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6. Exponential & Logarithmic Functions
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Algebra 2

6.02 Applications of exponential functions

Lesson
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Exponential growth and decay

Exponential functions can be classified as exponential growth functions or exponential decay functions based on the value of the base.

\displaystyle f\left(x\right)=ab^x
\bm{a}
The y-intercept
\bm{b}
The base
Exponential growth

Increasing by a base, b, where b>1

Exponential decay

Decreasing by a base, b, where 0<b<1

Growth factor

The base of an exponential growth function.

Decay factor

The base of an exponential decay function.

Exponential functions can also be expressed in terms of their constant percent rate of change.

\displaystyle f\left(x\right)=a\left(1\pm r\right)^x
\bm{a}
The initial value
\bm{r}
The growth or decay rate
Growth rate

The fixed percent by which an exponential function increases.

In the form f\left(x\right)=a\left(1+r\right)^x, r is a growth rate, and the constant percent rate of change is positive.

Decay rate

The fixed percent by which an exponential function decreases.

In the form f\left(x\right)=a\left(1-r\right)^x, r is a decay rate, and the constant percent rate of change is negative.

Exploration

Here are three exponential functions represented in three different ways:

Function 1:

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Function 2:

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Function 3:

Reagan decided to get a part-time job to begin saving for a new car. He was hired by a company that pays \$21\,500 in the first year, and he will receive a 5\% raise each year after that.

  1. Which of the functions would you model using y=ab^x and why?
  2. Which of the functions would you model using y=a\left(1+r\right)^x and why?
  3. In what type of situations would you prefer knowing the growth factor, b, over the growth rate, r?

We can determine the key features of an exponential function from its equation or graph:

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  • f\left(x\right)=ab^x, b>1

  • f\left(x\right)=a\left(1+r\right)^x, r>0

  • The graph is increasing

  • y approaches a minimum value of 0

  • The domain is \left(-\infty, \infty\right)

  • The range is \left(0, \infty\right)

  • The y-intercept is at \left(0,\, a\right)

  • The horizontal asymptote is y=0

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  • f\left(x\right)=ab^x, 0<b<1

  • f\left(x\right)=a\left(1-r\right)^x, r>0

  • The graph is decreasing

  • y approaches a minimum value of 0

  • The domain is \left(-\infty, \infty\right)

  • The range is \left( -\infty, 0\right)

  • The y-intercept is at \left(0,\, a\right)

  • The horizontal asymptote is y=0

Examples

Example 1

A frog population, F, has been following the model F(n)=480(0.952)^n, where n is the number of years.

a

What is the initial population of frogs?

Worked Solution
Create a strategy

The initial value will be when time is 0.

Apply the idea

When n=0, F=480, so the population was 480.

Reflect and check

Since negative values for time do not make sense, the y-intercept will be the leftmost point on the graph. In this case, we refer to it as the initial value.

b

What is the decay rate? Explain what it means in this context.

Worked Solution
Create a strategy

We can find the decay rate using that for y=ab^x, b=1-r has decay rate of r.

Apply the idea

Since b=0.952:

\displaystyle b\displaystyle =\displaystyle 1-r
\displaystyle 0.952\displaystyle =\displaystyle 1-r
\displaystyle r\displaystyle =\displaystyle 1-0.952
\displaystyle r\displaystyle =\displaystyle 0.048

This means that the decay rate is 4.8\%. The frog population is declining at 4.8\% per year.

c

Find F(3), interpret what this means in the context.

Worked Solution
Create a strategy

We can substitute n=3, and then evaluate using a calculator rounding to the nearest whole number.

Apply the idea
\displaystyle F(n)\displaystyle =\displaystyle 480(0.952)^n
\displaystyle F(3)\displaystyle =\displaystyle 480(0.952)^3
\displaystyle F(3)\displaystyle \approx\displaystyle 414.1446758
\displaystyle F(3)\displaystyle \approx\displaystyle 414

After 3 years, there would be 414 frogs remaining.

d

Using technology, graph the frog population for 0 <n< 120, and describe the end behavior.

Worked Solution
Create a strategy

We can input the equation and then adjust the view using the zoom so we can see from n=0 to n=120.

Apply the idea

A good window setting is 0<n<120 and 0<F(n)<480.

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Based on the graph, we can see that the frog population is decreasing over time. The end behavior of the function shows that as the number of years n increases, the frog population F(n) approaches 0 but never reaches it. This indicates that the frog population will continue to decrease.

Reflect and check

By utilizing technology to monitor frog populations, we observe a declining trend over time, with the numbers approaching zero without actually reaching it. If this trend persists, it's important to note that in reality, a population cannot include a fraction of a frog. Therefore, the actual number of frogs would ultimately drop to zero.

e

Use your graph to determine approximately how many years it will take for the population to reach half of the original population.

Worked Solution
Create a strategy

We want to find half of 480 on the vertical axis and then slide over to find where the graph reaches that value.

Apply the idea

Half of 480 is \dfrac{480}{2}=240.

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Using our original scale we can see this is between n=10 and n=15, so we can zoom in to get a better estimation.

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From the zoomed in graph we can see that it is just after 14 years where the population reaches half of the original value.

Reflect and check

We can solve for the exponent using a graph, table of values, or writing both sides of the equation with the same base if possible.

Example 2

The population of rabbits in Lincoln county can be modeled by an exponential growth function. Conservationists have been measuring the population since 2015. After the first year, there were 46 rabbits. After the third year, there were 66 rabbits. After the fifth year, there were 95 rabbits, as shown in the table below.

Year201420152016201720182019
Population466695
a

Estimate the growth rate to one decimal place.

Worked Solution
Create a strategy

To estimate the growth rate, we will create data points from the given information. If we let x=0 represent 2015, then x=1 will be the first year. The associated data point is (1,46). The data point for the third year is (3,66).

These are not consecutive outputs, so we need to set up an equation and solve for r.

Apply the idea

To get from the first year to the third year, we multiplied the population by the growth factor twice. The equation would be 46\cdot b \cdot b = 66 which simplifies to 46b^2=66.

\displaystyle 46b^2\displaystyle =\displaystyle 66Given equation
\displaystyle b^2\displaystyle =\displaystyle \dfrac{66}{46}Division property of equality
\displaystyle b\displaystyle \approx\displaystyle 1.198Approximate by taking the square root of both sides

This represents the growth factor, so now we must solve for the growth rate.

\begin{aligned} 1.198 &= 1+r \\ r &= 0.198 \end{aligned}

Next, we can convert to a percentage and round it to one decimal place.

\begin{aligned} r&=19.8\% \\ &= 20\% \end{aligned}

b

Estimate the initial population.

Worked Solution
Create a strategy

The initial population will be the number of rabbits present in 2015. Now that we know the growth rate, we can use the number of rabbits in the first year, 2016, to help us solve for the initial population.

Apply the idea

To set up the equation, we begin by substituting the known values r=0.2 from part (a), x=1, and y=46 into the equation y=a(1+r)^x.

\displaystyle y\displaystyle =\displaystyle a\left(1+r\right)^xExponential growth equation
\displaystyle 46\displaystyle =\displaystyle a\left(1+0.2\right)^1Substitute y=46, r=0.2, and x=1
\displaystyle 46\displaystyle =\displaystyle a\left(1.2\right)Evaluate the addition and exponent
\displaystyle \dfrac{46}{1.2}\displaystyle =\displaystyle aDivision property of equality
\displaystyle 38\displaystyle \approx\displaystyle aApproximate by evaluating the division

The initial population of rabbits in 2015 was approximately 38.

c

Write the equation that models this situation.

Worked Solution
Create a strategy

Since we discussed the percentage growth, we will use the growth rate form of the equation, y=a\left(1+r\right)^x.

Apply the idea

We found r=20\%, a=38 in parts (a) and (b), so we can simply plug them into the formula.

y=38\left(1+0.2\right)^x

Reflect and check

Because we rounded the percentage in part (a), the values will not be exact when we check our answers. But since we are working with rabbits and we cannot have a decimal of a rabbit, it makes sense to round to the nearest whole rabbit.

Year 1: y=38(1+0.2)^1=45.6 \approx 46 rabbits

Year 2: y=38(1+0.2)^2=54.72 \approx 55 rabbits

Year 3: y=38(1+0.2)^3=65.664\approx66 rabbits

These answers match the information given in the problem.

Example 3

The graph shows the exponential decay of Plutonium-236, highlighting a rapid decrease in its quantity, measured in grams, over time, measured in years, due to its radioactive decay.

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a

Estimate the remaining grams of a Plutonium-236 sample after 8 years, given its exponential decay.

Worked Solution
Create a strategy

To estimate the remaining grams of Plutonium-236 after 8 years, we can use the graph to observe the trend of exponential decay. By locating the point on the graph corresponding to x = 8 (8 years), we can estimate the value of y (remaining grams of Plutonium-236).

Apply the idea

Looking on the graph where x = 8, we can see that the exponential decay curve is close to the y-axis gridline between 6 and 12. By estimating the position of the curve at x = 8, we can approximate the remaining grams of Plutonium-236.

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By zooming in, at x = 8, the curve appears to be very close to gridlines of 6 and 7 but slightly below the midpoint of 6 and 7 on the y-axis. Considering that the middle point between 6 and 7 is 6.5, we can estimate that the remaining grams of Plutonium-236 after 8 years is approximately 6.5 grams.

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Reflect and check

An alternative way to find the remaining grams of Plutonium-236 after 8 years is to first determine the decay factor by examining the trend shown in the graph and then finding the equation of the function. Once we have the equation, we can substitute x = 8 to find the remaining grams.

Looking at the graph, we can see that the curve goes through the points (0, 64) and (1, 48). We can use these points to find the decay factor (r). The exponential decay function is in the form y = a(1 - r)^x, where a is the initial amount and r is the decay factor.

Since the initial amount at x = 0 is 64, the equation becomes y = 64(1 - r)^x. Using the point (1, 48), we can solve for r:

\displaystyle 48\displaystyle =\displaystyle 64(1 - r)Substitute x = 1 and y = 48
\displaystyle 0.75\displaystyle =\displaystyle 1 - rDivide by 64 on both sides
\displaystyle r\displaystyle =\displaystyle 0.25Subtract 0.75 from both sides

Now that we have found the decay factor to be 0.25, we can write the equation of the function as y = 64(1 - 0.25)^x.

To find the remaining grams after 8 years we can substitute into the equation and use a calculator to evaluate:

\displaystyle y\displaystyle =\displaystyle 64(1 - 0.25)^8Substitute x = 8
\displaystyle y\displaystyle \approx\displaystyle 6.41Evaluate the expression

Using this alternative method, we can estimate that the remaining grams of Plutonium-236 after 8 years is approximately 6.41 grams which is close to our initial estimate of 6.5 grams based on the graph.

b

What is the end behavior of the function and explain what that means in the context of the scenario.

Worked Solution
Create a strategy

Observe the graph to determine the end behavior of the function as x approaches positive infinity. Then, explain the meaning of this behavior in the context of Plutonium-236's radioactive decay.

Apply the idea

As x approaches positive infinity, the value of the function approaches 0, which means the quantity of Plutonium-236 always decreases as time goes on. In the context of the scenario, this indicates that over time, the amount of Plutonium-236 will continue to decrease due to its radioactive decay, eventually becoming negligible.

Reflect and check

Let's verify the end behavior of the function by using the equation 64(1-0.25)^x and plugging in large values of x using a calculator. Evaluating the function for large x values will help us understand how the function behaves as time goes on.

For instance, let's calculate the value of the function for x = 100, x = 500, and x = 1000:

\displaystyle 64(1-0.25)^{100}\displaystyle \approx\displaystyle 1.84 \cdot 10^{-7}Evaluate for x = 100
\displaystyle 64(1-0.25)^{500}\displaystyle \approx\displaystyle 1.28 \cdot 10^{-35}Evaluate for x = 500
\displaystyle 64(1-0.25)^{1000}\displaystyle \approx\displaystyle 1.64 \cdot 10^{-63}Evaluate for x = 1000

As we can see, as x increases, the value of the function approaches 0. This is a horizontal asymptote of y=0. This confirms the end behavior we observed earlier, where the amount of Plutonium-236 will continue to decrease due to radioactive decay, eventually becoming negligible.

Idea summary

Exponential functions can also be expressed in terms of their constant percent rate of change.

\displaystyle f\left(x\right)=a\left(1\pm r\right)^x
\bm{a}
The initial value
\bm{r}
The growth or decay rate

This means that for f(x)=ab^x:

  • if b>1, the growth rate will be r=b-1.
  • if 0<b<1, the decay rate will be r=1-b.

Compound interest

A common application of exponential growth is compound interest, where interest applies to the current balance, not the initial investment or loan.

We can write the compound interest formula as:

\displaystyle A=P\left(1+r\right)^n
\bm{A}
closing balance or future value of the investment
\bm{P}
principal (starting) amount
\bm{r}
periodic interest rate, as a decimal
\bm{n}
number of time periods

The compounding period is the length of time between interest payments. Here are some possible periods:

Number of period per yearLength of time
Annually11 \text{ year}
Semi-annually26 \text{ months}
Quarterly43 \text{ months}
Monthly121 \text{ month}
Bi-weekly262 \text{ weeks}
Weekly521 \text{ week}
Daily3651 \text{ day}

For example, for an investment that earns a yearly interest rate of 6\%, compounding quarterly. Since there are 4 quarters in a year:\begin{aligned}\text{Quarterly interest rate} &=6\% \text{ per } 4\text{ quarters}\\ &=\dfrac{0.06}{4}\\&=0.015\\&=1.5\%\end{aligned}

It is also important to match the total number of periods to the duration of the investment. To do this, we multiply the number of years by the number of compounding periods per year.

If interest is calculated quarterly for 5 years, there will have been: 4\text{ quarters per year}\cdot 5\text{ years}=20\text{ time periods}

Examples

Example 4

Valerie borrows \$1250 at an annual rate of 7.6\% compounding annually.

a

Which of these expressions represents the amount Valerie must repay after 15 years, assuming that she hasn't paid anything back?

A
1250 \cdot \left(1+0.076\right) \cdot 15
B
1250 \cdot \left(1+0.076\right) \cdot \dfrac{15}{12}
C
1250 \cdot \left(1+0.076\right)^{\frac{15}{12}}
D
1250 \cdot \left(1+0.076\right)^{15}
Worked Solution
Create a strategy

Use the compound interest formula: A=P\left(1+r\right)^{n}

Apply the idea
  • P=1250
  • r=\dfrac{7.6}{100}=0.076
  • n=15
\displaystyle A\displaystyle =\displaystyle P\left(1+r\right)^{n}Write the formula
\displaystyle =\displaystyle 1250 \left(1+0.076\right)^{15}Substitute the known values

The correct answer is option D.

b

How much will Valerie owe after fifteen years?

Worked Solution
Create a strategy

We can use the expression for the amount owed after 15 years, which we found in the first part of the question, and then evaluate it.

Apply the idea
\displaystyle A\displaystyle =\displaystyle 1250 \left(1+0.076\right)^{15}Expression from part (a)

Now, we can evaluate this expression:

\displaystyle A\displaystyle \approx\displaystyle 3750.54Evaluate

Therefore, Valerie will owe approximately \$3750.54 after fifteen years.

Reflect and check

Let's analyze the amount of interest Valerie has to pay over the fifteen years. We can calculate the interest by subtracting the initial amount borrowed from the total amount owed after fifteen years.

\displaystyle \text{Interest}\displaystyle =\displaystyle 3750.54 - 1250Subtract the initial amount borrowed
\displaystyle =\displaystyle \$2500.54Evaluate

The interest generated on the loan over the fifteen years is \$2500.54. This means that Valerie has to pay an extra \$2500.54 in interest over the course of the loan. As a result, it is essential for Valerie to consider this additional cost when making decisions about the loan and her financial situation.

It is always important to consider the interest that accumulates over time when taking out a loan, as it can significantly impact the total amount that needs to be repaid.

Example 5

A \$9450 investment earns interest at an annual rate of 2.6\% compounded monthly over 14 years.

a

What is the value of the investment after 14 years?

Worked Solution
Create a strategy

Use the compound interest formula: A=P\left(1+r\right)^{n}

Apply the idea

We are given that:

P=9450, r=\dfrac{0.026}{12} and n= 14\cdot 12 \text{ months}=168\text{ months}

\displaystyle A\displaystyle =\displaystyle P\left(1+r\right)^{n}
\displaystyle =\displaystyle 9450 \left(1+\dfrac{0.026}{12}\right) ^{168}Substitute the values
\displaystyle =\displaystyle \$13\,593.90Evaluate

The future value is \$13\,593.90

b

The following is a graph of the investment over time.

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Estimate how much the investment will be worth in 15 years.

Worked Solution
Create a strategy

To estimate how much the investment will be worth in 15 years, we can look at the point on the graph where x=15, and estimate the value of y.

Apply the idea
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Looking at the graph at x=15, we can see that the y-value is approximately y=14\,000.

The estimated investment in 15 years is \$14\,000.

Reflect and check

To improve the accuracy of our estimations, we can use technology to evaluate the exponential function at x=15 and at y=16\,000. Let's use the compound interest formula:

A=P\left(1+\dfrac{r}{n}\right)^{nt}

where A is the final amount, P is the principal amount, r is the annual interest rate (as a decimal), n is the number of times interest is compounded per year, and t is the number of years.

\displaystyle A\displaystyle =\displaystyle 9450\left(1+\dfrac{0.026}{12}\right)^{12(15)}Substitute P=9450, r=0.026, n=12, and t=15
\displaystyle A\displaystyle \approx\displaystyle 13,951.58Calculate the value using technology

The investment will be worth approximately \$13,951.58 in 15 years.

c

Estimate how long it will take the investment to be worth \$16\,000.

Worked Solution
Create a strategy

Locate the point on the graph where y=16\,000 and determine the corresponding x value.

Apply the idea
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At y=16\,000, the corresponding x-value is very close to x=20.

In just over 20 years, the investment will be worth \$16\,000.

Reflect and check

Let's solve for a more precise time that it will take for the investment to reach \$16,000.

\displaystyle 16,000\displaystyle =\displaystyle 9450\left(1+\dfrac{0.026}{12}\right)^{12t}Substitute P=9450, A=16\,000, r=0.026, and n=12
\displaystyle t\displaystyle \approx\displaystyle 20.27Solve for t using technology

It will take approximately 20.27 years for the investment to be worth \$16,000.

By using technology to evaluate the exponential function, we have found a more accurate approximation for the time it takes for the investment to reach \$16,000.

Idea summary

We can calculate compound interest with the formula:

\displaystyle A=P\left(1+r\right)^n
\bm{A}
closing balance or future value of the investment
\bm{P}
principal amount
\bm{r}
interest rate, as a decimal
\bm{n}
number of time periods

If interest is compounded daily, weekly, bi-weekly, monthly, quarterly, or half-yearly, then we need to convert the interest rate and number of time periods to the same units.

Outcomes

A2.F.2

The student will investigate and analyze characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions algebraically and graphically.

A2.F.2a

Determine and identify the domain, range, zeros, and intercepts of a function presented algebraically or graphically, including graphs with discontinuities.

A2.F.2c

Determine the intervals on which the graph of a function is increasing, decreasing, or constant.

A2.F.2f

For any value, x, in the domain of f, determine f(x) using a graph or equation. Explain the meaning of x and f(x) in context, where applicable.

A2.F.2g

Describe the end behavior of a function.

A2.F.2h

Determine the equations of any vertical and horizontal asymptotes of a function using a graph or equation (rational, exponential, and logarithmic).

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