Chapter 1
Section 8-6
Exponential and Logarithmic
Functions, Applications, and Models
8-6-1
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Exponential and Logarithmic
Functions, Applications, and Models
• Exponential Functions and Applications
• Logarithmic Functions and Applications
• Exponential Models in Nature
8-6-2
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Exponential Function
An exponential function with base b, where
b > 0 and b  1, is a function of the form
f ( x)  b ,
x
where x is any real number.
8-6-3
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Example: Exponential Function
(b > 1)
x
2x
–1
1/2
0
1
1
2
2
4
f ( x)  2
x
y
x
The x-axis is the horizontal asymptote of each graph.
8-6-4
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Example: Exponential Function
(0 < b < 1)
x
(1/2) x
–2
4
–1
2
0
1
1
1/2
y
x
1
f ( x)   
2
x
The x-axis is the horizontal asymptote of each graph.
8-6-5
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Graph of f ( x )  b
x
1. The graph always will contain the point (0, 1).
2. When b > 1 the graph will rise from left to right.
When 0 < b < 1, the graph will fall from left to
right.
3. The x-axis is the horizontal asymptote.
4. The domain is (, ) and the range is (0, ).
8-6-6
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Exponential with Base e
e  2.718281828.
f ( x)  e x
y
x
8-6-7
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Compound Interest Formula
Suppose that a principal of P dollars is invested at
an annual interest rate r (in percent, expressed as
a decimal), compounded n times per year. Then
the amount A accumulated after t years is given
by the formula
nt
 r
A  P 1   .
 n
8-6-8
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Example: Compound Interest Formula
Suppose that $2000 dollars is invested at an
annual rate of 8%, compounded quarterly. Find
the total amount in the account after 6 years if no
withdrawals are made.
Solution
 r
A  P 1  
 n
nt
 .08 
A  2000 1 

4 

4(6)
8-6-9
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Example: Compound Interest Formula
Solution (continued)
A  2000 1.02 
24
A  2000 1.60844  3216.88
There would be $3216.88 in the account at
the end of six years.
8-6-10
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Continuous Compound Interest Formula
Suppose that a principal of P dollars is invested at
an annual interest rate r (in percent, expressed as
a decimal), compounded continuously. Then the
amount A accumulated after t years is given by
the formula
A  Pe .
rt
8-6-11
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Example: Continuous Compound Interest
Formula
Suppose that $2000 dollars is invested at an
annual rate of 8%, compounded continuously.
Find the total amount in the account after 6 years
if no withdrawals are made.
Solution
A  Pe
rt
A  2000e
.08(6)
8-6-12
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Example: Compound Interest Formula
Solution (continued)
A  2000e.48
A  2000(1.61607)  3232.14
There would be $3232.14 in the account at
the end of six years.
8-6-13
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Exponential Growth Formula
The continuous compound interest formula is an
example of an exponential growth function. In
situations involving growth or decay of a quantity, the
amount present at time t can often be approximated by
a function of the form
A(t )  A0e ,
kt
where A0 represents the amount present at time t = 0,
and k is a constant. If k > 0, there is exponential
growth; if k < 0, there is exponential decay.
8-6-14
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Definition of log b x
For b > 0, b  1, if
b  x then logb x.
y
8-6-15
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Exponential and Logarithmic Equations
Exponential
Equation
3  81
4
10  10, 000
4
43  1/ 64
3 1
0
Logarithmic
Equation
4  log3 81
4  log10 10, 000
3  log 4 (1/ 64)
0  log3 1
8-6-16
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Logarithmic Function
A logarithmic function with base b,
where b > 0 and b  1, is a function of the
form
g ( x)  logb x, where x  0.
8-6-17
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Graph of g ( x)  log b x
The graph of y = log b x can be found by
interchanging the roles of x and y in the
function f (x) = bx. Geometrically, this is
accomplished by reflecting the graph of
f (x) = bx about the line y = x.
The y-axis is called the vertical asymptote
of the graph.
8-6-18
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Example: Logarithmic Functions
g ( x)  log 2 x
g ( x)  log1/ 2 x
y
y
x
x
8-6-19
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Graph of g ( x)  log b x
1. The graph always will contain the point (1, 0).
2. When b > 1 the graph will rise from left to right.
When 0 < b < 1, the graph will fall from left to
right.
3. The y-axis is the vertical asymptote.
4. The domain is (0, ) and the range is (, ).
8-6-20
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Natural Logarithmic Function
ln x  loge x
g(x) = ln x, called the natural logarithmic function,
is graphed below.
y
g ( x)  ln x
x
8-6-21
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Natural Logarithmic Function
The expression ln ek is the exponent to which
the base e must be raised in order to obtain ek.
There is only one such number that will do
this, and it is k. Thus for all real numbers k,
ln e  k.
k
8-6-22
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Example: Doubling Time
Suppose that a certain amount P is invested at an
annual rate of 5% compounded continuously. How
long will it take for the amount to double (doubling
time)?
Solution
rt
A  Pe
.05t
2P  Pe
.05t
2e
Sub in 2P for A (double).
Divide by P.
8-6-23
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Example: Doubling Time
Solution (continued)
ln 2  ln e
.05t
ln 2  .05t
ln 2
t
 13.9
.05
Take ln of both sides.
Simplify.
Divide by .05
Therefore, it would take about 13.9 years for
the initial investment P to double.
8-6-24
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Exponential Models in Nature
Radioactive materials disintegrate according
to exponential decay functions. The halflife of a quantity that decays exponentially is
the amount of time it takes for any initial
amount to decay to half its initial value.
8-6-25
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Example: Half-Life
Carbon 14 is a radioactive form of carbon that is
found in all living plants and animals. After a plant
or animal dies, the radiocarbon disintegrates. The
amount of carbon 14 present after t years is
modeled by the exponential equation
y  y0e
.0001216t
a) What is the half-life of carbon 14?
b) If an initial sample contains 1 gram of carbon 14,
how much will be left in 10,000 years?
8-6-26
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Example: Half-Life
Solution
a)
y  y0e
.0001216t
1
.0001216t
y0  y0e
2
1
ln    .0001216t
2
t  5700
The half-life of carbon
14 is about 5700 years.
8-6-27
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Example: Half-Life
Solution
b)
y  1e
.0001216(10000)
y  .30
There will be about .30 grams remaining.
8-6-28
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