Exponential Growth and
Decay
Section 3.5
Objectives
• Solve word problems requiring
exponential models.
Find the time required for an
investment of $5000 to grow
to $6800 at an interest rate
of 7.5% compounded
quarterly.
Formula needed:
nt
r

A  P 1  
n

Find the time required for an investment of $5000 to
grow to $6800 at an interest rate of 7.5% compounded
quarterly.
 
.075

6800  5000 1 



4 
6800 
.075 
 1 

5000 
4 
4t
 4t 
1.36  1.01875
 4t 
Take the logarithm
of both sides
ln1.36  ln1.01875
 4t 
ln1.36  4t ln(1.01875)
ln1.36
t
4 ln(1.01875)
t  4.138112586
Change to
exponential form
log1.01875(1.36)  4t
log1.01875(1.36)
t
4
Change of base
 log(1.36) 


log1.01875(1.36) log(1.01875) 
t 

4
4
It will take approximately 4.138112586 years.
The population of a certain city was
292000 in 1998, and the observed
relative growth rate is 2% per year.
• Find a function that models the
population after t years.
• Find the projected population in
the year 2004.
• In what year will the population
reach 365004?
The population of a certain city was
292000 in 1998, and the observed
relative growth rate is 2% per year.
• Find a function that models the
population after t years.
Populations are assumed to have continuous growth. Thus
we use the formula
A  Pe
rt
We are going to let t = 0 represent the year 1998 and t is the
number of years since 1998.
A  292000e .02t
This model will give us the population t years after 1998.
The population of a certain city was
292000 in 1998, and the observed
relative growth rate is 2% per year.
• Find the projected population in
the year 2004.
We use the model from the first part of the question
A  292000e .02t
Now we plug in for t. We need to know what t will
represent the year 2004. We find that using a
subtraction problem. 2004-1998 = 6 tells us that 2004
is 6 years since 1998 and thus our t is 6.
A  292000e .02*6 
A  329229.0807
This is not the answer. People must be whole, thus we must round this
number. Either 329229 people or 329230 people would be considered
correct.
The population of a certain city was
292000 in 1998, and the observed
relative growth rate is 2% per year.
• In what year will the population
reach 365004?
We use the model from the first part of the question
A  292000e .02t
Now we plug in 365004 for A and solve for t.
365004  292000e .02t
365004
 e .02t
292000
continued on next screen
The population of a certain city was
292000 in 1998, and the observed
relative growth rate is 2% per year.
• In what year will the population
reach 365004? take a logarithm of both sides
of the equation
change to
logarithmic form
 365004 
ln
 .02t

 292000 
 365004 
ln
 292000   t
.02
t  11.15772551
OR
 365004 
.02t


ln

ln
e

 292000 
 365004 
ln
 (.02t ) ln e

 292000 
 365004 
ln
 (.02t )

 292000 
 365004 
ln
 292000   t
.02
t  11.15772551
continued on next screen
The population of a certain city was
292000 in 1998, and the observed
relative growth rate is 2% per year.
• In what year will the population
reach 365004?
We still don’t have the answer to the question. We know that
11.15772551 years after 1998 the population will reach 365004. We
need to know what year that is. We find this by adding
1998 + 11.15772551 = 2009.15772551
Now do we round up to 2010 or keep it at 2009. We must always keep it at
the lower year (no matter what the decimal). Any decimal part of a year is
still during that year. That the answer to the question is in the year 2009,
the population will reach 365004.
The count in a bacteria culture was 600 after 15 minutes and
16054 after 35 minutes. Assume that growth can be modeled
rt
exponentially by a function of the form
A
(
t
)

Pe
where t is in minutes.
• Find the relative growth rate.
We do not know a starting bacteria count, but the growth rate can be
figured from any point in time to any other point in time as long as we
use the passage of time as the t value.
We will use 600 as the starting population only for the purpose of this
part of the problem. We then have 16054 bacteria 20 minutes later
(the amount of time between 15 and 35 minutes). With these numbers
we can solve for r.
continued on next screen
The count in a bacteria culture was 600 after 15 minutes and
16054 after 35 minutes. Assume that growth can be modeled
rt
exponentially by a function of the form
A
(
t
)

Pe
where t is in minutes.
• Find the relative growth rate.
r *20 
16054  600e
16054
 e r *20 
600
 16054 
ln
  r * 20
 600 
 16054 
ln

 600   r
20
We can now use this r for the other parts of this problem. It will be
important tat we do not round our r value in future parts of the problem.
The count in a bacteria culture was 600 after 15 minutes and
16054 after 35 minutes. Assume that growth can be modeled
rt
exponentially by a function of the form
A
(
t
)

Pe
where t is in minutes.
• What was the initial size of the culture?
We know the relative growth rate r. We also have two different
populations after two different periods of time. We can use either one
for our population A after time t. We will need to solve for P.
I will use (15, 600) as the population after time t.
600  Pe
 16054

 ln

600 *15 

 20





600
 16054

 ln

600

*15 
 20





P
e
P  51.00083003
Since bacteria must be whole, the
initial population must be 51 bacteria.
This means that out model for future
parts of the problem will be
A(t )  51e
 16054
 ln
600 *t

 20








The count in a bacteria culture was 600 after 15 minutes and
16054 after 35 minutes. Assume that growth can be modeled
rt
exponentially by a function of the form
A
(
t
)

Pe
where t is in minutes.
• Find the doubling period in minutes.
Doubling time is how long it takes for the number of bacteria to get to
be two times as many as the initial population. This means that we need
to plug in an ending population of 2*51 or 102.
102  51e
102
e
51
2e
 16054
 ln
600 *t

 20


 16054
 ln
600 *t

 20


 16054
 ln
600 *t

 20




















16054
ln
600 * t
ln2 
20
ln2
t
 16054 
 ln

600 

 20





t  4.217784019
The doubling time
is approximately
4.21778 minutes.
The count in a bacteria culture was 600 after 15 minutes and
16054 after 35 minutes. Assume that growth can be modeled
rt
exponentially by a function of the form
A
(
t
)

Pe
where t is in minutes.
• Find the population after 110 minutes.
For this one we just plug 110 into the model for t.
A(t )  51e
 16054

 ln

600 *110

 20





A(t )  3617813887
The count in a bacteria culture was 600 after 15 minutes and
16054 after 35 minutes. Assume that growth can be modeled
rt
exponentially by a function of the form
A
(
t
)

Pe
where t is in minutes.
• When will the population reach 15000?
For this one, we plug in 15000 for A(t) and solve the t.
15000  51e
15000
e
51
 16054
 ln
600 *t

 20


 16054
 ln
600 *t

 20


 15000 
ln

 51 
ln












16054
600 *t
20
 15000 
ln

 51 
t 
  16054  
 ln

  600  


20




t  34.58688146
The population will reach 15000 in
approximately 34.58688 minutes.
The half-life of strontium-90 is 28 years. Suppose we have
a 80 mg sample.
• Find a function that models the mass m(t) remaining
after t years.
In order to make a model, we need an initial amount and a decay
rate. We are given the initial amount is 80 mg. We will have to
calculate r. In order to do this, we will use the definition of
half-life. Half-life is the number of years it takes to get from
one amount to half of that amount. Thus for our problem, we
know that in 28 years we will have 40 mg (half of 80). Now we
plug all this into the continuous decay formula (same as
continuous growth formula).
A  Pe rt
40  80e
r *28
40
 e r *28
80
1
 e r *28
2
1
ln  r * 28
2
1
ln
2 r
28
Model:
A  80e
 ln(. 5) 
*t 

 28

It will be important tat we do not round our r value in future parts of
the problem.
The half-life of strontium-90 is 28 years. Suppose we have
a 80 mg sample.
• How much of the sample will remain after 100
years?
Model:
A  80e
 ln(. 5) 
*t 

 28

We will use the model found in the previous part of the problem
and plug in 100 for the t.
A  80e
 ln(. 5 )

*100

 28

A  6.729500963
In 100 years, there will be 6.729500963 grams
The half-life of strontium-90 is 28 years. Suppose we have
a 80 mg sample.
• How long will it take the sample to decay to a mass
of 20 mg?
For this part of the problem, we will use the
model found in part 1. We will plug in 20 for A
and solve for t.
20  80e
 ln(. 5 ) 
*t 

 28

 ln(. 5 ) 
*t 
28


20
 e
80
.25  e
 ln(. 5 ) 
*t 

 28

ln(.5)
ln(.25) 
*t
28
Model:
A  80e
 ln(. 5) 
*t 

28


ln(.25)
t
 ln(.5) 


 28 
t  56
It will take 56 years to get down to
20 mg. You could also have done this
just using the definition of half-life.
A wooden artifact from an
ancient tomb contains 35% of the
carbon-14 that is present in living
trees. How long ago was the
artifact made? (The half-life of
carbon-14 is 5730 years.)
For this problem we will need to find a model to represent the
decay of carbon-14. This will require that we find r. Once we have
r and the model, we will be able to solve the problem.
It is assumed that a living tree has 100% of the carbon-14 in living
plants. Once cut down, the amount of carbon-14 starts decreasing
(decaying) based on the half-life. This means that we can use the
half-life information with the staring point 100. The half-life
information tells us that we have 50% of the carbon-14 after 5730
years.
continued on next screen
A wooden artifact from an ancient tomb contains 35% of the
carbon-14 that is present in living trees. How long ago was the
artifact made? (The half-life of carbon-14 is 5730 years.)
Finding r using the half-life information
50  100e r *5730
50
 e r *5730
100
.5  e r *5730
ln(.5)  r * 5730
ln(.5)
r
5730
This means that our
model is
A(t )  100e
 ln(. 5 ) 
*t 

 5730 
continued on next screen
A wooden artifact from an ancient tomb contains 35% of the
carbon-14 that is present in living trees. How long ago was the
artifact made? (The half-life of carbon-14 is 5730 years.)
We use the model from the previous page. Since the ending amount
of carbon-14 is 35%, we will plug 35 in for A(t) – the amount after
time t. Now we solve for t.
35  100e
 ln(. 5 ) 
*t 

 5730 
 ln(. 5 )

*t 

35
 e  5730 
100
.35  e
 ln(. 5 ) 
*t 

 5730 
ln(.35) 
ln(.5)
*t
5730
ln(.35)
t
 ln(.5) 


5730


t  8678.50428
This means that our wooden
artifact is approximately
8678.50428 years old.
An infectious strain of bacteria increases
in number at a relative growth rate of
190% per hour. When a certain critical
number of bacteria are present in the
bloodstream, a person becomes ill. If a
single bacterium infects a person, the
critical level is reached in 24 hours. How
long will it take for the critical level to be
reached if the same person is infected
with 10 bacteria?
continued on next screen
An infectious strain of bacteria increases in number at a relative growth rate
of 190% per hour. When a certain critical number of bacteria are present in
the bloodstream, a person becomes ill. If a single bacterium infects a
person, the critical level is reached in 24 hours. How long will it take for the
critical level to be reached if the same person is infected with 10 bacteria?
In order to answer this question, we have to know the relative growth rate
and the number of bacteria that is classified as the critical level. In the
problem, we are give the relative growth rate (r = 1.9). We know that if
you start with 1 bacteria, you will have the critical level number in 24
hours. Thus we can use a model to find the critical level which we need to
answer the question.
A  1e
1.9*24 
A  63654392070000000000
Thus we need to find out how long it will take to reach this critical
number 63654392070000000000. We will plug in this number for A
and 10 for P and solve for t
continued on next screen
An infectious strain of bacteria increases in number at a relative growth rate
of 190% per hour. When a certain critical number of bacteria are present in
the bloodstream, a person becomes ill. If a single bacterium infects a
person, the critical level is reached in 24 hours. How long will it take for the
critical level to be reached if the same person is infected with 10 bacteria?
Thus we need to find out how long it will take to reach this critical
number 63654392070000000000. We will plug in this number for A
We will plug in 10 for P since we are asked how long it takes to get to
the critical level if we start with 10 bacteria. Now we solve for t.
63654392070000000000  10e 1.9*t 
63654392070000000000
 e 1.9*t 
10
63654392070000000000
ln
 1.9t
10
 63654392070000000000 
ln

10

 t
1.9
t  22.78811311
It will take 22.7881131 hours to reach the critical level or
for a person to become ill.