after what y-value does the exponential function appear to surpass the linear function?

Learning Objectives

In this section, you will:

• Evaluate exponential functions.
• Notice the equation of an exponential function.
• Use compound involvement formulas.
• Evaluate exponential functions with base of operations e.

India is the second most populous country in the world with a population of most[latex]\,ane.25\,[/latex]billion people in 2013. The population is growing at a rate of nearly[latex]\,1.two%\,[/latex]each year^[1] . If this charge per unit continues, the population of India will exceed Communist china'south population by the year[latex]\,2031.[/latex]When populations grow apace, we often say that the growth is "exponential," meaning that something is growing very speedily. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, nosotros will have a expect 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 instance, in the equation[latex]\,f\left(ten\correct)=3x+four,[/latex]the slope tells us the output increases past three each time the input increases by i. The scenario in the India population example is dissimilar because we accept a percent change per unit of measurement time (rather than a constant alter) 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 of america doubled from 2009 to 2011. In 2011, 2.v% of the population was vegan, adhering to a diet that does non include any animal products—no meat, poultry, fish, dairy, or eggs. If this charge per unit continues, vegans volition make up 10% of the U.South. population in 2015, 40% in 2019, and lxxx% in 2021.

What exactly does it mean to grow exponentially? What does the give-and-take double accept in common with percent increase? People toss these words around errantly. Are these words used correctly? The words certainly appear oftentimes in the media.

• Percent change refers to a change based on a pct of the original amount.
• Exponential growth refers to an increase based on a abiding multiplicative charge per unit of change over equal increments of time, that is, a percent increment of the original amount over time.
• Exponential decay refers to a decrease based on a abiding multiplicative charge per unit of alter over equal increments of fourth dimension, that is, a percent decrease of the original amount over time.

For usa to gain a articulate understanding of exponential growth, let us dissimilarity exponential growth with linear growth. We will construct two functions. The outset function is exponential. We will starting time with an input of 0, and increase each input by 1. Nosotros will double the corresponding consecutive outputs. The second part is linear. We will outset with an input of 0, and increment each input by 1. Nosotros will add together two to the respective consecutive outputs. See (Figure).

[latex]x[/latex]	[latex]f\left(ten\right)={2}^{10}[/latex]	[latex]chiliad\left(x\right)=2x[/latex]
0	1	0
i	2	2
2	4	4
3	8	6
4	sixteen	viii
5	32	10
6	64	12

From (Figure) nosotros tin infer that for these two functions, exponential growth dwarfs linear growth.

• Exponential growth refers to the original value from the range increases by the aforementioned percentage over equal increments institute in the domain.
• Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain.

Evidently, the deviation between "the same pct" and "the same amount" is quite meaning. For exponential growth, over equal increments, the constant multiplicative charge per unit of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant condiment rate of change over equal increments resulted in adding ii to the output whenever the input was increased past 1.

The general form of the exponential office is[latex]\,f\left(10\right)=a{b}^{x},\,[/latex]where[latex]\,a\,[/latex]is whatever nonzero number,[latex]\,b\,[/latex]is a positive real number non equal to 1.

• If[latex]\,b>ane,[/latex]the function grows at a rate proportional to its size.
• If[latex]\,0<b<ane,[/latex] the part decays at a rate proportional to its size.

Permit's look at the function[latex]\,f\left(x\correct)={two}^{10}\,[/latex]from our example. We will create a table ((Figure)) to determine the respective outputs over an interval in the domain from[latex]\,-three\,[/latex]to[latex]\,three.[/latex]

[latex]10[/latex]	[latex]-3[/latex]	[latex]-ii[/latex]	[latex]-i[/latex]	[latex]0[/latex]	[latex]1[/latex]	[latex]2[/latex]	[latex]3[/latex]
[latex]f\left(x\right)={2}^{x}[/latex]	[latex]{2}^{-3}=\frac{1}{8}[/latex]	[latex]{2}^{-two}=\frac{1}{4}[/latex]	[latex]{ii}^{-ane}=\frac{i}{2}[/latex]	[latex]{2}^{0}=ane[/latex]	[latex]{two}^{1}=2[/latex]	[latex]{2}^{2}=four[/latex]	[latex]{ii}^{iii}=8[/latex]

Let the states examine the graph of[latex]\,f\,[/latex]by plotting the ordered pairs we notice on the table in (Figure), and and then make a few observations.

Figure one.

Let's ascertain the beliefs of the graph of the exponential role[latex]\,f\left(x\correct)={2}^{x}\,[/latex]and highlight some its cardinal characteristics.

• the domain is[latex]\,\left(-\infty ,\infty \right),[/latex]
• the range is[latex]\,\left(0,\infty \right),[/latex]
• as[latex]\,x\to \infty ,f\left(x\right)\to \infty ,[/latex]
• as [latex]\,x\to -\infty ,f\left(x\right)\to 0,[/latex]
• [latex]\,f\left(10\correct)\,[/latex]is always increasing,
• the graph of[latex]\,f\left(x\correct)\,[/latex]will never touch the ten-centrality because base 2 raised to any exponent never has the result of zero.
• [latex]\,y=0\,[/latex]is the horizontal asymptote.
• the y-intercept is 1.

Exponential Part

For whatever real number[latex]\,10,[/latex]an exponential function is a part with the course

[latex]f\left(ten\right)=a{b}^{x}[/latex]

where

• [latex]\,a\,[/latex]is a non-zero real number chosen the initial value and
• [latex]\,b\,[/latex]is any positive real number such that[latex]\,b\ne 1.[/latex]
• The domain of[latex]\,f\,[/latex]is all real numbers.
• The range of[latex]\,f\,[/latex]is all positive existent numbers if[latex]\,a>0.[/latex]
• The range of[latex]\,f\,[/latex]is all negative existent numbers if[latex]\,a<0.[/latex]
• The y-intercept is[latex]\,\left(0,a\right),[/latex]and the horizontal asymptote is[latex]\,y=0.[/latex]

Identifying Exponential Functions

Which of the following equations are non exponential functions?

• [latex]f\left(x\right)={4}^{iii\left(x-2\right)}[/latex]
• [latex]thousand\left(x\right)={x}^{three}[/latex]
• [latex]h\left(x\correct)={\left(\frac{1}{3}\right)}^{x}[/latex]
• [latex]j\left(x\right)={\left(-ii\right)}^{x}[/latex]

Try It

Which of the following equations represent exponential functions?

• [latex]f\left(x\right)=ii{x}^{two}-3x+ane[/latex]
• [latex]g\left(x\right)={0.875}^{x}[/latex]
• [latex]h\left(x\right)=1.75x+ii[/latex]
• [latex]j\left(10\right)={1095.6}^{-2x}[/latex]

Show Solution

[latex]g\left(10\right)={0.875}^{x}\,[/latex]and[latex]j\left(x\correct)={1095.6}^{-2x}\,[/latex]stand for exponential functions.

Evaluating Exponential Functions

Remember that the base of an exponential office must be a positive real number other than[latex]\,1.[/latex]Why do we limit the base [latex]b\,[/latex]to positive values? To ensure that the outputs will be existent numbers. Discover what happens if the base is not positive:

• Let[latex]\,b=-9\,[/latex]and[latex]\,x=\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.

Why do nosotros limit the base to positive values other than [latex]1?[/latex]Because base of operations [latex]1\,[/latex]results in the constant role. Observe what happens if the base is [latex]1:[/latex]

• Allow[latex]\,b=ane.\,[/latex]Then[latex]\,f\left(x\right)={i}^{x}=1\,[/latex]for whatsoever value of[latex]\,x.[/latex]

To evaluate an exponential function with the form[latex]\,f\left(x\correct)={b}^{x},[/latex]we only substitute [latex]10\,[/latex]with the given value, and calculate the resulting power. For example:

Let [latex]\,f\left(x\right)={2}^{x}.\,[/latex]What is [latex]f\left(three\correct)?[/latex]

[latex]\brainstorm{array}{lll}f\left(ten\right)\hfill & ={2}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & ={2}^{3}\text{ }\hfill & \text{Substitute }x=three.\hfill \\ \hfill & =viii\text{ }\hfill & \text{Evaluate the power}\text{.}\hfill \end{array}[/latex]

To evaluate an exponential function with a course other than the basic form, information technology is important to follow the order of operations. For case:

Let[latex]\,f\left(x\right)=30{\left(2\right)}^{ten}.\,[/latex]What is[latex]\,f\left(3\right)?[/latex]

[latex]\begin{array}{lll}f\left(ten\right)\hfill & =30{\left(2\correct)}^{x}\hfill & \hfill \\ f\left(three\right)\hfill & =30{\left(2\correct)}^{3}\hfill & \text{Substitute }10=3.\hfill \\ \hfill & =thirty\left(eight\right)\text{ }\hfill & \text{Simplify the ability 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 exist wrong:

[latex]f\left(3\right)=30{\left(two\right)}^{3}\ne {sixty}^{3}=216,000[/latex]

Evaluating Exponential Functions

Let [latex]\,f\left(ten\right)=5{\left(iii\right)}^{x+1}.\,[/latex]Evaluate[latex]\,f\left(2\right)\,[/latex]without using a calculator.

Try It

Let[latex]f\left(x\right)=8{\left(i.2\right)}^{x-5}.\,[/latex]Evaluate[latex]\,f\left(3\right)\,[/latex]using a estimator. Round to four decimal places.

Show Solution

[latex]5.5556[/latex]

Defining Exponential Growth

Because the output of exponential functions increases very rapidly, the term "exponential growth" is often used in everyday linguistic communication to describe anything that grows or increases chop-chop. Still, exponential growth can be defined more precisely in a mathematical sense. If the growth charge per unit is proportional to the amount present, the function models exponential growth.

Exponential Growth

A function that models exponential growth grows by a rate proportional to the amount nowadays. For any real number[latex]\,x\,[/latex]and any positive real numbers[latex]\,a \,[/latex]and[latex]\,b\,[/latex]such that[latex]\,b\ne 1,[/latex]an exponential growth function has the grade

[latex]\text{ }f\left(x\right)=a{b}^{x}[/latex]

where

• [latex]a\,[/latex]is the initial or starting value of the role.
• [latex]b\,[/latex]is the growth cistron or growth multiplier per unit of measurement[latex]\,10[/latex].

In more full general terms, we have an exponential function, in which a abiding base of operations is raised to a variable exponent. To differentiate between linear and exponential functions, permit's consider 2 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[latex]\,A\left(x\right)=100+50x.\,[/latex]Company B has 100 stores and expands by increasing the number of stores by fifty% each year, so its growth can be represented past the function [latex]\,B\left(x\right)=100{\left(i+0.5\right)}^{x}.[/latex]

A few years of growth for these companies are illustrated in (Figure).

Year, [latex]x[/latex]	Stores, Company A	Stores, Company B
[latex]0[/latex]	[latex]100+50\left(0\right)=100[/latex]	[latex]100{\left(1+0.5\right)}^{0}=100[/latex]
[latex]one[/latex]	[latex]100+50\left(ane\right)=150[/latex]	[latex]100{\left(1+0.5\correct)}^{1}=150[/latex]
[latex]ii[/latex]	[latex]100+50\left(two\right)=200[/latex]	[latex]100{\left(one+0.5\right)}^{two}=225[/latex]
[latex]3[/latex]	[latex]100+50\left(three\right)=250[/latex]	[latex]100{\left(i+0.5\right)}^{3}=337.five[/latex]
[latex]x[/latex]	[latex]A\left(x\right)=100+50x[/latex]	[latex]B\left(x\right)=100{\left(1+0.5\correct)}^{ten}[/latex]

The graphs comparing the number of stores for each visitor over a five-year period are shown in (Figure). We tin can meet that, with exponential growth, the number of stores increases much more rapidly than with linear growth.

Figure two. The graph shows the numbers of stores Companies A and B opened over a v-year catamenia.

Detect that the domain for both functions is[latex]\,\left[0,\infty \correct),[/latex]and the range for both functions is[latex]\,\left[100,\infty \right).\,[/latex]Afterward year one, Company B always has more stores than Company A.

At present we volition turn our attending to the function representing the number of stores for Company B,[latex]\,B\left(x\right)=100{\left(i+0.5\right)}^{x}.\,[/latex]In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and[latex]\,ane+0.5=i.5\,[/latex]represents the growth gene. Generalizing further, we can write this function as[latex]\,B\left(x\right)=100{\left(one.5\right)}^{x},[/latex]where 100 is the initial value,[latex]\,one.five\,[/latex]is chosen the base, and[latex]\,x\,[/latex]is called the exponent.

Evaluating a Real-World Exponential Model

At the commencement of this section, nosotros learned that the population of India was almost[latex]\,1.25\,[/latex]billion in the year 2013, with an almanac growth rate of near[latex]\,1.ii%.\,[/latex]This situation is represented by the growth office[latex]\,P\left(t\right)=i.25{\left(i.012\right)}^{t},[/latex] where [latex]\,t\,[/latex] is the number of years since[latex]\,2013.\,[/latex]To the nearest thousandth, what will the population of Bharat be in[latex]\,\text{2031?}[/latex]

Try It

The population of Communist china was about one.39 billion in the year 2013, with an annual growth rate of about[latex]\,0.six%.\,[/latex]This situation is represented by the growth office[latex]\,P\left(t\right)=1.39{\left(one.006\right)}^{t},[/latex] where [latex]\,t\,[/latex] is the number of years since[latex]\,2013.[/latex]To the nearest thousandth, what will the population of People's republic of china exist for the year 2031? How does this compare to the population prediction we made for Bharat in (Effigy)?

Show Solution

About[latex]\,i.548\,[/latex]billion people; by the year 2031, India's population will exceed China'due south by nearly 0.001 billion, or i million people.

Finding Equations of Exponential Functions

In the previous examples, we were given an exponential role, which nosotros then evaluated for a given input. Sometimes we are given data near an exponential function without knowing the function explicitly. We must use the information to first write the form of the role, then determine the constants[latex]\,a\,[/latex]and[latex]\,b,[/latex]and evaluate the function.

How To

Given two information points, write an exponential model.

1. If one of the data points has the form[latex]\,\left(0,a\right),[/latex] and then[latex]\,a\,[/latex]is the initial value. Using[latex]\,a,[/latex] substitute the second point into the equation[latex]\,f\left(x\correct)=a{\left(b\right)}^{x},[/latex] and solve for[latex]\,b.[/latex]
2. If neither of the data points have the grade[latex]\,\left(0,a\right),[/latex] substitute both points into two equations with the class[latex]\,f\left(x\right)=a{\left(b\right)}^{x}.\,[/latex]Solve the resulting system of two equations in ii unknowns to find[latex]\,a\,[/latex]and[latex]\,b.[/latex]
3. Using the[latex]\,a\,[/latex]and[latex]\,b\,[/latex]found in the steps in a higher place, write the exponential function in the form[latex]\,f\left(ten\right)=a{\left(b\right)}^{x}.[/latex]

Writing an Exponential Model When the Initial Value Is Known

In 2006, fourscore deer were introduced into a wild fauna refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function[latex]\,N\left(t\right)\,[/latex]representing the population[latex]\,\left(N\correct)\,[/latex]of deer over time[latex]\,t.[/latex]

Try It

A wolf population is growing exponentially. In 2011,[latex]\,129\,[/latex]wolves were counted. By[latex]\,\text{2013,}\,[/latex]the population had reached 236 wolves. What 2 points can be used to derive an exponential equation modeling this situation? Write the equation representing the population[latex]\,Northward\,[/latex]of wolves over time[latex]\,t.[/latex]

Testify Solution

[latex]\left(0,129\right)\,[/latex]and[latex]\,\left(ii,236\right);\,\,\,N\left(t\right)=129{\left(\text{1}\text{.3526}\right)}^{t}[/latex]

Writing an Exponential Model When the Initial Value is Non Known

Detect an exponential function that passes through the points[latex]\,\left(-2,vi\right)\,[/latex]and[latex]\,\left(2,1\right).[/latex]

Endeavour It

Given the two points[latex]\,\left(1,3\right)\,[/latex]and[latex]\,\left(2,4.5\correct),[/latex]find the equation of the exponential function that passes through these two points.

Prove Solution

[latex]f\left(x\correct)=2{\left(1.5\right)}^{ten}[/latex]

Do two points ever decide a unique exponential part?

Yes, provided the ii points are either both above the x-axis or both beneath the 10-axis and take dissimilar x-coordinates. But go along in mind that we likewise demand to know that the graph is, in fact, an exponential function. Not every graph that looks exponential actually is exponential. We need to know the graph is based on a model that shows the aforementioned percent growth with each unit increase in[latex]\,ten,[/latex] which in many existent world cases involves time.

How To

Given the graph of an exponential office, write its equation.

1. First, identify 2 points on the graph. Choose the y-intercept as ane of the ii points whenever possible. Try to cull points that are as far autonomously as possible to reduce round-off mistake.
2. If one of the data points is the y-intercept[latex]\,\left(0,a\right)[/latex], so[latex]\,a\,[/latex]is the initial value. Using[latex]\,a,[/latex] substitute the second bespeak into the equation[latex]\,f\left(x\correct)=a{\left(b\correct)}^{ten},[/latex] and solve for[latex]\,b.[/latex]
3. If neither of the data points accept the form[latex]\,\left(0,a\correct),[/latex] substitute both points into two equations with the class[latex]\,f\left(x\right)=a{\left(b\correct)}^{ten}.\,[/latex]Solve the resulting organisation of 2 equations in 2 unknowns to find[latex]\,a\,[/latex]and[latex]\,b.[/latex]
4. Write the exponential function,[latex]\,f\left(ten\correct)=a{\left(b\right)}^{x}.[/latex]

Writing an Exponential Role Given Its Graph

Observe an equation for the exponential function graphed in (Figure).

Effigy 5.

Endeavor It

Detect an equation for the exponential function graphed in (Effigy).

Effigy 6.

Show Solution

[latex]f\left(x\right)=\sqrt{2}{\left(\sqrt{2}\right)}^{10}.\,[/latex]Answers may vary due to round-off error. The answer should exist very close to[latex]\,1.4142{\left(ane.4142\right)}^{x}.[/latex]

How To

Given two points on the curve of an exponential function, apply a graphing calculator to observe the equation.

1. Press [STAT].
2. Clear any existing entries in columns L1 or L2.
3. In L1, enter the ten-coordinates given.
4. In L2, enter the corresponding y-coordinates.
5. Printing [STAT] once again. Cursor right to CALC, coil downward to ExpReg (Exponential Regression), and printing [ENTER].
6. The screen displays the values of a and b in the exponential equation[latex]\,y=a\cdot {b}^{x}[/latex].

Using a Graphing Figurer to Find an Exponential Function

Utilize a graphing calculator to find the exponential equation that includes the points[latex]\,\left(2,24.8\right)\,[/latex]and[latex]\,\left(v,198.4\correct).[/latex]

Attempt It

Use a graphing calculator to detect the exponential equation that includes the points (3, 75.98) and (6, 481.07).

Show Solution

[latex]y\approx 12\cdot {one.85}^{x}[/latex]

Applying the Chemical compound-Involvement Formula

Savings instruments in which earnings are continually reinvested, such equally mutual funds and retirement accounts, employ compound involvement. The term compounding refers to interest earned non only on the original value, merely on the accumulated value of the account.

The annual pct rate (April) of an account, also chosen the nominal charge per unit, 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 yr. In fact, when involvement is compounded more one time a year, the constructive interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.

Nosotros tin calculate the compound involvement using the chemical compound interest formula, which is an exponential function of the variables time[latex]\,t,[/latex] principal[latex]\,P,[/latex] APR[latex]\,r,[/latex] and number of compounding periods in a twelvemonth[latex]\,northward:[/latex]

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

For example, observe (Figure), which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases equally the compounding frequency increases.

Frequency	Value after 1 twelvemonth
Annually	$1100
Semiannually	$1102.50
Quarterly	$1103.81
Monthly	$1104.71
Daily	$1105.16

The Compound Interest Formula

Compound interest tin be calculated using the formula

[latex]A\left(t\correct)=P{\left(1+\frac{r}{northward}\correct)}^{nt}[/latex]

where

• [latex]A\left(t\right)\,[/latex]is the business relationship value,
• [latex]t\,[/latex]is measured in years,
• [latex]P\,[/latex]is the starting amount of the account, ofttimes called the principal, or more generally present value,
• [latex]r\,[/latex]is the annual percentage rate (April) expressed equally a decimal, and
• [latex]n\,[/latex]is the number of compounding periods in ane year.

Computing Compound Interest

If we invest $3,000 in an investment account paying iii% interest compounded quarterly, how much will the account be worth in 10 years?

Attempt Information technology

An initial investment of $100,000 at 12% involvement is compounded weekly (use 52 weeks in a year). What volition the investment exist worth in 30 years?

Using the Compound Interest Formula to Solve for the Primary

A 529 Plan is a college-savings program that allows relatives to invest money to pay for a child's future college tuition; the account grows tax-free. Lily wants to gear up upwardly 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 vi% compounded semi-annually (twice a twelvemonth). To the nearest dollar, how much volition Lily need to invest in the account now?

Effort It

Refer to (Effigy). 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. (Effigy) shows that the increase from annual to semi-almanac 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 $one invested at 100% interest for 1 twelvemonth, compounded at various frequencies, listed in (Effigy).

Frequency	[latex]A\left(t\correct)={\left(1+\frac{ane}{n}\right)}^{due north}[/latex]	Value
Annually	[latex]{\left(i+\frac{ane}{1}\right)}^{one}[/latex]	$ii
Semiannually	[latex]{\left(1+\frac{i}{ii}\right)}^{2}[/latex]	$ii.25
Quarterly	[latex]{\left(1+\frac{1}{four}\right)}^{iv}[/latex]	$two.441406
Monthly	[latex]{\left(i+\frac{1}{12}\right)}^{12}[/latex]	$two.613035
Daily	[latex]{\left(1+\frac{1}{365}\right)}^{365}[/latex]	$2.714567
Hourly	[latex]{\left(one+\frac{ane}{\text{8760}}\right)}^{\text{8760}}[/latex]	$2.718127
In one case per minute	[latex]{\left(i+\frac{1}{\text{525600}}\right)}^{\text{525600}}[/latex]	$two.718279
In one case per 2nd	[latex]{\left(ane+\frac{1}{31536000}\correct)}^{31536000}[/latex]	$two.718282

These values appear to exist budgeted a limit as[latex]\,n\,[/latex]increases without spring. In fact, equally[latex]\,north\,[/latex]gets larger and larger, the expression[latex]\,{\left(one+\frac{one}{n}\right)}^{n}\,[/latex]approaches a number used so frequently in mathematics that it has its own name: the letter[latex]\,e.\,[/latex]This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.

The Number due east

The alphabetic character east represents the irrational number

[latex]{\left(i+\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 of operations e, we use the approximation,[latex]\,east\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.

Using a Calculator to Discover Powers of east

Calculate[latex]\,{e}^{3.fourteen}.\,[/latex]Circular to five decimal places.

Show Solution

On a computer, press the button labeled[latex]\,\left[{eastward}^{10}\right].\,[/latex]The window shows[latex]\,\left[e^(\text{ }\right].\,[/latex]Type[latex]\,3.xiv\,[/latex]and then shut parenthesis,[latex]\,\left[)\right].\,[/latex]Printing [ENTER]. Rounding to[latex]\,5\,[/latex]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. Information technology is not used to find powers of[latex]\,due east.[/latex]

Endeavor It

Apply a figurer to find[latex]\,{due east}^{-0.five}.\,[/latex]Round to 5 decimal places.

Show Solution

[latex]{east}^{-0.5}\approx 0.60653[/latex]

Investigating Continuous Growth

Then far we have worked with rational bases for exponential functions. For about real-world phenomena, however, e is used as the base of operations for exponential functions. Exponential models that utilize[latex]\,e\,[/latex]equally the base are called continuous growth or decay models. We see these models in finance, computer science, and nigh of the sciences, such as physics, toxicology, and fluid dynamics.

The Continuous Growth/Disuse Formula

For all real numbers[latex]\,t,[/latex]and all positive numbers[latex]\,a\,[/latex]and[latex]\,r,[/latex]continuous growth or decay is represented by the formula

[latex]A\left(t\right)=a{e}^{rt}[/latex]

where

• [latex]a\,[/latex]is the initial value,
• [latex]r\,[/latex]is the continuous growth rate per unit fourth dimension,
• and[latex]\,t\,[/latex]is the elapsed time.

If[latex]\,r>0\,[/latex], so the formula represents continuous growth. If[latex]\,r<0\,[/latex], and then the formula represents continuous decay.

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form

[latex]A\left(t\right)=P{e}^{rt}[/latex]

where

• [latex]P\,[/latex]is the principal or the initial invested,
• [latex]r\,[/latex]is the growth or involvement rate per unit of measurement time,
• and [latex]t\,[/latex]is the menstruum or term of the investment.

How To

Given the initial value, rate of growth or decay, and time[latex]\,t,[/latex] solve a continuous growth or disuse function.

1. Use the information in the trouble to decide[latex]\,a[/latex], the initial value of the function.
2. Use the information in the problem to determine the growth rate[latex]\,r.[/latex]
   1. If the trouble refers to continuous growth, and then[latex]\,r>0.[/latex]
   2. If the trouble refers to continuous decay, then[latex]\,r<0.[/latex]
3. Utilise the information in the trouble to make up one's mind the fourth dimension[latex]\,t.[/latex]
4. Substitute the given data into the continuous growth formula and solve for[latex]\,A\left(t\correct).[/latex]

Calculating Continuous Growth

A person invested $i,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one yr?

Try It

A person invests $100,000 at a nominal 12% interest per twelvemonth compounded continuously. What will be the value of the investment in 30 years?

Show Solution

$3,659,823.44

Calculating Continuous Decay

Radon-222 decays at a continuous charge per unit of 17.3% per twenty-four hours. How much volition 100 mg of Radon-222 decay to in 3 days?

Try It

Using the data in (Figure), how much radon-222 volition remain after one year?

Bear witness Solution

3.77E-26 (This is calculator notation for the number written as[latex]\,3.77×{x}^{-26}\,[/latex]in scientific notation. While the output of an exponential office is never zippo, this number is so close to zero that for all practical purposes we can accept nix equally the respond.)

Central Equations

definition of the exponential part	[latex]f\left(10\correct)={b}^{ten}\text{, where }b>0, b\ne 1[/latex]
definition of exponential growth	[latex]f\left(x\right)=a{b}^{x},\text{ where }a>0,b>0,b\ne 1[/latex]
compound interest formula	[latex]\begin{array}{l}A\left(t\right)=P{\left(one+\frac{r}{n}\right)}^{nt} ,\text{ where}\hfill \\ A\left(t\right)\text{ is the account value at time }t\hfill \\ t\text{ is the number of years}\hfill \\ P\text{ is the initial investment, often chosen the master}\hfill \\ r\text{ is the annual percent rate (April), or nominal charge per unit}\hfill \\ n\text{ is the number of compounding periods in i yr}\hfill \end{array}[/latex]
continuous growth formula	[latex]A\left(t\right)=a{e}^{rt},\text{ where}[/latex] [latex]t[/latex]is the number of unit time periods of growth [latex]a[/latex]is the starting corporeality (in the continuous compounding formula a is replaced with P, the principal) [latex]e[/latex]is the mathematical constant,[latex] \text{ }e\approx 2.718282[/latex]

Key Concepts

• An exponential office is defined as a function with a positive abiding other than[latex]\,one\,[/latex]raised to a variable exponent. Come across (Figure).
• A function is evaluated by solving at a specific value. Meet (Figure) and (Effigy).
• An exponential model can be plant when the growth rate and initial value are known. Meet (Figure).
• An exponential model can exist found when the ii data points from the model are known. See (Effigy).
• An exponential model tin can be establish using ii data points from the graph of the model. Run across (Effigy).
• An exponential model tin can be establish using two information points from the graph and a calculator. Run across (Figure).
• The value of an account at any fourth dimension[latex]\,t\,[/latex]can exist calculated using the compound involvement formula when the primary, annual interest rate, and compounding periods are known. See (Figure).
• The initial investment of an account can be constitute using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. Come across (Figure).
• The number[latex]\,due east\,[/latex]is a mathematical constant often used as the base of real world exponential growth and disuse models. Its decimal approximation is[latex]\,east\approx 2.718282.[/latex]
• Scientific and graphing calculators have the key[latex]\,\left[{east}^{x}\right]\,[/latex]or[latex]\,\left[\mathrm{exp}\left(x\correct)\right]\,[/latex]for calculating powers of[latex]\,e.\,[/latex]See (Effigy).
• Continuous growth or disuse models are exponential models that apply[latex]\,e\,[/latex]as the base. Continuous growth and disuse models can exist institute when the initial value and growth or decay rate are known. Come across (Figure) and (Effigy).

Section Exercises

Verbal

Explain why the values of an increasing exponential function will somewhen overtake the values of an increasing linear function.

Testify Solution

Linear functions take a constant charge per unit of modify. Exponential functions increase based on a percent of the original.

Given a formula for an exponential function, is it possible to determine whether the role 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 respective exactly to the existent value."^[ii] Develop a reasonable statement for why the term nominal rate is used to describe the annual pct rate of an investment account that compounds involvement.

Show Solution

When interest is compounded, the percentage of involvement earned to principal ends up existence greater than the annual percentage rate for the investment account. Thus, the annual per centum rate does not necessarily correspond to the real involvement earned, which is the very definition of nominal.

Algebraic

For the following exercises, identify whether the statement represents an exponential part. Explain.

The average annual population increment of a pack of wolves is 25.

A population of bacteria decreases past a factor of[latex]\,\frac{1}{eight}\,[/latex]every[latex]\,24\,[/latex]hours.

Show Solution

exponential; the population decreases past a proportional rate.

The value of a coin drove has increased past[latex]\,3.25%\,[/latex]annually over the final[latex]\,20\,[/latex]years.

For each preparation session, a personal trainer charges his clients[latex]\,\text{\$}5\,[/latex]
less than the previous training session.

Show Solution

not exponential; the accuse decreases by a constant amount each visit, and so the argument represents a linear office. .

The height of a projectile at time[latex]\,t\,[/latex]is represented by the part[latex]\,h\left(t\right)=-iv.9{t}^{two}+18t+40.[/latex]

For the following exercises, consider this scenario: For each year[latex]\,t,[/latex]the population of a forest of trees is represented by the part[latex]\,A\left(t\right)=115{\left(1.025\right)}^{t}.\,[/latex]In a neighboring forest, the population of the same type of tree is represented by the part[latex]\,B\left(t\correct)=82{\left(1.029\correct)}^{t}.\,[/latex](Round answers to the nearest whole number.)

Which wood's population is growing at a faster rate?

Show Solution

The wood represented by the function[latex]\,B\left(t\right)=82{\left(1.029\right)}^{t}.[/latex]

Which woods had a greater number of trees initially? By how many?

Assuming the population growth models continue to stand for the growth of the forests, which forest volition have a greater number of trees after[latex]\,20\,[/latex]years? By how many?

Bear witness Solution

Afterwards[latex]\,t=20\,[/latex]years, forest A will have[latex]\,43\,[/latex]more trees than woods B.

Assuming the population growth models continue to represent the growth of the forests, which forest will accept a greater number of trees after[latex]\,100\,[/latex]years? By how many?

Talk over the higher up results from the previous 4 exercises. Assuming the population growth models continue to represent the growth of the forests, which forest volition have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model?

Show Solution

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than woods A and will remain that way as long every bit the population growth models concur. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other ecology and biological factors.

For the following exercises, decide whether the equation represents exponential growth, exponential decay, or neither. Explain.

[latex]y=300{\left(1-t\right)}^{five}[/latex]

[latex]y=220{\left(i.06\right)}^{x}[/latex]

Show Solution

exponential growth; The growth factor,[latex]\,1.06,[/latex] is greater than[latex]\,i.[/latex]

[latex]y=16.five{\left(1.025\right)}^{\frac{i}{x}}[/latex]

[latex]y=eleven,701{\left(0.97\right)}^{t}[/latex]

Show Solution

exponential decay; The decay cistron,[latex]\,0.97,[/latex] is between[latex]\,0\,[/latex]and[latex]\,1.[/latex]

For the following exercises, find the formula for an exponential role that passes through the two points given.

[latex]\left(0,6\right)\,[/latex]and[latex]\,\left(3,750\correct)[/latex]

[latex]\left(0,2000\right)\,[/latex]and[latex]\,\left(2,20\right)[/latex]

Show Solution

[latex]f\left(x\right)=2000{\left(0.ane\right)}^{ten}[/latex]

[latex]\left(-1,\frac{3}{2}\right)\,[/latex]and[latex]\,\left(3,24\right)[/latex]

[latex]\left(-two,6\correct)\,[/latex]and[latex]\,\left(3,1\right)[/latex]

Show Solution

[latex]f\left(x\right)={\left(\frac{1}{vi}\correct)}^{-\frac{3}{5}}{\left(\frac{i}{6}\correct)}^{\frac{10}{5}}\approx 2.93{\left(0.699\right)}^{x}[/latex]

[latex]\left(3,ane\right)\,[/latex]and[latex]\,\left(five,four\correct)[/latex]

For the following exercises, determine whether the table could correspond a function that is linear, exponential, or neither. If it appears to exist exponential, find a role that passes through the points.

[latex]x[/latex]	1	2	three	four
[latex]f\left(x\right)[/latex]	70	forty	10	-20

[latex]x[/latex]	i	2	3	four
[latex]h\left(x\right)[/latex]	70	49	34.iii	24.01

[latex]x[/latex]	one	two	3	4
[latex]thou\left(ten\right)[/latex]	80	61	42.nine	25.61

[latex]x[/latex]	i	2	three	4
[latex]f\left(x\correct)[/latex]	10	20	40	eighty

[latex]ten[/latex]	1	2	3	four
[latex]g\left(10\right)[/latex]	-3.25	two	7.25	12.5

For the following exercises, use the compound involvement formula,[latex]\,A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}.[/latex]

Later a certain number of years, the value of an investment account is represented by the equation[latex]\,10,250{\left(1+\frac{0.04}{12}\correct)}^{120}.\,[/latex]What is the value of the account?

What was the initial eolith fabricated to the account in the previous do?

Show Solution

[latex]$ten,250[/latex]

How many years had the account from the previous practise been accumulating involvement?

An account is opened with an initial eolith of $half-dozen,500 and earns[latex]\,3.6%\,[/latex]interest compounded semi-annually. What will the account exist worth in[latex]\,20\,[/latex]years?

Evidence Solution

[latex]$xiii,268.58[/latex]

How much more than would the account in the previous exercise take been worth if the interest were compounding weekly?

Solve the compound interest formula for the main,[latex]\,P[/latex].

Evidence Solution

[latex]P=A\left(t\right)\cdot {\left(1+\frac{r}{northward}\correct)}^{-nt}[/latex]

Utilise the formula found in the previous exercise to calculate the initial deposit of an account that is worth[latex]\,$xiv,472.74\,[/latex]afterward earning[latex]\,5.5%\,[/latex]involvement compounded monthly for[latex]\,5\,[/latex]years. (Round to the nearest dollar.)

How much more would the account in the previous two exercises be worth if it were earning interest for[latex]\,5\,[/latex]more years?

Show Solution

[latex]$4,572.56[/latex]

Employ properties of rational exponents to solve the compound interest formula for the involvement rate,[latex]\,r.[/latex]

Use the formula found in the previous do to calculate the involvement charge per unit for an account that was compounded semi-annually, had an initial deposit of $ix,000 and was worth $xiii,373.53 afterwards x years.

Show Solution

[latex]iv%[/latex]

Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded monthly, had an initial deposit of $5,500, and was worth $38,455 after 30 years.

For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.

[latex]y=3742{\left(e\right)}^{0.75t}[/latex]

Show Solution

continuous growth; the growth rate is greater than[latex]\,0.[/latex]

[latex]y=150{\left(e\right)}^{\frac{iii.25}{t}}[/latex]

[latex]y=2.25{\left(east\right)}^{-2t}[/latex]

Show Solution

continuous decay; the growth rate is less than[latex]\,0.[/latex]

Suppose an investment account is opened with an initial eolith of[latex]\,$12,000\,[/latex]earning[latex]\,7.2%\,[/latex]interest compounded continuously. How much will the account exist worth later on[latex]\,30\,[/latex]years?

How much less would the account from Do 42 exist worth after[latex]\,30\,[/latex]years if it were compounded monthly instead?

Bear witness Solution

[latex]$669.42[/latex]

Numeric

For the following exercises, evaluate each office. Circular answers to four decimal places, if necessary.

[latex]f\left(x\right)=2{\left(5\correct)}^{x},[/latex] for[latex]\,f\left(-3\right)[/latex]

[latex]f\left(ten\right)=-{iv}^{2x+3},[/latex] for[latex]\,f\left(-1\right)[/latex]

Show Solution

[latex]f\left(-1\right)=-4[/latex]

[latex]f\left(ten\right)={e}^{x},[/latex] for[latex]\,f\left(3\right)[/latex]

[latex]f\left(x\right)=-2{eastward}^{x-ane},[/latex] for[latex]\,f\left(-1\correct)[/latex]

Show Solution

[latex]f\left(-i\right)\approx -0.2707[/latex]

[latex]f\left(x\right)=2.7{\left(4\right)}^{-x+i}+ane.v,[/latex] for[latex]f\left(-2\correct)[/latex]

[latex]f\left(10\correct)=ane.2{e}^{2x}-0.3,[/latex] for[latex]\,f\left(3\right)[/latex]

Show Solution

[latex]f\left(3\right)\approx 483.8146[/latex]

[latex]f\left(10\right)=-\frac{3}{ii}{\left(three\right)}^{-10}+\frac{iii}{2},[/latex] for[latex]\,f\left(two\right)[/latex]

Engineering science

For the following exercises, utilise a graphing calculator to find the equation of an exponential role given the points on the curve.

[latex]\left(0,iii\right)\,[/latex]and[latex]\,\left(3,375\correct)[/latex]

Testify Solution

[latex]y=3\cdot {five}^{ten}[/latex]

[latex]\left(3,222.62\correct)\,[/latex]and[latex]\,\left(10,77.456\correct)[/latex]

[latex]\left(20,29.495\right)\,[/latex]and[latex]\,\left(150,730.89\right)[/latex]

Show Solution

[latex]y\approx eighteen\cdot {ane.025}^{10}[/latex]

[latex]\left(5,two.909\correct)\,[/latex]and[latex]\,\left(13,0.005\right)[/latex]

[latex]\left(11,310.035\right)\,[/latex] and [latex]\left(25,356.3652\right)[/latex]

Show Solution

[latex]y\approx 0.two\cdot {1.95}^{x}[/latex]

Extensions

The annual percentage yield (APY) of an investment account is a representation of the bodily interest rate earned on a compounding account. It is based on a compounding period of one year. Show that the APY of an account that compounds monthly tin can be establish with the formula[latex]\,\text{APY}={\left(one+\frac{r}{12}\right)}^{12}-1.[/latex]

Echo the previous practise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a office[latex]\,I\left(n\right)\,[/latex]for the APY of any account that compounds[latex]\,north\,[/latex]times per year.

Show Solution

[latex]\text{APY}=\frac{A\left(t\right)-a}{a}=\frac{a{\left(1+\frac{r}{365}\right)}^{365\left(1\right)}-a}{a}=\frac{a\left[{\left(1+\frac{r}{365}\correct)}^{365}-ane\right]}{a}={\left(1+\frac{r}{365}\right)}^{365}-one;[/latex][latex]I\left(north\correct)={\left(1+\frac{r}{n}\right)}^{n}-1[/latex]

Remember that an exponential function is any equation written in the grade[latex]\,f\left(x\correct)=a\cdot {b}^{x}\,[/latex]such that[latex] a [/latex]and[latex] b [/latex]are positive numbers and[latex] b\ne 1. [/latex]Whatsoever positive number[latex] b [/latex]can be written as[latex] b={e}^{n} [/latex]for some value of[latex] n[/latex]. Use this fact to rewrite the formula for an exponential function that uses the number[latex] east [/latex]as a base.

In an exponential disuse role, the base of the exponent is a value between 0 and ane. Thus, for some number[latex]\,b>i,[/latex] the exponential disuse office can be written every bit[latex]\,f\left(10\right)=a\cdot {\left(\frac{1}{b}\right)}^{x}.\,[/latex]Utilize this formula, along with the fact that[latex]\,b={due east}^{northward},[/latex] to testify that an exponential decay function takes the course[latex]\,f\left(x\right)=a{\left(due east\right)}^{-nx}\,[/latex]for some positive number[latex]\,n\,[/latex].

Show Solution

Let[latex]\,f\,[/latex]be the exponential decay role[latex]\,f\left(10\right)=a\cdot {\left(\frac{1}{b}\right)}^{10}\,[/latex]such that[latex]\,b>i.\,[/latex]And so for some number[latex]\,n>0,[/latex][latex]f\left(ten\correct)=a\cdot {\left(\frac{1}{b}\right)}^{10}=a{\left({b}^{-i}\right)}^{x}=a{\left({\left({east}^{n}\right)}^{-1}\right)}^{10}=a{\left({e}^{-n}\correct)}^{10}=a{\left(e\correct)}^{-nx}.[/latex]

The formula for the amount[latex]\,A\,[/latex]in an investment account with a nominal interest rate[latex]\,r\,[/latex]at any time[latex]\,t\,[/latex]is given by[latex]\,A\left(t\right)=a{\left(due east\right)}^{rt},[/latex]where[latex]\,a\,[/latex]is the amount of primary initially deposited into an business relationship that compounds continuously. Prove that the percentage of interest earned to chief at any time[latex]\,t\,[/latex]can be calculated with the formula[latex]\,I\left(t\right)={due east}^{rt}-1.[/latex]

Real-World Applications

The trick population in a certain region has an annual growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to exist in the year 2020?

Bear witness Solution

[latex]47,622\,[/latex]fox

A scientist begins with 100 milligrams of a radioactive substance that decays exponentially. After 35 hours, 50mg of the substance remains. How many milligrams will remain afterwards 54 hours?

In the twelvemonth 1985, a house was valued at $110,000. Past the year 2005, the value had appreciated to $145,000. What was the annual growth charge per unit between 1985 and 2005? Assume that the value continued to grow by the aforementioned percentage. What was the value of the house in the yr 2010?

Testify Solution

[latex]ane.39%;\,[/latex][latex]$155,368.09[/latex]

A motorcar was valued at $38,000 in the year 2007. Past 2013, the value had depreciated to $11,000 If the motorcar's value continues to drop by the same percentage, what will information technology be worth by 2017?

Jamal wants to salve $54,000 for a down payment on a habitation. How much will he need to invest in an account with 8.2% APR, compounding daily, in lodge to attain his goal in 5 years?

Show Solution

[latex]$35,838.76[/latex]

Kyoko has $10,000 that she wants to invest. Her bank has several investment accounts to choose from, all compounding daily. Her goal is to have $15,000 by the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? (Hint: solve the compound interest formula for the involvement charge per unit.)

Alyssa opened a retirement business relationship with 7.25% APR in the year 2000. Her initial deposit was $xiii,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?

Show Solution

[latex]$82,247.78;\,[/latex][latex]$449.75[/latex]

An investment business relationship with an annual interest rate of 7% was opened with an initial deposit of $4,000 Compare the values of the business relationship after ix years when the interest is compounded annually, quarterly, monthly, and continuously.

Glossary

annual percentage charge per unit (APR)

the yearly interest rate earned by an investment account, also chosen nominal rate

compound interest

involvement earned on the total rest, not just the chief

exponential growth

a model that grows past a charge per unit proportional to the amount present

nominal charge per unit

the yearly interest rate earned past an investment business relationship, also chosen almanac percentage charge per unit

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Source: https://courses.lumenlearning.com/suny-osalgebratrig/chapter/exponential-functions/

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