4.1: Exponential Functions
Defining an Exponential Function
A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, \(2.5\%\) of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up \(10\%\) of the U.S. population in 2015, \(40\%\) in 2019, and \(80\%\) in 2050.
What exactly does it mean to grow exponentially ? What does the word double have in common with percent increase ? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.
- Percent change refers to a change based on a percent of the original amount.
- Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
- Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time.
For us to gain a clear understanding of exponential growth , let us contrast exponential growth with linear growth . We will construct two functions. The first function is exponential. We will start with an input of \(0\), and increase each input by \(1\). We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of \(0\), and increase each input by \(1\). We will add \(2\) to the corresponding consecutive outputs (Table \(\PageIndex{1}\)).
From Table \(\PageIndex{1}\) we can infer that for these two functions, exponential growth dwarfs linear growth.
- Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain.
- Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain.
| \(x\) | \(f(x)=2^x\) | \(g(x)=2x\) |
|---|---|---|
| 0 | 1 | 0 |
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 6 |
| 4 | 16 | 8 |
| 5 | 32 | 10 |
| 6 | 64 | 12 |
Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding \(2\) to the output whenever the input was increased by one.
The general form of the exponential function is \(f(x)=ab^x\), where \(a\) is any nonzero number, \(b\) is a positive real number not equal to \(1\).
- If \(b>1\),the function grows at a rate proportional to its size.
- If \(0<b<1\), the function decays at a rate proportional to its size.
Let’s look at the function \(f(x)=2^x\) from our example. We will create a table (Table \(\PageIndex{2}\)) to determine the corresponding outputs over an interval in the domain from \(−3\) to \(3\).
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(f(x)=2^x\) | \(2^{−3}=\dfrac{1}{8}\) | \(2^{−2}=\dfrac{1}{4}\) | \(2^{−1}=\dfrac{1}{2}\) | \(2^0=1\) | \(2^1=2\) | \(2^2=4\) | \(2^3=8\) |
Let us examine the graph of \(f\) by plotting the ordered pairs from Table \(\PageIndex{2}\) and then make a few observations \(\PageIndex{1}\).
Let’s define the behavior of the graph of the exponential function \(f(x)=2^x\) and highlight some its key characteristics.
- the domain is \((−\infty,\infty)\),
- the range is \((0,\infty)\),
- as \(x\rightarrow \infty\), \(f(x)\rightarrow \infty\),
- as \(x\rightarrow −\infty\), \(f(x)\rightarrow 0\),
- \(f(x)\) is always increasing,
- the graph of \(f(x)\) will never touch the x -axis because base two raised to any exponent never has the result of zero.
- \(y=0\) is the horizontal asymptote.
- the y -intercept is \(1\).
For any real number \(x\),an exponential function is a function with the form
\[f(x)=ab^x\]
where
- \(a\) is a non-zero real number called the initial value and
- \(b\) is any positive real number such that \(b≠1\).
- The domain of \(f\) is all real numbers.
- The range of \(f\) is all positive real numbers if \(a>0\).
- The range of \(f\) is all negative real numbers if \(a<0\).
- The y -intercept is \((0,a)\),and the horizontal asymptote is \(y=0\).
Which of the following equations are not exponential functions?
- \(f(x)=4^{3(x−2)}\)
- \(g(x)=x^3\)
- \(h(x)=\left(\dfrac{1}{3}\right)^x\)
- \(j(x)=(−2)^x\)
Solution
By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, \(g(x)=x^3\) does not represent an exponential function because the base is an independent variable. In fact, \(g(x)=x^3\) is a power function.
Recall that the base \(b\) of an exponential function is always a positive constant, and \(b≠1\). Thus, \(j(x)={(−2)}^x\) does not represent an exponential function because the base, \(−2\), is less than \(0\).
Defining Exponential Growth
Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.
A function that models exponential growth grows by a rate proportional to the amount present. For any real number \(x\) and any positive real numbers \(a\) and \(b\) such that \(b≠1\),an exponential growth function has the form
\[f(x)=ab^x\]
where
- \(a\) is the initial or starting value of the function.
- \(b\) is the growth factor or growth multiplier per unit \(x\).
In more general terms, we have an exponential function , in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has \(100\) stores and expands by opening \(50\) new stores a year, so its growth can be represented by the function \(A(x)=100+50x\). Company B has \(100\) stores and expands by increasing the number of stores by \(50\%\) each year, so its growth can be represented by the function \(B(x)=100{(1+0.5)}^x\).
A few years of growth for these companies are illustrated in Table \(\PageIndex{3}\).
| Year, \(x\) | Stores, Company A | Stores, Company B |
|---|---|---|
| \(0\) | \(100+50(0)=100\) | \(100{(1+0.5)}^0=100\) |
| \(1\) | \(100+50(1)=150\) | \(100{(1+0.5)}^1=150\) |
| \(2\) | \(100+50(2)=200\) | \(100{(1+0.5)}^2=225\) |
| \(3\) | \(100+50(3)=250\) | \(100{(1+0.5)}^3=337.5\) |
| \(x\) | \(A(x)=100+50x\) | \(B(x)=100{(1+0.5)}^x\) |
The graphs comparing the number of stores for each company over a five-year period are shown in Figure \(\PageIndex{2}\). We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.
Notice that the domain for both functions is \([0,\infty)\),and the range for both functions is \([100,\infty)\). After year 1, Company B always has more stores than Company A.
Now we will turn our attention to the function representing the number of stores for Company \(B\), \(B(x)=100{(1+0.5)}^x\). In this exponential function, \(100\) represents the initial number of stores, \(0.50\) represents the growth rate, and \(1+0.5=1.5\) represents the growth factor. Generalizing further, we can write this function as \(B(x)=100{(1.5)}^x\),where \(100\) is the initial value, \(1.5\) is called the base , and \(x\) is called the exponent .
Finding Equations of Exponential Functions
In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants \(a, a\) and \(b, b\),and evaluate the function.
- If one of the data points has the form \((0,a)\), then \(a\) is the initial value. Using \(a\), substitute the second point into the equation \(f(x)=a{(b)}^x\), and solve for \(b\).
- If neither of the data points have the form \((0,a)\), substitute both points into two equations with the form \(f(x)=a{(b)}^x\). Solve the resulting system of two equations in two unknowns to find \(a\) and \(b\).
- Using the \(a\) and \(b\) found in the steps above, write the exponential function in the form \(f(x)=a{(b)}^x\).
In 2006, \(80\) deer were introduced into a wildlife refuge. By 2012, the population had grown to \(180\) deer. The population was growing exponentially. Write an algebraic function \(N(t)\) representing the population \((N)\) of deer over time \(t\).
Solution
We let our independent variable \(t\) be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, \(a=80\). We can now substitute the second point into the equation \(N(t)=80b^t\) to find \(b\):
\[\begin{align*} N(t)&= 80b^t\\ 180&= 80b^6 \qquad \text{Substitute using point } (6, 180)\\ \dfrac{9}{4}&= b^6 \qquad \text{Divide and write in lowest terms}\\ b&= {\left (\dfrac{9}{4} \right )}^{\tfrac{1}{6}} \qquad \text{Isolate b using properties of exponents}\\ b&\approx 1.1447 \qquad \text{Round to 4 decimal places} \end{align*}\]
Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.
The exponential model for the population of deer is \(N(t)=80{(1.1447)}^t\). (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)
We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure \(\PageIndex{3}\) passes through the initial points given in the problem, \((0, 80)\) and \((6, 180)\). We can also see that the domain for the function is \([0,\infty)\),and the range for the function is \([80,\infty)\).
A wolf population is growing exponentially. In 2011, \(129\) wolves were counted. By 2013, the population had reached \(236\) wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population \(N\) of wolves over time \(t\).
- Answer
-
\((0,129)\) and \((2,236)\); \(N(t)=129{(1.3526)}^t\)
Find an exponential function that passes through the points \((−2,6)\) and \((2,1)\).
Solution
Because we don’t have the initial value, we substitute both points into an equation of the form \(f(x)=ab^x\), and then solve the system for \(a\) and \(b\).
- Substituting \((−2,6)\) gives \(6=ab^{−2}\)
- Substituting \((2,1)\) gives \(1=ab^2\)
Use the first equation to solve for \(a\) in terms of \(b\):
\[\begin{align*} 6&= ab^{-2}\\ \dfrac{6}{b^{-2}}&= a \qquad \text{Divide}\\ a&= 6b^2 \qquad \text{Use properties of exponents to rewrite the denominator} \end{align*}\]
Substitute a in the second equation, and solve for \(b\):
\[\begin{align*} 1&= ab^{2}\\ 1&= 6b^2 b^2\\ &= 6b^4 \qquad \text{Substitute a}\\ b&= \left (\dfrac{1}{6} \right )^{\tfrac{1}{4}} \qquad \text{Round 4 decimal places rewrite the denominator}\\ b&\approx 0.6389 \end{align*}\]
Use the value of \(b\) in the first equation to solve for the value of \(a\):
\[\begin{align*} a&= 6b^{2}\\ &\approx 6(0.6389)^2 \\ &\approx 2.4492 \end{align*}\]
Thus, the equation is \(f(x)=2.4492{(0.6389)}^x\).
We can graph our model to check our work. Notice that the graph in Figure \(\PageIndex{4}\) passes through the initial points given in the problem, \((−2, 6)\) and \((2, 1)\). The graph is an example of an exponential decay function.
- First, identify two points on the graph. Choose the \(y\)-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.
- If one of the data points is the \(y\)-intercept \((0,a)\), then \(a\) is the initial value. Using \(a\), substitute the second point into the equation \(f(x)=a{(b)}^x\), and solve for \(b\)
- If neither of the data points have the form \((0,a)\), substitute both points into two equations with the form \(f(x)=a{(b)}^x\). Solve the resulting system of two equations in two unknowns to find \(a\) and \(b\).
- Write the exponential function, \(f(x)=a{(b)}^x\).
Applying the Compound-Interest Formula
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest . The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate , is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.
We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time \(t\), principal \(P\), \(APR\) \(r\), and number of compounding periods in a year \(n\):
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt} \nonumber\]
For example, observe Table \(\PageIndex{4}\), which shows the result of investing \($1,000\) at \(10\%\) for one year. Notice how the value of the account increases as the compounding frequency increases.
| Frequency | Value after \(1\) year |
|---|---|
| Annually | \($1100\) |
| Semiannually | \($1102.50\) |
| Quarterly | \($1103.81\) |
| Monthly | \($1104.71\) |
| Daily | \($1105.16\) |
Compound interest can be calculated using the formula
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt}\]
where
- \(A(t)\) is the account value,
- \(t\) is measured in years,
- \(P\) is the starting amount of the account, often called the principal, or more generally present value,
- \(r\) is the annual percentage rate (APR) expressed as a decimal, and
- \(n\) is the number of compounding periods in one year.
If we invest \($3,000\) in an investment account paying \(3\%\) interest compounded quarterly, how much will the account be worth in \(10\) years?
Solution
Because we are starting with \($3,000\), \(P=3000\). Our interest rate is \(3\%\), so \(r = 0.03\). Because we are compounding quarterly, we are compounding \(4\) times per year, so \(n=4\). We want to know the value of the account in \(10\) years, so we are looking for \(A(10)\),the value when \(t = 10\).
\[\begin{align*} A(t)&= P{\left (1+\dfrac{r}{n} \right )}^{nt} \qquad \text{Use the compound interest formula}\\ A(10)&= 3000{\left (1+\dfrac{0.03}{4} \right )}^{(4)\cdot (10)} \qquad \text{Substitute using given values}\\ &\approx \$4045.05 \qquad \text{Round to two decimal places} \end{align*}\]
The account will be worth about \($4,045.05\) in \(10\) years.
Evaluating Functions with Base \(e\)
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table \(\PageIndex{5}\) shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.
Examine the value of \($1\) invested at \(100\%\) interest for \(1\) year, compounded at various frequencies, listed in Table \(\PageIndex{5}\).
| Frequency | \(A(t)={\left (1+\dfrac{1}{n} \right )}^n\) | Value |
|---|---|---|
| Annually | \({\left (1+\dfrac{1}{1} \right )}^1\) | \($2\) |
| Semiannually | \({\left (1+\dfrac{1}{2} \right )}^2\) | \($2.25\) |
| Quarterly | \({\left (1+\dfrac{1}{4} \right )}^4\) | \($2.441406\) |
| Monthly | \({\left (1+\dfrac{1}{12} \right )}^{12}\) | \($2.613035\) |
| Daily | \({\left (1+\dfrac{1}{365} \right )}^{365}\) | \($2.714567\) |
| Hourly | \({\left (1+\dfrac{1}{8760} \right )}^{8760}\) | \($2.718127\) |
| Once per minute | \({\left (1+\dfrac{1}{525600} \right )}^{525600}\) | \($2.718279\) |
| Once per second | \({\left (1+\dfrac{1}{31536000} \right )}^{31536000}\) | \($2.718282\) |
These values appear to be approaching a limit as \(n\) increases without bound. In fact, as \(n\) gets larger and larger, the expression \({\left (1+\dfrac{1}{n} \right )}^n\) approaches a number used so frequently in mathematics that it has its own name: the letter \(e\). This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.
The letter \(e\) represents the irrational number
\[{\left (1+\dfrac{1}{n} \right )}^n\]
as \(n\) increases without bound
The letter \(e\) is used as a base for many real-world exponential models. To work with base \(e\), we use the approximation, \(e≈2.718282\). The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.
Calculate \(e^{3.14}\). Round to five decimal places.
Solution
On a calculator, press the button labeled \([e^x]\). The window shows \([e {}^( ]\). Type \(3.14\) and then close parenthesis, \([)]\). Press [ENTER]. Rounding to \(5\) decimal places, \(e^{3.14}≈23.10387\). Caution: Many scientific calculators have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of \(e\).