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Copyright © 2007 Pearson Education, Inc.
Slide 5-1
Chapter 5: Inverse, Exponential and
Logarithmic Functions
5.1
5.2
5.3
5.4
5.5
Inverse Functions
Exponential Functions
Logarithms and Their Properties
Logarithmic Functions
Exponential and Logarithmic Equations and
Inequalities
5.6 Further Applications and Modeling with
Exponential and Logarithmic Functions
Copyright © 2007 Pearson Education, Inc.
Slide 5-2
5.6 Further Applications and Modeling with
Exponential and Logarithmic Functions
• Physical Science Applications: A(t )  A0e kt
– A0 is some initial quantity
– t represents time
– k > 0 represents the growth constant, and k < 0
represents the decay constant
Copyright © 2007 Pearson Education, Inc.
Slide 5-3
5.6 Exponential Decay Function Involving
Radioactive Isotopes
Example Nuclear energy derived from radioactive isotopes
can be used to supply power to space vehicles. Suppose that
the output of the radioactive power supply for a certain
satellite is given by the function defined by
y  40e
.004t
,
where y is measured in watts and t is the time in days.
(a) What is the initial output of the power supply?
(b) After how many days will the output be reduced to
35 watts?
(c) After how many days will the output be half of its initial
amount? (That is, what is its half-life?)
Copyright © 2007 Pearson Education, Inc.
Slide 5-4
5.6 Exponential Decay Function Involving
Radioactive Isotopes
Solution
(a) Let t = 0 and evaluate y.
y  40e.004( 0)  40e0  40
The initial output is 40 watts.
(b) Let y = 35 and solve for t.
35  40e .004t
35
 e .004t
40
35
ln  ln e .004t
40
35
ln 35
40
ln  .004t  t 
 33.4 days
40
 .004
Copyright © 2007 Pearson Education, Inc.
Slide 5-5
5.6 Exponential Decay Function Involving
Radioactive Isotopes
(c) Because the initial amount is 40, the half-life is
the value of t for which y  12 (40)  20.
20  40e .004t
20
 e .004t
40
20
ln  ln e .004t
40
ln .5
ln .5  .004t  t 
 173 days
 .004
Copyright © 2007 Pearson Education, Inc.
Slide 5-6
5.6 Exponential Decay Function
Involving Radioactive Isotopes
The half-life can be obtained from the graph of
.004t
by noting that when t = x = 173,
y  40e
y  20 = 12 (40).
Copyright © 2007 Pearson Education, Inc.
Slide 5-7
5.6 Age of a Fossil using
Carbon-14 Dating
Example Carbon-14 is a radioactive form of carbon found
in all living plants and animals. After a plant or animal dies,
the radiocarbon disintegrates. Scientists determine the age of
the remains by comparing the amount of carbon-14 present
with the amount found in living plants and animals. The
amount of carbon-14 present after t years is given by
1
.
A(t )  A0e kt , with k  (ln 2)5700
Find the half-life.
Solution
Copyright © 2007 Pearson Education, Inc.
1
.
Let A(t )   12  A0 and k  (ln 2)5700
1
A0  A0e (ln 2 )(1 / 5700) t Divide by A0.
2
1
 e (ln 2 )(1 / 5700) t
2
Slide 5-8
5.6 Age of a Fossil using
Carbon-14 Dating
1
ln  ln e (ln 2 )(1 / 5700) t
2
ln 2
(ln 1  ln 2)  
t
5700
5700

(ln 1  ln 2)  t
ln 2
5700
ln 2  t
ln 2
5700  t
Take the ln of both sides.
ln ex = x and quotient rule
for logarithms
Isolate t.
Distribute and use the fact
that ln1 = 0.
The half-life is 5700 years.
Copyright © 2007 Pearson Education, Inc.
Slide 5-9
5.6 Finding Half-life
Example Radium-226, which decays according to A(t )  A0e kt ,
has a half-life of about 1612 years. Find k. How long does it
take a 10-gram sample to decay to 6 grams?
Solution
The half-life tells us that A(1612) = (½)A0.
1
A0  A0e k (1612)
2
1
 e 1612k
2
1
ln  ln e 1612k
2
1
ln  1612k
2
ln 12
k
or k  .00043
 1612
Copyright © 2007 Pearson Education, Inc.
Slide 5-10
5.6 Finding Half-life
Thus, radium-226 decays according to the equation
A(t )  A0e .00043t . Now let A(t) = 6 and A0 = 10 to
find t.
6  10e .00043t
.6  e .00043t
ln .6  ln e .00043t
ln .6  .00043t
ln .6
t
 .00043
t  1188 years
Copyright © 2007 Pearson Education, Inc.
Slide 5-11
5.6 Financial Applications
Example How long will it take $1000 invested at
6%, compounded quarterly, to grow to $2700?
Solution
and n = 4.
Find t when A = 2700, P = 1000, r = .06,
.06 
2700  10001 

4 

2.7  1.0154t
4t
log 2.7  log 1.0154t
log 2.7  4t log 1.015
log 2.7
 t  t  16.678 or 16 34 years
4 log 1.015
Copyright © 2007 Pearson Education, Inc.
Slide 5-12
5.6 Amortization Payments
•
A loan of P dollars at interest i per period may
be amortized in n equal periodic payments of
R dollars made at the end of each period, where
P
R
.
n
1  (1  i ) 


i


•
The total interest I that will be paid during the
term of the loan is
I  nR  P.
Copyright © 2007 Pearson Education, Inc.
Slide 5-13
5.6 Using Amortization to Finance an
Automobile
Example You agree to pay $24,000 for a used SUV. After
a down payment of $4000, the balance will be paid off in
36 equal monthly payments at 8.5% interest per year. Find the
amount of each payment. How much interest will you pay
over the life of the loan?
Solution
P  24000  4000  20000, i  .085 / 12  .007083, n  36
20000
R
 631.35
36
1  (1  .007083) 


.
007083


The monthly payment wi ll be $631.35. The total interest paid
will be 36($631.35)  $20,000  $2728.60.
Copyright © 2007 Pearson Education, Inc.
Slide 5-14