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Slide 4.3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
OBJECTIVES
Logarithmic Function
Learn to define logarithmic functions.
Learn to evaluate logarithms.
Learn to find the domains of logarithmic functions.
Learn to graph logarithmic functions.
Learn to use logarithms to evaluate exponential equations.
SECTION 4.3
1
2
3
4
5
Slide 4.3- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
DEFINITION OF THELOGARITHMIC FUNCTION
For x > 0, a > 0, and a ≠ 1,
y loga x if and only if x ay .
The function f (x) = loga x, is called the logarithmic function with base a.
The logarithmic function is the inverse function of the exponential function.
Slide 4.3- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1Converting from Exponential to Logarithmic Form
a. 43 64
Write each exponential equation in logarithmic form.
b. 1
2
4
1
16c. a 2 7
Solution
a. 43 64 log4 64 3
b. 1
2
4
1
16 log1 2
1
164
c. a 2 7 loga 7 2
Slide 4.3- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2Converting from Logarithmic Form to Exponential Form
a. log3 243 5
Write each logarithmic equation in exponential form.
b. log2 5 x c. loga N x
Solution
a. log3 243 5 243 35
b. log2 5 x 5 2x
c. loga N x N ax
Slide 4.3- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3 Evaluating Logarithms
a. log5 25
Find the value of each of the following logarithms.
b. log2 16 c. log1 3 9
d. log7 7 e. log6 1 f. log4
1
2
Solution
a. log5 25 y 25 5y or 52 5y y 2
b. log2 16 y 16 2y or 24 2y y 4
Slide 4.3- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3 Evaluating Logarithms
Solution continued
d. log7 7 y 7 7y or 71 7y y 1
e. log6 1 y 1 6y or 60 6y y 0
f. log4
1
2y
1
24 y or 2 1 22 y y
1
2
c. log1 3 9 y 9 1
3
y
or 32 3 y y 2
Slide 4.3- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Using the Definition of Logarithm
a. log5 x 3
Solve each equation.
b. log3
1
27y
c. logz 1000 3 d. log2 x2 6x 10 1
Solutiona. log5 x 3
x 5 3
x 1
53 1
125The solution set is
1
125
.
Slide 4.3- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Using the Definition of Logarithm
Solution continued
b. log3
1
27y
1
273y
3 3 3y
3 y
c. logz 1000 3
1000 z3
103 z3
10 z
The solution set is 3 .
The solution set is 10 .
Slide 4.3- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Using the Definition of Logarithm
Solution continuedd. log2 x2 6x 10 1
x2 6x 10 21 2
x2 6x 8 0
x 2 x 4 0
x 2 0 or x 4 0
x 2 or x 4
The solution set is 2, 4 .
Slide 4.3- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
DOMAIN OF LOGARITHMIC FUNCTION
Domain of f –1(x) = loga x is (0, ∞)
Range of f –1(x) = loga x is (–∞, ∞)
Recall thatDomain of f (x) = ax is (–∞, ∞)Range of f (x) = ax is (0, ∞)
Since the logarithmic function is the inverse of the exponential function,
Thus, the logarithms of 0 and negative numbers are not defined.
Slide 4.3- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Finding the Domain
a. f x log3 2 x Find the domain of each function.
2 x 0
2 x
b. f x log3
x 2
x 1
Solution
a. Domain of a logarithmic function must be positive, that is,
The domain of f is (–∞, 2).
Slide 4.3- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Finding the Domain
Solution continuedx 2
x 1 0b. Domain must be positive, that is,
The domain of f is (–∞, –1) U (2, ∞).
Set numerator = 0 and denominator = 0.x – 2 = 0 x + 1 = 0x = 2 x = –1
Create a sign graph.
Slide 4.3- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
BASIC PROPERTIES OF LOGARITHMS
1. loga a = 1.
2. loga 1 = 0.
3. loga ax = x, for any real number x.4. aloga x x, for any x 0.
For any base a > 0, with a ≠ 1,
Slide 4.3- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Sketching a Graph
Sketch the graph of y = log3 x.Solution by plotting points.
Make a table of values.
x y = log3 x (x, y)
3–3 = 1/27 –3 (1/27, –3)
3–2 = 1/9 –2 (1/9, –2)
3–3 = 1/3 –1 (1/3, –1)
30 = 1 0 (1, 0)
31 = 3 1 (3, 1)
32 = 9 2 (9, 2)
Slide 4.3- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Sketching a Graph
Solution continued
Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log3 x.
Slide 4.3- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Sketching a Graph
Solution by using the inverse function
Graph y = f (x) = 3x
Reflect the graph of y = 3x in the line y = x to obtain the graph of y = f –1(x) = log3 x
Slide 4.3- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
Domain (0, ∞) Range (–∞, ∞)
Domain (–∞, ∞) Range (0, ∞)
x-intercept is 1 No y-intercept
y-intercept is 1 No x-intercept
x-axis (y = 0) is the horizontal asymptote
y-axis (x = 0) is the vertical asymptote
Slide 4.3- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
Is one-to-one, that is, logau = logav if and only if u = v
Is one-to-one , that is, au = av if and only if u = v
Increasing if a > 1 Decreasing if 0 < a < 1
Increasing if a > 1 Decreasing if 0 < a < 1
Slide 4.3- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
GRAPHS OF LOGARITHMIC FUNCTIONS
f (x) = loga x (0 < a < 1)f (x) = loga x (a > 1)
Slide 4.3- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using Transformations
Start with the graph of f (x) = log3 x and use transformations to sketch the graph of each function.
a. f x log3 x 2
c. f x log3 x
b. f x log3 x 1
d. f x log3 x
State the domain and range and the vertical asymptote for the graph of each function.
Slide 4.3- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using Transformations
Solution
Shift up 2Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0
a. f x log3 x 2
Slide 4.3- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using Transformations
Solution continued
Shift right 1Domain (1, ∞)Range (–∞, ∞)Vertical asymptote x = 1
b. f x log3 x 1
Slide 4.3- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using Transformations
Solution continued
Reflect graph of y = log3 x in the x-axis Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0
c. f x log3 x
Slide 4.3- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Using Transformations
Solution continued
Reflect graph of y = log3 x in the y-axis Domain (∞, 0)Range (–∞, ∞)Vertical asymptote x = 0
d. f x log3 x
Slide 4.3- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
COMMON LOGARITHMS
1. log 10 = 1.
2. log 1 = 0.
3. log 10x = x4. 10log x x
The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x = log10 x. Thus,
y = log x if and only if x = 10 y.
Applying the basic properties of logarithms
Slide 4.3- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Using Transformations to Sketch a Graph
Sketch the graph of y 2 log x 2 .Solution
Graph f (x) = log x and shift it right 2 units.
f x log x f x log x 2
Slide 4.3- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Using Transformations to Sketch a Graph
Solution continued
y log x 2 Reflect in x-axis
y 2 log x 2 Shift up 2
Slide 4.3- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
NATURAL LOGARITHMS
1. ln e = 1
2. ln 1 = 0
3. log ex = x4. eln x x
The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = loge x. Thus,
y = ln x if and only if x = e y.
Applying the basic properties of logarithms
Slide 4.3- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 9 Evaluating the Natural Logarithm
Evaluate each expression.
a. ln e4 b. ln1
e2.5 c. ln 3
Solution
a. ln e4 4
b. ln1
e2.5 ln e 2.5 2.5
(Use a calculator.)c. ln 3 1.0986123
Slide 4.3- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
NEWTON’S LAW OF COOLING
Newton’s Law of Cooling states that
where T is the temperature of the object at time t, Ts is the surrounding temperature, and T0 is the value of T at t = 0.
T Ts T0 Ts e kt ,
Slide 4.3- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 McDonald’s Hot Coffee
The local McDonald’s franchise has discovered that when coffee is poured from a pot whose contents are at 180ºF into a noninsulated pot in the store where the air temperature is 72ºF, after 1 minute the coffee has cooled to 165ºF. How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at 125ºF?
Slide 4.3- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 McDonald’s Hot Coffee
Use Newton’s Law of Cooling with T0 = 180 and Ts = 72 to obtain
Solution
We have T = 165 and t = 1.
T 72 180 72 e kt
T 72 108e kt
165 72 108e k
93
108e k
ln93
108
k
k 0.1495317
Slide 4.3- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 10 McDonald’s Hot Coffee
Substitute this value for k.
Solution continued
Solve for t when T = 125.
T 72 108e 0.1495317t
125 72 108e 0.1495317t
125 72
108e 0.1495317t
ln53
108
0.1495317t
t 1
0.1495317ln
53
108
t 4.76
The employee should wait about 5 minutes.
Slide 4.3- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
GROWTH AND DECAY MODEL
A is the quantity after time t.A0 is the initial (original) quantity (when t = 0).r is the growth or decay rate per period.t is the time elapsed from t = 0.
A A0ert
Slide 4.3- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 11 Chemical Toxins in a Lake
In a large lake, one-fifth of the water is replaced by clean water each year. A chemical spill deposits 60,000 cubic meters of soluble toxic waste into the lake.
a. How much of this toxin will be left in the lake after four years?
b. How long will it take for the toxic chemical in the lake to be reduced to 6000 cubic meters?
Slide 4.3- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 11 Chemical Toxins in a Lake
A A0ert
A 60,000e
1
5
t
One-fifth (1/5) of the water in the lake is replaced by clean water every year, the decay rate for the toxin is r = –1/5 and A0 = 60,000. So,
Solution
where A is the amount of toxin (in cubic meters) after t years.
Slide 4.3- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 11 Chemical Toxins in a Lake
A 60,000e
1
5
4
A 26,959.74 cm3
a. Substitute t = 4.
Solution continued
b. Substitute A = 6000 and solve for t.
6000 60,000e
1
5
t
1
10e
1
5
t
0.1 e
1
5
t
ln 0.1 1
5t
t 5 ln 0.1 t 11.51 years