Learning activity 3.3: Solving problems involving exponential functions

Exponential functions in the real world

Previously, you learned about the properties of exponential functions with equations ${y = a}^{x} .$

In this learning activity, you will learn to use exponential functions to model and solve real-world problems involving exponential growth or decay. You will learn to solve problems using exponential equations as well as the graphs of exponential functions.

Exponential functions are applicable in many real-world situations. The most common applications are population growth, exponential decay and compound interest.

For instance, the following is a graph of the growth of the solar power photovoltaic global capacity from 1995 to 2012.

On the vertical axis, the dependent variable, is the capacity of solar power. This is measured in gigawatts. To familiarize yourself with a gigawatt, explore your preferred internet search engine and enter the terms “how much power is 1 gigawatt” and “United States government.” You may consider other search terms to find resources to help your understanding.

You probably noticed that the shape of the preceding graph represents an exponential function.

The graph can be used to find information about this situation. For example, to estimate the capacity of solar power for the year 2024, extend the curve and find the year 2024 on the horizontal axis. From here follow the vertical line up to the place where it intersects the curve. From this intersection point, draw a horizontal line over to the vertical axis to locate the corresponding value. It is approximately 5,800 gigawatts, as you can notice on the following graph:

You can also estimate when the capacity of solar power is expected to reach 3,000 gigawatts. Find the number 3,000 on the vertical axis. Trace a horizontal line across until it intersects the curve. From this point draw a vertical line down to the horizontal axis and find the corresponding value. It is expected that in 2,022, the capacity of solar power should reach 3,000 gigawatts.

Notebook

You can use your notebook to work through the examples as your learning. Compare your work with the suggested answers to check your understanding.

Example 1

Now you can try finding information when given a graph of a situation modelled by an exponential function.

Amy invests $2,000 at a rate of 7% compounded annually. The following graph represents the growth of her investment:

Use the graph to answer the following questions. When you’re finished, compare your answers to the suggested ones.

How much money will Amy have after 22 years?

Find 22 years on the horizontal axis , then move your finger up to the graph. Locate the corresponding value on the vertical axis. It is approximately $9,000.

So, after 22 years, Amy will have about $9,000.

How long will it take for Amy’s investment to double in value?

When doubled, $2,000 becomes $4,000. Find the number 4 ($4,000) on the vertical axis. Draw a horizontal line across to the graph. From here, indicate a vertical line down to the horizontal axis to find the corresponding value. It is approximately 10 years.

Therefore, Amy’s investment will double in approximately 10 years.

How long will it take for Amy’s investment to triple in value?

When multiplied by 3, Amy’s $2,000 becomes $6,000. Find the number 6 ($6,000) on the vertical axis indicate a horizontal line across to the curve. Then, draw a vertical line down to the horizontal axis to find the corresponding value. It is approximately 16 years until Amy has $6,000.

Therefore, Amy’s investment will triple in approximately sixteen years.

Example 2

At 9:00 a.m., Gina finds out that they have been promoted to vice-president of sales at their office. The news about Gina’s promotion continues to spread this way throughout the company.

To determine what type of function is represented by this situation, use a table to generate data and then create the corresponding graph.

To construct a table of values, let $t$ represent the number of half-hour intervals and let $n$ represent the total number of people who have heard about the promotion at the end of each half-hour.

In order to determine the type of function:

Complete the table beside on the number of people who have heard the rumour in your notebook (the first two entries have been completed for you).

t	n (number of people who have heard the rumour)
0	1
1	2
2	4
3	8
4	16
5	32
6	64
7	128

Use the table we just examined to generate a graph.

What type of function does this graph represent?

As shown by both the table and the graph, the spread of this rumour can be modelled by an exponential function.

It is important to note that the ratios of the differences should be constant for a table of values to represent an exponential graph. You can review this concept in Learning Activity 3.2.

Exponential growth

As you learned earlier, exponentials or “powers” were previously defined as a mathematical operation of repeated multiplication. Repeated multiplication occurs frequently in problems that involve growth.

Let us examine the following graphs from earlier:

Notice that the two graphs rise from left to right. Each graph represents a situation of exponential growth. As the values along the horizontal axis increase, the values along the vertical axis increase slowly at first and then increase extremely rapidly.

The graph of the basic exponential function ${y = a}^{x}$ increases when the value of the base is greater than 1 therefore, the equations of functions that represent exponential growth all have a base that is greater than 1.

Think

Many real-world applications represent exponential growth. Think of examples that would be modelled by exponential growth and then compare them with some suggested examples:

the spread of a virus or rumour
the growth of bacteria
the growth of the population of a town or city, and
the growth of an investment

In the Minds On section of this learning activity, you estimated values from a given graph by drawing (or following) dotted lines. As was mentioned, this estimation method is not a very precise way of solving a problem.

Exponential model and growth

Suppose a quantity $y$ doubles every $T$ units of time. If $y_{0}$ is the initial amount, then the quantity y after t units of time is given by $y = y_{0} {(2)}^{\frac{t}{T}}$

Example 1

Let $t = time$ in days. At $t = 0$ , initially there were 20 bacteria. Suppose that the bacteria doubles every 100 hours. Give the exponential model for the bacteria as a function of $t$ .

Initially at $t = 0$

Initially at $t = 100$

Initially at $t = 200$

Initially at $t = 300$

Number of bacteria $= 20$

Number of bacteria $= 20 (2)$

Number of bacteria $= 20 {(2)}^{2}$

Number of bacteria $= 20 {(2)}^{3}$

An exponential model for this situation is $y = 20 {(2)}^{\frac{t}{100}}$ .

Example 2

At $t = 0$ , 500 bacteria are in petri dish and this amount triples every 15 days.

Give an exponential model for the situation.
How many bacteria are in the dish after 45 days.

Solution:

If $y_{0} = 500$ is the initial amount, bacteria triples and $T = 15$ is unit of time, then

Exponential model is $y = {500 (3)}^{\frac{t}{15}}$
Substitute $t = 50$ in the exponential model to get bacteria in the dish after 45 days, which gives

$y = {500 (3)}^{\frac{45}{15}}$

$= {500 (3)}^{3}$

$= 13,500$

Therefore, after 45 days the total number of bacteria in the dish are 13,500.

In the following exercise, you’ll use the equation as well as the graph of the exponential function to solve problems involving exponential growth.

Notebook

Example 3

A start-up social media company launched a new app that allows users to connect to each other via virtual reality (VR) devices. After number of users had grown substantially since its debut. The user growth can be modelled by the equation ${u (t) = 5 (1.83)}^{t}$ , where $u$ is the number of users in thousands and $t$ is the number of years since the app’s launch.

Complete the table of values beside in your notebook. Approximate your values to one decimal place.

t	${u (t) = 5 (1.83)}^{t}$
0	5
1	9.2
2	16.7
3	30.6
4	56.1
5	102.6
6	187.8
7	343.7
8	628.9
9	1,150.9

Using your notebook, and the table of values, graph the equation for 0 ≤ t ≤ 10.

To graph the equation, you will need to decide on an appropriate scale and appropriate labels. You are interested in finding the number of users at a particular time; therefore, time is the input (that is, the known variable).

Since t is the input variable, it is plotted along the horizontal axis. Since u is the output variable, it is plotted along t a grid. Draw a smooth curve to connect the points, as shown in the following graph:

What was the initial number of users at the app’s launch?

The initial year for which you have data is the app’s launch date, which is represented by $t = 0$ .

Substitute $t = 0$ in the equation and then solve for u:

${u (0) = 5 (1.83)}^{(0)}$

Recall from the exponent laws: ${1.83}^{0} = 1$

$u (0) = 5 (1)$

$u (0) = 5$

The number of users was 5,000 at launch.

Estimate the number of users 12 years after the app’s launch.

Substitute $t = 12$ in the equation and then solve for $u$ :

$u (12) = {5 (1.83)}^{(12)}$

$u (12) = 5 (1,410.63)$

$u (12) = 7,053.14$

$u (12) = 7.05 m i l l i o n$

If the current growth rate holds, then 12 years after the app’s launch, there should be approximately 7.05 million users.

According to the model, when will the number of users reach 500,000?

Substitute $u = 500$ in the equation and then solve for $t$ :

$500 = {5 (1.83)}^{t}$ Divide each side by 5

$100 = {1.83}^{t}$

From the table of values, you can determine that $u = 500$ lies between $t = 7$ and $t = 8$ , so the nearest whole number is $t = 7$ .

Therefore, seven years after its launch, the app should have 500,000 users.

Another method is to use trial and error to find a value of $t$ so that ${(1.83)}^{t} = 100$ .

Exponential decay

When the graph of an exponential function decreases—that is, when the value of $y$ decreases as the value of $x$ increases—then the situation represents exponential decay. As you have learned, the graph of the basic exponential function ${y = a}^{x}$ decreases when the value of the base is between 0 and 1. Therefore, the equations of functions that represent exponential decay all have a base that is between 0 and 1.

Explore this!

Explore the following video to identify the graphs of exponential growth and functions.

Many real-world applications represent exponential decay.

Notebook

Can you think of examples that would be modelled by exponential decay? Make a list of a few examples in your notebook, then compare them with the suggested examples.

the amount of prescription drug in your body
the amount of caffeine in your body
radioactive decay of chemical substances
depreciation of a value of an object, such as a car

Exponential model and exponential decay

Example

Suppose that the half-life of a certain radioactive substance is 10 days and the initial amount of substance is 10g. Find the amount of substance remaining after 30 days.

Solution

Let $t$ be time in days

Initially, when, $t = 0$

when, $t = 10$

when, $t = 20$

when, $t = 30$

Amount of substance $= 10 g$

Amount of substance $= 5 g$

Amount of substance $= 2.5 g ($ Half of $5 g)$

Amount of substance $= 1.25 g ($ Half of $5 g)$

The exponential model in this situation is $y = {10 (\frac{1}{2})}^{\frac{t}{10}}$ .

Try it!

Unsold rechargeable batteries on store shelves lose their charge, even though they are not in use. This is why you should fully charge a rechargeable battery before its first use. A cellphone battery loses 2% of its charge every day, when not in use.

The percent charge, $C$ , that remains in the battery after $t$ days can be modelled by the equation ${C (t) = 100 (0.98)}^{t}$ .

Use the equation to estimate the percent charge remaining after:

25 days

Substitute $t = 25$ into the equation and solve:

$\begin{array}{l} C (t) = 100 {(0.98)}^{t} \\ C (25) = 100 {(0.98)}^{25} \\ C (25) = 60.3% \end{array}$

50 days

Substitute $t = 50$

${C (50) = 100 (0.98)}^{(50)}$

$C (50) = 36.4%$

75 days

Substitute $t = 75$

${C (75) = 100 (0.98)}^{(75)}$

$C (75) = 22.0%$

100 days

Substitute $t = 100$

${C (100) = 100 (0.98)}^{(100)}$

$C (100) = 13.3%$

Use the equation ${C (t) = 100 (0.98)}^{t}$ to determine the time it will take for the battery to lose half its charge.

The half-life of 100% is 50%.

Substitute $C = 50$ : $50 = 100 {(0.98)}^{t}$

Divide each side by $100$ : $0.5 = {(0.98)}^{t}$

Use trial and error to find a value of $t$ .

From earlier, you know that when $t = 25$ , the charge remaining is 60.4% and when $t = 50$ the charge remaining is 36.4%.

Try different values of $t$ between 25 and 50 to find the one that gives the closest value to 50%. The solution is $t = 34$ because ${(0.98)}^{34} = 0.50$ .

Therefore, the battery will have half its charge after 34 days.

Notebook

Using your notebook, graph the following function.

Self-check

Rate your level of understanding from 1 (I am still confused) to 5 (I fully understand this concept) based on your results from the questions you just completed.

I am able to:

Statement	1	2	3	4	5
Identify and solve problems involving exponential growth
Identify and solve problems involving exponential decay

If there are any criteria where you rated your level of understanding a 3 or below, you should review the concepts before moving on to the next learning activity.

Math journal

At the end of the course, you will fine-tune 8 entries (two from each unit) from your math journal and submit them as your “Culminating assessment - Math journal” (Opens in new window)

Take two pictures; one representation of exponential decay and the other of exponential growth in the real world and add these to your math journal. Explain how you know why they represent exponential functions and how you know which is decay and which is growth.

Once you are comfortable with the success criteria, complete the questions below to assess your progress.

Notebook

Check your understanding

Now you have the opportunity to check your understanding of real-world applications of exponential functions.

Problem 1:

Problem 1: Population growth

With the advent of modern medicine and modern farming, for many, there is plenty to eat and people do not die from disease as often as they did in the past. As a result, the human population has been growing at an amazing rate.

This population explosion has caused many people to move to cities in order to find housing and employment. For example, the population of Vancouver in 1940 was approximately 19,000 people. Today, the greater Vancouver area has more than 2.4 million people.

This question explores a similar situation in the fictional town of Mathville.

In 1958, the population of the town of Mathville was 3,000. From 1958 onward, the population doubled every 6 years.

Press the Question 1 tab to begin.

Question 1

Complete a table like the following detailing Mathville’s population growth since 1958 in your notebook:

Year	Six-year interval	Population
1958	0	3,000
1964	1	6,000
1970	2	12,000
1976	3	24,000
1982	4	48,000
1988	5	96,000
1994	6	192,000
2000	7	384,000
2006	8	768,000
2012	9	1,536,000

Question 2

In your notebook graph the data.

To graph the data on graph paper, choose an appropriate scale for each axis. For the horizontal axis, let one square represent one six-year interval, starting with 1958. Label the horizontal axis Time (Six-year intervals from 1958).

For the vertical axis, let one square represent 300,000. To avoid writing the final three zeros, use 300, 600, 1,200, and so on and label the vertical axis Population (in thousands).

Plot the points and then draw a smooth curve to connect them, as shown in the following graph:

Question 3

Using your graph, estimate the population in 2009.

Notice from the table that the year 2009 falls between 2006 and 2012. This represents the interval 8.5 along the horizontal axis. Place your finger halfway between 8 and 9 on the horizontal axis. Move your finger up the graph and then find the corresponding value on the vertical axis. It is 1,100. The dotted lines drawn on the graph illustrate how to estimate the population at this time. Note that the vertical scale represents the population in thousands, which means that 1,100 becomes 1,100,000.

Using the following graph, the estimated population in 2009 is 1,100,000.

Question 4

Use your graph to estimate when the population will reach 5 million.

To determine when the population will reach five million, find five million on the vertical axis. It is the value 5,000. Place your finger here and then move horizontally to touch the graph. Find the corresponding value on the horizontal axis. The dotted lines on the following graph illustrate this procedure.

The value is approximately 10 and $\frac{2}{3}$ intervals.

This corresponds to $(10 \times 6) + (\frac{2}{3} \times 6) = 64$ years since 1958 which would be the year 2022.

So the population of Mathville will reach five million in 2022.

Problem 2

Coffee, tea, cola and chocolate contain caffeine. After caffeine enters your body, it is processed by the liver, which breaks it down into other substances that are excreted from the body. When you consume caffeine, the percent, $P$ left in your body is represented by the function ${P (n) = 100 (0.87)}^{n}$ , where $n$ is the elapsed time in hours.

Complete a table of values like the one besides:

n	P
0	100
2	75.7
4	57.3
6	43.4
8	32.8
10	24.8
12	18.8
14	14.2
16	10.8
18	8.2
20	6.2
22	4.7
24	3.5

In your notebook, graph the table of values for caffeine consumption.

Use the equation to estimate the percent of caffeine remaining in your body after:

1 hour

Substitute $n = 1$

${P (1) = 100 (0.87)}^{1}$

$P (1) = 87 %$

After one hour, the amount of caffeine remaining is 87%.

9 hours

Substitute $n = 9$

${P (9) = 100 (0.87)}^{9}$

$P (9) = 28.6 %$

After nine hours, the amount of caffeine remaining is 28.6%.

15 hours

Substitute $n = 15$

${P (15) = 100 (0.87)}^{15}$

$P (15) = 12.4 %$

After 15 hours, the amount of caffeine remaining is 12.4%.

Use the equation to determine how long it will take for the percent of caffeine to drop by 50%.

${50 = 100 (0.87)}^{n}$

${0.5 = (0.87)}^{n}$

Use trial and error. From the values in the table, try $n = 5$ .

${(0.87)}^{5} = 0.5$

It takes approximately five hours for the caffeine to drop to 50%.