Module 4.1_Examples
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Module 4.1- Exponential Functions Examples
Example 1- The court-imposed fine for the city of Yonkers
You may recall that the city council defied a district court-order to desegregate housing in Yonkers. So, Judge Sand imposed a fine against the city starting at $100 on August 2nd and doubling daily until officials obeyed the court order.
Previously we modeled the relationship between the number of days since August 2
nd
(when the court-order was issued) and the the fine for the city of Yonkers. Here again is our model.
y
=
100
⋅
2
t
Let's identify how our model fits the definition of an exponential function given above.
t
=
¿
number of days since August 2
nd
and this is
x
in
f
(
x
)=
a
⋅
b
x
.
y
=
¿
the fine (in dollars) imposed on the city of Yonkers and this is
f
(
x
)
in
f
(
x
)=
a
⋅
b
x
.
The initial value is
a
=
$
100
. And, the conditions that
a
is a real number and
a≠
0
are met.
The growth factor is
b
=
2
. And the conditions that
b
is a positive real number and
b≠
1
are
met.
Example 2- Repopulating a wildlife refuge
Prior to 2012, the deer population in a wildlife refuge area had been hunted to elimination. In 2014, 80 deer were reintroduced into the wildlife refuge area. By 2020, the population had grown to 180 deer. If this population grows exponentially, find an exponential model for the deer population in this wildlife refuge area and use it to predict the deer population in 2025.
Solution:
We define our input variable to be
t
, the number of years after 2014, and the output variable to be
N
, the number of deer in the wildlife refuge area
t
years after 2014. Then our input-output pairs must have the form
(
t ,N
)
.
In 2014,
t
=
0
and
N
=
80
. So, the first input-output pair is
(
0,80
)
.
In 2020,
t
=
6
and
N
=
180
. So, the second input-output pair is
(
6,180
)
.
Since we are given that the deer population is growing exponentially, we will use the model
N
(
t
)=
a
⋅
b
t
. To find the model for the deer population, we need to find values for parameter
a
(the initial value) and the parameter
b
(the growth factor).
Notice that by choosing our input variable
t
to be the number of years after 2014, we "gave" ourselves the initial value for the function. So, we know that the parameter
a
=
80
.
So, we can substitute
a
=
80
into
N
(
t
)=
a
⋅
b
t
.
N
(
t
)=
80
⋅
b
t
We have used the first input-output pair to obtain the parameter
a
. Now, to find the parameter
b
we'll substitute the second input-output pair into
N
(
t
)=
80
⋅
b
t
.
The second input-output pair is
(
6,180
)
.
This tells us that
t
=
6
, and
N
(
6
)=
180
. Making the substitutions gives us ...
180
=
80
⋅
b
6
Dividing both sides by 80, gives us ...
b
6
=
180
80
Taking the 6th root of both sides, gives us ...
b
=
6
√
180
80
=
(
180
80
)
(
1
6
)
Entering the expression
(
180
80
)
(
1
6
)
into our graphing calculators gives us ...
b≈
1.1447
So the deer population is modeled by the exponential function
N
(
t
)=
80
(
1.1447
)
t
.
In 2025,
t
=
2025
−
2014
=
11
.
N
(
11
)=
80
(
1.1447
)
11
≈
354
.
In 2025 we predict that population of deer in the wildlife refuge area will be 354.
Example 3- The growth rate for the population of deer
Find the growth rate for the population of deer from the previous example. Then interpret the growth rate in context.
Solution:
Recall that the growth factor is
b
=
1
+
r
where
r
is the growth rate in decimal form.
Substituting
b≈
1.1447
into
b
=
1
+
r
and solving for
r
, gives us
r
=
0.1447
. This is the growth rate in decimal form.
So the grow rate is 14.47%. This means that the deer population in the wildlife refuge area is increasing 14.47% each year.
Example 4- Find the exponential function passing through two points
Find a formula for an exponential function passing through the points
(−
2,6
)
and
(
2,1
)
.
Solution:
Since we don’t have the initial value, we will take a general approach that will work for any function form with unknown parameters. We will substitute both of the given input-
output pairs into the function form
y
=
a
⋅
b
x
solve for the unknown parameters,
a
and
b
.
All input-output pairs have the form
(
x , y
)
.
Equation 1: Substituting
(−
2,6
)
into
y
=
a
⋅
b
x
gives us
6
=
a
⋅
b
−
2
.
Equation 2: Substituting
(
2,1
)
into
y
=
a
⋅
b
x
gives us
1
=
a
⋅
b
2
.
Equation 1:
6
=
a
⋅
b
−
2
Equation 2:
1
=
a
⋅
b
2
First we solve
Equation 1
for
a
. Multiplying both sides by
b
2
yields
a
=
6
b
2
.
Substituting
a
=
6
b
2
into
Equation 2
yields
1
=(
6
b
2
)
⋅
b
2
.
Now, we just solve for
b
.
1
=(
6
b
2
)
⋅
b
2
1
=
6
b
4
b
4
=
1
6
b
=
4
√
1
6
(Note: since we will use
b
to find
a
, we avoid round-off error in our calculations by not rounding
b
yet.)
To find
a
, we substitute
b
=
4
√
1
6
into
a
=
6
b
2
.
a
=
6
(
4
√
1
6
)
2
=
6
⋅
√
1
6
Now, we are ready to round.
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b
=
4
√
1
6
≈
0.6389
a
=
6
⋅
√
1
6
≈
2.449 5
Finally, we pull it altogether and substitute the values for
a
and
b
into the exponential function
f
(
x
)=
a
⋅
b
x
to get the following function.
f
(
x
)=
2.4495
⋅
(
0.6389
)
x
Example 5- Find the equation of an exponential function from its graph
Find the equation of the exponential function given in the graph below. Do not round the initial value nor the growth factor.
Solution:
The initial value for the function is not clear in this graph, so we will instead work using two clearly identifiable points. There are three clearly identifiable points: (−1,1), (1,2), and (3,4). Two points are sufficient to find the equation for a standard exponential function, so we can choose any two of these points. Let's work with (1,2) and (3,4). This was an arbitrary choice. Choosing any two points will yield the same exponential function (as long as we perform the calculations correctly).
Substituting (1,2) into
y
=
a
⋅
b
x
yields
2
=
a
⋅
b
1
.
Substituting (3,4) into
y
=
a
⋅
b
x
yields
4
=
a
⋅
b
3
.
Equation 1:
2
=
a
⋅
b
Equation 2:
4
=
a
⋅
b
3
Solving Equation 1 for
a
yields
a
=
2
b
.
a
=
2
b
Substituting
a
=
2
b
into Equation 2 yields
4
=
(
2
b
)
⋅
b
3
.
Now, we just solve for
b
.
4
=
(
2
b
)
⋅
b
3
4
=
2
⋅
b
2
2
=
b
2
b
=
±
2
But, by the definition of an exponential function,
b
>
0
.
b
=
√
2
Substituting
b
=
√
2
into
a
=
2
b
yields
a
=
2
√
2
.
a
=
2
√
2
=
√
2
So the equation of the exponential function given in the graph is
f
(
x
)=
√
2
⋅
(
√
2
)
x
Example 6- Radioactive Isotope
Bismuth-210 is an isotope that radioactively decays by about 13% each day. So, each day 13% of
the remaining Bismuth-210 transforms into another atom (polonium-210 in this case). Initially we have 150 mg of Bismuth-210. How much remains after one week? If necessary, round to four
decimal places.
Solution:
The decrease is 13% per day, so
r
=−
0.13
, and the number of milligrams of Bismuth-210
remaining after
t
days is modeled by the exponential function,
g
(
t
)=
a
⋅
b
t
, where:
o
a
=
150
mg, and
o
b
=
1
+(−
0.13
)=
0.87
.
So,
g
(
t
)=
150
⋅
(
0.87
)
t
.
Since one week is 7 days, we evaluate
g
(
t
)
for
t
=
7
.
g
(
7
)=
150
⋅
(
0.87
)
7
⟶
g
(
7
)
≈
56.5882
mg
After 7 days, approximately 56.5882mg of Bismuth-210 remains.
Example 7- Exponential decay given in increments of 4 hours
A scientist begins with 100 mg of a radioactive substance. After 4 hours, it has decayed to 80 mg. Assume the substance continues to decay exponentially at this rate.
Find an exponential model in the form
f
(
t
)=
a
⋅
b
t
for the amount of substance remaining after
t
hours.
Identify the growth factor and hourly percent decrease.
How much of the substance remains after 15 hours?
Solution:
We begin by identifying the exponentional model
f
(
t
)=
a
⋅
b
t
for the amount of substance remaining after
t
hours.
The initial amount is 100 mg.
The amount remaining after 4 hours is 80 mg.
newvalue
previous value
=
80
100
=
0.8
This means we multiply
100
(
0.8
)
to get the amount of the substance remaining after 4 hours. Then we multiply by 0.8 again to get
100
(
0.8
)
2
which is the amount of substance remaining after another 4 hours, and so on and so forth. Therefore we can model the exponential decay of the radioactive substance as follows.
f
(
t
)=
100
⋅
(
0.8
)
t
4
f
(
t
)=
100
⋅
(
0.8
)
1
4
∙t
f
(
t
)=
100
⋅
(
0.8
1
4
)
t
(applied the power rule for exponents)
f
(
t
)
≈
100
⋅
(
0.9457
)
t
Now we can identify the growth factor and hourly percent decrease.
The exponential function
f
(
t
)
≈
100
⋅
(
0.9457
)
t
is in the form
f
(
t
)=
a
⋅
b
t
.
So the
growth factor is
b
=
0.9457
.
Also, we know that
b
=
1
+
r
where
r
is the hourly percent decrease in decimal form.
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Substituting
b
=
0.9457
into
b
=
1
+
r
, gives us
0.9457
=
1
+
r
.
Solving this equation for
r
yields
r
=−
0.0543
.
So the
hourly percent decrease is 5.43%
. This means the amount of the radioactive substance decreases 5.43% each hour.
Finally, we can use our model to determine the amount of the substance remaining after 15 hours.
f
(
15
)
≈
100
⋅
(
0.9457
)
15
f
(
15
)
≈
43.28
mg.
After 15 hours, approximately 43.28 mg of the radioactive substance remains.
Example 8- Exponential growth given in increments of five years
In 2014, the median annual household income in the United States was $55,613. In 2019, the median household income in the United States was $69,560.
Find an exponential model
h
(
t
)=
a
⋅
b
t
for the median annual household income in the United States
t
years after 2014.
Identify the growth factor and hourly percent increase.
If the median annual household income in the U.S. continued to grow at this rate, what would it be in 2030? Does this seem reasonable to you? Why or why not?
Solution:
We begin by finding an exponential model,
h
(
t
)
, for the median annual household income in the United States
t
years after 2014.
Initially, the median household income was $55,610. Five years later, the median household income was $69,560.
newvalue
previous value
=
69,560
$
55,610
≈
1.25085
This means we multiply
$
55,610
(
1.25085
)
to get the median household income $69,560 (with a little round off error) in 2029 (5 years later). Then we multiply by 1.25085 again to get
$
55,610
(
1.25085
)
2
which is the predicted median household income in another 5 years, and so on and so forth. Therefore we can model the exponential growth of the median household income as follows.
g
(
t
)=
55,610
⋅
(
1.25085
)
t
5
g
(
t
)=
55,610
⋅
(
1.25085
)
1
5
∙ t
g
(
t
)=
55,610
⋅
(
1.25085
1
5
)
t
g
(
t
)
≈
55,610
⋅
(
1.04578
)
t
Now we're ready to identify the growth factor and hourly percent decrease.
The exponential function
g
(
t
)
≈
55,610
⋅
(
1.04578
)
t
is in the form
g
(
t
)=
a
⋅
b
t
.
So the
growth factor is
b
=
1.04578
.
Also, we know that
b
=
1
+
r
where
r
is the hourly percent decrease in decimal form.
Substituting
b
=
1.04578
into
b
=
1
+
r
, gives us
1.04578
=
1
+
r
.
Solving this equation for
r
yields
r
=
0.04578
.
So the
yearly percent increase is 4.578%
. This means the median household income in the United States was increasing approximately 4.578% each year.
Now we can use our model to predict the annual household income in the U.S. in the year 2030. We'll also answer the question, "Does this seem reasonable to you? Why or why not?"
Since 2014 is the initial year, we calculate
t
=
2030
−
2014
=
16
years in 2030.
g
(
16
)
≈
55,610
⋅
(
1.04578
)
16
g
(
16
)
≈$
113,815
So, if we assume that the median household income continues to grow exponentially, we would expect the median household income to be $113,815 in 2030.
Example 9- Compounding monthly
A certificate of deposit (CD) is a type of savings account offered by banks. A CD typically offers a higher interest rate in return for a fixed length of time you leave your money invested with the bank. If a bank offers a 24 month CD with an annual interest rate of 1.8% compounded monthly,
how much will a $1000 investment grow to over those 24 months?
Solution:
First, we must notice that the interest rate is an annual rate, but it is compounded monthly. This means that the interest is calculated and added to the account monthly. To find the monthly interest rate, we divide the annual rate of 1.8% by 12 since there are 12 months in a year. Each month we will earn
1.8%
12
=
0.15%
interest. From this, we can set
up an exponential function, with:
our initial deposit of $1000
a growth rate of
1.8%
12
, and
our input
m,
the number of months after our initial deposit.
A
(
m
)=
1000
(
1
+
0.018
12
)
m
After 24 months, the amount in the account will be
A
(
24
)=
1000
(
1
+
0.018
12
)
24
=
1,036.63
.
So, the initial investment of $1,000 will grow to $1,036.63 after 24 months.
Example 10- Suspending loan payments
Suppose that three years ago, the world shut down due to a pandemic, and you owed $2,000 on a credit card that charges 21% annual interest compounded quarterly. Further suppose that the credit card company allowed you to skip making payments for the past three years with no late fees, but the interest continued to compound. What is your credit card balance after three years?
Solution:
Since the interest is "compounded quarterly," we will use the compound-interest exponential model.
A
(
t
)
=
p
(
1
+
r
n
)
n∙ t
We have:
p
=
$
2,000
(the initial loan amount)
r
=
0.21
(the annual interest rate of 21% in decimal form)
n
=
4
(quarterly means 4 compoundings in
one
year)
t
=
3
(the number of years the interest is compounded)
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Substituting these values into the compound-interest model gives us the following.
A
(
3
)=
2000
(
1
+
0.21
4
)
4
∙
(
3
)
=
2000
(
1.0015
)
12
≈
3,695.69
So, after three years of making no payment, the balance on credit card is $3,695.69.
Example 11- Solving for the initial amount,
p
A 529 plan is a college savings plan in which a relative can invest money to save for a child’s eventual college tuition. The funds in a 529 plan grow tax free. Lily wants to make a one-time investment in a 529 account for her new granddaughter, and she wants the investment to grow to $40,000 over 18 years. She believes the account will earn a 6% annual interest rate compounded semi-annually (twice per year). How much money should Lily invest in the account now so that her investment will grow to $40,000 in 18 years?
Solution:
Since, the interest is compounded, we will use the compound-interest form of an exponential function.
A
(
t
)
=
p
(
1
+
r
n
)
n∙ t
p
=
?
(We do not know the initial value, i.e. the principal investment
p
. In fact, to answer the question, we must find the initial value
p
.)
r
=
0.06
(The annual interest rate is 6%.)
n
=
2
(The interest is compounded semi-annually, i.e. twice per year).
t
=
18
(Lily wants to reach her investment goal in 18 years.)
A
(
18
)=
$
40,000
(Lily wants the principal investment to grow to $40,000.)
Substituting the know values into the exponential function,
A
(
t
)
=
p
(
1
+
r
n
)
n∙ t
, gives us.
$
40,000
=
p
(
1
+
0.06
2
)
2
⋅
(
18
)
$
40,000
=
p
(
2.8983
)
p
=
$
40,000
2,8983
=
$
13,801
To grow her investment $40,000 in 18 years, Lily will need to invest $13,801 now.
Example 12- Continuous decay
Radon-222 decays at a continuous rate of 17.3% per day. Given 100mg of Radon-222, how much Radon-222 will remain in 3 days?
Solution:
Since we are given a continuous decay rate, we will use the continuous growth formula. Also, since the substance is decaying, we know the decay rate will be negative. So,
r
=−
0.173
.
So, we need to substitute
a
=
100
mg
,
r
=−
0.173
, and
t
=
3
days into the continuous growth formula
f
(
t
)=
a
⋅
e
rt
.
f
(
3
)=
100
⋅
e
−
0.173
⋅
(
3
)
f
(
3
)=
100
⋅
e
−
0.519
f
(
3
)
≈
59.5115
mg
So, if we begin with 100mg of Radon-222, in the 3 days, 59.5115mg will remain.