Skip to main content

Section 3.1 Exponential Growth and Decay

Linear functions have constant average rate of change and model many important phenomena. In other settings, it is natural for a quantity to change at a rate that is proportional to the amount of the quantity present. For instance, whether you put $\(100\) or $\(100000\) or any other amount in a mutual fund, the investment's value changes at a rate proportional the amount present. We often measure that rate in terms of the annual percentage rate of return.

Suppose that a certain mutual fund has a \(10\)% annual return. If we invest $\(100\text{,}\) after \(1\) year we still have the original $\(100\text{,}\) plus we gain \(10\)% of $\(100\text{,}\) so

\begin{equation*} 100 \xrightarrow{\text{year } 1} 100 + 0.1(100) = 1.1(100)\text{.} \end{equation*}

If we instead invested $\(100000\text{,}\) after \(1\) year we again have the original $\(100000\text{,}\) but now we gain \(10\)% of $\(100000\text{,}\) and thus

\begin{equation*} 100000 \xrightarrow{\text{year } 1} 100000 + 0.1(100000) = 1.1(100000)\text{.} \end{equation*}

We therefore see that regardless of the amount of money originally invested, say \(P\text{,}\) the amount of money we have after \(1\) year is \(1.1P\text{.}\)

If we repeat our computations for the second year, we observe that

\begin{equation*} 1.1(100) \xrightarrow{\text{year } 2} 1.1(100) + 0.1(1.1(100) = 1.1(1.1(100)) = 1.1^2 (100)\text{.} \end{equation*}

The ideas are identical with the larger dollar value, so

\begin{equation*} 1.1(100000) \xrightarrow{\text{year } 2} 1.1(100000) + 0.1(1.1(100000) = 1.1(1.1(100000)) = 1.1^2 (100000)\text{,} \end{equation*}

and we see that if we invest \(P\) dollars, in \(2\) years our investment will grow to \(1.1^2 P\text{.}\)

Of course, in \(3\) years at \(10\)%, the original investment \(P\) will have grown to \(1.1^3 P\text{.}\) Here we see a new kind of pattern developing: annual growth of \(10\)% is leading to powers of the base \(1.1\text{,}\) where the power to which we raise \(1.1\) corresponds to the number of years the investment has grown. We often call this phenomenon exponential growth.

Preview Activity 3.1.1.

Suppose that at age \(20\) you have $\(20000\) and you can choose between one of two ways to use the money: you can invest it in a mutual fund that will, on average, earn \(8\)% interest annually, or you can purchase a new automobile that will, on average, depreciate \(12\)% annually. Let's explore how the $\(20000\) changes over time.

Let \(I(t)\) denote the value of the $\(20000\) after \(t\) years if it is invested in the mutual fund, and let \(V(t)\) denote the value of the automobile \(t\) years after it is purchased.

  1. Determine \(I(0)\text{,}\) \(I(1)\text{,}\) \(I(2)\text{,}\) and \(I(3)\text{.}\)

  2. Note that if a quantity depreciates \(12\)% annually, after a given year, \(88\)% of the quantity remains. Compute \(V(0)\text{,}\) \(V(1)\text{,}\) \(V(2)\text{,}\) and \(V(3)\text{.}\)

  3. Based on the patterns in your computations in (a) and (b), determine formulas for \(I(t)\) and \(V(t)\text{.}\)

  4. Use Desmos to define \(I(t)\) and \(V(t)\text{.}\) Plot each function on the interval \(0 \le t \le 20\) and record your results on the axes in Figure 3.1.1, being sure to label the scale on the axes. What trends do you observe in the graphs? How do \(I(20)\) and \(V(20)\) compare?

    Figure 3.1.1. Blank axes for plotting \(I\) and \(V\text{.}\)

Subsection 3.1.1 Exponential functions of form \(f(t) = ab^t\)

In Preview Activity 3.1.1, we encountered the functions \(I(t)\) and \(V(t)\) that had the same basic structure. Each can be written in the form \(g(t) = ab^t\) where \(a\) and \(b\) are positive constants and \(b \ne 1\text{.}\) Based on our earlier work with transformations, we know that the constant \(a\) is a vertical scaling factor, and thus the main behavior of the function comes from \(b^t\text{,}\) which we call an “exponential function”.

Definition 3.1.2.

Let \(b\) be a real number such that \(b \gt 0\) and \(b \ne 1\text{.}\) We call the function defined by

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

an exponential function with base \(b\text{.}\)

For an exponential function \(f(t) = b^t\text{,}\) we note that \(f(0) = b^0 = 1\text{,}\) so an exponential function of this form always passes through \((0,1)\text{.}\) In addition, because a positive number raised to any power is always positive (for instance, \(2^{10} = 1096\) and \(2^{-10} = \frac{1}{2^{10}} = \frac{1}{2096}\)), the output of an exponential function is also always positive. In particular, \(f(t) = b^t\) is never zero and thus has no \(x\)-intercepts.

Because we will be frequently interested in functions such as \(I(t)\) and \(V(t)\) with the form \(ab^t\text{,}\) we will also refer to functions of this form as “exponential”, understanding that technically these are vertical stretches of exponential functions according to Definition 3.1.2. In Preview Activity 3.1.1, we found that \(I(t) = 20000(1.08)^t\) and \(V(t) = 20000(0.88)^t\text{.}\) It is natural to call \(1.08\) the “growth factor” of \(I\) and similarly \(0.88\) the growth factor of \(V\text{.}\) In addition, we note that these values stem from the actual growth rates: \(0.08\) for \(I\) and \(-0.12\) for \(V\text{,}\) the latter being negative because value is depreciating. In general, for a function of form \(f(t) = ab^t\text{,}\) we call \(b\) the growth factor. Moreover, if \(b = 1+r\text{,}\) we call \(r\) the growth rate. Whenever \(b \gt 1\text{,}\) we often say that the function \(f\) is exhibiting “exponential growth”, wherease if \(0 \lt b \lt 1\text{,}\) we say \(f\) exhibits “exponential decay”.

We explore the properties of functions of form \(f(t) = ab^t\) further in Activity 3.1.2.

Activity 3.1.2.

In Desmos, define the function \(g(t) = ab^t\) and create sliders for both \(a\) and \(b\) when prompted. Click on the sliders to set the minimum value for each to \(0.1\) and the maximum value to \(10\text{.}\) Note that for \(g\) to be an exponential function, we require \(b \ne 1\text{,}\) even though the slider for \(b\) will allow this value.

  1. What is the domain of \(g(t) = ab^t\text{?}\)

  2. What is the range of \(g(t) = ab^t\text{?}\)

  3. What is the \(y\)-intercept of \(g(t) = ab^t\text{?}\)

  4. How does changing the value of \(b\) affect the shape and behavior of the graph of \(g(t) = ab^t\text{?}\) Write several sentences to explain.

  5. For what values of the growth factor \(b\) is the corresponding growth rate positive? For which \(b\)-values is the growth rate negative?

  6. Consider the graphs of the exponential functions \(p\) and \(q\) provided in Figure 3.1.3. If \(p(t) = ab^t\) and \(q(t) = cd^t\text{,}\) what can you say about the values \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) (beyond the fact that all are positive and \(b \ne 1\) and \(d \ne 1\))? For instance, can you say a certain value is larger than another? Or that one of the values is less than \(1\text{?}\)

    Figure 3.1.3. Graphs of exponential functions \(p\) and \(q\text{.}\)

Subsection 3.1.2 Determining formulas for exponential functions

To better understand the roles that \(a\) and \(b\) play in an exponential function, let's compare exponential and linear functions. In Table 3.1.4 and Table 3.1.5, we see output for two different functions \(r\) and \(s\) that correspond to equally spaced input.

Table 3.1.4. Data for the function \(r\text{.}\)
\(t\) \(0\) \(3\) \(6\) \(9\)
\(r(t)\) \(12\) \(10\) \(8\) \(6\)
Table 3.1.5. Data for the function \(s\text{.}\)
\(t\) \(0\) \(3\) \(6\) \(9\)
\(s(t)\) \(12\) \(9\) \(6.75\) \(5.0625\)

In Table 3.1.4, we see a function that exhibits constant average rate of change since the change in output is always \(\triangle r = -2\) for any change in input of \(\triangle t = 3\text{.}\) Said differently, \(r\) is a linear function with slope \(m = -\frac{2}{3}\text{.}\) Since its \(y\)-intercept is \((0,12)\text{,}\) the function's formula is \(y = r(t) = 12 - \frac{2}{3}t\text{.}\)

In contrast, the function \(s\) given by Table 3.1.5 does not exhibit constant average rate of change. Instead, another pattern is present. Observe that if we consider the ratios of consecutive outputs in the table, we see that

\begin{equation*} \frac{9} {12}= \frac{3}{4}, \frac{6.75}{9} = 0.75 = \frac{3}{4}, \text{ and } \frac{5.0625}{6.75} = 0.75 = \frac{3}{4}\text{.} \end{equation*}

So, where the differences in the outputs in Table 3.1.4 are constant, the ratios in the outputs in Table 3.1.5 are constant. The latter is a hallmark of exponential functions and may be used to help us determine the formula of a function for which we have certain information.

If we know that a certain function is linear, it suffices to know two points that lie on the line to determine the function's formula. It turns out that exponential functions are similar: knowing two points on the graph of a function known to be exponential is enough information to determine the function's formula. In the following example, we show how knowing two values of an exponential function enables us to find both \(a\) and \(b\) exactly.

Example 3.1.6.

Suppose that \(p\) is an exponential function and we know that \(p(2) = 11\) and \(p(5) = 18\text{.}\) Determine the exact values of \(a\) and \(b\) for which \(p(t) = ab^t\text{.}\)


Since we know that \(p(t) = ab^t\text{,}\) the two data points give us two equations in the unknowns \(a\) and \(b\text{.}\) First, using \(t = 2\text{,}\)

\begin{equation} ab^2 = 11\text{,}\label{eq-growth-p-2}\tag{3.1.1} \end{equation}

and using \(t = 5\) we also have

\begin{equation} ab^5 = 18\text{.}\label{eq-growth-p-5}\tag{3.1.2} \end{equation}

Because we know that the quotient of outputs of an exponential function corresponding to equally-spaced inputs must be constant, we thus naturally consider the quotient \(\frac{18}{11}\text{.}\) Using Equation (3.1.1) and Equation (3.1.2), it follows that

\begin{equation*} \frac{18}{11} = \frac{ab^5}{ab^2}\text{.} \end{equation*}

Simplifying the fraction on the right, we see that \(\frac{18}{11} = b^3 \text{.}\) Solving for \(b\text{,}\) we find that \(b = \sqrt[3]{\frac{18}{11}}\) is the exact value of \(b\text{.}\) Substituting this value for \(b\) in Equation (3.1.1), it then follows that \(a \left( \sqrt[3]{\frac{18}{11}} \right)^2 = 11\text{,}\) so \(a = \frac{11}{\left( \frac{18}{11} \right)^{2/3}} \text{.}\) Therefore,

\begin{equation*} p(t) = \frac{11}{\left( \frac{18}{11} \right)^{2/3}} \left( \sqrt[3]{\frac{18}{11}} \right)^t \approx 7.9215 \cdot 1.1784^t\text{,} \end{equation*}

and a plot of \(y = p(t)\) confirms that the function indeed passes through \((2,11)\) and \((5,18)\) as shown in Figure 3.1.7.

Figure 3.1.7. Plot of \(p(t) = ab^t\) that passes through \((2,11)\) and \((5,18)\text{.}\)
Activity 3.1.3.

The value of an automobile is depreciating. When the car is \(3\) years old, its value is $\(12500\text{;}\) when the car is \(7\) years old, its value is $\(6500\text{.}\)

  1. Suppose the car's value \(t\) years after its purchase is given by the function \(V(t)\) and that \(V\) is exponential with form \(V(t) = ab^t\text{,}\) what are the exact values of \(a\) and \(b\text{?}\)

  2. Using the exponential model determined in (a), determine the purchase value of the car and estimate when the car will be worth less than $\(1000\text{.}\)

  3. Suppose instead that the car's value is modeled by a linear function \(L\) and satisfies the values stated at the outset of this activity. Find a formula for \(L(t)\) and determine both the purchase value of the car and when the car will be worth $\(1000\text{.}\)

  4. Which model do you think is more realistic? Why?

Subsection 3.1.4 Summary

  • We say that a function is exponential whenever its algebraic form is \(f(t) = ab^t\) for some positive constants \(a\) and \(b\) where \(b \ne 1\text{.}\) (Technically, the formal definition of an exponential function is one of form \(f(t) = b^t\text{,}\) but in our everyday usage of the term “exponential” we include vertical stretches of these functions and thus allow \(a\) to be any positive constant, not just \(a = 1\text{.}\))

  • To determine the formula for an exponential function of form \(f(t) = ab^t\text{,}\) we need to know two pieces of information. Typically this information is presented in one of two ways.

    • If we know the amount, \(a\text{,}\) of a quantity at time \(t = 0\) and the rate, \(r\text{,}\) at which the quantity grows or decays per unit time, then it follows \(f(t) = a(1+r)^t\text{.}\) In this setting, \(r\) is often given as a percentage that we convert to a decimal (e.g., if the quantity grows at a rate of \(7\)% per year, we set \(r = 0.07\text{,}\) so \(b = 1.07\)).

    • If we know any two points on the exponential function's graph, then we can set up a system of two equations in two unknowns and solve for both \(a\) and \(b\) exactly. In this situation, it is useful to consider the quotient of the two known outputs, as demonstrated in Example 3.1.6.

  • Exponential functions of the form \(f(t) = ab^t\) (where \(a\) and \(b\) are both positive and \(b \ne 1\)) exhibit the following important characteristics:

    • The domain of any exponential function is the set of all real numbers and the range of any exponential function is the set of all positive real numbers.

    • The \(y\)-intercept of the exponential function \(f(t) = ab^t\) is \((0,a)\) and the function has no \(x\)-intercepts.

    • If \(b \gt 1\text{,}\) then the exponential function is always increasing and always increases at an increasing rate. If \(0 \lt b \lt 1\text{,}\) then the exponential function is always decreasing and always decreases at an increasing rate.

Exercises 3.1.5 Exercises


Grinnell Glacier in Glacier National Park in Montana covered about \(142\) acres in 2007 and was found to be shrinking at about \(4.4\)% per year. 2 

See Exercise 34 on p. 146 of Connally's Functions Modeling Change.
  1. Let \(G(t)\) denote the area of Grinnell Glacier in acres in year \(t\text{,}\) where \(t\) is the number of years since 2007. Find a formula for \(G(t)\) and define the function in Desmos.

  2. How many acres of ice were in the glacier in 1997? In 2012? What does the model predict for 2022?

  3. How many total acres of ice were lost from 2007 to 2012?

  4. What was the average rate of change of \(G\) from 2007 to 2012? Write a sentence to explain the meaning of this number and include units on your answer. In addition, how does this compare to the average rate of change of \(G\) from 2012 to 2017?

  5. How would you you describe the overall behavior of \(G\text{,}\) and thus what is happening to the Grinnell Glacier?


Consider the exponential function \(f\) whose graph is given by Figure 3.1.12. Note that \(f\) passes through the two noted points exactly.

Figure 3.1.12. A plot of the exponential function \(f\text{.}\)
  1. Determine the values of \(a\) and \(b\) exactly.

  2. Determine the average rate of change of \(f\) on the intervals \([2,7]\) and \([7,12]\text{.}\) Which average rate is greater?

  3. Find the equation of the linear function \(L\) that passes through the points \((2,20)\) and \((7,5)\text{.}\)

  4. Which average rate of change is greater? The average rate of change of \(f\) on \([0,2]\) or the average rate of change of \(L\) on \([0,2]\text{?}\)


A cup of hot coffee is brought outside on a cold winter morning in Winnipeg, Manitoba, where the surrounding temperature is \(0\) degrees Fahrenheit. A temperature probe records the coffee's temperature (in degrees Fahrenheit) every minute and generates the data shown in Table 3.1.13.

Table 3.1.13. The temperature, \(F\text{,}\) of the coffee at time \(t\text{.}\)
\(t\) \(0\) \(2\) \(4\) \(6\) \(8\) \(10\)
\(F(t)\) \(175\) \(129.64\) \(96.04\) \(71.15\) \(52.71\) \(39.05\)
  1. Assume that the data in the table represents the overall trend of the behavior of \(F\text{.}\) Is \(F\) linear, exponential, or neither? Why?

  2. Is it possible to determine an exact formula for \(F\text{?}\) If yes, do so and justify your formula; if not, explain why not.

  3. What is the average rate of change of \(F\) on \([4,6]\text{?}\) Write a sentence that explains the practical meaning of this value in the context of the overall exercise.

  4. How do you think the data would appear if instead of being in a regular coffee cup, the coffee was contained in an insulated mug?


The amount (in milligrams) of a drug in a person's body following one dose is given by an exponential decay function. Let \(A(t)\) denote the amount of drug in the body at time \(t\) in hours after the dose was taken. In addition, suppose you know that \(A(3) = 22.7\) and \(A(6) = 15.2\text{.}\)

  1. Find a formula for \(A\) in the form \(A(t) = ab^t\text{,}\) where you determine the values of \(a\) and \(b\) exactly.

  2. What is the size of the initial dose the person was given?

  3. How much of the drug remains in the person's body \(8\) hours after the dose was taken?

  4. Estimate how long it will take until there is less than \(1\) mg of the drug remaining in the body.

  5. Compute the average rate of change of \(A\) on the intervals \([3,5]\text{,}\) \([5,7]\text{,}\) and \([7,9]\text{.}\) Write at least one careful sentence to explain the meaning of the values you found, including appropriate units. Then write at least one additional sentence to explain any overall trend(s) you observe in the average rate of change.

  6. Plot \(A(t)\) on an appropriate interval and label important points and features of the graph to highlight graphical interpretations of your answers in (b), (c), (d), and (e).