2. Exponential and Logarithmic Functions
Chapter 4

3. 4.1 Inverse Functions
Determine whether a function is one-to-one, and if it is, find a formula for its inverse.
Simplify expressions of the type (f f 1)(x) and (f 1 f)(x).

4. Inverses
When we go from an output of a function back to its input or inputs, we get an inverse relation. When that relation is a function, we have an inverse function.
Interchanging the first and second coordinates of each ordered pair in a relation produces the inverse relation.
Example: Consider the relation g given by
g = {(2, 4), (3, 4), (8, 5).
Solution: The inverse of the relation is
{(4, 2), (4, 3), (5, 8)}.

5. Inverse Relation
If a relation is defined by an equation, interchanging the variables produces an equation of the inverse relation.
Example: Find an equation for the inverse of the relation:
y = x2  2x.
Solution: We interchange x and y and obtain an equation of the inverse: x = y2  2y.
Graphs of a relation and its inverse are always reflections of each other across the line y = x.

6. Inverses of Functions
If the inverse of a function f is also a function, it is named f 1 and read “f-inverse.” The negative 1 in f 1 is not an exponent. This does not mean the reciprocal of f. f 1(x) is not equal to

7. One-to-One Functions
A function f is one-to-one if different inputs have different outputs.
That is,
if a  b
then
f(a)  f(b).
A function f is one-to-one if when the outputs are the same, the inputs are the same.
That is,
if f(a) = f(b)
then
a = b.

8. Properties of One-to-One Functions and Inverses
If a function is one-to-one, then its inverse is a function.
The domain of a one-to-one function f is the range of the inverse f 1.
The range of a one-to-one function f is the domain of the inverse f 1.
A function that is increasing over its domain or is decreasing over its domain is a one-to-one function.

9. Horizontal-Line Test
If it is possible for a horizontal line to intersect the graph of a function more than once, then the function is not one-to-one and its inverse is not a function.

10. Horizontal-Line Test Graph f(x) = 3x + 4.
Example: From the graph at the left, determine whether the function is one-to-one and thus has an inverse that is a function.
Solution: No horizontal line intersects the graph more than once, so the function is one-to-one. It has an inverse that is a function.
f(x) = 3x + 4

11. Horizontal-Line Test Graph f(x) = x2  2.
Example: From the graph at the left, determine whether the function is one-to-one and thus has an inverse that is a function.
Solution: There are many horizontal lines that intersect the graph more than once. The inputs 1 and 1 have the same output, 1. Thus the function is not one-to-one. The inverse is not a function.

12. Obtaining a Formula for an Inverse
If a function f is one-to-one, a formula for its inverse can generally be found as follows:
Replace f(x) with y.
Interchange x and y.
Solve for y.
Replace y with f 1(x).

13. Example
Determine whether the function f(x) = 3x  2 is one-to-one, and if it is, find a formula for f 1(x).
Solution: The graph is that of a line and passes the horizontal-line test. Thus it is one-to-one and its inverse is a function.
1. Replace f(x) with y: y = 3x  2
2. Interchange x and y: x = 3y  2
3. Solve for y:
x + 2 = 3y
4. Replace y with f 1(x):
f 1(x) =

14. Example Graph f(x) = 3x  2 and f 1(x) =
using the same set of axes. Then compare the two graphs.
Solution: The solutions of the inverse function can be found from those of the original function by interchanging the first and second coordinates of each ordered pair. The graph f 1 is a reflection of the graph f across the line y = x.

15. Solution 7 3 4 2 2 5 1
f(x) = 3x  2 x 2 4 1 2 1 5
f 1(x) = x

16. Inverse Functions and Composition
If a function f is one-to-one, then f 1 is the unique function such that each of the following holds:
for each x in the domain of f, and
for each x in the domain of f 1.

17. Example
Given that f(x) = 7x  2, use composition of functions to show that f 1(x) = (x + 2)/7.
Solution:

18. Restricting a Domain
When the inverse of a function is not a function, the domain of the function can be restricted to allow the inverse to be a function. In such cases, it is convenient to consider “part” of the function by restricting the domain of f(x). If the domain is restricted, then its inverse is a function.

19. 4.2 Exponential Functions and Graphs
Graph exponential equations and functions.
Solve applied problems involving exponential functions and their graphs.

20. Exponential Function
The function f(x) = ax, where x is a real number, a > 0 and a  1, is called the exponential function, base a.
The base needs to be positive in order to avoid the complex numbers that would occur by taking even roots of negative numbers.
Examples:

21. Graphing Exponential Functions
To graph an exponential function, follow the steps listed:
1. Compute some function values and list the results in a table.
2. Plot the points and connect them with a smooth curve. Be sure to plot enough points to determine how steeply the curve rises.

22. Example Graph the exponential function y = f(x) = 3x. (3, 1/27) 1/27
(2, 1/9)
1/9 2
(1, 1/3)
1/3 1
(3, 27) 27 3 9 1
y = f(x) = 3x
(2, 9) 2
(1, 3)
(0, 1)
(x, y) x

23. Example Graph the exponential function . (3, 1/27) 1/27 3 (2, 1/9) 1/9
(1, 1/3)
1/3 1
(3, 27) 27 3 9
(2, 9) 2
(1, 3) 1
(0, 1)
(x, y) x

24. Example
Graph y = 3x + 2.
The graph is the graph of y = 3x shifted to left 2 units.
243 3 81 2 27 1 9
1/3 y 1 2 3 x

25. Example
Graph y = 4  3x
The graph is a reflection of the graph of y = 3x across the x-axis, followed by a reflection across the y-axis and then a shift up of units.
3.96 3
3.88 2
3.67 1 5
23 y 1 2 3 x

26. The Number e
e  …
Find each value of ex, to four decimal places, using the ex key on a calculator.
a) e4 b) e0.25
c) e2 d) e1
a) b) c) d)

27. Graphs of Exponential Functions, Base e
Graph f(x) = ex.
7.389 2
2.718 1
0.368
0.135
f(x) 1 2 x

28. Example Graph f(x) = 2  e3x. 1.99 2 1.95 1 18.09 401.43 f(x) 1 21 2 x

29. Example Graph f(x) = ex+2. 0.135 4 0.368 3 20.086 1 7.389 2.718 f(x)1 2 x

30. 4.3 Logarithmic Functions and Graphs
Graph logarithmic functions.
Convert between exponential and logarithmic equations.
Find common and natural logarithms with and without using a calculator.
Change logarithm bases

31. Logarithmic Functions
These functions are inverses of exponential functions.
Graph: x = 3y.
1. Choose values for y.
2. Compute values for x.
3. Plot the points and connect them with a smooth curve.
* Note that the curve does not touch or cross the y-axis.

32. Logarithmic Functions continued
Graph: x = 3y
(1/27, 3) 3
1/27
(1/9, 2) 2
1/9
(1/3, 1) 1
1/3 2 1 y
(9, 2) 9
(3, 1) 3
(1, 0)
(x, y)
x = 3y

33. Logarithmic Function, Base a
We define y = loga x as that number y such that x = ay, where x > 0 and a is a positive constant other than 1.
We read loga x as “the logarithm, base a, of x.”

34. Finding Certain Logarithms
Find each of the following logarithms.
a) log b) log
c) log d) log
a) The exponent to which we raise 2 to obtain 16 is 4; thus log2 16 = 4.
b) The exponent to which we raise 10 to obtain 1000 is 3; thus log = 3.
c) The exponent we raise 16 to get 4 is ½, so log16 4 = ½.
d) We have The exponent to which we raise 10
to get is 3, so log = 3.

35. Logarithms loga 1 = 0 and loga a = 1, for any logarithmic base a.
Convert each of the following to a logarithmic equation.
a) 25 = 5x
log5 25 = x
b) ew = 30
loge 30 = w
The exponent is the logarithm.
The base remains the same.

36. Example Convert each of the following to an exponential equation.
a) log7 343 = log7 343 = = 343
b) logb R = 12
logb R = 12 b12 = R
The logarithm is the exponent.
The base remains the same.

37. Example
Find each of the following common logarithms on a calculator. Round to four decimal places.
a) log 723,456
b) log
c) log (4)
Does not exist
ERR: nonreal ans
log (4)
4.6108 
log
5.8594
log 723,456
Rounded
Readout
Function Value

38. Natural Logarithms
Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus,
ln x means loge x.
ln 1 = 0 and ln e = 1, for the logarithmic base e.

39. Example
Find each of the following natural logarithms on a calculator. Round to four decimal places.
a) ln 723,456
b) ln
c) ln (4)
Does not exist
ERR: nonreal ans
ln (4)   ln
ln 723,456
Rounded
Readout
Function Value

40. Changing Logarithmic Bases
The Change-of-Base Formula
For any logarithmic bases a and b, and any positive number M,

41. Example Find log6 8 using common logarithms.
Solution: First, we let a = 10, b = 6, and M = Then we substitute into the change-of-base formula:

42. Example We can also use base e for a conversion.
Find log6 8 using natural logarithms.
Solution: Substituting e for a, 6 for b and 8 for M, we have

43. Graphs of Logarithmic Functions
Graph: y = f(x) = log6 x.
Select y.
Compute x. 2
1/36 1
1/6 3
216 2 36 1 6 y
x,or 6y

44. Example
Graph each of the following. Describe how each graph can be obtained from the graph of y = ln x. Give the domain and the vertical asymptote of each function.
a) f(x) = ln (x  2)
b) f(x) = 2  ln x
c) f(x) = |ln (x + 1)|

45. Graph f(x) = ln (x  2)
The graph is a shift 2 units right. The domain is the set of all real numbers greater than 2. The line x = 2 is the vertical asymptote.
1.099 5
0.693 4 3
0.693
2.5
1.386
2.25
f(x) x

46. Graph f(x) = 2  ln x
The graph is a vertical shrinking, followed by a reflection across the x-axis, and then a translation up 2 units. The domain is the set of all positive real numbers. The y-axis is the vertical asymptote.
1.598 5
1.725 3 2 1
2.173
0.5
2.576
0.1
f(x) x

47. Graph f(x) = |ln (x + 1)|
The graph is a translation 1 unit to the left. Then the absolute value has the effect of reflecting negative outputs across the x-axis. The domain is the set of all real numbers greater than 1. The line x = 1 is the vertical asymptote.
1.946 6
1.386 3
0.693 1
0.5
f(x) x

48. Application: Walking Speed
In a study by psychologists Bornstein and Bornstein, it was found that the average walking speed w, in feet per second, of a person living in a city of population P, in thousands, is given by the function
w(P) = 0.37 ln P

49. Application: Walking Speed continued
The population of Philadelphia, Pennsylvania, is 1,517,600. Find the average walking speed of people living in Philadelphia.
Since 1,517,600 = thousand, we substitute for P, since P is in thousands:
w(1517.6) = 0.37 ln
 2.8 ft/sec.
The average walking speed of people living in Philadelphia is about 2.8 ft/sec.

50. 4.4 Properties of Logarithmic Functions
Convert from logarithms of products, powers, and quotients to expressions in terms of individual logarithms, and conversely.
Simplify expressions of the type loga ax and

51. Logarithms of Products
The Product Rule
For any positive numbers M and N and any logarithmic base a,
loga MN = loga M + loga N.
(The logarithm of a product is the sum of the logarithms of the factors.)

52. Example
Express as a single logarithm:
Solution:

53. Logarithms of Powers The Power Rule
For any positive number M, any logarithmic base a, and any real number p,
loga Mp = p loga M.
(The logarithm of a power of M is the exponent times the logarithm of M.)

54. Examples
Express as a product.
Express as a product.

55. Logarithms of Quotients
The Quotient Rule
For any positive numbers M and N, and any logarithmic base a, .
(The logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.)

56. Examples Express as a difference of logarithms.
Express as a single logarithm.

57. Applying the Properties
Express in terms of sums and differences of logarithms.

58. Example
Express as a single logarithm.

59. Final Properties The Logarithm of a Base to a Power
For any base a and any real number x,
loga ax = x.
(The logarithm, base a, of a to a power is the power.)
A Base to a Logarithmic Power
For any base a and any positive real number x,
(The number a raised to the power loga x is x.)

60. Examples Simplify. a) loga a6 b) ln e8 Solution: a) loga a6 = 6

61. 4.5 Solving Exponential and Logarithmic Equations
Solve exponential and logarithmic equations.

62. Solving Exponential Equations
Equations with variables in the exponents, such as
3x = 40 and 53x = 25, are called exponential equations.
Base-Exponent Property
For any a > 0, a  1,
ax = ay  x = y.

63. Example Solve: . Write each side with the same base.
Since the bases are the same number, 5, we can use the base-exponent property and set the exponents equal:
Check: 52x  3 = 125
52(3)  3 ? 125
53 ? 125
125 = 125 True
The solution is 3.

64. Graphical Solution
We will use the Intersect method. We graph y1 = and y2 = 53

65. Another Property Property of Logarithmic Equality
For any M > 0, N > 0, a > 0, and a  1,
loga M = loga N  M = N.
Solve: 2x = 50
This is an exact answer. We cannot simplify further, but we can approximate using a calculator.
x 
We can check by finding  50.

66. Graphical Solution
We will use the Intersect method. We graph y1 = 2x and y2 = 50

67. Example
Solve: e0.25w = 12
The solution is about 9.94.

68. Solving Logarithmic Equations
Equations containing variables in logarithmic expressions, such as log2 x = 16 and log x + log (x + 4) = 1, are called logarithmic equations.
To solve logarithmic equations algebraically, we first try to obtain a single logarithmic expression on one side and then write an equivalent exponential equation.

69. Example
Solve: log4 x = 3
Check:
log4 x = 3
The solution is

70. Graphical Solution
We use the change of base formula and graph the equations
y1 =
y2 = 3

71. Example Solve: Check: For x = 3: For x = 3:
Negative numbers do not have real-number logarithms. The solution is 3.

72. Example
Solve:
The value 6 checks and is the solution.

73. 4.6 Applications and Models: Growth and Decay
Solve applied problems involving exponential growth and decay and compound interest.
Find models involving exponential and logarithmic functions.

74. Population Growth
The function P(t) = P0ekt, k > 0 can model many kinds of population growths.
In this function:
P0 = population at time 0,
P = population after time,
t = amount of time,
k = exponential growth rate.
The growth rate unit must be the same as the time unit.

75. Example
Population Growth of the United States. In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau)
Find the exponential growth function.
What will the population be in 2020?
After how long will the population be double what it was in 1990?

76. Solution
At t = 0 (1990), the population was about 249 million. We substitute 249 for P0 and 0.08 for k to obtain the exponential growth function.
P(t) = 249e0.08t
In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t:
P(3) = 249e0.08(3) = 249e0.24  317.
The population will be approximately 317 million in 2020.

77. Solution continued We are looking for the doubling time T.
ln 2 = ln e0.08T (Taking the natural logarithm on both sides)
ln 2 = 0.08T (ln ex = x)
= T
8.7 T
The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.

78. Interest Compound Continuously
The function P(t) = P0ekt can be used to calculate interest that is compounded continuously.
In this function:
P0 = amount of money invested,
P = balance of the account,
t = years,
k = interest rate compounded continuously.

79. Example
Suppose that $2000 is deposited into an IRA at an interest rate k, and grows to $ after 12 years.
What is the interest rate?
Find the exponential growth function.
What will the balance be after the first 5 years?
How long did it take the $2000 to double?

80. Solution
At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is
P(t) = 2000ekt. We know that P(12) = $ We then substitute and solve for k:
$ = 2000e12k
The interest rate is about 9%.

81. Solution continued The exponential growth function is
P(t) = 2000e0.09t.
The balance after 5 years is
P(5) = 2000e0.09(5)
= 2000e0.45
 $

82. Solution continued
To find the doubling time T, set P(T) = 2  P0 = $4000 and solve for T.
4000 = 2000e0.09T
2 = e0.09T
ln 2 = ln e0.09T
ln 2 = 0.09T
= T
7.7  T
The original investment of $2000 doubled in about 7.7 years.

83. Growth Rate and Doubling Time
The growth rate k and the doubling time T are related by
kT = ln 2 or
* The relationship between k and T does not depend on P0.

84. Example
A certain town’s population is doubling every years. What is the exponential growth rate?
Solution:

85. Models of Limited Growth
In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value.
One model of such growth is
which is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

86. Exponential Decay
Decay, or decline, of a population is represented by the function P(t) = P0ekt, k > 0.
In this function:
P0 = initial amount of the substance,
P = amount of the substance left after time,
t = time,
k = decay rate.
The half-life is the amount of time it takes for half of an amount of substance to decay.

87. Graphs

88. Example
Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined?
Solution: We know (from Example 5 in our book), that the function for carbon dating is
P(t) = P0e t.
If the charcoal has lost 7.3% of its carbon-14 from its initial amount P0, then 92.7%P0 is the amount present.

89. Example continued
To find the age of the charcoal, we solve the equation for t :
The charcoal was about 632 years old.