4.1: Exponential Functions
- Refresher Topics
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Systems
- Solving 2x2 Systems of Linear Equations - Substitution Method
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Systems
In this section, you will:
- Evaluate exponential functions.
- Find the equation of an exponential function.
- Use compound interest formulas.
- Evaluate exponential functions with base \(e\).
When the original iteration of this textbook was written (2013), India was the second most populous country in the world with a population of about \(1.25\) billion people. At that time, the population was growing at a rate of about \(1.2\%\) each year. If that rate continued, the population of India would have exceeded China's population by the year \(2031\). Interestingly, the rate of growth for India's population actually sped up! In 2025, India's population was \( 1.42 \) billion while China's was \( 1.41 \) billion. When populations grow rapidly, we often say that the growth is "exponential," meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions , which model this kind of rapid growth.
Identifying Exponential Functions
When exploring linear growth, we observed a constant rate of change - a constant number by which the output increased for each unit increase in input. For example, in the equation \(f(x)=3x+4\), the slope tells us the output increases by \( 3 \) each time the input increases by \( 1 \). We call this the Slope Addition Property and it is unique to linear functions. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people.
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 continued, vegans would have made up \( 10\% \) of the U.S. population in 2015, \( 40\% \) in 2019, and \( 80\% \) in 2021.
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. See Table \( \PageIndex{ 1 } \).
| \(x\) | \(f(x)= 2^x\) | \(g(x)=2x\) |
|---|---|---|
| 0 | 1 | 0 |
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 6 |
| 4 | 16 | 8 |
| 5 | 32 | 10 |
| 6 | 64 | 12 |
| 7 | 128 | 14 |
| 8 | 256 | 16 |
| 9 | 512 | 18 |
| 10 | 1024 | 20 |
From Table \( \PageIndex{ 1 } \) we can infer that for these two functions, exponential growth dwarfs linear growth.
- Exponential growth means the original value from the range increases by the same percentage over equal increments in the domain.
- Linear growth means the original value from the range increases by the same amount over equal increments in the domain.
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. We call this the Base Multiplier Property , and it is unique to exponential functions. 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. As mentioned previously, we call this the Slope Addition Property .
The general form of the exponential function is \(f(x) = a b^x \), 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 we observe on the table in Figure \( \PageIndex{ 1 } \), and then make a few observations.
Figure \( \PageIndex{ 1 } \)
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 \(x \to \infty , f(x) \to \infty \),
- as \(x \to − \infty , f(x) \to 0\),
- \(f(x)\) 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\[f(x)=a b^x, \nonumber \]where
- \(a\) is a non-zero real number called the initial value .
- \(b\) is any positive real number such that \(b \neq 1\).
- 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?
- \(f(x)= 4^{3( x−2 )}\)
- \(g(x)= x^3\)
- \(h(x)= \left( \frac{1}{3} \right)^x\)
- \(j(x)= ( −2 )^x\)
- Solution
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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 \(b \neq 1\). Thus, \(j(x)= ( −2 )^x\) does not represent an exponential function because the base, \(−2\), is less than \(0\).
Which of the following equations represent exponential functions?
- \(f(x)=2 x^2 −3x+1\)
- \(g(x)= 0.875^x\)
- \(h(x)=1.75x+2\)
- \(j(x)= 1095.6^{-2x}\)
Evaluating Exponential Functions
Recall that the base of an exponential function must be a positive real number other than \(1\). Why do we limit the base \(b\) to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:
- Let \(b=−9\) and \(x= \frac{1}{2} \). Then \(f(x)=f\left( \frac{1}{2} \right)= ( −9 )^{1/2} =\sqrt{−9} \), which is not a real number.
Why do we limit the base to positive values other than \(1\)? Because base \(1\) results in the constant function. Observe what happens if the base is \(1\):
- Let \(b=1\). Then \(f(x)= 1^x =1\) for any value of \(x\).
To evaluate an exponential function with the form \(f(x)= b^x \), we simply substitute \(x\) with the given value, and calculate the resulting power. For example, let \(f(x)= 2^x \). What is \(f(3)\)?\[ \begin{array}{rrclcl} & f( x ) & = & 2^x & & \\[6pt] \implies & f( 3 ) & = & 2^3 & \quad & \left( \text{substituting} \right) \\[6pt] & & = & 8 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]To evaluate an exponential function with a form other than the basic form, it is important to follow the Order of Operations. For example, let \(f(x)=30( 2 )^x \). What is \(f(3)\)?\[ \begin{array}{rrclcl} & f( x ) & = & 30(2)^x & & \\[6pt] \implies & f( 3 ) & = & 30(2)^3 & \quad & \left( \text{substituting} \right) \\[6pt] & & = & 30(8) & \quad & \left( \text{exponents before multiplication} \right) \\[6pt] & & = & 240 & \quad & \left( \text{multiplication} \right) \\[6pt] \end{array} \nonumber \]Note that if the Order of Operations were not followed, the result would be incorrect:\[f(3) = 30( 2 )^3 \neq 60^3 = 216,000. \nonumber \]
Let \(f( x )=5 ( 3 )^{x+1} \). Evaluate \(f( 2 )\) without using a calculator.
- Solution
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Follow the Order of Operations. Be sure to pay attention to the parentheses.\[ \begin{array}{rrclcl} & f(x) & = & 5(3)^{x + 1} & & \\[6pt] \implies & f(2) & = & 5(3)^{2 + 1} & \quad & \left( \text{substituting} \right) \\[6pt] & & = & 5(3)^{3} & \quad & \left( \text{simplifying} \right) \\[6pt] & & = & 5(27) & \quad & \left( \text{simplifying} \right) \\[6pt] & & = & 135 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]
Let \(f( x )=8 ( 1.2 )^{x−5} \). Evaluate \(f( 3 )\) using a calculator. Round to four decimal places.
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 \(b \neq 1\), an exponential growth function has the form\[f(x)=a \, b^x, \nonumber \]where
- \(a\) is the initial or starting value of the function and
- \(b\) 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.
Figure \( \PageIndex{ 2 } \): The graph shows the numbers of stores Companies A and B opened over a five-year period.
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 .
At the beginning of this section, we learned that the population of India was about \(1.25\) billion in the year 2013, with an annual growth rate of about \(1.2\%\). This situation is represented by the growth function \(P(t)=1.25 ( 1.012 )^t \), where \(t\) is the number of years since 2013. To the nearest thousandth, and using this model, what will the population of India be in 2031?
- Solution
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To estimate the population in 2031, we evaluate the model for \(t=18\), because 2031 is \(18\) years after 2013. Rounding to the nearest thousandth,\[P(18)=1.25 ( 1.012 )^{18} \approx 1.549. \nonumber \]There will be about 1.549 billion people in India in the year 2031.
The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about \(0.6\%\). This situation is represented by the growth function \(P(t)=1.39 ( 1.006 )^t \), where \(t\) is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example \( \PageIndex{ 3 } \)?
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 \(a\) and \(b\), and evaluate the function.
Given two data points, write an exponential model.
- 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 exponential function \(N(t)\) representing the population \(( N )\) of deer over time \(t\).
- Solution
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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 \(N(t)=80 b^t\) to find \(b\):\[ \begin{array}{rrclcl} & N(t) & = & 80b^t & & \\[6pt] \implies & 180 & = & 80b^6 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & \dfrac{9}{4} & = & b^6 & \quad & \left( \text{dividing both sides by }80 \right) \\[6pt] \implies & \left(\dfrac{9}{4}\right)^{1/6} & = & b & \quad & \left( \text{raising both sides to the }\frac{1}{6}\text{ power} \right) \\[6pt] \implies & b & \approx & 1.1447 & \quad & \left( \text{rounding to four decimal places} \right) \\[6pt] \end{array} \nonumber \] NOTES: When we took the sixth root of both sides, we did not bother writing \( b = \pm \left( \frac{9}{4} \right)^{1/6} \). This is because the base of an exponential function is never negative. Moreover, do not round any intermediate calculations. Doing so leads to error propagation.
The exponential model for the population of deer is \(N(t) \approx 80 ( 1.1447 )^t \). (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 \([80, \infty )\).
Figure \( \PageIndex{ 3 } \): Graph showing the population of deer over time, \(N(t) \approx =80 ( 1.1447 )^t \), \(t\) years after 2006
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 \(t\).
Find an exponential function that passes through the points \(( −2,6 )\) and \(( 2,1 )\).
- Solution
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Because we don't have the initial value, we substitute both points into an equation of the form \(f(x)=a \, b^x \), and then solve the system for \(a\) and \(b\).
- Substituting \(( −2,6 )\) gives \(6=a \, b^{−2}\)
- Substituting \(( 2,1 )\) gives \(1=a \, b^2\)
Use the first equation to solve for \(a\) in terms of \(b\):\[ \begin{array}{rrclcl} & 6 & = & a \, b^{-2} & & \\[6pt] \implies & 6b^2 & = & a & \quad & \left( \text{multiplying both sides by }b^2 \right) \\[6pt] \end{array} \nonumber \]Substitute \(a\) in the second equation, and solve for \(b\):\[ \begin{array}{rrclcl} & 1 & = & a \, b^{2} & & \\[6pt] \implies & 1 & = & 6 b^2 b^2 & \quad & \left( \text{substituting }a = 6 b^2 \right) \\[6pt] \implies & 1 & = & 6 b^4 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \implies & \dfrac{1}{6} & = & b^4 & \quad & \left( \text{dividing both sides by }6 \right) \\[6pt] \implies & \left(\dfrac{1}{6}\right)^{1/4} & = & b & \quad & \left( \text{raise both sides to the }\frac{1}{4}\text{ power} \right) \\[6pt] \end{array} \nonumber \]Now that we have \( b \), we go back to the first equation to find \( a \):\[ a = 6 b^2 = 6 \left(\left( \dfrac{1}{6} \right)^{1/4}\right)^2 = 6 \left( \dfrac{1}{6} \right)^{2/4} = 6 \left( \dfrac{1}{6} \right)^{1/2} = \dfrac{6}{6^{1/2}} = 6^{1 - 1/2} = 6^{1/2} = \sqrt{6}. \nonumber \]Thus, the equation is\[ f(x) = \sqrt{6} \left( \left( \dfrac{1}{6} \right)^{1/4} \right)^x = \sqrt{6} \left( \dfrac{1}{6} \right)^{x/4}.\nonumber \]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.
Figure \( \PageIndex{ 4 } \): The graph of \(f(x)=\sqrt{6} \left( \frac{1}{6} \right)^{x/4}\) models exponential decay.
Given the two points \(( 1,3 )\) and \(( 2,4.5 )\), find the equation of the exponential function that passes through these two points.
Do two points always determine a unique exponential function?
Yes, provided the two points are either both above the \( x \)-axis or both below the \( x \)-axis and have different \( x \)-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in \(x\), which in many real world cases involves time.
Find an equation for the exponential function graphed in Figure \( \PageIndex{ 5 } \).
Figure \( \PageIndex{ 5 } \)
- Solution
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We can choose the \( y \)-intercept of the graph, \(( 0,3 )\), as our first point. This gives us the initial value, \(a=3\). Next, choose a point on the curve some distance away from \(( 0,3 )\) that has integer coordinates. One such point is \((2,12)\).\[ \begin{array}{rrclcl} & y & = & a b^x & \quad & \left( \text{writing the general form} \right) \\[6pt] \implies & y & = & 3 b^x & \quad & \left( \text{substituting }a = 3 \right) \\[6pt] \implies & 12 & = & 3 b^2 & \quad & \left( \text{substituting }x = 2\text{ and }y = 12 \right) \\[6pt] \implies & 4 & = & b^2 & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt] \implies & \pm 2 & = & b & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \implies & 2 & = & b & \quad & \left( \text{exponential bases are positive} \right) \\[6pt] \end{array} \nonumber \]Substitute \(a\) and \(b\) into the standard form to yield the equation \(f(x)=3 (2)^x \).
Find an equation for the exponential function graphed in Figure \( \PageIndex{ 6 } \).
Figure \( \PageIndex{ 6 } \)
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\[A(t)=P \left( 1 + \dfrac{r}{n} \right)^{nt}, \nonumber \]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
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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 \(t=10\).\[ \begin{array}{rrclcl} & A(t) & = & P\left( 1 + \dfrac{r}{n} \right)^{nt} & & \\[6pt] \implies & A(10) & = & 3000 \left( 1 + \dfrac{0.03}{4} \right)^{4 \cdot 10} & \quad & \left( \text{substituting known values} \right) \\[6pt] & & \approx & \$4,045.05 & \quad & \left( \text{rounding to two decimal places} \right) \\[6pt] \end{array} \nonumber \]The account will be worth about $4,045.05 in 10 years.
An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?
A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child's future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?
- Solution
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The nominal interest rate is 6%, so \(r=0.06\). Interest is compounded twice a year, so \(k=2\).
We want to find the initial investment, \(P\), needed so that the value of the account will be worth $40,000 in \(18\) years. Substitute the given values into the compound interest formula, and solve for \(P\).\[ \begin{array}{rrclcl} & A(t) & = & P\left( 1 + \dfrac{r}{n} \right)^{nt} & & \\[6pt] \implies & 40,000 & = & P \left( 1 + \dfrac{0.06}{2} \right)^{2 \cdot 18} & \quad & \left( \text{substituting known values} \right) \\[6pt] \implies & 40,000 & = & P \left( 1.03 \right)^{36} & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & \dfrac{40,000}{1.03^{36}} & = & P & \quad & \left( \text{isolating }P \right) \\[6pt] \implies & P & \approx & \$13,801 & \quad & \left( \text{rounding to the nearest dollar} \right) \\[6pt] \end{array} \nonumber \]Lily will need to invest $13,801 to have $40,000 in 18 years.
Refer to Example \( \PageIndex{ 8 } \). To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?
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 | \(A(n)= \left( 1+ \frac{1}{n}\right)^n\) | Value |
|---|---|---|
| Annually | \(\left( 1 + \frac{1}{1} \right)^{1}\) | $2 |
| Semiannually | \(\left( 1 + \frac{1}{2} \right)^{2}\) | $2.25 |
| Quarterly | \(\left( 1 + \frac{1}{4} \right)^{4}\) | $2.441406 |
| Monthly | \(\left( 1 + \frac{1}{12} \right)^{12}\) | $2.613035 |
| Daily | \(\left( 1 + \frac{1}{365} \right)^{365}\) | $2.714567 |
| Hourly | \(\left( 1 + \frac{1}{8760} \right)^{8760}\) | $2.718127 |
| Once per minute | \(\left( 1 + \frac{1}{525600} \right)^{525600}\) | $2.718279 |
| Once per second | \(\left( 1 + \frac{1}{31536000} \right)^{31536000}\) | $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 \(\left( 1 + \frac{1}{n} \right)^n\) 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.
The letter \( e \) represents the irrational number\[\left( 1 + \dfrac{1}{n} \right)^n, \text{ as } n \text{ increases without bound.} \nonumber \]
The letter \( e \) is used as a base for many real-world exponential models. To work with base \( e \), we use the approximation, \(e \approx 2.718282\). 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
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On a calculator, press the button labeled \( \mathrm{[e^x]}\). The window shows \(\mathrm{e^(}\). Type \(3.14\) and then close parenthesis, \(\mathrm{)}\). Press \(\mathrm{[ENTER]}\). Rounding to \(5\) decimal places, \(e^{3.14} \approx 23.10387\).
Caution: Many scientific calculators have an \(\mathrm{Exp}\) button, which is used to enter numbers in scientific notation. It is not used to find powers of \(e\).
Use a calculator to find \(e^{−0.5} \). Round to five decimal places.
Investigating Continuous Growth
So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, \( e \) is used as the base for exponential functions. Exponential models that use \(e\) as the base are called continuous growth or decay models . We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.
For all real numbers \(t\), and all positive numbers \(a\) and \(r\), continuous growth or decay is represented by the formula\[A(t)=a e^{rt}, \nonumber \]where
- \(a\) is the initial value,
- \(r\) is the continuous growth rate per unit time,
- and \(t\) is the elapsed time.
If \(r > 0\) , then the formula represents continuous growth. If \(r < 0\) , then the formula represents continuous decay.
For business applications, the continuous growth formula is called the continuous compounding formula and takes the form\[A(t)=P e^{rt}, \nonumber \]where
- \(P\) is the principal or the initial invested,
- \(r\) is the growth or interest rate per unit time,
- and \(t\) is the period or term of the investment.
A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?
- Solution
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Since the account is growing in value, this is a continuous compounding problem with growth rate \(r=0.10\). The initial investment was $1,000, so \(P=1000\). We use the continuous compounding formula to find the value after \(t=1\) year:\[ \begin{array}{rclcl} A(t) & = & Pe^{rt} & & \\[6pt] & = & 1000 e^{0.1} & \quad & \left( \text{substituting known values} \right) \\[6pt] & \approx & 1105.17 & \quad & \left( \text{rounding to the nearest hundredth} \right) \\[6pt] \end{array} \nonumber \]The account is worth $1,105.17 after one year.
A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?
Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?
- Solution
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Since the substance is decaying, the rate, \(17.3\%\), is negative. So, \(r=−0.173\). The initial amount of radon-222 was \(100\) mg, so \(a=100\). We use the continuous decay formula to find the value after \(t=3\) days:\[ \begin{array}{rclcl} A(t) & = & a e^{rt} & & \\[6pt] & = & 100 e^{-0.173(3)} & \quad & \left( \text{substituting known values} \right) \\[6pt] & \approx & 59.5115 & \quad & \left( \text{rounding to four decimal places} \right) \\[6pt] \end{array} \nonumber \]So approximately 59.5115 mg of radon-222 will remain.
Using the data in Example \( \PageIndex{ 11 } \), how much radon-222 will remain after one year?
Access these online resources for additional instruction and practice with exponential functions.