Transcript Ch 5 Inverse, Exponential and Logarithmic Functions
Chapter 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions 5.3 Logarithms and Their Properties 5.4 Logarithmic Functions 5.5 Exponential and Logarithmic Equations and Inequalities
5.6 Further Applications and Modeling with Exponential and Logarithmic Functions Slide 5-2
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5.6 Further Applications and Modeling with Exponential and Logarithmic Functions • Physical Science Applications:
A
(
t
)
A
0
e kt
–
A
0 is some initial quantity –
t
represents time –
k
> 0 represents the growth constant, and
k
< 0 represents the decay constant
Slide 5-3
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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
40
e
.
004
t
, 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?)
Slide 5-4
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5.6 Exponential Decay Function Involving Radioactive Isotopes
Solution
(a) Let
t
= 0 and evaluate
y
.
y
40
e
.
004 ( 0 ) 40
e
0 40 The initial output is 40 watts.
(b) Let
y
= 35 and solve for
t
.
35 40
e
.
004
t
35 40
e
.
004
t
ln 35 40 ln
e
.
004
t
ln 35 40 .
004
t
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t
ln 35 40 .
004 33 .
4 days
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
1 2 ( 40 ) 20 .
20 40
e
.
004
t
20 40
e
.
004
t
ln 20 40 ln
e
.
004
t
ln .
5 .
004
t
t
ln .
5 .
004 173 days
Slide 5-6
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5.6 Exponential Decay Function Involving Radioactive Isotopes The half-life can be obtained from the graph of
y
40
e
.
004
t
by noting that when
t
=
x
= 173,
y
20 = 2 1 ( 40 ).
Slide 5-7
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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 A
(
t
)
A
0
e
kt
, with
k
years is given by (ln 2 ) 5700 .
Find the half-life.
Solution
Let
A
(
t
) 2
A
0 and
k
(ln 2 ) 5700 .
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1 2
A
0
A
0
e
(ln 2 )( 1 / 5700 )
t
1 2
e
(ln 2 )( 1 / 5700 )
t
Divide by
A
0 .
Slide 5-8
5.6 Age of a Fossil using Carbon-14 Dating ln 1 2 ln
e
(ln 2 )( 1 / 5700 )
t
(ln 1 ln 2 ) ln 2 5700
t
5700 (ln 1 ln 2 ln 2 )
t
5700 ln ln 2 2 5700
t t
Take the ln of both sides.
ln
e x
=
x
and quotient rule for logarithms Isolate
t
.
Distribute and use the fact that ln1 = 0.
The half-life is 5700 years.
Slide 5-9
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5.6 Finding Half-life
Example
Radium-226, which decays according to
A
(
t
) has a half-life of about 1612 years. Find
k
. How long does it
A
0
e
kt
, take a 10-gram sample to decay to 6 grams?
Solution
The half-life tells us that
A
(1612) = (½)
A
0 .
1 2
A
0 1
A
0
e
k e
1612
k
( 1612 ) 2 1 ln 2 1 ln 2 ln
e
1612
k
1612
k k
ln 1 2 1612 or
k
.
00043 Copyright © 2007 Pearson Education, Inc.
Slide 5-10
5.6 Finding Half-life Thus, radium-226 decays according to the equation
A
(
t
)
A
0
e
.
00043
t
.
Now let
A
(
t
) = 6 and
A
0 = 10 to find
t
.
6 10
e
.
00043
t
.
6
e
.
00043
t
ln .
6 ln
e
.
00043
t
ln .
6 .
00043
t
ln .
6 .
00043
t 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, 2700 1000 1 .
06 4
t
4 2 .
7 1 .
015 4
t
log 2 .
7 log 2 .
7 log 1 .
015 4
t
4
t
log 1 .
015 log 2 .
7 4 log 1 .
015
t
t
16 .
678 or 16 3 4 years 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
R
1
P
( 1
i i
)
n
.
The total interest
I
that will be paid during the term of the loan is
I
nR
P
.
Slide 5-13
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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
R
1 ( 1 20000 .
007083 ) 36 .
007083 631 .
35 The monthly payment wi ll be $631.35.
The total interest paid will be 36($631.35
) $ 20,000 $2728.60.
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Slide 5-14