Over Which Interval Does The Exponential Show A More Rapid Rate Of Change?

Learning Objectives

In this department, you will:

• Evaluate exponential functions.
• Find the equation of an exponential function.
• Employ compound interest formulas.
• Evaluate exponential functions with base e.

India is the 2nd about populous state in the globe with a population of almost[latex]\,ane.25\,[/latex]billion people in 2013. The population is growing at a rate of about[latex]\,one.two%\,[/latex]each year^[1] . If this rate continues, the population of India volition exceed Communist china's population by the year[latex]\,2031.[/latex]When populations abound speedily, we oftentimes say that the growth is "exponential," meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a expect at exponential functions, which model this kind of rapid growth.

Identifying Exponential Functions

When exploring linear growth, nosotros observed a constant rate of modify—a constant number by which the output increased for each unit increment in input. For example, in the equation[latex]\,f\left(ten\correct)=3x+four,[/latex]the slope tells us the output increases by three each fourth dimension the input increases by 1. The scenario in the India population example is dissimilar because we have a percent modify per unit fourth dimension (rather than a constant change) in the number of people.

Defining an Exponential Function

A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.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 rate continues, vegans will make upwards 10% of the U.S. population in 2015, 40% in 2019, and eighty% in 2021.

What exactly does it mean to grow exponentially? What does the word double have in common with percent increase? People toss these words around errantly. Are these words used correctly? The words certainly announced frequently in the media.

• Percent alter refers to a change based on a per centum of the original amount.
• Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
• Exponential decay refers to a decrease based on a constant multiplicative rate of modify over equal increments of time, that is, a percent decrease of the original amount over time.

For us to gain a clear agreement of exponential growth, permit us contrast exponential growth with linear growth. Nosotros will construct ii functions. The first function is exponential. Nosotros volition kickoff with an input of 0, and increment each input by 1. We volition double the corresponding sequent outputs. The second function is linear. Nosotros will start with an input of 0, and increase each input by ane. We will add 2 to the corresponding consecutive outputs. Encounter (Figure).

[latex]ten[/latex]	[latex]f\left(x\right)={2}^{10}[/latex]	[latex]g\left(x\right)=2x[/latex]
0	1	0
1	ii	2
2	4	4
3	8	6
4	xvi	8
v	32	ten
vi	64	12

From (Figure) we can infer that for these ii functions, exponential growth dwarfs linear growth.

• Exponential growth refers to the original value from the range increases past the same percentage over equal increments found 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.

Apparently, the difference between "the aforementioned percent" and "the same amount" is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by 1. For linear growth, the abiding additive rate of change over equal increments resulted in calculation 2 to the output whenever the input was increased by one.

The general form of the exponential function is[latex]\,f\left(x\right)=a{b}^{x},\,[/latex]where[latex]\,a\,[/latex]is any nonzero number,[latex]\,b\,[/latex]is a positive real number not equal to ane.

• If[latex]\,b>one,[/latex]the office grows at a charge per unit proportional to its size.
• If[latex]\,0<b<1,[/latex] the function decays at a rate proportional to its size.

Permit's look at the function[latex]\,f\left(x\right)={2}^{x}\,[/latex]from our example. We will create a table ((Figure)) to decide the corresponding outputs over an interval in the domain from[latex]\,-iii\,[/latex]to[latex]\,3.[/latex]

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

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

Figure 1.

Permit'due south define the behavior of the graph of the exponential role[latex]\,f\left(x\correct)={2}^{10}\,[/latex]and highlight some its key characteristics.

• the domain is[latex]\,\left(-\infty ,\infty \correct),[/latex]
• the range is[latex]\,\left(0,\infty \right),[/latex]
• as[latex]\,10\to \infty ,f\left(x\correct)\to \infty ,[/latex]
• every bit [latex]\,ten\to -\infty ,f\left(10\right)\to 0,[/latex]
• [latex]\,f\left(x\right)\,[/latex]is ever increasing,
• the graph of[latex]\,f\left(x\right)\,[/latex]volition never touch the x-axis because base 2 raised to whatsoever exponent never has the result of zero.
• [latex]\,y=0\,[/latex]is the horizontal asymptote.
• the y-intercept is one.

Exponential Function

For whatsoever existent number[latex]\,x,[/latex]an exponential part is a function with the form

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

where

• [latex]\,a\,[/latex]is a non-nothing real number called the initial value and
• [latex]\,b\,[/latex]is any positive real number such that[latex]\,b\ne ane.[/latex]
• The domain of[latex]\,f\,[/latex]is all existent numbers.
• The range of[latex]\,f\,[/latex]is all positive real numbers if[latex]\,a>0.[/latex]
• The range of[latex]\,f\,[/latex]is all negative real numbers if[latex]\,a<0.[/latex]
• The y-intercept is[latex]\,\left(0,a\correct),[/latex]and the horizontal asymptote is[latex]\,y=0.[/latex]

Identifying Exponential Functions

Which of the following equations are not exponential functions?

• [latex]f\left(10\right)={4}^{3\left(10-2\right)}[/latex]
• [latex]g\left(x\right)={ten}^{3}[/latex]
• [latex]h\left(x\right)={\left(\frac{1}{three}\right)}^{x}[/latex]
• [latex]j\left(x\right)={\left(-2\correct)}^{x}[/latex]

Try It

Which of the following equations represent exponential functions?

• [latex]f\left(10\correct)=two{x}^{2}-3x+i[/latex]
• [latex]g\left(x\right)={0.875}^{x}[/latex]
• [latex]h\left(x\right)=one.75x+two[/latex]
• [latex]j\left(x\right)={1095.half dozen}^{-2x}[/latex]

Show Solution

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

Evaluating Exponential Functions

Retrieve that the base of an exponential part must be a positive real number other than[latex]\,ane.[/latex]Why practice we limit the base of operations [latex]b\,[/latex]to positive values? To ensure that the outputs volition be real numbers. Discover what happens if the base of operations is not positive:

• Let[latex]\,b=-9\,[/latex]and[latex]\,ten=\frac{1}{2}.\,[/latex]Then[latex]\,f\left(10\right)=f\left(\frac{1}{2}\right)={\left(-nine\correct)}^{\frac{1}{2}}=\sqrt{-ix},[/latex]which is not a real number.

Why do we limit the base to positive values other than [latex]1?[/latex]Considering base [latex]1\,[/latex]results in the constant function. Discover what happens if the base is [latex]i:[/latex]

• Permit[latex]\,b=1.\,[/latex]Then[latex]\,f\left(ten\right)={1}^{x}=1\,[/latex]for any value of[latex]\,x.[/latex]

To evaluate an exponential part with the form[latex]\,f\left(x\right)={b}^{ten},[/latex]we merely substitute [latex]ten\,[/latex]with the given value, and calculate the resulting power. For case:

Let [latex]\,f\left(x\right)={ii}^{10}.\,[/latex]What is [latex]f\left(iii\right)?[/latex]

[latex]\begin{array}{lll}f\left(10\correct)\hfill & ={ii}^{ten}\hfill & \hfill \\ f\left(3\right)\hfill & ={2}^{3}\text{ }\hfill & \text{Substitute }ten=3.\hfill \\ \hfill & =8\text{ }\hfill & \text{Evaluate the power}\text{.}\hfill \stop{array}[/latex]

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

Let[latex]\,f\left(10\right)=30{\left(two\right)}^{x}.\,[/latex]What is[latex]\,f\left(3\correct)?[/latex]

[latex]\brainstorm{array}{lll}f\left(ten\right)\hfill & =thirty{\left(2\right)}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & =30{\left(2\correct)}^{3}\hfill & \text{Substitute }x=3.\hfill \\ \hfill & =thirty\left(viii\correct)\text{ }\hfill & \text{Simplify the power first}\text{.}\hfill \\ \hfill & =240\hfill & \text{Multiply}\text{.}\hfill \finish{array}[/latex]

Annotation that if the order of operations were not followed, the result would be wrong:

[latex]f\left(3\correct)=xxx{\left(2\correct)}^{3}\ne {60}^{3}=216,000[/latex]

Evaluating Exponential Functions

Let [latex]\,f\left(ten\right)=5{\left(3\right)}^{10+ane}.\,[/latex]Evaluate[latex]\,f\left(two\correct)\,[/latex]without using a computer.

Try Information technology

Let[latex]f\left(x\right)=8{\left(1.2\right)}^{10-five}.\,[/latex]Evaluate[latex]\,f\left(3\right)\,[/latex]using a calculator. Round to four decimal places.

Show Solution

[latex]5.5556[/latex]

Defining Exponential Growth

Because the output of exponential functions increases very quickly, the term "exponential growth" is ofttimes used in everyday language to describe annihilation that grows or increases rapidly. However, exponential growth can be divers more precisely in a mathematical sense. If the growth rate is proportional to the amount nowadays, the function models exponential growth.

Exponential Growth

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

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

where

• [latex]a\,[/latex]is the initial or starting value of the function.
• [latex]b\,[/latex]is the growth factor or growth multiplier per unit[latex]\,x[/latex].

In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, permit'due south consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can exist represented by the function[latex]\,A\left(x\correct)=100+50x.\,[/latex]Visitor B has 100 stores and expands past increasing the number of stores past l% each yr, so its growth can be represented by the office [latex]\,B\left(10\correct)=100{\left(1+0.five\right)}^{x}.[/latex]

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

Year, [latex]10[/latex]	Stores, Company A	Stores, Company B
[latex]0[/latex]	[latex]100+50\left(0\right)=100[/latex]	[latex]100{\left(ane+0.5\right)}^{0}=100[/latex]
[latex]1[/latex]	[latex]100+l\left(1\right)=150[/latex]	[latex]100{\left(one+0.5\right)}^{1}=150[/latex]
[latex]2[/latex]	[latex]100+50\left(two\correct)=200[/latex]	[latex]100{\left(ane+0.five\correct)}^{two}=225[/latex]
[latex]three[/latex]	[latex]100+l\left(3\correct)=250[/latex]	[latex]100{\left(1+0.5\right)}^{3}=337.five[/latex]
[latex]x[/latex]	[latex]A\left(x\correct)=100+50x[/latex]	[latex]B\left(x\right)=100{\left(1+0.5\right)}^{ten}[/latex]

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

Figure 2. The graph shows the numbers of stores Companies A and B opened over a five-year catamenia.

Discover that the domain for both functions is[latex]\,\left[0,\infty \right),[/latex]and the range for both functions is[latex]\,\left[100,\infty \right).\,[/latex]After twelvemonth 1, Company B always has more than stores than Company A.

Now we will plough our attending to the function representing the number of stores for Company B,[latex]\,B\left(ten\correct)=100{\left(i+0.5\right)}^{x}.\,[/latex]In this exponential role, 100 represents the initial number of stores, 0.fifty represents the growth charge per unit, and[latex]\,1+0.5=ane.5\,[/latex]represents the growth gene. Generalizing further, we can write this function as[latex]\,B\left(x\right)=100{\left(1.five\right)}^{ten},[/latex]where 100 is the initial value,[latex]\,ane.five\,[/latex]is chosen the base of operations, and[latex]\,x\,[/latex]is called the exponent.

Evaluating a Real-Globe Exponential Model

At the beginning of this section, we learned that the population of India was about[latex]\,1.25\,[/latex]billion in the year 2013, with an almanac growth rate of nigh[latex]\,1.2%.\,[/latex]This situation is represented by the growth function[latex]\,P\left(t\correct)=1.25{\left(1.012\correct)}^{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 exist in[latex]\,\text{2031?}[/latex]

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The population of People's republic of china was about ane.39 billion in the year 2013, with an annual growth charge per unit of almost[latex]\,0.6%.\,[/latex]This situation is represented by the growth function[latex]\,P\left(t\right)=ane.39{\left(1.006\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 Red china exist for the year 2031? How does this compare to the population prediction we made for India in (Figure)?

Evidence Solution

Nigh[latex]\,1.548\,[/latex]billion people; by the twelvemonth 2031, Bharat'southward population will exceed China's by nigh 0.001 billion, or 1 million people.

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 part without knowing the role explicitly. We must use the information to first write the grade of the part, and then determine the constants[latex]\,a\,[/latex]and[latex]\,b,[/latex]and evaluate the function.

How To

Given two data points, write an exponential model.

1. If one of the information points has the form[latex]\,\left(0,a\right),[/latex] then[latex]\,a\,[/latex]is the initial value. Using[latex]\,a,[/latex] substitute the 2d point into the equation[latex]\,f\left(x\right)=a{\left(b\right)}^{ten},[/latex] and solve for[latex]\,b.[/latex]
2. If neither of the data points take the form[latex]\,\left(0,a\right),[/latex] substitute both points into ii equations with the form[latex]\,f\left(x\right)=a{\left(b\right)}^{x}.\,[/latex]Solve the resulting system of two equations in two unknowns to discover[latex]\,a\,[/latex]and[latex]\,b.[/latex]
3. Using the[latex]\,a\,[/latex]and[latex]\,b\,[/latex]found in the steps above, write the exponential office in the form[latex]\,f\left(10\right)=a{\left(b\right)}^{x}.[/latex]

Writing an Exponential Model When the Initial Value Is Known

In 2006, eighty deer were introduced into a wildlife refuge. Past 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\right)\,[/latex]of deer over time[latex]\,t.[/latex]

Endeavor It

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

Show Solution

[latex]\left(0,129\right)\,[/latex]and[latex]\,\left(2,236\right);\,\,\,N\left(t\correct)=129{\left(\text{one}\text{.3526}\right)}^{t}[/latex]

Writing an Exponential Model When the Initial Value is Non Known

Find an exponential role that passes through the points[latex]\,\left(-2,6\right)\,[/latex]and[latex]\,\left(ii,i\right).[/latex]

Endeavour Information technology

Given the ii points[latex]\,\left(1,3\right)\,[/latex]and[latex]\,\left(2,four.5\right),[/latex]discover the equation of the exponential function that passes through these 2 points.

Show Solution

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

Do 2 points always determine a unique exponential function?

Yes, provided the two points are either both above the 10-axis or both below the x-axis and take dissimilar x-coordinates. But keep in heed that we besides need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same pct growth with each unit increase in[latex]\,x,[/latex] which in many real world cases involves time.

How To

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

1. Commencement, identify two points on the graph. Choose the y-intercept equally one of the 2 points whenever possible. Try to cull points that are equally far apart as possible to reduce circular-off fault.
2. If one of the data points is the y-intercept[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(10\correct)=a{\left(b\right)}^{x},[/latex] and solve for[latex]\,b.[/latex]
3. If neither of the information points have the form[latex]\,\left(0,a\right),[/latex] substitute both points into two equations with the form[latex]\,f\left(x\right)=a{\left(b\right)}^{x}.\,[/latex]Solve the resulting organisation of two equations in 2 unknowns to find[latex]\,a\,[/latex]and[latex]\,b.[/latex]
4. Write the exponential function,[latex]\,f\left(10\right)=a{\left(b\right)}^{10}.[/latex]

Writing an Exponential Function Given Its Graph

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

Figure 5.

Try It

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

Effigy 6.

Prove Solution

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

How To

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

1. Press [STAT].
2. Clear any existing entries in columns L1 or L2.
3. In L1, enter the x-coordinates given.
4. In L2, enter the respective y-coordinates.
5. Printing [STAT] again. Cursor correct to CALC, gyre down to ExpReg (Exponential Regression), and press [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 Estimator to Discover an Exponential Function

Employ a graphing calculator to find the exponential equation that includes the points[latex]\,\left(2,24.8\right)\,[/latex]and[latex]\,\left(v,198.iv\right).[/latex]

Try It

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

Evidence Solution

[latex]y\approx 12\cdot {1.85}^{10}[/latex]

Applying the Chemical compound-Interest Formula

Savings instruments in which earnings are continually reinvested, such every bit common funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, only on the accumulated value of the business relationship.

The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly involvement charge per unit earned past an investment account. The term nominal is used when the compounding occurs a number of times other than one time per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up existence greater than the nominal rate! This is a powerful tool for investing.

We tin can calculate the compound interest using the chemical compound interest formula, which is an exponential part of the variables time[latex]\,t,[/latex] chief[latex]\,P,[/latex] Apr[latex]\,r,[/latex] and number of compounding periods in a year[latex]\,n:[/latex]

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

For case, observe (Effigy), which shows the effect of investing $1,000 at ten% for one twelvemonth. Notice how the value of the account increases equally the compounding frequency increases.

Frequency	Value afterwards 1 yr
Annually	$1100
Semiannually	$1102.50
Quarterly	$1103.81
Monthly	$1104.71
Daily	$1105.16

The Compound Interest Formula

Compound interest can be calculated using the formula

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

where

• [latex]A\left(t\right)\,[/latex]is the account value,
• [latex]t\,[/latex]is measured in years,
• [latex]P\,[/latex]is the starting corporeality of the business relationship, oftentimes called the principal, or more than mostly present value,
• [latex]r\,[/latex]is the annual pct rate (April) expressed every bit a decimal, and
• [latex]due north\,[/latex]is the number of compounding periods in one yr.

Computing 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?

Endeavor Information technology

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?

Using the Chemical 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'southward future college tuition; the business relationship grows revenue enhancement-free. Lily wants to set upward a 529 account for her new granddaughter and wants the account to abound to $40,000 over 18 years. She believes the account will earn half dozen% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the business relationship now?

Try It

Refer to (Figure). To the nearest dollar, how much would Lily need to invest if the business relationship is compounded quarterly?

Evaluating Functions with Base of operations e

Equally we saw earlier, the amount earned on an account increases as the compounding frequency increases. (Figure) shows that the increase from annual to semi-almanac compounding is larger than the increment from monthly to daily compounding. This might lead us to enquire whether this pattern will continue.

Examine the value of $one invested at 100% involvement for 1 year, compounded at various frequencies, listed in (Effigy).

Frequency	[latex]A\left(t\correct)={\left(1+\frac{1}{n}\right)}^{due north}[/latex]	Value
Annually	[latex]{\left(1+\frac{one}{1}\right)}^{1}[/latex]	$ii
Semiannually	[latex]{\left(ane+\frac{1}{2}\right)}^{2}[/latex]	$ii.25
Quarterly	[latex]{\left(1+\frac{1}{4}\right)}^{4}[/latex]	$2.441406
Monthly	[latex]{\left(one+\frac{1}{12}\correct)}^{12}[/latex]	$two.613035
Daily	[latex]{\left(one+\frac{1}{365}\right)}^{365}[/latex]	$2.714567
Hourly	[latex]{\left(1+\frac{1}{\text{8760}}\right)}^{\text{8760}}[/latex]	$2.718127
Once per minute	[latex]{\left(1+\frac{ane}{\text{525600}}\right)}^{\text{525600}}[/latex]	$two.718279
Once per second	[latex]{\left(1+\frac{1}{31536000}\right)}^{31536000}[/latex]	$two.718282

These values announced to exist approaching a limit as[latex]\,due north\,[/latex]increases without bound. In fact, as[latex]\,n\,[/latex]gets larger and larger, the expression[latex]\,{\left(1+\frac{1}{northward}\correct)}^{n}\,[/latex]approaches a number used and so frequently in mathematics that it has its own name: the letter[latex]\,eastward.\,[/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 eastward

The alphabetic character e represents the irrational number

[latex]{\left(1+\frac{1}{n}\right)}^{n},\text{as}\,n\,\text{increases without bound}[/latex]

The letter eastward is used as a base for many real-world exponential models. To piece of work with base e, we employ the approximation,[latex]\,east\approx ii.718282.\,[/latex]The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who showtime investigated and discovered many of its properties.

Using a Calculator to Find Powers of due east

Calculate[latex]\,{e}^{iii.14}.\,[/latex]Round to five decimal places.

Show Solution

On a figurer, printing the button labeled[latex]\,\left[{east}^{10}\right].\,[/latex]The window shows[latex]\,\left[e^(\text{ }\correct].\,[/latex]Type[latex]\,3.14\,[/latex]and then close parenthesis,[latex]\,\left[)\right].\,[/latex]Press [ENTER]. Rounding to[latex]\,5\,[/latex]decimal places,[latex]\,{e}^{iii.14}\approx 23.10387.\,[/latex]Caution: Many scientific calculators accept an "Exp" push button, which is used to enter numbers in scientific notation. It is not used to discover powers of[latex]\,e.[/latex]

Try Information technology

Utilise a calculator to find[latex]\,{e}^{-0.v}.\,[/latex]Circular to v decimal places.

Show Solution

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

Investigating Continuous Growth

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

The Continuous Growth/Decay 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{eastward}^{rt}[/latex]

where

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

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

For concern applications, the continuous growth formula is chosen the continuous compounding formula and takes the grade

[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 interest rate per unit of measurement time,
• and [latex]t\,[/latex]is the period 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 decay function.

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

Computing Continuous Growth

A person invested $1,000 in an business relationship earning a nominal 10% per twelvemonth compounded continuously. How much was in the business relationship at the end of one year?

Try Information technology

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?

Show Solution

$3,659,823.44

Calculating Continuous Decay

Radon-222 decays at a continuous rate of 17.three% per mean solar day. How much volition 100 mg of Radon-222 decay to in three days?

Attempt It

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

Show Solution

3.77E-26 (This is calculator annotation for the number written as[latex]\,3.77×{10}^{-26}\,[/latex]in scientific note. While the output of an exponential function is never cypher, this number is so close to cipher that for all practical purposes we tin can accept zero every bit the respond.)

Key Equations

definition of the exponential function	[latex]f\left(x\right)={b}^{x}\text{, where }b>0, b\ne ane[/latex]
definition of exponential growth	[latex]f\left(x\right)=a{b}^{ten},\text{ where }a>0,b>0,b\ne 1[/latex]
chemical compound interest formula	[latex]\begin{assortment}{50}A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} ,\text{ where}\hfill \\ A\left(t\correct)\text{ is the business relationship value at fourth dimension }t\hfill \\ t\text{ is the number of years}\hfill \\ P\text{ is the initial investment, often chosen the primary}\hfill \\ r\text{ is the annual percent rate (APR), or nominal rate}\hfill \\ n\text{ is the number of compounding periods in one year}\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 amount (in the continuous compounding formula a is replaced with P, the chief) [latex]e[/latex]is the mathematical constant,[latex] \text{ }due east\approx 2.718282[/latex]

Cardinal Concepts

• An exponential function is defined as a function with a positive constant other than[latex]\,1\,[/latex]raised to a variable exponent. See (Figure).
• A function is evaluated by solving at a specific value. Run across (Figure) and (Figure).
• An exponential model can be constitute when the growth rate and initial value are known. See (Effigy).
• An exponential model can be found when the two information points from the model are known. Encounter (Figure).
• An exponential model can be establish using 2 data points from the graph of the model. See (Figure).
• An exponential model can be constitute using ii data points from the graph and a calculator. See (Figure).
• The value of an account at whatsoever fourth dimension[latex]\,t\,[/latex]can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See (Effigy).
• The initial investment of an account can be plant using the chemical compound involvement formula when the value of the account, annual involvement rate, compounding periods, and life span of the account are known. See (Effigy).
• The number[latex]\,e\,[/latex]is a mathematical constant often used as the base of real world exponential growth and disuse models. Its decimal approximation is[latex]\,e\approx 2.718282.[/latex]
• Scientific and graphing calculators have the cardinal[latex]\,\left[{east}^{x}\right]\,[/latex]or[latex]\,\left[\mathrm{exp}\left(x\correct)\correct]\,[/latex]for calculating powers of[latex]\,e.\,[/latex]See (Effigy).
• Continuous growth or disuse models are exponential models that utilise[latex]\,e\,[/latex]equally the base. Continuous growth and disuse models tin exist constitute when the initial value and growth or decay rate are known. See (Figure) and (Effigy).

Section Exercises

Verbal

Explain why the values of an increasing exponential office will eventually overtake the values of an increasing linear part.

Show Solution

Linear functions have a constant rate of change. Exponential functions increase based on a percentage of the original.

Given a formula for an exponential part, is it possible to determine whether the function grows or decays exponentially only by looking at the formula? Explain.

The Oxford Dictionary defines the word nominal every bit a value that is "stated or expressed but not necessarily respective exactly to the existent value."^[2] 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.

Show Solution

When involvement is compounded, the percentage of interest earned to main ends up being greater than the annual percent rate for the investment account. Thus, the annual per centum charge per unit does not necessarily correspond to the existent interest earned, which is the very definition of nominal.

Algebraic

For the post-obit exercises, identify whether the statement represents an exponential function. Explain.

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

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

Show Solution

exponential; the population decreases past a proportional rate.

The value of a coin drove has increased by[latex]\,3.25%\,[/latex]annually over the last[latex]\,twenty\,[/latex]years.

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

Show Solution

not exponential; the accuse decreases past a constant amount each visit, so the statement represents a linear role. .

The height of a projectile at time[latex]\,t\,[/latex]is represented past the function[latex]\,h\left(t\right)=-4.ix{t}^{ii}+18t+xl.[/latex]

For the post-obit exercises, consider this scenario: For each year[latex]\,t,[/latex]the population of a woods of copse is represented by the function[latex]\,A\left(t\correct)=115{\left(i.025\right)}^{t}.\,[/latex]In a neighboring woods, the population of the same type of tree is represented by the function[latex]\,B\left(t\correct)=82{\left(i.029\correct)}^{t}.\,[/latex](Round answers to the nearest whole number.)

Which forest'southward population is growing at a faster rate?

Show Solution

The forest represented by the part[latex]\,B\left(t\correct)=82{\left(one.029\correct)}^{t}.[/latex]

Which forest had a greater number of trees initially? Past how many?

Assuming the population growth models continue to stand for the growth of the forests, which wood will have a greater number of copse afterward[latex]\,20\,[/latex]years? By how many?

Evidence Solution

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

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

Hash out the higher up results from the previous four exercises. Bold the population growth models proceed to represent the growth of the forests, which forest will 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 volition vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, just considering wood B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that manner as long as the population growth models concord. 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, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.

[latex]y=300{\left(ane-t\right)}^{5}[/latex]

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

Bear witness Solution

exponential growth; The growth factor,[latex]\,1.06,[/latex] is greater than[latex]\,1.[/latex]

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

[latex]y=11,701{\left(0.97\right)}^{t}[/latex]

Evidence Solution

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

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

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

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

Show Solution

[latex]f\left(10\right)=2000{\left(0.one\correct)}^{x}[/latex]

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

[latex]\left(-2,6\right)\,[/latex]and[latex]\,\left(3,1\right)[/latex]

Show Solution

[latex]f\left(x\right)={\left(\frac{1}{6}\right)}^{-\frac{three}{5}}{\left(\frac{one}{half dozen}\right)}^{\frac{x}{v}}\approx 2.93{\left(0.699\right)}^{x}[/latex]

[latex]\left(3,1\right)\,[/latex]and[latex]\,\left(5,four\right)[/latex]

For the following exercises, make up one's mind whether the table could stand for a function that is linear, exponential, or neither. If it appears to be exponential, discover a part that passes through the points.

[latex]x[/latex]	ane	2	3	four
[latex]f\left(x\right)[/latex]	lxx	40	ten	-20

[latex]ten[/latex]	one	2	3	iv
[latex]h\left(x\right)[/latex]	70	49	34.3	24.01

[latex]x[/latex]	1	ii	iii	4
[latex]m\left(x\right)[/latex]	lxxx	61	42.ix	25.61

[latex]x[/latex]	1	ii	3	4
[latex]f\left(x\correct)[/latex]	10	20	40	80

[latex]x[/latex]	ane	2	iii	four
[latex]one thousand\left(x\correct)[/latex]	-iii.25	two	7.25	12.5

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

After a sure number of years, the value of an investment business relationship is represented past the equation[latex]\,10,250{\left(1+\frac{0.04}{12}\right)}^{120}.\,[/latex]What is the value of the account?

What was the initial deposit made to the account in the previous exercise?

Prove Solution

[latex]$10,250[/latex]

How many years had the account from the previous exercise been accumulating interest?

An business relationship is opened with an initial deposit of $vi,500 and earns[latex]\,3.half-dozen%\,[/latex]interest compounded semi-annually. What will the business relationship be worth in[latex]\,20\,[/latex]years?

Show Solution

[latex]$13,268.58[/latex]

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

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

Testify Solution

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

Use the formula found in the previous exercise to summate the initial deposit of an account that is worth[latex]\,$14,472.74\,[/latex]after earning[latex]\,5.5%\,[/latex]involvement compounded monthly for[latex]\,5\,[/latex]years. (Round to the nearest dollar.)

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

Bear witness Solution

[latex]$4,572.56[/latex]

Utilise backdrop of rational exponents to solve the compound interest formula for the interest rate,[latex]\,r.[/latex]

Apply 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 $nine,000 and was worth $thirteen,373.53 after 10 years.

Show Solution

[latex]four%[/latex]

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

For the following exercises, make up one's mind 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 charge per unit is greater than[latex]\,0.[/latex]

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

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

Show Solution

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

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

How much less would the business relationship from Exercise 42 exist worth after[latex]\,xxx\,[/latex]years if it were compounded monthly instead?

Show Solution

[latex]$669.42[/latex]

Numeric

For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.

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

[latex]f\left(x\right)=-{4}^{2x+3},[/latex] for[latex]\,f\left(-one\right)[/latex]

Evidence Solution

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

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

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

Show Solution

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

[latex]f\left(x\correct)=2.7{\left(four\right)}^{-x+one}+1.5,[/latex] for[latex]f\left(-2\right)[/latex]

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

Testify Solution

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

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

Technology

For the following exercises, apply a graphing calculator to detect the equation of an exponential function given the points on the curve.

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

Bear witness Solution

[latex]y=3\cdot {5}^{x}[/latex]

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

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

Prove Solution

[latex]y\approx 18\cdot {i.025}^{x}[/latex]

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

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

Show Solution

[latex]y\approx 0.ii\cdot {i.95}^{x}[/latex]

Extensions

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

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

Show Solution

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

Think that an exponential function is whatever equation written in the form[latex]\,f\left(x\right)=a\cdot {b}^{ten}\,[/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]tin be written equally[latex] b={e}^{n} [/latex]for some value of[latex] n[/latex]. Apply this fact to rewrite the formula for an exponential office that uses the number[latex] e [/latex]as a base.

In an exponential decay function, the base of the exponent is a value between 0 and 1. Thus, for some number[latex]\,b>i,[/latex] the exponential disuse office tin can be written as[latex]\,f\left(x\right)=a\cdot {\left(\frac{1}{b}\right)}^{x}.\,[/latex]Use this formula, along with the fact that[latex]\,b={e}^{n},[/latex] to show that an exponential decay role takes the form[latex]\,f\left(10\correct)=a{\left(e\right)}^{-nx}\,[/latex]for some positive number[latex]\,n\,[/latex].

Evidence Solution

Let[latex]\,f\,[/latex]be the exponential decay function[latex]\,f\left(x\right)=a\cdot {\left(\frac{1}{b}\correct)}^{x}\,[/latex]such that[latex]\,b>ane.\,[/latex]Then for some number[latex]\,north>0,[/latex][latex]f\left(x\right)=a\cdot {\left(\frac{i}{b}\correct)}^{x}=a{\left({b}^{-ane}\right)}^{x}=a{\left({\left({e}^{n}\right)}^{-one}\right)}^{x}=a{\left({e}^{-n}\right)}^{x}=a{\left(e\right)}^{-nx}.[/latex]

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

Real-World Applications

The trick population in a certain region has an annual growth rate of nine% per year. In the yr 2012, there were 23,900 pull a fast one on counted in the area. What is the fob population predicted to be in the twelvemonth 2020?

Show 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 after 54 hours?

In the year 1985, a business firm was valued at $110,000. By the yr 2005, the value had appreciated to $145,000. What was the almanac growth charge per unit between 1985 and 2005? Assume that the value continued to grow by the same percent. What was the value of the firm in the year 2010?

Show Solution

[latex]1.39%;\,[/latex][latex]$155,368.09[/latex]

A car was valued at $38,000 in the twelvemonth 2007. By 2013, the value had depreciated to $eleven,000 If the automobile'southward value continues to driblet by the same percentage, what will it be worth by 2017?

Jamal wants to relieve $54,000 for a down payment on a home. How much volition he need to invest in an account with 8.2% APR, compounding daily, in society to reach his goal in 5 years?

Show Solution

[latex]$35,838.76[/latex]

Kyoko has $10,000 that she wants to invest. Her depository financial institution has several investment accounts to cull from, all compounding daily. Her goal is to have $15,000 past the fourth dimension she finishes graduate school in half dozen years. To the nearest hundredth of a percent, what should her minimum almanac interest rate exist in order to reach her goal? (Hint: solve the compound interest formula for the interest rate.)

Alyssa opened a retirement business relationship with 7.25% April in the year 2000. Her initial deposit was $13,500. How much will the account exist worth in 2025 if interest compounds monthly? How much more would she brand if interest compounded continuously?

Testify Solution

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

An investment account with an annual interest charge per unit of vii% was opened with an initial eolith of $4,000 Compare the values of the account after 9 years when the involvement is compounded annually, quarterly, monthly, and continuously.

Glossary

annual percentage rate (April)

the yearly interest rate earned by an investment account, likewise called nominal rate

compound interest

interest earned on the total balance, not simply the main

exponential growth

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

nominal charge per unit

the yearly interest rate earned by an investment account, besides called annual percentage charge per unit

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