2. CHAPTER 5: Exponential and Logarithmic Functions
5.1 Inverse Functions
5.2 Exponential Functions and Graphs
5.3 Logarithmic Functions and Graphs
5.4 Properties of Logarithmic Functions
5.5 Solving Exponential and Logarithmic Equations
5.6 Applications and Models: Growth and Decay; and Compound Interest
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3. 5.3 Logarithmic Functions and Graphs
Find common logarithms and natural logarithms with and without a calculator.
Convert between exponential and logarithmic equations.
Change logarithmic bases.
Graph logarithmic functions.
Solve applied problems involving logarithmic functions.
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4. Logarithmic Functions
These functions are inverses of exponential functions. We can draw the graph of the inverse of an exponential function by interchanging x and y.
To Graph: x = 2y.
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.
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Example
Graph: x = 2y.
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Example (continued)
This curve looks like the graph of y = 2x reflected across the line y = x, as we would expect for an inverse. The inverse of y = 2x is x = 2y.
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7. 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.”
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8. Finding Certain Logarithms - Example
Find each of the following logarithms.
a) log10 10,000 b) log c) log2 8
d) log9 3 e) log6 1 f) log8 8
Solution:
a) The exponent to which we raise 10 to obtain 10,000 is 4; thus log10 10,000 = 4.
b) We have
The exponent to which we raise 10 to get 0.01 is –2, so log = –2.
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Example (continued)
c) log2 8: The exponent to which we raise 2 to get 8 is 3, so log2 8 = 3.
d) log9 3: The exponent to which we raise 9 to get 3 is 1/2; thus log9 3 = 1/2.
e) log6 1: 1 = 60. The exponent to which we raise 6 to get 1 is 0, so log6 1 = 0.
f) log8 8: 8 = 81. The exponent to which we raise 8 to get 8 is 4, so log8 8 = 1.
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Logarithms
loga 1 = 0 and loga a = 1, for any logarithmic base a.
A logarithm is an exponent!
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Example
Convert each of the following to a logarithmic equation.
a) 16 = 2x b) 10–3 = c) et = 70
The exponent is the logarithm.
a) 16 = 2x log216 = x
The base remains the same.
Solution:
b) 10–3 = g log = –3
c) et = 70 g log e 70 = t
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Example
Convert each of the following to an exponential equation.
a) log 2 32= 5 b) log a Q= 8 c) x = log t M
The logarithm is the exponent.
a) log 2 32 = = 32
The base remains the same.
Solution:
b) log a Q = 8 g a8 = Q
c) x = log t M g tx = M
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Example
Find each of the following common logarithms on a calculator. Round to four decimal places.
a) log 645,778 b) log c) log (3)
Solution:
Function Value Readout Rounded
a) log 645,
b) log –4.6216
c) log (–3) Does not exist.
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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.
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Example
Find each of the following natural logarithms on a calculator. Round to four decimal places.
a) ln 645,778 b) ln c) log (5)
d) ln e e) ln 1
Solution:
Function Value Readout Rounded
a) ln 645,
b) ln –
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Example (continued)
Solution:
Function Value Readout Rounded
c) ln (–5) Does not exist.
b) ln e
c) ln
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17. Changing Logarithmic Bases
The Change-of-Base Formula
For any logarithmic bases a and b, and any positive number M,
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Example
Find log5 8 using common logarithms.
Solution:
First, we let a = 10, b = 5, and M = 8. Then we substitute into the change-of-base formula:
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Example
We can also use base e for a conversion.
Find log5 8 using natural logarithms.
Solution:
Substituting e for a, 6 for b and 8 for M, we have
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20. Graphs of Logarithmic Functions - Example
Graph: y = f (x) = log5 x.
Solution: Method 1
y = log5 x is equivalent to x = 5y. Select y and compute x.
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Example (continued)
Graph: y = f (x) = log5 x.
Solution: Method 2
Use a graphing calculator. First change bases.
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Example (continued)
Graph: y = f (x) = log5 x.
Solution: Method 3
Calculators which graph inverses automatically.
Begin with y1 = 5x, the graphs of both y1 and its inverse y2 = log 5 x will be drawn.
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23. Copyright © 2009 Pearson Education, Inc.
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 + 3)
b) f (x) = 3  ln x
c) f (x) = |ln (x – 1)|
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Example (continued)
a) f (x) = ln (x + 3)
The graph is a shift 3 units left. The domain is the set of all real numbers greater than –3, (–3, ∞). The line x = –3 is the vertical asymptote.
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Example (continued)
b) f (x) = 3 – ln x
The graph is a vertical shrinking of y = ln x, followed by a reflection across the x-axis and a translation up 3 units. The domain is the set of all positive real numbers, (0, ∞). The y-axis is the vertical asymptote.
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Example (continued)
c) f (x) = |ln (x – 1)|
The graph is a translation of y = ln x, right 1 unit. The effect of the absolute is to reflect the negative output across the x-axis. The domain is the set of all positive real numbers greater than 1, (1, ∞). The line x =1 is the vertical asymptote.
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Application
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
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Example
a. The population of Hartford, Connecticut, is 124,848. Find the average walking speed of people living in Hartford.
The population of San Antonio, Texas, is 1,236,249. Find the average walking speed of people living in San Antonio.
Graph the function.
d. A sociologist computes the average walking speed in a city to the approximately 2.0 ft/sec. Use this information to estimate the population of the city.
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Example (continued)
Solution:
a. Since P is in thousands and 124,848 = thousand, we substitute for P:
w( ) = 0.37 ln
 1.8 ft/sec.
The average walking speed of people living in Hartford is about 1.8 ft/sec.
b. Substitute for P:
w( ) = 0.37 ln
 2.7 ft/sec.
The average walking speed of people living in San Antonio is about 2.7 ft/sec.
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Example (continued)
c. Graph with a viewing window [0, 600, 0, 4] because inputs are very large and outputs are very small by comparison.
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Example (continued)
d. To find the population for which the walking speed is 2.0 ft/sec, we substitute 2.0 for w(P), 2.0 = 0.37 ln P , and solve for P.
Use the Intersect method.
Graph y1 = 0.37 ln x and y2 = 2.
In a city with an average walking speed of 2.0 ft/sec, the population is about thousand or 194,500.
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