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Exponential Functions

Evaluate exponential functions.

Exponential Functions

Find the value of a function (exponential). (IA 3.5.3)
Graph exponential functions. (IA 10.2.1)

Example

Evaluate the function $f(x)={3}^{x}$ for the given values

Try it.

ⓐ $f(2)$
ⓑ $f(-1)$
ⓒ $f(2h)$

Solution

ⓐ Replace x with 2 and find the value of the function $f(2)={3}^{2}=9$
ⓑ Replace x with -1 and find the value of the function $f(2)={3}^{-1}=\frac{1}{3}$
ⓒ Replace x with 2h and simplify if possible $f(2)={3}^{2h}$

Find the value of an exponential function.

Try it.

Evaluate the function $f(x)={(\frac{3}{2})}^{x}$ for the given values.

ⓐ $f(2)$
ⓑ $f(-2)$
ⓒ $f(a)$

Try it.

We also find the value of the function when we solve application problems involving exponential functions.
Medicare Premiums. The monthly Medicare Part B health-care premium for most beneficiaries ages 65 and older has increased significantly since 1975. The monthly premium has increased from about $7 in 1975 to $110.50 in 2011 (Source: Centers for Medicare and Medicaid Services). The following exponential function models the premium increases:
$M(x)=7{(1.080)}^{x}$ where x is the number of years since 1975.
Estimate the monthly Medicare Part B premium in 1985, in 1992, and in 2002. (Note that x is the number of years since 1975, so for 1985, x=10.) Round to the nearest dollar.

Try it.

We can find Compound Interest using $A=P{(1+\frac{r}{n})}^{nt}$ ,
Where A is the amount of money, P is the principal, t is the number of years, r is the interest rate, and n is the number of times the interest was compounded per year.
Suppose that $960 is invested at 7% interest, compounded semiannually.

ⓐ Find the function for the amount to which the investment grows after t years.
ⓑ Find the amount of money in the account at t=1, 6, 10, 15, and 20 years.

Condensed — the full section is in OpenStax Precalculus 2e.

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. 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.

Condensed — the full section is in OpenStax Precalculus 2e.

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 $f(x)=f(\frac{1}{2})={(-9)}^{\frac{1}{2}}=\sqrt{-9},$ 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 $f(3)?$

\[\begin{array}{lll}f(x) & ={2}^{x} & \\ f(3) & ={2}^{3}\ & \text{Substitute }x=3. \\ & =8\ & \text{Evaluate the power}\text{.}\end{array}\]

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 $f(3)?$

\[\begin{array}{lll}f(x) & =30{(2)}^{x} & \\ f(3) & =30{(2)}^{3} & \text{Substitute }x=3. \\ & =30(8)\ & \text{Simplify the power first}\text{.} \\ & =240 & \text{Multiply}\text{.}\end{array}\]

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

\[f(3)=30{(2)}^{3}\ne {60}^{3}=216,000\]

Example

Try it.

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.

\[\begin{array}{lll}f(x) & =5{(3)}^{x+1} & \\ f(2) & =5{(3)}^{2+1} & \text{Substitute }x=2. \\ & =5{(3)}^{3} & \text{Add the exponents}. \\ & =5(27) & \text{Simplify the power}\text{.} \\ & =135 & \text{Multiply}\text{.}\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

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.

Example

Try it.

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 $f(x)=a{b}^{x},$ 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 $b:$

Substitute $a$ in the second equation, and solve for $b:$

Use the value of $b$ in the first equation to solve for the value of $a:$

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 passes through the initial points given in the problem, $(-2,\ \text{6})$ and $(2,\ \text{1}).$ The graph is an example of an exponential decay function.

Condensed — the full section is in OpenStax Precalculus 2e.

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 $n:$

\[A(t)=P{(1+\frac{r}{n})}^{nt}\]

For example, observe , 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

Example

Try it.

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 $t\ =\ 10.$

\[\begin{array}{llllll}A(t) & =P{(1+\frac{r}{n})}^{nt} & \text{Use the compound interest formula}. \\ A(10) & =3000{(1+\frac{0.03}{4})}^{4⋅10}\begin{array}{llll} & & & \end{array} & \text{Substitute using given values}. \\ & \approx \text{\$}4045.05 & \text{Round to two decimal places}.\end{array}\]

The account will be worth about $4,045.05 in 10 years.

Condensed — the full section is in OpenStax Precalculus 2e.

Evaluating Functions with Base

As we saw earlier, the amount earned on an account increases as the compounding frequency increases. 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 .

Frequency	$A(n)={(1+\frac{1}{n})}^{n}$	Value
Annually	${(1+\frac{1}{1})}^{1}$	$2
Semiannually	${(1+\frac{1}{2})}^{2}$	$2.25
Quarterly	${(1+\frac{1}{4})}^{4}$	$2.441406
Monthly	${(1+\frac{1}{12})}^{12}$	$2.613035
Daily	${(1+\frac{1}{365})}^{365}$	$2.714567
Hourly	${(1+\frac{1}{\text{8760}})}^{\text{8760}}$	$2.718127
Once per minute	${(1+\frac{1}{\text{525600}})}^{\text{525600}}$	$2.718279
Once per second	${(1+\frac{1}{31536000})}^{31536000}$	$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 ${(1+\frac{1}{n})}^{n}$ 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.

Example

Try it.

Calculate ${e}^{3.14}.$ Round to five decimal places.

Solution

On a calculator, press the button labeled $[{e}^{x}].$ The window shows $[e\wedge (\ ].$ Type $3.14$ and then close parenthesis, $[)].$ Press [ENTER]. Rounding to $5$ decimal places, ${e}^{3.14}\approx 23.10387.$ 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 $e.$

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.

Example

Try it.

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

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:

\[\begin{array}{llllll}A(t) & =P{e}^{rt} & \text{Use the continuous compounding formula}. \\ & =1000{(e)}^{0.1}\begin{array}{llll} & & & \end{array} & \text{Substitute known values for }P,r,\ \text{and }t. \\ & \approx 1105.17 & \text{Use a calculator to approximate}.\end{array}\]

The account is worth $1,105.17 after one year.

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

definition of the exponential function	$f(x)={b}^{x}\text{, where }b>0,b\ne 1$
definition of exponential growth	$f(x)=a{b}^{x},\ \text{where }a>0,b>0,b\ne 1$
compound interest formula	$\begin{array}{l}A(t)=P{(1+\frac{r}{n})}^{nt},\ \text{where} \\ A(t)\ \text{is the account value at time }t \\ t\ \text{is the number of years} \\ P\ \text{is the initial investment, often called the principal} \\ r\ \text{is the annual percentage rate (APR), or nominal rate} \\ n\ \text{is the number of compounding periods in one year}\end{array}$
continuous growth formula	$A(t)=a{e}^{rt},\ \text{where}$ $t$ is the number of unit time periods of growth $a$ is the starting amount (in the continuous compounding formula a is replaced with P, the principal) $e$ is the mathematical constant, $e\approx 2.718282$

Key Concepts

An exponential function is defined as a function with a positive constant other than $1$ raised to a variable exponent. See .
A function is evaluated by solving at a specific value. See and .
An exponential model can be found when the growth rate and initial value are known. See .
An exponential model can be found when the two data points from the model are known. See .
An exponential model can be found using two data points from the graph of the model. See .
An exponential model can be found using two data points from the graph and a calculator. See .
The value of an account at any time $t$ can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See .
The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See .
The number $e$ is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is $e\approx 2.718282.$
Scientific and graphing calculators have the key $[{e}^{x}]$ or $[\exp (x)]$ for calculating powers of $e.$ See .
Continuous growth or decay models are exponential models that use $e$ as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. ⓐ $f(2)$
2. ⓑ $f(-1)$
3. ⓒ $f(2h)$
Sýna svarið
- ⓐ Replace x with 2 and find the value of the function $f(2)={3}^{2}=9$
- ⓑ Replace x with -1 and find the value of the function $f(2)={3}^{-1}=\frac{1}{3}$
- ⓒ Replace x with 2h and simplify if possible $f(2)={3}^{2h}$
We also find the value of the function when we solve application problems involving exponential functions.
Medicare Premiums. The monthly Medicare Part B health-care premium for most beneficiaries ages 65 and older has increased significantly since 1975. The monthly premium has increased from about $7 in 1975 to $110.50 in 2011 (Source: Centers for Medicare and Medicaid Services). The following exponential function models the premium increases:
$M(x)=7{(1.080)}^{x}$ where x is the number of years since 1975.
Estimate the monthly Medicare Part B premium in 1985, in 1992, and in 2002. (Note that x is the number of years since 1975, so for 1985, x=10.) Round to the nearest dollar.
We can find Compound Interest using $A=P{(1+\frac{r}{n})}^{nt}$ ,
Where A is the amount of money, P is the principal, t is the number of years, r is the interest rate, and n is the number of times the interest was compounded per year.
Suppose that $960 is invested at 7% interest, compounded semiannually.
1. ⓐ Find the function for the amount to which the investment grows after t years.
2. ⓑ Find the amount of money in the account at t=1, 6, 10, 15, and 20 years.
Graph the exponential function $f(x)={2}^{x}$ by making a table.
$x$ $y=f(x)$
Graph the exponential function $f(x)={(\frac{1}{2})}^{x}$ by making a table.
$x$ $y=f(x)$

How does it compare with the graph of $f(x)={2}^{x}$ ?
Graph $f(x)={3}^{x}$ , $f(x)={4}^{x}$ , $f(x)={2.5}^{x}$ in the same viewing window using a graphing calculator or program. What is the relationship between the base a and the shape of the graph?
Graph $f(x)={0.2}^{x}$ , $f(x)={0.4}^{x}$ , $f(x)={0.7}^{x}$ in the same viewing window using a graphing calculator or program. What is the relationship between the base a and the shape of the graph?
Fill in the Properties of Exponential Function.
$f(x)={a}^{x},a>0,a\ne 1$
Is it continuous?
Is it one-to-one?
Domain
Range
Increasing if
Decreasing if
Asymptotes
Intercepts
Graph the exponential function $f(x)={e}^{x}$ by making a table.
$x$ $y=f(x)$

What is the domain of $f(x)$ ?
What is the range of $f(x)$ ?
Which of the following equations are not exponential functions?
- $f(x)={4}^{3(x-2)}$
- $g(x)={x}^{3}$
- $h(x)={(\frac{1}{3})}^{x}$
- $j(x)={(-2)}^{x}$
Sýna svarið

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 $b\ne 1.$ 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?
- $f(x)=2{x}^{2}-3x+1$
- $g(x)={0.875}^{x}$
- $h(x)=1.75x+2$
- $j(x)={1095.6}^{-2x}$
Sýna svarið

$g(x)={0.875}^{x}$ and $j(x)={1095.6}^{-2x}$ represent exponential functions.
Let $f(x)=5{(3)}^{x+1}.$ Evaluate $f(2)$ without using a calculator.

Sýna svarið

Follow the order of operations. Be sure to pay attention to the parentheses.
\[\begin{array}{lll}f(x) & =5{(3)}^{x+1} & \\ f(2) & =5{(3)}^{2+1} & \text{Substitute }x=2. \\ & =5{(3)}^{3} & \text{Add the exponents}. \\ & =5(27) & \text{Simplify the power}\text{.} \\ & =135 & \text{Multiply}\text{.}\end{array}\]
Let $f(x)=8{(1.2)}^{x-5}.$ Evaluate $f(3)$ using a calculator. Round to four decimal places.

Sýna svarið

$5.5556$
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, what will the population of India be in $\text{2031?}$

Sýna svarið

To estimate the population in 2031, we evaluate the models for $t=18,$ because 2031 is $18$ years after 2013. Rounding to the nearest thousandth,
\[P(18)=1.25{(1.012)}^{18}\approx 1.549\]
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 ?

Sýna svarið

About $1.548$ billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.
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 $t.$

Sýna svarið

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 $b:$
\[\begin{array}{llllll}N(t) & =80{b}^{t} & \\ \ 180 & =80{b}^{6} & \text{Substitute using point }(6,180). \\ \ \ \frac{9}{4} & ={b}^{6} & \text{Divide and write in lowest terms}. \\ \ \ \ \ b & ={(\frac{9}{4})}^{\frac{1}{6}} & \text{Isolate }b\ \text{using properties of exponents}. \\ \ \ \ \ b & \approx 1.1447\begin{array}{llll} & & & \end{array} & \text{Round to 4 decimal places}.\end{array}\]
NOTE: 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 passes through the initial points given in the problem, $(0,80)$ and $(\text{6},180).$ 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 $\text{2013,}$ 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 $t.$

Sýna svarið

$(0,129)$ and $(2,236);\ \ N(t)=129{(\text{1}\text{.3526})}^{t}$
Find an exponential function that passes through the points $(-2,6)$ and $(2,1).$
Sýna svarið
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 $b:$

Substitute $a$ in the second equation, and solve for $b:$
Use the value of $b$ in the first equation to solve for the value of $a:$

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 passes through the initial points given in the problem, $(-2,\ \text{6})$ and $(2,\ \text{1}).$ The graph is an example of an exponential decay function.
Given the two points $(1,3)$ and $(2,4.5),$ find the equation of the exponential function that passes through these two points.

Sýna svarið

$f(x)=2{(1.5)}^{x}$
Find an equation for the exponential function graphed in .

Sýna svarið

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 $(2,12).$
\[\begin{array}{lllll}\ y=a{b}^{x} & \text{Write the general form of an exponential equation}. \\ \ y=3{b}^{x} & \text{Substitute the initial value 3 for }a. \\ 12=3{b}^{2} & \text{Substitute in 12 for }y\ \text{and 2 for }x. \\ \ 4={b}^{2} & \text{Divide by 3}. \\ \ b=\pm 2\begin{array}{llll} & & & \end{array} & \text{Take the square root}.\end{array}\]
Because we restrict ourselves to positive values of $b,$ we will use $b=2.$ 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 .

Sýna svarið

$f(x)=\sqrt{2}{(\sqrt{2})}^{x}.$ Answers may vary due to round-off error. The answer should be very close to $1.4142{(1.4142)}^{x}.$
Use a graphing calculator to find the exponential equation that includes the points $(2,24.8)$ and $(5,198.4).$

Sýna svarið

Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4.
Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values $a=6.2$ and $b=2$ will be displayed. The exponential equation is $y=6.2⋅{2}^{x}.$
Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).

Sýna svarið

$y\approx 12⋅{1.85}^{x}$
If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?

Sýna svarið

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 $t\ =\ 10.$
\[\begin{array}{llllll}A(t) & =P{(1+\frac{r}{n})}^{nt} & \text{Use the compound interest formula}. \\ A(10) & =3000{(1+\frac{0.03}{4})}^{4⋅10}\begin{array}{llll} & & & \end{array} & \text{Substitute using given values}. \\ & \approx \text{\$}4045.05 & \text{Round to two decimal places}.\end{array}\]
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?

Sýna svarið

about $3,644,675.88
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?

Sýna svarið

The nominal interest rate is 6%, so $r=0.06.$ Interest is compounded twice a year, so $n=2.$
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 $P.$
\[\begin{array}{llllll}\ \ \ \ A(t) & =P{(1+\frac{r}{n})}^{nt} & \text{Use the compound interest formula}. \\ 40,000 & =P{(1+\frac{0.06}{2})}^{2(18)}\begin{array}{llll} & & & \end{array} & \text{Substitute using given values }A\text{, }r,n\text{, and }t. \\ 40,000 & =P{(1.03)}^{36} & \text{Simplify}. \\ \frac{40,000}{{(1.03)}^{36}} & =P & \text{Isolate }P. \\ \ \ \ \ \ P & \approx \text{\$}13,801 & \text{Divide and round to the nearest dollar}.\end{array}\]
Lily will need to invest $13,801 to have $40,000 in 18 years.
Refer to . To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?

Sýna svarið

$13,693
Calculate ${e}^{3.14}.$ Round to five decimal places.

Sýna svarið

On a calculator, press the button labeled $[{e}^{x}].$ The window shows $[e\wedge (\ ].$ Type $3.14$ and then close parenthesis, $[)].$ Press [ENTER]. Rounding to $5$ decimal places, ${e}^{3.14}\approx 23.10387.$ 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 $e.$
Use a calculator to find ${e}^{-0.5}.$ Round to five decimal places.

Sýna svarið

${e}^{-0.5}\approx 0.60653$
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?

Sýna svarið

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:
\[\begin{array}{llllll}A(t) & =P{e}^{rt} & \text{Use the continuous compounding formula}. \\ & =1000{(e)}^{0.1}\begin{array}{llll} & & & \end{array} & \text{Substitute known values for }P,r,\ \text{and }t. \\ & \approx 1105.17 & \text{Use a calculator to approximate}.\end{array}\]
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?

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$3,659,823.44
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?

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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:
\[\begin{array}{llllll}A(t) & =a{e}^{rt} & \text{Use the continuous growth formula}. \\ & =100{e}^{-0.173(3)}\begin{array}{llll} & & & \end{array} & \text{Substitute known values for }a,r,\ \text{and }t. \\ & \approx 59.5115 & \text{Use a calculator to approximate}.\end{array}\]
So 59.5115 mg of radon-222 will remain.
Using the data in , how much radon-222 will remain after one year?

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3.77E-26 (This is calculator notation for the number written as $3.77\times {10}^{-26}$ in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)
Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function.

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Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.
Given a formula for an exponential function, is it possible to determine whether the function grows or decays exponentially just by looking at the formula? Explain.
The Oxford Dictionary defines the word nominal as a value that is “stated or expressed but not necessarily corresponding exactly to the real value.”Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina. Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest.

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When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal.
The average annual population increase of a pack of wolves is 25.
A population of bacteria decreases by a factor of $\frac{1}{8}$ every $24$ hours.

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exponential; the population decreases by a proportional rate. .
The value of a coin collection has increased by $3.25\%$ annually over the last $20$ years.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\neg,\ \wedge,\ \vee,\ \Rightarrow,\ \Leftrightarrow

not, and, or, implies, iff
Logical connectives.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Exponential Functions

Evaluate exponential functions.
Find the equation of an exponential function.
Use compound interest formulas.
Evaluate exponential functions with base
Find the value of a function (exponential). (IA 3.5.3)
Graph exponential functions. (IA 10.2.1)

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Prófaðu þitt eigið

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Meira í Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

Frequency	\(A(n)={(1+\frac{1}{n})}^{n}\)	Value
Annually	\({(1+\frac{1}{1})}^{1}\)	$2
Semiannually	\({(1+\frac{1}{2})}^{2}\)	$2.25
Quarterly	\({(1+\frac{1}{4})}^{4}\)	$2.441406
Monthly	\({(1+\frac{1}{12})}^{12}\)	$2.613035
Daily	\({(1+\frac{1}{365})}^{365}\)	$2.714567
Hourly	\({(1+\frac{1}{\text{8760}})}^{\text{8760}}\)	$2.718127
Once per minute	\({(1+\frac{1}{\text{525600}})}^{\text{525600}}\)	$2.718279
Once per second	\({(1+\frac{1}{31536000})}^{31536000}\)	$2.718282

definition of the exponential function	\(f(x)={b}^{x}\text{, where }b>0,b\ne 1\)
definition of exponential growth	\(f(x)=a{b}^{x},\ \text{where }a>0,b>0,b\ne 1\)
compound interest formula	\(\begin{array}{l}A(t)=P{(1+\frac{r}{n})}^{nt},\ \text{where} \\ A(t)\ \text{is the account value at time }t \\ t\ \text{is the number of years} \\ P\ \text{is the initial investment, often called the principal} \\ r\ \text{is the annual percentage rate (APR), or nominal rate} \\ n\ \text{is the number of compounding periods in one year}\end{array}\)
continuous growth formula	\(A(t)=a{e}^{rt},\ \text{where}\) \(t\) is the number of unit time periods of growth \(a\) is the starting amount (in the continuous compounding formula a is replaced with P, the principal) \(e\) is the mathematical constant, \(e\approx 2.718282\)

\(x\)	\(y=f(x)\)

\(x\)	\(y=f(x)\)

\(x\)	\(y=f(x)\)