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Section 3.4 Gist of Exponential Functions

In this lesson we learn about a family of functions called exponential. Just like linear functions, exponentials have their or own distinct numerical patterns, unique graphs and general formulas.

Subsection 3.4.1 Exponential Patterns in Data

When looking at data, linear functions will have a pattern of adding the same number repeatedly (a positive or negative constant) for constant changes in input. This leads to the fact that linear functions have constant rate of change (slope).

Input Output
\(0\) \(5\)
\({}+5\rightarrow\) \(5\) \(6.25\) \(\leftarrow{}+1.25\)
\({}+5\rightarrow\) \(10\) \(7.5\) \(\leftarrow{}+1.25\)
\({}+2\cdot5\rightarrow\) \(20\) \(10\) \(\leftarrow{}+2\cdot1.25\)
\({}+3\cdot5\rightarrow\) \(35\) \(13.75\) \(\leftarrow{}+3\cdot1.25\)
\({}+5\rightarrow\) \(40\) \(15\) \(\leftarrow{}+1.25\)

The table shows \(1.25\) units added to the output for every \(5\) unit change in the input.

In turn, exponentials have a pattern of repeated multiplication by the same number called a growth factor. This leads to the fact that exponential functions have constant percent change for constant changes in input. Multiplying is interpreted as a percent change.

Input Output
\(0\) \(5\)
\({}+5\rightarrow\) \(5\) \(6.25\) \(\leftarrow{}\times1.25\)
\({}+5\rightarrow\) \(10\) \(7.8125\) \(\leftarrow{}\times1.25\)
\({}+5\rightarrow\) \(20\) \(12.207\) \(\leftarrow{}\times1.25\)
\({}+5\rightarrow\) \(35\) \(23.842\) \(\leftarrow{}\times1.25\)
\({}+5\rightarrow\) \(40\) \(29.802\) \(\leftarrow{}\times1.25\)

The table shows the output increases by \(25 \%\) (multiplication by \(1.25\)) for every \(5\) unit change in the input.

Two standard ways of communicating exponential information are through the measures of half-life and doubling time.

The half-life is defined as the time it takes to reduce a quantity by \(50 \%\) or by \(\frac{1}{2}\text{.}\)

The doubling-time is defined as the time it takes for something to double.

The units of half-life or doubling-time are therefore time units, like hours, days or even years.

Since half-life and doubling times are assumed to be constants, anything that has a half-life or doubling-time is therefore modeled exponentially. In other words, only exponential functions have a half-life or a doubling time.

A certain pain relieving drug has a half-life of about \(3\) hours. If a person has this drug in their system, how long will it take for the drug to reduce to \(10 \%\) of the drug in the body?

Solution

We use an exponential model \(ab^{t}\) because of the repeated \(50 \%\) loss. The problem states it takes \(3\) hours to reduce by \(50 \%\text{.}\) Therefore the hourly growth factor, \(b\) occurs \(3\) times for a total of \(50 \%\) loss. In math we say

\begin{equation*} b^{3}=0.5 \end{equation*}

and solving for \(b\) we get

\begin{align*} b^{3} \amp=0.5\\ \left(b^{3}\right)^{1/3} \amp=(0.5)^{1/3}\\ b \amp=(0.5)^{1/3}\text{.} \end{align*}

Because the problem does not tell us exactly how much of the drug was in the system to begin with, for simplicity we will imagine that beginning amount was \(1\)mg. Now, knowing the hourly percent loss, we can write a formula

\begin{align*} D(t) \amp= ab^t\\ \amp= 1 \left((0.5)^{1/3}\right)^t\\ \amp= (0.5)^{t/3} \end{align*}

where \(D(t)\) is the amount of drug in the body in mg and \(t\) is time in hours.

We use the formula to determine how many hours it takes to reduce to \(10 \%\text{.}\)

In this example we solve the equation

\begin{equation*} (0.5)^{t/3}=0.1 \end{equation*}

graphically.

Figure 3.4.4.

The cost of renting an apartment in a growing town has increased by \(21.4 \%\) over the last \(5\) years. If this trend continues, in how many years will rental prices double?

Solution

We use an exponential model \(ab^{t}\) because of the repeated \(21.4 \%\) growth every \(5\) years. The annual growth factor \(b\) is found by realizing it occurs \(5\) times for a total growth of \(21.4 \%\text{.}\) This means that

\begin{equation*} b^{5}=1.214 \end{equation*}

and solving this equation we get

\begin{align*} b^{5} \amp=1.214\\ (b^{5})^{\frac{1}{5}} \amp=(1.214)^{\frac{1}{5}}\text{.} \end{align*}

Our model is therefore \(ab^{\frac{t}{5}}\) where \(a\) is the current cost of renting an appartment. We want to find the time it takes for the cost of renting to double. Since the cost of renting an apartment now is \(a\) dollars, then we want to know when the price will be \(2a\text{.}\)

In math we would write

\begin{equation*} ab^{\frac{t}{5}}=2a \end{equation*}

but since \(a\) cancels on both sides we can just write

\begin{equation*} b^{\frac{t}{5}}=2\text{.} \end{equation*}

Solving this equation for \(t\) gives us the doubling time.

In this example we solve the equation graphically

Figure 3.4.6.

If the current rate of increase continues, rental prices will double in approximately \(18\) years.

Subsection 3.4.2 Graph of the Exponential Function

Consider the difference graphically between the tables of data in the previous examples. In the linear table 3.4.1 we are adding \(1.25\) to the output every \(5\) units of input. In the exponential table 3.4.2 we are multiplying the output by \(1.25\) (i.e. a \(25 \%\) increase) for every \(5\) units of input.

Figure 3.4.8. Adding \(1.25\) every \(5\) units

Plotting the ordered pairs for the linear model gives us a graph of a straight line. The graph has both horizontal and vertical intercepts.

Figure 3.4.10. Multiplying by \(1.25\) every \(5\) units

Plotting the ordered pairs for the exponential model gives us a graph of a smooth curve. In this example the curve is concave up.

Consider the graphs of the two exponential functions.

Figure 3.4.11. Constant percent decrease
Figure 3.4.12. Constant percent increase

Notice that as you move along the \(x\)-axis, each graph tends to drop closer and closer to the \(x\)-axis. The decreasing exponential approaches the \(x\)-axis as you move to the right while the increasing exponential approaches the \(x\)-axis as you move to the left.

In fact, the farther you move right or left, the closer each graph gets to the \(x\)-axis. In truth neither graph will ever reach the \(x\)-axis.

The x-axis is the horizontal line \(y = 0\) and the graphs are “tending towards \(y = 0\)” as the inputs move to the right or left.

We say that the \(x\)-axis is a horizontal asymptote with equation \(y = 0\text{.}\) It means that the graph is getting closer and closer to the \(x\)-axis in at least one direction.

Subsection 3.4.3 General Form of Exponential Formulas

Patterns for linear and exponential models are found in language too.

For instance statements like, “An hourly rate of \(\$ 15\) per hour” or “A cost of \(\$ 2\) per square foot” are both providing information as linear patterns. The statements indicate some constant amount is added (or subtracted) to the output for each increase in the input.

On the other hand statements like, “An annual decrease of \(15 \%\)” or “...is expected to rise by \(2.1 \%\) per person”, are language examples of exponential patterns. These statements indicate the output is changing by the same multiple (percent) for each change in input.

By counting how many times we add or how many times we multiply, we can interpret a pattern in language into a math pattern, usually called a formula.

The forms of linear and exponential formulas are unique. They each have their own characteristics.

Linear formulas may look like

\begin{equation*} y = mx + b \end{equation*}

or

\begin{equation*} ax+by=k \end{equation*}

or

\begin{equation*} y=m(x-h)+k \end{equation*}

There are many forms, but most likely the first one is most familiar.

The formula for an exponential function may look like \(f(x) = ab^{x}\) or \(f(x)=ae^{kx}\text{.}\)

\(ab^{x}\) is called the discrete form of the exponential and \(ae^{kx}\) is called the continuous form.

There is also the financial exponential model for compounded interest

\begin{equation*} f(x)=\left(1+\frac{r}{n}\right)^{(n \cdot t)} \end{equation*}

In the linear function \(f(t)=mt+b\) the \(t\) is a counter. It counts how many times the number \(m\) is repeatedly added.

If a car is traveling at \(50\) miles per hour, then we can write a linear equation to calculate the distance the car travels as a function of time, \(t\) hours.

\begin{equation*} f(t)=50t \end{equation*}

means that each hour we add another \(50\) miles to the distance traveled by the car. After \(6\) hours of driving, the vehicle has covered

\begin{equation*} f(6)=50+50+50+50+50+50=50*6=300 \end{equation*}

\(300\) miles.

If the vehicle had already covered \(60\) miles when we started counting, then the equation could be written as

\begin{equation*} f(t)=60+50t \end{equation*}

which simply says we are adding to the \(60\) miles that were already covered.

The reason there are at least two forms of the exponential formula is there are at least two ways to derive or create an exponential model.

In practice, when we observe or measure a constant percent change we typically use the information collected to write a formula directly from the data. The result is the discrete form of the exponential formula.

We can measure the increase in a population over a period of time and directly apply our findings into a discrete exponential equation to model the population growth.

A population of \(500\) grows by \(8 \%\) every \(10\) years.

\(P(t)=500(1.08)^{\frac{t}{10}}\)

Notice how we went from an observation about the population growth directly to an equation without any algebra. We literally translated information from one language, English, to another, Math.

However, with just one algebra step we can get an equation to model the annual percent increase.

\begin{gather*} P(t)=500(1.08)^{\frac{t}{10}}\\ P(t)=500(1.08^{\frac{1}{10}})^{t}\\ P(t) \approx 500(1.007726)^{t} \end{gather*}

In this example \(a=500\) is the initial size of the population when the measuring began and \(b \approx 1.007726\) is the annual growth factor (i.e. the number that is multiplied repeatedly) representing an annual percent growth of about \(0.7726 \%\) each year.

Another way to arrive at an exponential model is through higher level calculus called differential equations. Of course we will not get into this subject, but here is a tiny little glimpse.

The annual rate of growth of a population is proportional to the size of the population. The bigger a population is, the faster it grows.

In calculus this simple statement looks like

\begin{equation*} \frac{dP}{dt}=kP \end{equation*}

where \(t\) measures time, like years, and \(\frac{dP}{dt}\) is called the instantaneous rate of change. It is similar to the average rate of change \(\frac{\Delta P}{\Delta t}\) we study in this course.

This equation can be solved using calculus, but instead of the solution being a number like \(t=7\) the solution is another formula.

Specifically, the solution is

\begin{equation*} P(t)=ae^{kt} \end{equation*}

If a solution to a differential equation is

\begin{equation*} P(t)=300e^{0.007726t} \end{equation*}

then as a scientist or engineer, you must recognize the solution as the continuous form of exponential growth. You should also know the equation is stating an initial population size of \(300\) and a continuous annual growth of \(0.7726 \%\text{.}\)

But, most importantly, you have to know the continuous \(0.7726 \%\) growth is NOT the actual growth of the population. Continuous information must be converted into its discrete equivalent in order to get practical use from the solution.

Since

\begin{equation*} b=e^{0.007726} \approx 1.007756 \end{equation*}

we find the continuous growth represents an actual annual percent growth of about \(0.7756 \%\text{.}\)

In the “Compounding and the Number e” activity we learned that frequent, low interest charges or payments can cost or make more money over time than charging a larger interest rate just once.

For instance, consider a \(\$ 2300\) investment that earns \(8 \%\) annual interest.

Waiting to the end of each year to receive an \(8 \%\) increase will result in

\begin{equation*} 2300\left(1+\frac{0.08}{1}\right)^{1 \cdot 1} \approx \$ 2484 \end{equation*}

at the end of one year.

Over a period of \(5\) years the same investment would earn

\begin{equation*} 2300\left(1+\frac{0.08}{1}\right)^{1 \cdot 5} \approx \$ 3379.45 \end{equation*}

However, by compounding the same \(15 \%\)interest monthly, with \(12\) months in a year, the investment would collect

\begin{equation*} 2300\left(1+\frac{0.08}{12}\right)^{12 \cdot 1} \approx \$ 2490.90 \end{equation*}

with

\begin{equation*} 2300\left(1+\frac{0.08}{12}\right)^{12 \cdot 5} \approx \$ 3426.65 \end{equation*}

over \(5\) years.

Subsection 3.4.4 Conclusion

Test your understanding of this chapter with the following exercises.

Student Learning Outcome: Define, distinguish and apply exponential functions.

Student Learning Outcome: Sketch the graph of an exponential function.

Student Learning Outcome: Identify a horizontal asymptote from a graph, numerical model, verbal description or equation.

Student Learning Outcome: Distinguish growth vs. decay given any model (verbal, numerical, graphical, equation).

Student Learning Outcome: Determine percent change of an exponential function.

Student Learning Outcome: Model real-world situations algebraically using exponential functions.

Identify the meaning of the parameters of an exponential function.

Exercises 3.4.5 Exercises

Formulas for Exponential Functions

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Evaluating Exponential Functions

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Building a Formula

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Exponential vs. Linear

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Additional Problems

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