )( ) =
2
2
2ln ln(( )( ) = +
( ) − ( ) =( ) = ( ) − ( )−
ln ln
ln ln
ln ln
x y
x y
x x
no
2
2
21f
tte: This problem is the same as
example 4 in thee instruction
manual but solved differently.
Booth solutions are correct.
ln ln lnx x( ) − ( ) =22( )
3.. A e
e
xx
x
B e
x
x
.
ln ln
.
2 1
2 1
3
1
1
2 1 02 1
12
2
+
+
=
( ) = ( )+ =
= −
= −
xx
x
e
e e
x
x
x
=
( ) = ( )( ) + =
= − ( )
= −( )
0
3 02
2 3 0
3 2
2
ln ln
ln
ln
ln33
0 2 5
1 2 54 2
42
2
0 2 5
c x
e exx
x
D
x
. ln
.
ln
= +( )
== +
− =
= − = −
+( )
lln ln
lnln
x
x
e e
x e
x e or
x
( ) + ( ) =( ) =
=
=
=
( )
5 6
5 6
5
51
5 6
6
6
55
5 6
5 6 0
2 3 0
6
2
2
e
A e e
y y
y y
Letx x4. . − = −
− + =
−( ) −( ) =
.
ln ln
ln
y e
y
y
e
e
x
x
x
x
=
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
ln ln
ln
.
y
y
e
e
x
B e e
x
x
x x
− =
=
=
( ) = ( )= ( )
+ =
3 0
3
3
3
3
2 7 42 Let y e
y y
y y
y
y
x=
+ − =
−( ) +( ) =− =
=
.
2 7 4 0
2 1 4 0
2 1 0
2 1
2
yy
e
e
x
x
x
x
x
=
=
( ) == ( ) − ( )= − ( )=
1212
12
1 2
0 2
ln ln
ln ln
ln
( )
−− ( )
+ =
= −
= −
( ) = ( )
ln
ln ln –
2
4 0
4
4
4
y
y
e
e
x
x
error:
no soluution
Lesson Practice 6BLesson 6B1. y e
x e
x
x
y
=
= ( )
+
+
1
1
ln
reverse variables
(( ) = ( )( ) = +
( ) − =
( ) = ( ) −
+
−
ln
ln
ln
ln
e
x y
x y
x x
y
y 1
1
1
1
1f
2. ==
= ( )
( ) =
e
x e
x e
x
x
y
y
2
2
2ln ln
ln
reverse variables
( )(( ) =
( ) =( ) = ( )
=+ ( )
y
x y
x x
A e
e
x
x
22
2
2
1
3
ln
ln
.
ln
–
ln
f
3.++ ( )( ) = ( )+ ( ) = ( )
= ( ) − ( )
=
ln ln
ln ln
ln ln
ln
3 2
3 2
2 3
2
x
x
x (33
1 2 21
1 2 2
1 2 2
)B e e
e e
x xx
x x
x x
.
ln ln
+ −
+ −
=
( ) = ( )+ = −
− + = −223
3
3 5 1
3 5 1
4
2
2
2
− = −=
+ +( ) = −( )
+ + = −
+
xx
c x x x
x x x
x x
. ln ln
++ =
+( ) +( ) =+ =
= −
=
=
4 0
2 2 02 0
2
23
2 3
x xx
x
D x
e ex
l x
. ln
n
( )( )
22
2
2 3 7
2 3 7
2 7
3
3
2
2
2
=
=
( ) + = ( )
+ =
− +
e
x e
A x x
y y
y y
4. . ln ln
33 0
2 1 3 0
2 1 0
2 1
=
( )( ) =
= ( )
− =
=
=
y y
y x
y
Y
y
– –
ln Let
11212
3 0
3
3
12
ln
ln
ln
ln
x
e e
x e
y
y
x
e
x
( ) =
=
=
− =
=
( ) =
( )
xx
x x
e
x e
B e e
y y
y y
y y
( )=
=
=
=
− =
( ) −( ) =
3
3
2
2
2
2
2
2 0
2 0
. Lett
undefined
y e
y
e
e
x
x
x
x
=
=
=
( ) = ( )= ( )
ln ln
ln
0
0
0
0 ln ln
ln
y
y
e
e
x
x
x
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
y = x
= ln(x) – 1y = ex + 12.7
f−1 x( )
Lesson 6B1. y e
x e
x
x
y
=
= ( )
+
+
1
1
ln
reverse variables
(( ) = ( )( ) = +
( ) − =
( ) = ( ) −
+
−
ln
ln
ln
ln
e
x y
x y
x x
y
y 1
1
1
1
1f
2. ==
= ( )
( ) =
e
x e
x e
x
x
y
y
2
2
2ln ln
ln
reverse variables
( )(( ) =
( ) =( ) = ( )
=+ ( )
y
x y
x x
A e
e
x
x
22
2
2
1
3
ln
ln
.
ln
–
ln
f
3.++ ( )( ) = ( )+ ( ) = ( )
= ( ) − ( )
=
ln ln
ln ln
ln ln
ln
3 2
3 2
2 3
2
x
x
x (33
1 2 21
1 2 2
1 2 2
)B e e
e e
x xx
x x
x x
.
ln ln
+ −
+ −
=
( ) = ( )+ = −
− + = −223
3
3 5 1
3 5 1
4
2
2
2
− = −=
+ +( ) = −( )
+ + = −
+
xx
c x x x
x x x
x x
. ln ln
++ =
+( ) +( ) =+ =
= −
=
=
4 0
2 2 02 0
2
23
2 3
x xx
x
D x
e ex
l x
. ln
n
( )( )
22
2
2 3 7
2 3 7
2 7
3
3
2
2
2
=
=
( ) + = ( )
+ =
− +
e
x e
A x x
y y
y y
4. . ln ln
33 0
2 1 3 0
2 1 0
2 1
=
( )( ) =
= ( )
− =
=
=
y y
y x
y
Y
y
– –
ln Let
11212
3 0
3
3
12
ln
ln
ln
ln
x
e e
x e
y
y
x
e
x
( ) =
=
=
− =
=
( ) =
( )
xx
x x
e
x e
B e e
y y
y y
y y
( )=
=
=
=
− =
( ) −( ) =
3
3
2
2
2
2
2
2 0
2 0
. Lett
undefined
y e
y
e
e
x
x
x
x
=
=
=
( ) = ( )= ( )
ln ln
ln
0
0
0
0 ln ln
ln
y
y
e
e
x
x
x
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
f (x) = 2 ln(x)–1
x2y = e
y = x
LeSSon pRAcTIce 6B - LeSSon pRAcTIce 6c
SoLUTIonScALcULUS 271
Lesson 6B1. y e
x e
x
x
y
=
= ( )
+
+
1
1
ln
reverse variables
(( ) = ( )( ) = +
( ) − =
( ) = ( ) −
+
−
ln
ln
ln
ln
e
x y
x y
x x
y
y 1
1
1
1
1f
2. ==
= ( )
( ) =
e
x e
x e
x
x
y
y
2
2
2ln ln
ln
reverse variables
( )(( ) =
( ) =( ) = ( )
=+ ( )
y
x y
x x
A e
e
x
x
22
2
2
1
3
ln
ln
.
ln
–
ln
f
3.++ ( )( ) = ( )+ ( ) = ( )
= ( ) − ( )
=
ln ln
ln ln
ln ln
ln
3 2
3 2
2 3
2
x
x
x (33
1 2 21
1 2 2
1 2 2
)B e e
e e
x xx
x x
x x
.
ln ln
+ −
+ −
=
( ) = ( )+ = −
− + = −223
3
3 5 1
3 5 1
4
2
2
2
− = −=
+ +( ) = −( )
+ + = −
+
xx
c x x x
x x x
x x
. ln ln
++ =
+( ) +( ) =+ =
= −
=
=
4 0
2 2 02 0
2
23
2 3
x xx
x
D x
e ex
l x
. ln
n
( )( )
22
2
2 3 7
2 3 7
2 7
3
3
2
2
2
=
=
( ) + = ( )
+ =
− +
e
x e
A x x
y y
y y
4. . ln ln
33 0
2 1 3 0
2 1 0
2 1
=
( )( ) =
= ( )
− =
=
=
y y
y x
y
Y
y
– –
ln Let
11212
3 0
3
3
12
ln
ln
ln
ln
x
e e
x e
y
y
x
e
x
( ) =
=
=
− =
=
( ) =
( )
xx
x x
e
x e
B e e
y y
y y
y y
( )=
=
=
=
− =
( ) −( ) =
3
3
2
2
2
2
2
2 0
2 0
. Lett
undefined
y e
y
e
e
x
x
x
x
=
=
=
( ) = ( )= ( )
ln ln
ln
0
0
0
0 ln ln
ln
y
y
e
e
x
x
x
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
Lesson 6B1. y e
x e
x
x
y
=
= ( )
+
+
1
1
ln
reverse variables
(( ) = ( )( ) = +
( ) − =
( ) = ( ) −
+
−
ln
ln
ln
ln
e
x y
x y
x x
y
y 1
1
1
1
1f
2. ==
= ( )
( ) =
e
x e
x e
x
x
y
y
2
2
2ln ln
ln
reverse variables
( )(( ) =
( ) =( ) = ( )
=+ ( )
y
x y
x x
A e
e
x
x
22
2
2
1
3
ln
ln
.
ln
–
ln
f
3.++ ( )( ) = ( )+ ( ) = ( )
= ( ) − ( )
=
ln ln
ln ln
ln ln
ln
3 2
3 2
2 3
2
x
x
x (33
1 2 21
1 2 2
1 2 2
)B e e
e e
x xx
x x
x x
.
ln ln
+ −
+ −
=
( ) = ( )+ = −
− + = −223
3
3 5 1
3 5 1
4
2
2
2
− = −=
+ +( ) = −( )
+ + = −
+
xx
c x x x
x x x
x x
. ln ln
++ =
+( ) +( ) =+ =
= −
=
=
4 0
2 2 02 0
2
23
2 3
x xx
x
D x
e ex
l x
. ln
n
( )( )
22
2
2 3 7
2 3 7
2 7
3
3
2
2
2
=
=
( ) + = ( )
+ =
− +
e
x e
A x x
y y
y y
4. . ln ln
33 0
2 1 3 0
2 1 0
2 1
=
( )( ) =
= ( )
− =
=
=
y y
y x
y
Y
y
– –
ln Let
11212
3 0
3
3
12
ln
ln
ln
ln
x
e e
x e
y
y
x
e
x
( ) =
=
=
− =
=
( ) =
( )
xx
x x
e
x e
B e e
y y
y y
y y
( )=
=
=
=
− =
( ) −( ) =
3
3
2
2
2
2
2
2 0
2 0
. Lett
undefined
y e
y
e
e
x
x
x
x
=
=
=
( ) = ( )= ( )
ln ln
ln
0
0
0
0 ln ln
ln
y
y
e
e
x
x
x
− =
=
=
( ) = ( )= ( )
2 0
2
2
2
2
Lesson Practice 6CLesson Practice 6C1. y x
x y
=
=
2
2
2
2 reverse variabless( )
=
± =
( ) = ±
=
( ) = ( )=
−
x y
x y
x x
e e
e e
x
x
x
2
2
2
4
2
1
4
4
f
2.
ln ln
1114
3 1 1
3 13 1
1
3 1 1
x
x
e ex e
x e
x e
x
=
−( ) ==
− == +
= +
−( )3. ln
ln
33
7 10 0
7 10 0
5
2
2
4. e e l y e
y y
y y
x x x− + = =
− + =
−( )
et
−−( ) =− =
=
=
( ) = ( )= ( )
2 0
5 0
5
5
5
5
ln ln
ln
y
y
e
e
x
x
x ln ln
ln
ln
y
y
e
e
x
x
x
x
− =
=
=
( ) = ( )= ( )
( ) =
2 0
2
2
2
2
225. lln ln
ln
x y x
y y
y y
y y
y
x
( ) = ( )
=
− =
( ) −( ) ==
( )
let 2
2
2
2 0
2 0
0
==
=
==
− =
=
( ) =( )0
1
2 0
2
20
0
e e
x ex
y
y
x
e
xln
ln
lnxx
x x
x x
x x
e
x e
e e
e e
e e
( )=
=
− + =
−( ) −( ) =− = −
2
2
2 3 2 0
1 2 0
1 0
6.
22 0
1 21 2
0 2
=
= =
= ( ) = ( )= ( )
e ex x
x
x x
ln ln
,ln
y = 2x2
f (x) = –1
y = x
± x
2
LeSSon pRAcTIce 6c - LeSSon pRAcTIce 6D
SoLUTIonS cALcULUS272
Lesson Practice 6C1. y x
x y
=
=
2
2
2
2 reverse variabless( )
=
± =
( ) = ±
=
( ) = ( )=
−
x y
x y
x x
e e
e e
x
x
x
2
2
2
4
2
1
4
4
f
2.
ln ln
1114
3 1 1
3 13 1
1
3 1 1
x
x
e ex e
x e
x e
x
=
−( ) ==
− == +
= +
−( )3. ln
ln
33
7 10 0
7 10 0
5
2
2
4. e e l y e
y y
y y
x x x− + = =
− + =
−( )
et
−−( ) =− =
=
=
( ) = ( )= ( )
2 0
5 0
5
5
5
5
ln ln
ln
y
y
e
e
x
x
x ln ln
ln
ln
y
y
e
e
x
x
x
x
− =
=
=
( ) = ( )= ( )
( ) =
2 0
2
2
2
2
225. lln ln
ln
x y x
y y
y y
y y
y
x
( ) = ( )
=
− =
( ) −( ) ==
( )
let 2
2
2
2 0
2 0
0
==
=
==
− =
=
( ) =( )0
1
2 0
2
20
0
e e
x ex
y
y
x
e
xln
ln
lnxx
x x
x x
x x
e
x e
e e
e e
e e
( )=
=
− + =
−( ) −( ) =− = −
2
2
2 3 2 0
1 2 0
1 0
6.
22 0
1 21 2
0 2
=
= =
= ( ) = ( )= ( )
e ex x
x
x x
ln ln
,ln
Lesson Practice 6DLesson Practice 6D1. e
e
x
x
x
2 2
2 2
5
5
2 2
+
+
=
( ) = ( )+ =
ln ln
lln
ln
ln
5
2 5 2
5 22
2 5 3
2
2
( )= ( ) −
=( ) −
+ = =
x
x
e e let y ex x x2.
yy y
y y
y y
y
y
y
ex
2
2
5 3
2 5 3 0
2 1 3 0
2 1 0
2 1
12
+ =
+ − =
−( ) +( ) =− =
=
=
==
=
= − ( )= − ( )= − ( )
+
12
12
1 2
0 2
2
x
x
x
x
y
ln
ln ln
ln
ln
( )
33 0
3
3
3
2 22
=
= −
= −
= −( )
+( ) =+
y
e
x
x
e
x
x
ln
lnln
undefined
3.(( )
=
+ =
= −
+( ) + ( ) =+( )( ) =
e
x e
x e
x
x
2
2
2
2
2
1 4 3
4 1 3
4. ln ln
ln
lln
l
ln
4 4 3
4 4
4 4
44
4 4 3
3
3
3
x
e e
x e
x e
x e
x
+( ) ==
+ =
= −
= −
+( )
5. nn
– –
2 4 2
2 4
2 4
42
0 1
1
11
2
2
2
3
3
x
x e
x e
x e
x y
e
e
−( ) =− =
= +
= +
6.
33
3
3
3
e
Finding
y e
x ex y
y x
x
y
the inverse
=
=( ) =
=( )
ln
ln33
31f − ( ) = ( )
x xln
LeSSon pRAcTIce 6D - LeSSon pRAcTIce 7A
SoLUTIonScALcULUS 273
Lesson Practice 6D1. e
e
x
x
x
2 2
2 2
5
5
2 2
+
+
=
( ) = ( )+ =
ln ln
lln
ln
ln
5
2 5 2
5 22
2 5 3
2
2
( )= ( ) −
=( ) −
+ = =
x
x
e e let y ex x x2.
yy y
y y
y y
y
y
y
ex
2
2
5 3
2 5 3 0
2 1 3 0
2 1 0
2 1
12
+ =
+ − =
−( ) +( ) =− =
=
=
==
=
= − ( )= − ( )= − ( )
+
12
12
1 2
0 2
2
x
x
x
x
y
ln
ln ln
ln
ln
( )
33 0
3
3
3
2 22
=
= −
= −
= −( )
+( ) =+
y
e
x
x
e
x
x
ln
lnln
undefined
3.(( )
=
+ =
= −
+( ) + ( ) =+( )( ) =
e
x e
x e
x
x
2
2
2
2
2
1 4 3
4 1 3
4. ln ln
ln
lln
l
ln
4 4 3
4 4
4 4
44
4 4 3
3
3
3
x
e e
x e
x e
x e
x
+( ) ==
+ =
= −
= −
+( )
5. nn
– –
2 4 2
2 4
2 4
42
0 1
1
11
2
2
2
3
3
x
x e
x e
x e
x y
e
e
−( ) =− =
= +
= +
6.
33
3
3
3
e
Finding
y e
x ex y
y x
x
y
the inverse
=
=( ) =
=( )
ln
ln33
31f − ( ) = ( )
x xln–1 1
y = e 3x
(1, 20)
Lesson Practice 6D1. e
e
x
x
x
2 2
2 2
5
5
2 2
+
+
=
( ) = ( )+ =
ln ln
lln
ln
ln
5
2 5 2
5 22
2 5 3
2
2
( )= ( ) −
=( ) −
+ = =
x
x
e e let y ex x x2.
yy y
y y
y y
y
y
y
ex
2
2
5 3
2 5 3 0
2 1 3 0
2 1 0
2 1
12
+ =
+ − =
−( ) +( ) =− =
=
=
==
=
= − ( )= − ( )= − ( )
+
12
12
1 2
0 2
2
x
x
x
x
y
ln
ln ln
ln
ln
( )
33 0
3
3
3
2 22
=
= −
= −
= −( )
+( ) =+
y
e
x
x
e
x
x
ln
lnln
undefined
3.(( )
=
+ =
= −
+( ) + ( ) =+( )( ) =
e
x e
x e
x
x
2
2
2
2
2
1 4 3
4 1 3
4. ln ln
ln
lln
l
ln
4 4 3
4 4
4 4
44
4 4 3
3
3
3
x
e e
x e
x e
x e
x
+( ) ==
+ =
= −
= −
+( )
5. nn
– –
2 4 2
2 4
2 4
42
0 1
1
11
2
2
2
3
3
x
x e
x e
x e
x y
e
e
−( ) =− =
= +
= +
6.
33
3
3
3
e
Finding
y e
x ex y
y x
x
y
the inverse
=
=( ) =
=( )
ln
ln33
31f − ( ) = ( )
x xln
1
y = ln( )x
3
Lesson Practice 7ALesson Practice 5A1. lim
lim
x
x
x x
x
→
→
−( ) =
−
3
3
2
2
3 +
llim lim
lim
x x
x
x
x x
→ →
→
+ =
− + =
− −( ) = −
3 3
27 2 7 42
3
9 3 3 9
2. −− = −
− +
+= − +
+=
→
→−
4 1
3 3
2 2
0 0 30 2
320
2
2
23.
4.
lim
lim
t
z
t t
t
z 1
z
c
2
π
+
−= − +
−= −
( )( )→
2
2 14 2
12
3
4
5. lim tansinθ
θθ oos
π
π cos π4
π
θ
θ
( ) =
=
−
⋅
=
→
limtan
sin4
34
41
22
22
( )( ) ( )
−− = −
−
−
=
→
124
2
9
33 26. lim
x
x
x x
x
When 3, the functio
2
nn is undefined.
we get,
Factoring
xxlim→
+( )3
3 xx
x x
xx
x x
x
x
−( )−( ) =
+( )= =
+ +
→
→−
3
3
3 63
2
5
3
3
lim
lim
2
7. 663
1
5
3
1
x
x x
x
x x
x
x
+=
+( ) +( )+( ) = −
+ +
→−
→−
lim
lim
2 3
3
2
8. 661
2 311
x
x xx
limit
x
+( ) =
+( ) +( )+( ) =
→−lim
does not eexist
sin 2 cos9.
10.
lim
lim
θθ θ
→( ) + ( )( ) =
+ ( ) =0
0 2 1 2
θθθ θ
→( ) − ( ) +( ) =
( ) − + =
03
4 1 0 3 7
4 sec sin
TeST 4 - TeST 6
SoLUTIonScALcULUS 427
Test 41.
2.
f
f
f
x x x
B
x
( ) = −
−( ) = −( ) − −( )= +=
2
2
2
1 1 2 1
1 23
:
(( ) = −
−( ) = −( ) − −( )= +
( ) =
x x
c f x x x
x x
X x
2
2 22
2
4 2
2
2
2
2
:
3. f −−
+( ) = +( ) − +( )
= + + − −
(
2
2
2 2 2
2
2 2
x
D x h x h x h
x xh h x h
x
: f
f4. )) = − ( ) =
( )( ) = −( )
= −( )=
x x x x
c x x x
x x
2
2
2
2 3
2
3 2
3
;
:
g
g f g
xx x
A x x
y x
x y
2 6
3
3
3
−
( ) = −
= −
= − (
5. : f
switch variables))− = −
− + =
= −
( ) = −−
x y
x y
y x
x x
c A
3
3
3
31f
6. : is the sine ffunction.
B is a non-vertical line, so it is
a fuunction.
c is a parabola with the "c" shape
so itt fails the vertical line test.
D is the naturall log function.
of inverse7.
8.
B definition
c x
:
: f (( ) =
=
= ( )
=
= ±
2
2
2
2
2
2
2
2
2
x
y x
x y
x y
y x
switch variables
ff − ( ) = ±1
2x x
This is a graph of a parabola with
thee "c" shape. It is not a function,
but its graphh is in quadrants
I and IV.
9. A c dc d
d c c
. =
=
( ) =
π
π
π
Test 5Test 51. B x: cosThe period of y is 2 ,
so the
= ( ) π
pperiod of y
will be half as large.
= ( )cos
:
2x
c2. ssec
sin , ,
( )π4
21
2
0 0
= = =
( ) = =
hypadj
c y When3. y π 2ππ,
3π, 4π etc
2 2 2π
θ θ θ
θ θ
= = =
= =
0 2
0
π
ππ π
π π
of fr
2
02
θ
θ
=
= , ,
4. A definition eequency
quadrant II, tangen
5. A : tan tan º( )34
135π =
tt is negative
tan
of ampli
( )34
1π = −
6. c Definition ttude
π
is the vertical shift
π
7.
8.
9.
B
B
c
–
csc
2
74( )== =
( ) = =
csc º –
sin , ,
x π π,
π
315 2
2 0 02
3
10. D x when
2252
192
, , , 2π, π ... π 10
are 21 solution
π
There ss.
Test 51. B x: cosThe period of y is 2 ,
so the
= ( ) π
pperiod of y
will be half as large.
= ( )cos
:
2x
c2. ssec
sin , ,
( )π4
21
2
0 0
= = =
( ) = =
hypadj
c y When3. y π 2ππ,
3π, 4π etc
2 2 2π
θ θ θ
θ θ
= = =
= =
0 2
0
π
ππ π
π π
of fr
2
02
θ
θ
=
= , ,
4. A definition eequency
quadrant II, tangen
5. A : tan tan º( )34
135π =
tt is negative
tan
of ampli
( )34
1π = −
6. c Definition ttude
π
is the vertical shift
π
7.
8.
9.
B
B
c
–
csc
2
74( )== =
( ) = =
csc º –
sin , ,
x π π,
π
315 2
2 0 02
3
10. D x when
2252
192
, , , 2π, π ... π 10
are 21 solution
π
There ss.
Test 6Test 6
1.
2.
c
D
ln ln ln ln ln
l
92
12
9 9 9 312( )
= ( ) = = ( ) = ( )( )nn ln
ln
ln
ln
x
x
e ex e
x e
D
x
( ) − ( ) =
=
=
=
=
4 2
42
4
4
4 2
2
2
( )
(
( )
3. 663
6 3
2
2
)= ( ) ( )
= −( )= −( )
ln – ln
ln
ln
4. B y x
x y switch vaariables( )
=
= −
+ =
( ) = +
−( )
−
e e
e y
e y
x e
x y
x
x
X
ln 2
1
2
2
2f
5. cc
A x
ln ln ln
ln ln
: ln
2 10 2 10
20 2 5
2
( ) + ( ) = ( )( )= ( ) = ( )
(6. )) − ( ) = −
− = −
− + =
−( ) −( ) =− =
5 4
5 4
5 4 0
1 4 0
1 0
2
2
ln x
y y
y y
y y
y
y ==
( ) ===
− =
=
( ) =( )
1
1
4 0
4
41
ln lnln
x
e ex e
y
y
xx ee e
x e
D definition
xln( )=
=
4
4
of an inverse
from
7.
lesson 4
8. B e e
e e
x e
x x
x x
2
2
3
3
2 3
=
( ) = ( )= ( ) +
ln ln
ln ln xx
x x
x
D x
x
e
( )= ( ) += ( )
( ) + ( ) =( ) =
2 3
3
2 7
2 7
ln
ln
ln ln
lnl
9.
nn
ln ln
2 7
7
7
3 1
3 1
2
2
1
1
3
x
x
x
e
x e
x e
A e
e
x
( )
−
−
=
=
=
=
( ) = ( )10.
−− ==
=
1 03 1
13
x
x
TeST 6 - TeST 7
SoLUTIonS cALcULUS428
Test 6
1.
2.
c
D
ln ln ln ln ln
l
92
12
9 9 9 312( )
= ( ) = = ( ) = ( )( )nn ln
ln
ln
ln
x
x
e ex e
x e
D
x
( ) − ( ) =
=
=
=
=
4 2
42
4
4
4 2
2
2
( )
(
( )
3. 663
6 3
2
2
)= ( ) ( )
= −( )= −( )
ln – ln
ln
ln
4. B y x
x y switch vaariables( )
=
= −
+ =
( ) = +
−( )
−
e e
e y
e y
x e
x y
x
x
X
ln 2
1
2
2
2f
5. cc
A x
ln ln ln
ln ln
: ln
2 10 2 10
20 2 5
2
( ) + ( ) = ( )( )= ( ) = ( )
(6. )) − ( ) = −
− = −
− + =
−( ) −( ) =− =
5 4
5 4
5 4 0
1 4 0
1 0
2
2
ln x
y y
y y
y y
y
y ==
( ) ===
− =
=
( ) =( )
1
1
4 0
4
41
ln lnln
x
e ex e
y
y
xx ee e
x e
D definition
xln( )=
=
4
4
of an inverse
from
7.
lesson 4
8. B e e
e e
x e
x x
x x
2
2
3
3
2 3
=
( ) = ( )= ( ) +
ln ln
ln ln xx
x x
x
D x
x
e
( )= ( ) += ( )
( ) + ( ) =( ) =
2 3
3
2 7
2 7
ln
ln
ln ln
lnl
9.
nn
ln ln
2 7
7
7
3 1
3 1
2
2
1
1
3
x
x
x
e
x e
x e
A e
e
x
( )
−
−
=
=
=
=
( ) = ( )10.
−− ==
=
1 03 1
13
x
x
Test 7Test 51. D Limit from the left = 1 and limit
