Lial/Hungerford/Holcomb/Mullins:
Mathematics with Applications 12e
Finite Mathematics with Applications 12e
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 1
Chapter 4
Exponential and Logarithmic
Functions
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 2
Section 4.1
Exponential Functions
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 3
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 4
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 5
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 6
Example:
Consider the function g ( x)  2 x.
(a) Rewrite the rule of g so that no minus signs appear in it.
Solution:
By the definition of negative exponents,
x
1 1
g ( x)  2 x  x    .
2 2
(b) Graph g(x).
Solution:
Either use a graphing calculator or graph by hand in the usual way,
as shown in Figure 4.3.
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 7
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 8
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 9
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 10
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 11
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 12
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 13
Figure 4.9
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 14
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 15
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 16
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 17
Section 4.2
Applications of Exponential
Functions
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 18
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 19
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 20
Example: Finance When money is placed in a bank account that pays
compound interest, the amount in the account grows
exponentially. Suppose such an account grows from $1000 to
$1316 in 7 years.
(a) Find a growth function of the form f (t )  y0b that gives
the amount in the account at time t years.
t
Solution:
The values of the account at time t  0 and t  7 are given; that is,
f (0)  1000 and f (7)  1316. Solve the first of these equations for y0 :
f (0)  1000
y0b 0  1000
Rule of f
y0  1000
b0  1
So the rule of f has the form f (t )  1000bt .
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 21
Example: Finance When money is placed in a bank account that pays
compound interest, the amount in the account grows
exponentially. Suppose such an account grows from $1000 to
$1316 in 7 years.
(a) Find a growth function of the form f (t )  yob that gives
the amount in the account at time t years.
t
Solution:
Now solve the equation f (7)  1316.
f (7)  1316
1000b 7  1316
Rule of f
b 7  1.316
Divide both sides by 1000.
b  1.316 
1/7
 1.04
Take the seventh root of each side.
So the rule of the function is f (t )  1000 1.04  .
t
(b) How much is in the account after 12 years?
Solution:
f (12)  1000 1.04   $1601.03.
ALWAYS LEARNING
12
Copyright © 2019 Pearson Education, Inc.
Slide 22
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 23
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 24
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 25
Section 4.3
Logarithmic Functions
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 26
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 27
Example: To find log 10,000, ask yourself, “To what exponent must 10
be raised to produce 10,000?” Since 104  10000, we see that
log 10,000  4. Similarly,
log1  0
because
100  1;
log.01  2
because
1
1
10  2 
 .01;
10 100
1
log 10 
2
because
101/2  10 .
ALWAYS LEARNING
2
Copyright © 2019 Pearson Education, Inc.
Slide 28
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 29
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 30
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 31
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 32
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 33
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 34
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 35
Figure 4.20
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 36
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 37
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 38
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 39
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 40
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 41
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 42
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 43
Section 4.4
Logarithmic and Exponential
Equations
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 44
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 45
Example: Solve log x  log( x  3)  log( x  1).
use the quotient property of logarithms to write the right side
Solution: First,
as a single logarithm:
log x  log( x  3)  log( x  1)
 x3
log(x)  log 
.
 x 1 
Quotient property of logarithms
The fact on the preceding slide now shows that
x3
x
x 1
Cross multiply.
x( x  1)  x  3
x2  x  x  3
Distributive Property
Add  x  3 to each side and combine like terms
x2  2 x  3  0
 x  3 x  1  0
x3
or
Factor.
x  1
Set each factor to 0 and solve for x.
Since log x is not defined when x  1, the only possible solution is
x  3.
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 46
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 47
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 48
ALWAYS LEARNING
Copyright © 2019 Pearson Education, Inc.
Slide 49