Transcript No Slide Title

CHAPTER 5:
Exponential and
Logarithmic Functions
5.1
5.2
5.3
5.4
5.5
Inverse Functions
Exponential Functions and Graphs
Logarithmic Functions and Graphs
Properties of Logarithmic Functions
Solving Exponential and Logarithmic
Equations
5.6 Applications and Models: Growth and Decay;
and Compound Interest
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Addison Wesley
5.3
Logarithmic Functions and Graphs
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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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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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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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Finding Certain Logarithms - Example
Find each of the following logarithms.
a) log10 10,000
b) log10 0.01
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.
1
1
b) We have 0.01 
 2  10 2.
100 10
The exponent to which we raise 10 to get 0.01 is –2,
so log10 0.01 = –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: 3  9  91 2. 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.
loga x  y  x  a y
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 = 0.001
c) et = 70
Solution:
a) 16 = 2x
The exponent is the logarithm.
log216 = x
The base remains the same.
b) 10–3 = 0.001 g log10 0.001 = –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
Solution:
a) log 2 32 = 5
The logarithm is the exponent.
25 = 32
The base remains the same.
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 0.0000239 c) log (3)
Solution:
Function Value
Readout
a) log 645,778
b) log 0.0000239
c) log (–3)
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Rounded
5.8101
–4.6216
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 0.0000239 c) log (5)
d) ln e
e) ln 1
Solution:
Function Value
Readout
a) ln 645,778
b) ln 0.0000239
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Rounded
13.3782
–10.6416
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Example (continued)
Solution:
Function Value
Readout
c) ln (–5)
Rounded
Does not exist.
b) ln e
1
c) ln 1
0
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Changing Logarithmic Bases
The Change-of-Base Formula
For any logarithmic bases a and b, and any positive
number M,
log a M
log b M 
.
log a b
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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:
log10 8
log 5 8 
log10 5
 1.2920
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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
log e 8
log 5 8 
log e 5
ln 8

 1.2920
ln 5
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Graphs of Logarithmic Functions - Example
Graph: y = f (x) = log5 x.
Solution:
y = log5 x is equivalent to x = 5y.
Select y and compute x.
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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)
1
b) f (x) = 3  ln x
2
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)
1
b) f (x) = 3  ln x
2
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 + 0.05.
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Example
a. The population of Savannah, Georgia, is 132,410.
Find the average walking speed of people living in
Savannah.
b. The population of Philadelphia, Pennsylvania, is
1,540,351. Find the average walking speed of
people living in Philadelphia.
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Example (continued)
Solution:
a. Since P is in thousands and 132,410 = 132.410
thousand, we substitute 132.410 for P:
w(132.410) = 0.37 ln 132.410 + 0.05
 1.9 ft/sec.
The average walking speed of people living in
Savannah is about 1.9 ft/sec.
b. Substitute 1540.351 for P:
w(1540.351) = 0.37 ln 1540.351 + 0.05
 2.8 ft/sec.
The average walking speed of people living in
Philadelphia is about 2.8 ft/sec.
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