Evaluate Exponential Functions
Learning Outcomes
- Identify the base of an exponential function and restrictions for its value.
- Evaluate exponential growth functions.
- Use the compound interest formula.
- Evaluate exponential functions with base e.
[latex]f(x)=ab^x[/latex]
where [latex]a[/latex] is any nonzero number and [latex]b\neq1[/latex] is a positive real number.- if [latex]b>1[/latex], the function grows at a rate proportional to its size.
- if [latex]0<b<1[/latex], the function decays at a rate proportional to its size.
- Consider a base of –9 and exponent of [latex]\frac{1}{2}[/latex]. Then [latex]f\left(x\right)=f\left(\frac{1}{2}\right)={\left(-9\right)}^{\frac{1}{2}}=\sqrt{-9}[/latex], which is not a real number.
- Consider a base of 1. Then [latex]f\left(x\right)={1}^{x}=1[/latex] for any value of x.
[latex]\begin{array}{llllllll}f\left(x\right)\hfill & ={2}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & ={2}^{3}\text{}\hfill & \text{Substitute }x=3. \hfill \\ \hfill & =8\text{}\hfill & \text{Evaluate the power}\text{.}\hfill \end{array}[/latex]
When evaluating an exponential function, it is important to follow the order of operations. For example: Let [latex]f\left(x\right)=30{\left(2\right)}^{x}[/latex]. What is [latex]f\left(3\right)[/latex]?[latex]\begin{array}{c}f\left(x\right)\hfill & =30{\left(2\right)}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & =30{\left(2\right)}^{3}\hfill & \text{Substitute }x=3.\hfill \\ \hfill & =30\left(8\right)\text{ }\hfill & \text{Simplify the power first}\text{.}\hfill \\ \hfill & =240\hfill & \text{Multiply}\text{.}\hfill \end{array}[/latex]
Note that if the order of operations were not followed, the result would be incorrect:[latex]f\left(3\right)=30{\left(2\right)}^{3}\ne {60}^{3}=216,000[/latex]
Example: Evaluating Exponential Functions
Let [latex]f\left(x\right)=5{\left(3\right)}^{x+1}[/latex]. Evaluate [latex]f\left(2\right)[/latex] without using a calculator.Answer: Follow the order of operations. Be sure to pay attention to the parentheses.
[latex]\begin{array}{c}f\left(x\right)\hfill & =5{\left(3\right)}^{x+1}\hfill & \hfill \\ f\left(2\right)\hfill & =5{\left(3\right)}^{2+1}\hfill & \text{Substitute }x=2.\hfill \\ \hfill & =5{\left(3\right)}^{3}\hfill & \text{Add the exponents}.\hfill \\ \hfill & =5\left(27\right)\hfill & \text{Simplify the power}\text{.}\hfill \\ \hfill & =135\hfill & \text{Multiply}\text{.}\hfill \end{array}[/latex]
Try It
Let [latex]f\left(x\right)=8{\left(1.2\right)}^{x - 5}[/latex]. Evaluate [latex]f\left(3\right)[/latex] using a calculator. Round to four decimal places.Answer: 5.5556
[ohm_question]73212[/ohm_question]A General Note: 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 [latex]b\ne 1[/latex], an exponential growth function has the form [latex-display]\text{ }f\left(x\right)=a{b}^{x}[/latex-display] where- a is the initial or starting value of the function.
- b is the growth factor or growth multiplier per unit x.
Tip for success
In mathematics, we carefully define terms, objects, and concepts so that there can be no ambiguity about what we mean. But such careful definition can be challenging to decipher. When reading definitions such as the one given above, it is helpful to write them out slowly by hand, perhaps more than once or twice, trying to explain each part to yourself as you go. That's how you'll build intuition for applying the term, object, or concept. Be sure to read the surrounding text as well, such as the paragraph below. It will help you understand the definition in more general terms.| 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 graph shows the numbers of stores Companies A and B opened over a five-year period.Example: Evaluating a Real-World Exponential Model
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 [latex]P\left(t\right)=1.25{\left(1.012\right)}^{t}[/latex] where t is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?Answer: 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,
[latex]P\left(18\right)=1.25{\left(1.012\right)}^{18}\approx 1.549[/latex]
There will be about 1.549 billion people in India in the year 2031.Try It
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 [latex]P\left(t\right)=1.39{\left(1.006\right)}^{t}[/latex] where t is the number of years since 2013. To the nearest thousandth, what will the population of China be in the year 2031? How does this compare to the population prediction we made for India in the previous example?Answer: 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.
[ohm_question]129476[/ohm_question]Using 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 compound interest using the compound interest formula which is an exponential function of the variables time t, principal P, APR r, and number of times compounded in a year n:[latex]A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}[/latex]
For example, observe the table below, 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 |
A General Note: The Compound Interest Formula
Compound interest can be calculated using the formula[latex]A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}[/latex]
where- A(t) is the accumulated value of the account
- 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
- n is the number of times compounded in a year
Example: Calculating Compound Interest
If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?Answer: 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.
[latex]\begin{array}{llllll}A\left(t\right)\hfill & =P\left(1+\frac{r}{n}\right)^{nt}\hfill & \text{Use the compound interest formula}. \\ A\left(10\right)\hfill & =3000\left(1+\frac{0.03}{4}\right)^{4\cdot 10}\hfill & \text{Substitute using given values}. \\ \text{ }\hfill & \approx 4045.05\hfill & \text{Round to two decimal places}.\end{array}[/latex]
The account will be worth about $4,045.05 in 10 years.Tip for success
When calculating the value of an exponential function such as the compound interest formula, be careful when entering your calculation into a calculator. Use as many parentheses as needed to ensure your intent is clear. The calculator will apply order of operations as it is typed, which can cause an incorrect calculation for your equation. Ex. To find the accumulated investment in the example above, you must calculate [latex]A\left(10\right)=3000\left(1+\frac{0.03}{4}\right)^{4\cdot 10}[/latex]. This would be entered in most scientific or graphing calculators as [latex]3000\left(1+(0.03/4)\right)\wedge\left(4\star10\right)[/latex].Try It
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?Answer: about $3,644,675.88
[ohm_question]2453[/ohm_question]Example: Using the Compound Interest Formula to Solve for the Principal
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?Answer: 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.
[latex]\begin{array}{c}A\left(t\right)\hfill & =P{\left(1+\frac{r}{n}\right)}^{nt}\hfill & \text{Use the compound interest formula}.\hfill \\ 40,000\hfill & =P{\left(1+\frac{0.06}{2}\right)}^{2\left(18\right)}\hfill & \text{Substitute using given values }A\text{, }r, n\text{, and }t.\hfill \\ 40,000\hfill & =P{\left(1.03\right)}^{36}\hfill & \text{Simplify}.\hfill \\ \frac{40,000}{{\left(1.03\right)}^{36}}\hfill & =P\hfill & \text{Isolate }P.\hfill \\ P\hfill & \approx 13,801\hfill & \text{Divide and round to the nearest dollar}.\hfill \end{array}[/latex]
Lily will need to invest $13,801 to have $40,000 in 18 years.Try It
Refer to the previous example. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?Answer: $13,693
Evaluating Exponential Functions with Base e
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below 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.| Frequency | [latex]A\left(t\right)={\left(1+\frac{1}{n}\right)}^{n}[/latex] | Value |
|---|---|---|
| Annually | [latex]{\left(1+\frac{1}{1}\right)}^{1}[/latex] | $2 |
| Semiannually | [latex]{\left(1+\frac{1}{2}\right)}^{2}[/latex] | $2.25 |
| Quarterly | [latex]{\left(1+\frac{1}{4}\right)}^{4}[/latex] | $2.441406 |
| Monthly | [latex]{\left(1+\frac{1}{12}\right)}^{12}[/latex] | $2.613035 |
| Daily | [latex]{\left(1+\frac{1}{365}\right)}^{365}[/latex] | $2.714567 |
| Hourly | [latex]{\left(1+\frac{1}{\text{8766}}\right)}^{\text{8766}}[/latex] | $2.718127 |
| Once per minute | [latex]{\left(1+\frac{1}{\text{525960}}\right)}^{\text{525960}}[/latex] | $2.718279 |
| Once per second | [latex]{\left(1+\frac{1}{31557600}\right)}^{31557600}[/latex] | $2.718282 |
understanding the number [latex]e[/latex]
The number [latex]e[/latex] is an irrational constant, like the number [latex]pi[/latex]. The number [latex]e[/latex] appears frequently in applications in the real world and is as important as numbers like [latex]0, 1, i, \text{ and } \pi[/latex], so it is apt that it was named in honor of one of the most prolific and important mathematicians to have lived, Leonhard Euler (pronounced like "oiler").A General Note: The Number [latex]e[/latex]
The letter e represents the irrational number[latex]{\left(1+\frac{1}{n}\right)}^{n},\text{as }n\text{ increases without bound}[/latex]
The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, [latex]e\approx 2.718282[/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.Example: Using a Calculator to Find Powers of [latex]e[/latex]
Calculate [latex]{e}^{3.14}[/latex]. Round to five decimal places.Answer: On a calculator, press the button labeled [latex]\left[{e}^{x}\right][/latex]. The window shows [e^( ]. Type 3.14 and then close parenthesis, [ e^(3.14) ]. Press [ENTER]. Rounding to 5 decimal places, [latex]{e}^{3.14}\approx 23.10387[/latex]. 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.
Try It
Use a calculator to find [latex]{e}^{-0.5}[/latex]. Round to five decimal places.Answer: [latex]{e}^{-0.5}\approx 0.60653[/latex]
[ohm_question]1495[/ohm_question]Licenses & Attributions
CC licensed content, Original
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- Question ID 1495. Authored by: WebWork-Rochester. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
- Question ID 2453. Authored by: Anderson,Tophe, mb Sousa,James. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
