Exponential Growth – Examples and Practice Problems

Exponential functions can be used to model population growth scenarios or other situations that follow patterns with growth at fixed rates. There are formulas that can be used to find solutions to most problems related to exponential growth.

Here, we will look at a summary of exponential growth and the formulas that can be used to solve these types of problems. In addition, we will look at several examples with answers of exponential growth in order to learn how to apply these formulas.

ALGEBRA
applications of exponential functions

Relevant for

Exploring examples of exponential growth.

See examples

ALGEBRA
applications of exponential functions

Relevant for

Exploring examples of exponential growth.

See examples

Summary of exponential growth

Exponential growth is a pattern of data that shows larger increases over time, creating the curve of an exponential function. For example, if a bacteria population starts with 2 in the first month, then with 4 in the second month, 16 in the third month, 256 in the fourth month, and so on, it means that the population grows exponentially with a power of 2 every month.

The following formula is used to model exponential growth. If a quantity grows by a fixed percentage at regular intervals, the pattern can be described by this function:

Exponential growth
y=a{{(1+r)}^x}

We recall that the original exponential function has the form y = a{{b}^x}. In the original growth formula, we have replaced b with 1+ r. So, in this formula we have:

  • a= initial value. This is the starting amount before growth.
  • r= growth rate. This is represented as a decimal.
  • x= time interval. This is the time that has passed.

Most naturally occurring events continually grow. For example, bacteria continue to grow over a 24-hour period. Bacteria don’t wait until the end of 24 hours to reproduce all at the same time.

To model the continuous growth that occurs naturally such as populations, bacteria, etc., we use the exponential ee can be thought of as a universal constant that represents growth possibilities using a continuous process. Furthermore, using e we can also represent growth measured periodically over time.

Therefore, if a quantity is continually growing with a fixed percentage, we can use the following formula to model this pattern:

Continuous Exponential Growth
A=A_{0}{{e}^{kt}}

In this formula we have:

  • A= final value. This is the amount after growth.
  • A_{0}= initial value. This is the amount before growth.
  • e= exponential. e is approximately equal to 2718…
  • k = continuous growth rate. It is also called the constant of proportionality.
  • atex t = elapsed time.

Exponential growth – Examples with answers

The following examples use the formulas detailed above and some variations to find the solution. It is recommended that you try to solve the exercises yourself before looking at the answer.

EXAMPLE 1

A population of bacteria grows according to the function f(x)=100{{e}^{0.02t}}, where t is measured in minutes. How many bacteria will there be after 4 hours (240 minutes)?

This is continuous growth, so we have the formula A=A_{0}{{e}^{kt}}. We can recognize the following data:

  • A_{0}=100
  • k=0.02
  • t=240

Therefore, we have:

f(240)=100{{e}^{0.02(240)}}\approx 12151

Therefore, there will be 12 151 bacteria after 4 hours.

EXAMPLE 2

A population of bacteria grows according to the function f(x)=100{{e}^{0.02t}}, where t is measured in minutes. When will the population reach 50 000?

Here, we have the same formula as the previous exercise, but now we have to find the time knowing the final quantity. We can recognize the following data:

  • A_{0}=100
  • k=0.02
  • A=50000

Therefore, we have:

50 000=100{{e}^{0.02t}}

 500={{e}^{0.02t}}

 \ln(500)=0.02t

t=\frac{\ln(500)}{0.02}

t\approx 310.73

Therefore, the population of bacteria will become 50 000 after 310.73 minutes.

EXAMPLE 3

We can model the population of a community with the formula A=10000({{e}^{0.005t}}). Here, A represents population and t represents time in years. What is the population after 10 years?

We already have a given formula: A=10000({{e}^{0.005t}}). We have to calculate the population using time t=10. Therefore, we substitute t=10 to get:

A=10000({{e}^{0.005(10)}})

 =10000({{e}^{0.05}})

=10000(1.0513)

=10513

Therefore, the population in the community after 10 years will be 10 513.

EXAMPLE 4

The population of a certain community was 10 000 in 1980. In 2000, it was found to have grown to 20 000. Form an exponential function to model the population of community P that changes through time t.

When we have continuous population growth, we can model the population with the general formula P=P_{0}({{e}^{\lambda t}}), where P_{0} represents the initial population, λ is the exponential growth constant and t is time.

Using the given information, we have to find the constant λ to complete the formula. Therefore, we have:

P=P_{0}({{e}^{\lambda t}})

20000=10000({{e}^{20 \lambda}})

 \frac{20000}{10000}={{e}^{20 \lambda}}

 2={{e}^{20 \lambda}}

 \ln(2)=20 \lambda

 \frac{\ln(2)}{20}=\lambda

 0.0347=\lambda

Thus, we can model the population growth of the community with the formula P=10000({{e}^{0.0347 t}}).

EXAMPLE 5

The population growth of a small city is modeled with the function P= P_{0}({{e}^{0.1234t}}). When did the population reach 37 500 if in 1980 the population was 12 500?

We can substitute the values in the formula with the given information:

P=P_{0}({{e}^{0.1234t}})

37500=12500({{e}^{0.1234t}})

Now, we have to solve for time:

37500=12500({{e}^{0.1234t}})

\frac{37500}{12500}=({{e}^{0.1234t}})

3=({{e}^{0.1234t}})

 \ln(3)=0.1234t

 \frac{\ln(3)}{0.1234}=t

8.9=t

Thus, the city’s population reached 37 500 in 1989.

EXAMPLE 6

One type of bacteria doubles every 5 minutes. Assuming we start with one bacterium, how many bacteria will we have at the end of 96 minutes?

We know that bacteria grow continuously, so we have to use the formula:

A=A_{0}({{e}^{kt}})

The bacteria doubles every 5 minutes, so after 5 minutes, we will have 2. We use this to find the value of k:

2=1({{e}^{5k}})

 \ln(2)=\ln({{e}^{5k}})

 \ln(2)=5k

k=\frac{\ln(2)}{5}

k=0.13863

Now, we form the equation using this value of k and solve using the time of 96 minutes:

A=A_{0}({{e}^{0.13863t}})

A=1({{e}^{0.13863(96)}})

A\approx602 248


Exponential growth – Practice problems

Practice using the exponential growth formulas with the following exercises. Solve the problems and select an answer. Check your answer to verify that you selected the correct one.

The population of a community was 12 500. After 20 years it was found that the population grew to 16 000. What will the population be after 50 years?

Choose an answer






One type of bacteria triples every 8 hours. Starting with 100 bacteria, how many will there be after 18 hours?

Choose an answer






One type of bacteria doubles every 6.5 hours. If there were 100 bacteria at the beginning, how many will there be after 1 and a half days?

Choose an answer







See also

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