Exponential Functions

Elizabeth Pople

27 Exponential Functions

Topics Covered:

In case you missed something in class, or just want to review a specific topic covered in this Module, here is a list of topics covered:

Define and Evaluate Exponential Functions

The functions we have studied so far do not give us a model for many naturally occurring phenomena. From the growth of populations and the spread of viruses to radioactive decay and compounding interest, the models are very different from what we have studied so far. These models involve exponential functions.

An exponential function is a function of the form f(x) = a^x where a > 0 and a ≠ 1.

Exponential Function

An exponential function, where a > 0 and a ≠ 1, is a function of the form

f(x) = a^x

Notice that in this function, the variable is the exponent. In our functions so far, the variables were the base.

This figure shows three functions: f of x equals negative 3x plus 4, which is marked as linear; f of x equals 2x squared plus 5x minus 3, which is marked as quadratic; and f of x equals 6 to the x power, which is marked exponential. For the functions marked linear and quadratic, x is the base. For the function marked exponential, x is the exponent for the base 6. Our definition says a ≠ 1. If we let a = 1, then f(x) = a^x becomes f(x) = 1^x. Since 1^x = 1 for all real numbers, f(x) = 1. This is the constant function.

Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base a to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:

$\displaystyle f(x)=(-4)^{x}$

$\displaystyle f\left(\frac{1}{2}\right)=(-4)^{\frac{1}{2}}$

$\displaystyle f\left(\frac{1}{2}\right)=\sqrt{-4}$ not a real number

In fact, f(x) = (−4)^x would not be a real number any time x is a fraction with an even denominator. So our definition requires a > 0.

Try it! – Identifying Exponential Functions

Which of the following equations are not exponential functions?

$\displaystyle f(x)=4^{3(x-2)}$
$\displaystyle g(x)=x^{3}$
$\displaystyle h(x)=\left(\frac{1}{3}\right)^{x}$
$\displaystyle f(x)=(-2)^{x}$

Solution

By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, g(x) = x³ does not represent an exponential function because the base is an independent variable. In fact, g(x) = x³ is a power function.

Recall that the base a 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.

2. Which of the following equations represent exponential functions?

f(x) = 2x² − 3x + 1
g(x) = 0.875^x
h(x) = 1.75^x + 2
j(x) = 1095.6^-2x

Solution

g(x) = 0.875^x and j(x) = 1095.6^-2x represent exponential functions.

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 a = 1. Then f(x) = 1^x = 1 for any value of x.

To evaluate an exponential function with the form f(x) = a^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)?

f(x) = 2^x

f(3) = 2³ Substitute x = 3.

= 8 Evaluate the power.

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)?

f(x) = 30(2)^x

f(3) = 30(2)³ Substitute x = 3.

= 30(8) Simplify the power first.

= 240 Multiply.

Note that if the order of operations were not followed, the result would be incorrect:

f(3) = 30(2)³ ≠ 60³ = 216,000

Try it! – Evaluating Exponential Functions

Let f(x) = 5(3)^{x + 1}. Evaluate f(2) without using a calculator.

Solution

Follow the order of operations. Be sure to pay attention to the parentheses.

f(x) = 5(3)^{x + 1}

f(2) = 5(3)^{2 + 1} Substitute x = 2.

= 5(3)³ Add the exponents.

= 5(27) Simplify the power.

= 135 Multiply.

2. Let f(x) = 8(1.2)^{x – 5}. Evaluate f(3) using a calculator. Round to four decimal places.

Solution

5.5556

Graphing Exponential Functions

By graphing a few exponential functions, we will be able to see their unique properties.

Try it!

On the same coordinate system graph f(x) = 2^x and g(x) = 3^x.

Solution

We will use point plotting to graph the functions. This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3).

2. Graph: g(x) = 5^x.

Solution

This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 5).

If we look at the graphs from the previous Example and Try Its, we can identify some of the properties of exponential functions.

The graphs of f(x) = 2^x and g(x) = 3^x, as well as the graphs of f(x) = 4^x and g(x) = 5^x, all have the same basic shape. This is the shape we expect from an exponential function where a > 1.

We notice, that for each function, the graph contains the point (0, 1). This make sense because a⁰ = 1 for any a.

The graph of each function, f(x) = a^x also contains the point (1, a). The graph of f(x) = 2^xcontained (1, 2) and the graph of g(x) = 3^x contained (1, 3). This makes sense as a¹ = a.

Notice too, the graph of each function f(x) = a^x also contains the point $\displaystyle \left(-1,\frac{1}{a}\right)$ . The graph of f(x) = 2^x contained $\displaystyle\left(-1,\frac{1}{2}\right)$ and the graph of g(x) = 3^x contained $\displaystyle\left(-1,\frac{1}{3}\right)$ . This makes sense as $\displaystyle a^{-1}=\frac{1}{a}$ .

What is the domain for each function? From the graphs we can see that the domain is the set of all real numbers. There is no restriction on the domain. We write the domain in interval notation as (−∞, ∞).

Look at each graph. What is the range of the function? The graph never hits the x-axis. The range is all positive numbers. We write the range in interval notation as (0, ∞).

Whenever a graph of a function approaches a line but never touches it, we call that line an asymptote. For the exponential functions we are looking at, the graph approaches the x-axis very closely but will never cross it, we call the line y = 0, the x-axis, a horizontal asymptote.

Properties of the Graph of f(x) = a^x when a > 1

Properties

Definition

Domain

(−∞, ∞)

Range

(0, ∞)

x–intercept

None

y–intercept

(0, 1)

Contains

(1, a), $\displaystyle\left(-1,\frac{1}{a}\right)$

Asymptote

x-axis, the line y = 0

This figure shows a curve that slopes upward from (negative 1, 1 over a) through (0, 1), up to (1, a).

Our definition of an exponential function f(x) = a^x says a > 0, but the examples and discussion so far has been about functions where a > 1. What happens when 0 < a < 1? The next example will explore this possibility.

Try it!

On the same coordinate system, graph $\displaystyle f(x)=\left(\frac{1}{2}\right)^{x}$ and $\displaystyle g(x)=\left(\frac{1}{3}\right)^{x}$ .

Solution

We will use point plotting to graph the functions. This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 2), (0, 1), and (1, 1 over 2). The second curve is marked in red and passes through the points (negative 1, 3), (0, 1), and (1, 1 over 3).

2. Graph: $\displaystyle f(x)=\left(\frac{1}{4}\right)^{x}$ .

Solution

This figure shows a curve that passes through (negative 1, 4), (0, 1) to a point just above (3, 0).

3. Graph: $\displaystyle g(x)=\left(\frac{1}{5}\right)^{x}$ .

Solution

This figure shows a curve that passes through (negative 1, 5), (0, 1) to a point just above (3, 0).

Now let’s look at the graphs from the previous Example and Try Its so we can now identify some of the properties of exponential functions where 0 < a < 1.

The graphs of $\displaystyle f(x)=\left(\frac{1}{2}\right)^{x}$ and $\displaystyle g(x)=\left(\frac{1}{3}\right)^{x}$ as well as the graphs of $\displaystyle f(x)=\left(\frac{1}{4}\right)^{x}$ and $\displaystyle g(x)=\left(\frac{1}{5}\right)^{x}$ all have the same basic shape. While this is the shape we expect from an exponential function where 0 < a < 1, the graphs go down from left to right while the previous graphs, when a > 1, went from up from left to right.

We notice that for each function, the graph still contains the point (0, 1). This makes sense because a⁰ = 1 for any a.

As before, the graph of each function, f(x) = a^x, also contains the point (1, a). The graph of $\displaystyle f(x)=\left(\frac{1}{2}\right)^{x}$ contained $\displaystyle\left(1,\frac{1}{2}\right)$ and the graph of $\displaystyle g(x)=\left(\frac{1}{3}\right)^{x}$ contained $\displaystyle\left(1,\frac{1}{3}\right)$ . This makes sense as a¹ = a.

Notice too that the graph of each function, f(x) = a^x, also contains the point $\displaystyle\left(-1,\frac{1}{a}\right)$ . The graph of $\displaystyle f(x)=\left(\frac{1}{2}\right)^{x}$ contained (−1, 2) and the graph of $\displaystyle g(x)=\left(\frac{1}{3}\right)^{x}$ contained (−1, 3). This makes sense as $\displaystyle a^{-1}=\frac{1}{a}$ .

We will summarize these properties in the chart below. Which also include when a > 1.

Properties of the Graph of f(x) = a^x

when a > 1

when 0 < a < 1

Domain

(−∞, ∞)

Domain

(−∞, ∞)

Range

(0, ∞)

Range

(0, ∞)

x-intercept

none

x-intercept

none

y-intercept

(0, 1)

y-intercept

(0,1 )

Contains

(1, a), $\displaystyle\left(-1,\frac{1}{a}\right)$

Contains

(1, a), $\displaystyle\left(-1,\frac{1}{a}\right)$

Asymptote

x-axis, the line y = 0

Asymptote

x-axis, the line y = 0

Basic shape

increasing

Basic shape

decreasing

This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).

It is important for us to notice that both of these graphs are one-to-one, as they both pass the horizontal line test. This means the exponential function will have an inverse. We will look at this later.

Transformations

When we graphed quadratic functions, we were able to graph using translation rather than just plotting points. Will that work in graphing exponential functions?

Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function f(x) = a^x without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.

The first transformation occurs when we add a constant c to the input of the parent function f(x) = a^x, giving us a horizontal shift c units in the opposite direction of the sign. For example, if we begin by graphing the parent function f(x) = 2^x, we can then graph two horizontal shifts alongside it, using c = 3 : the shift left, g(x) = 2^{x + 3}, and the shift right, h(x) = 2^{x – 3}. Both horizontal shifts are shown in the figure below.

Graph of three functions, g(x) = 2^(x+3) in blue, f(x) = 2^x in orange, and h(x)=2^(x-3). Each functions’ asymptotes are at y=0Note that each functions’ transformations are described in the text.

Try it!

On the same coordinate system graph f(x) = 2^x and g(x) = 2^{x + 1}.

Solution

We will use point plotting to graph the functions. This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1), (0, 2) and (1, 4).

2. On the same coordinate system, graph: f(x) = 2^x and g(x) = 2^{x – 1}.

Solution This figure shows the graphs of two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 1 power is marked in red and passes through the points (0, 1 over 2), (1, 1), and (2, 2).

Looking at the graphs of the functions f(x) = 2^x and g(x) = 2^x^{+ 1} in the last example, we see that adding one in the exponent caused a horizontal shift of one unit to the left. Recognizing this pattern allows us to graph other functions with the same pattern by translation.

The next transformation occurs when we add a constant d to the parent function f(x) = b^x, giving us a vertical shift d units in the same direction as the sign. For example, if we begin by graphing a parent function, f(x) = 2^x, we can then graph two vertical shifts alongside it, using d = 3 : the upward shift, g(x) = 2^x + 3 and the downward shift, h(x) = 2^x – 3. Both vertical shifts are shown in the figure below.

Graph of three functions, g(x) = 2^x+3 in blue with an asymptote at y=3, f(x) = 2^x in orange with an asymptote at y=0, and h(x)=2^x-3 with an asymptote at y=-3. Note that each functions’ transformations are described in the text.

Let’s now consider another situation that might be graphed more easily by translation, once we recognize the pattern.

Try it!

On the same coordinate system graph f(x) = 3^x and g(x) = 3^x – 2.

Solution

We will use point plotting to graph the functions. This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second curve is marked in red and passes through the points (negative 1, negative 5 over 3), (0, negative 1), and (1, 1).

2. On the same coordinate system, graph: f(x) = 3^x and g(x) = 3^x + 2

Solution
This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 1, 7 over 3), (0, 3) and (1, 5).

Looking at the graphs of the functions f(x) = 3^x and g(x) = 3^x – 2 in the last example, we see that subtracting 2 caused a vertical shift of down two units. Notice that the horizontal asymptote also shifted down 2 units. Recognizing this pattern allows us to graph other functions with the same pattern by translation.

For any constants c and d, the function f(x) = a^{x + c} + d shifts the parent function f(x) = a^x

vertically d units, in the same direction of the sign of d.
horizontally c units, in the opposite direction of the sign of c.
The y-intercept becomes (0, a^c + d).
The horizontal asymptote becomes y = d.
The range becomes (d, ∞).
The domain, (−∞, ∞) remains unchanged.

Try it! – Graphing a Shift of an Exponential Function

Graph f(x) = 2^{x + 1} − 3. State the domain, range, and asymptote.

Solution

We have an exponential equation of the form f(x) = b^{x + c} + d with b = 2, c = 1, and d = -3.

Draw the horizontal asymptote y = d, so draw y = -3.

Identify the shift as (-c, d) so the shift is (-1, -3).

Shift the graph of f(x) = b^x left 1 units and down 3 units. Graph of the function, f(x) = 2^(x+1)-3, with an asymptote at y=-3. Labeled points in the graph are (-1, -2), (0, -1), and (1, 1). The domain is (−∞, ∞); the range is (−3,∞); the horizontal asymptote is y = −3.

Stretch or Compress

While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function f(x) = a^x by a constant |c| > 0. For example, if we begin by graphing the parent function f(x) = 2^x, we can then graph the stretch, using c = 3, to get g(x) = 3(2)^x as shown in the figure on the left, and the compression, using $\displaystyle c=\frac{1}{3}$ , to get $\displaystyle h(x)=\frac{1}{3}(2)^{x}$ as shown in the figure on the right.

Two graphs where graph a is an example of vertical stretch and graph b is an example of vertical compression. — **(a)** *g(x) = 3(2)^x* stretches the graph of *f(x) = 2^x* vertically by a factor of 3, **(b)** *h(x) = $\displaystyle\frac{1}{3}\left(2\right)^{x}$* compresses the graph of *f(x) = 2^x* vertically by a factor of $\displaystyle \frac{1}{3}$ .

For any factor a > 0, the function f(x) = c(a)^x

is stretched vertically by a factor of c if |c| > 1.
is compressed vertically by a factor of c if |c| < 1.
has a y-intercept of (0, c).
has a horizontal asymptote at y = 0, a range of (0, ∞), and a domain of (−∞, ∞), which are unchanged from the parent function.

Try it! – Graphing the Stretch of an Exponential Function

Sketch a graph of $\displaystyle f(x)=4\left(\frac{1}{2}\right)^{x}$ . State the domain, range, and asymptote.

Solution

Before graphing, identify the behavior and key points on the graph.

Since $\displaystyle b=\frac{1}{2}$ is between zero and one, the left tail of the graph will increase without bound as x decreases, and the right tail will approach the x-axis as x increases.
Since a = 4, the graph of $\displaystyle f(x)=\left(\frac{1}{2}\right)^{x}$ will be stretched by a factor of 4.
Create a table of points as shown below.

xxx Points

x −3 −2 −1 0 1 2 3

$\displaystyle f(x)=4\left(\frac{1}{2}\right)^{x}$ 32 16 8 4 2 1 0.5
Plot the y-intercept, (0, 4) along with two other points. We can use (−1, 8) and (1, 2).

Draw a smooth curve connecting the points, as shown below. Graph of the function, f(x) = 4(1/2)^(x), with an asymptote at y=0. Labeled points in the graph are (-1, 8), (0, 4), and (1, 2). The domain is (−∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0.

Reflections

In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we multiply the parent function f(x) = a^x by -1, we get a reflection about the x-axis. When we multiply the input by -1, we get a reflection about the y-axis. For example, if we begin by graphing the parent function f(x) = 2^x, we can then graph the two reflections alongside it. The reflection about the x-axis, g(x) = -2^x, is shown in the figure on the left side and the reflection about the y-axis h(x) = 2^-x, is shown in the figure on the right side. Two graphs where graph a is an example of a reflection about the x-axis and graph b is an example of a reflection about the y-axis.

(a) g(x) = −2^x reflects the graph of f(x) = 2^x about the x-axis. (b) g(x) = 2^−x reflects the graph of f(x) = 2^x about the y-axis.

The function f(x) = −a^x

reflects the parent function f(x) = a^x about the x-axis.
has a y-intercept of (0, −1).
has a range of (−∞, 0).
has a horizontal asymptote at y = 0 and domain of (−∞, ∞), which are unchanged from the parent function.

The function f(x) = a^-x

reflects the parent function f(x) = a^x about the y-axis.
has a y-intercept of (0, 1), a horizontal asymptote at y = 0, a range of (0, ∞), and a domain of (−∞, ∞), which are unchanged from the parent function.

Try it! – Writing and Graphing the Reflection of an Exponential Function

Find and graph the equation for a function, g(x), that reflects $\displaystyle f(x)=\left(\frac{1}{4}\right)^{x}$ about the x-axis. State its domain, range, and asymptote.

Solution

Since we want to reflect the parent function $\displaystyle f(x)=\left(\frac{1}{4}\right)^{x}$ about the x-axis, we multiply f(x) by -1 to get, $\displaystyle g(x)=-\left(\frac{1}{4}\right)^{x}$ . Next we create a table of points as in the table below.

xxx

Points

x

−3

−2

−1

0

1

2

3

$\displaystyle g(x)=-\left(\frac{1}{4}\right)^{x}$

−64

−16

−4

−1

−0.25

−0.0625

−0.0156

Plot the y-intercept, (0, −1), along with two other points. We can use (−1, −4) and (1, −0.25).

Draw a smooth curve connecting the points: Graph of the function, g(x) = -(0.25)^(x), with an asymptote at y=0. Labeled points in the graph are (-1, -4), (0, -1), and (1, -0.25). The domain is (−∞, ∞); the range is (−∞, 0); the horizontal asymptote is y = 0.

Translations of the Parent Function *f(x) = b^x*
Translation	Form
Shift Horizontally c units to the left Vertically d units up	f(x) = b^{x + c} + d
Stretch and Compress Stretch if \|a\| > 1 Compression if 0 < \|a\| < 1	f(x) = ab^x
Reflect about the x-axis	f(x) = −b^x
Reflect about the y-axis	$\displaystyle f(x)=b^{-x}=\left(\frac{1}{b}\right)^{x}$
General equation for all translations	f(x) = ab^{x + c} + d

Translations of Exponential Functions

A translation of an exponential function has the form f(x) = ab^{x + c} + d

Where the parent function, y = b^x, b > 1, is

shifted horizontally c units to the left.
stretched vertically by a factor of |a| if |a| > 0.
compressed vertically by a factor of |a| if 0 < |a| < 1.
shifted vertically d units.
reflected about the x-axis when a < 0.

Note the order of the shifts, transformations, and reflections follow the order of operations.

Natural Base e

All of our exponential functions have had either an integer or a rational number as the base. We will now look at an exponential function with an irrational number as the base.

Before we can look at this exponential function, we need to define the irrational number, e. This number is used as a base in many applications in the sciences and business that are modeled by exponential functions. The number is defined as the value of $\displaystyle\left(1+\frac{1}{n}\right)^{n}$ as n gets larger and larger. We say, as n approaches infinity, or increases without bound. The table shows the value of $\displaystyle\left(1+\frac{1}{n}\right)^{n}$ for several values of n.

n

$\displaystyle\left(1+\frac{1}{n}\right)^{n}$

1

2

2.25

5

2.48832

10

2.59374246

100

2.704813829…

1,000

2.716923932…

10,000

2.718145927…

100,000

2.718268237…

1,000,000

2.718280469…

1,000,000,000

2.718281827…

e ≈ 2.718281827

The number e is like the number π in that we use a symbol to represent it because its decimal representation never stops or repeats. The irrational number e is called the natural base.

Natural Base e

The number e is defined as the value of $\displaystyle\left(1+\frac{1}{n}\right)^{n}$ , as n increases without bound. We say, as n approaches infinity,

e ≈ 2.718281827…

The exponential function whose base is e, f(x) = e^x is called the natural exponential function.

Natural Exponential Function

The natural exponential function is an exponential function whose base is e:

f(x) = e^x

The domain is (−∞, ∞) and the range is (0, ∞).

Let’s graph the function f(x) = e^x on the same coordinate system as g(x) = 2^x and h(x) = 3^x.Notice that the graph of f(x) = e^x is “between” the graphs of g(x) = 2^x and h(x) = 3^x. Does this make sense as 2 < e < 3?

Use Exponential Models in Applications

Exponential functions model many situations. If you own a bank account, you have experienced the use of an exponential function. There are two formulas that are used to determine the balance in the account when interest is earned. If a principal, P, is invested at an interest rate, r, for t years, the new balance, A, will depend on how often the interest is compounded. If the interest is compounded n times a year we use the formula $\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$ . If the interest is compounded continuously, we use the formula A = Pe^rt. These are the formulas for compound interest.

Compound Interest

For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:

$\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$ when compounded n times a year.

A = Pe^rt when compounded continuously.

As you work with the Interest formulas, it is often helpful to identify the values of the variables first and then substitute them into the formula.

Try it!

A total of $10,000 was invested in a college fund for a new grandchild. If the interest rate is 5%,how much will be in the account in 18 years by each method of compounding?

a. compound quarterly

b. compound monthly

c. compound continuously

Solution

Steps

Algebraic

A = ?

Identify the values of each variable in the formulas.

P = $10,000

Remember to express the percent as a decimal.

r = 0.05

t = 18 years

a.

Steps

Algebraic

For quarterly compounding, n = 4. There are 4 quarters in a year.

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

Substitute the values in the formula

$\displaystyle A=10,000\left(1+\frac{0.05}{4}\right)^{4\cdot 18}$

Compute the amount. Be careful to consider the order of operations as you enter the expression into your calculator.

A = $24,459.20

b.

Steps

Algebraic

For monthly compounding, n = 12. There are 12 months in a year.

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

Substitute the values in the formula.

$\displaystyle A=10,000\left(1+\frac{0.05}{12}\right)^{12\cdot 18}$

Compute the amount.

A = $24,550.08

c.

Steps

Algebraic

For compounding continuously,

A = Pe^rt

Substitute the values in the formula.

$\displaystyle A = 10,000e^{0.05\cdot 18}$

Compute the amount.

A = $24,596.03

2. Angela invested $15,000 in a savings account. If the interest rate is 4%, how much will be in the account in 10 years by each method of compounding?

a. compound quarterly

b. compound monthly

c. compound continuously

Solution

a. $22,332.96

b. $22,362.49

c. $22,377.37

Other topics that are modeled by exponential functions involve growth and decay. Both also use the formula A = Pe^rt we used for the growth of money. For growth and decay, generally we use A₀, as the original amount instead of calling it P, the principal. We see that exponential growth has a positive rate of growth and exponential decay has a negative rate of growth.

Exponential Growth and Decay

For an original amount, A₀, that grows or decays at a rate, r, for a certain time, t, the final amount, A, is:

A = A₀e^rt

Exponential growth is typically seen in the growth of populations of humans or animals or bacteria. Our next example looks at the growth of a virus.

Try it!

Chris is a researcher at the Center for Disease Control and Prevention and he is trying to understand the behavior of a new and dangerous virus. He starts his experiment with 100 of the virus that grows at a rate of 25% per hour. He will check on the virus in 24 hours. How many viruses will he find?

Solution

Steps

Algebraic

Identify the values of each variable in the formulas.

A = ?

Be sure to put the percent in decimal form.

A₀ = 100

Be sure the units match—the rate is per hour and the time is in hours.

r = 0.25/hour

t = 24 hours

Substitute the values in the formula: A = A₀e^rt.

$\displaystyle A=100e^{0.25\cdot 24}$

Compute the amount.

A = 40,342.88

Round to the nearest whole virus.

A = 40,343

The researcher will find 40,343 viruses.

2. Maria, a biologist is observing the growth pattern of a virus. She starts with 100 of the virus that grows at a rate of 10% per hour. She will check on the virus in 24 hours. How many viruses will she find?

Solution

She will find 1,102 viruses.

Access these online resources for additional instruction and practice with evaluating and graphing exponential functions.

Key Concepts

Properties of the Graph of f(x) = a^x

when a > 1

when 0 < a < 1

Domain

(−∞, ∞)

Domain

(−∞, ∞)

Range

(0, ∞)

Range

(0, ∞)

x-intercept

none

x-intercept

none

y-intercept

(0, 1)

y-intercept

(0, 1)

Contains

(1, a), $\displaystyle\left(-1,\frac{1}{a}\right)$

Contains

(1, a), $\displaystyle\left(-1,\frac{1}{a}\right)$

Asymptote

x-axis, the line y = 0

Asymptote

x-axis, the line y = 0

Basic shape

increasing

Basic shape

decreasing

This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). Compound Interest: For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is

$\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$ when compounded n times a year.
A = Pe^rt when compounded continuously.

Exponential Growth and Decay: For an original amount, A₀, that grows or decays at a rate, r, for a certain time, t, the final amount, A, is: A = A₀e^rt.

License

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NCB 0502 - Support for Corequisite Algebra Copyright © by Elizabeth Pople is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Define and Evaluate Exponential Functions

Graphing Exponential Functions

Transformations

Natural Base e

Use Exponential Models in Applications

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