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Exponential functions are useful for modeling a wide variety of real-life scenarios. This lesson will introduce the concepts of *exponential growth* and *exponential decay.* How they are used to model real-life situations will also be understood.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

While watching a documentary about ancient civilizations, Ali wondered how scientists could determine the age of an object belonging to such ancient civilizations.

Ali recalled that his uncle, Mr. Jones, is an archaeologist! Ali called and asked him how are the ages of ancient objects determined.

Carbon$-14$ is a substance present in organisms that, once they expire, begin to be released from an object's body at a slow rate. To date an object this way consists of measuring the amount of carbon$-14$ in a sample and comparing it to known values of different ages.

Well, carbon$-14$ has a half-life of $5730$ years. That means that after $5730$ years there will be half as much carbon$-14$ in the sample. In light of all this information, is it possible to write a function that can be used in carbon$-14$ dating?Explore

Consider the following exponential function.

$y=2⋅b_{x} $

How does its graph change when the value of $b$ changes? To visualize its effect, move the slider in the following applet.
Think about these two questions.

- What is the behavior of the function when $b>1$?
- What is the behavior of the function when $b<1$?

Discussion

For an exponential function $y=a(b)_{x},$ where $a>0$ and $b>1,$ the $y-$values increases as the $x-$values increases. Therefore, it can be classified as an increasing function. *exponential growth* functions.

$y=ab_{x}⇓Increasing whena>0andb>1 $

Such exponential functions are called
When a quantity *increases* by the same factor over equal intervals of time, it is said that such a quantity is in exponential growth. Exponential growth is modeled using exponential functions where $a>0$ and $b>1.$ *rate of growth*, in decimal form.

$y=a(b)_{t} $

In this form, $a$ is the initial amount, the base $b$ is the growth factor, and $t$ usually represents time. Like any other exponential function, $a$ also represents the $y-$intercept.
As shown, the greater the base $b,$ the faster the exponential function grows. Since the base $b$ is greater than $1,$ it can be written as the sum of $1$ and some positive number $r.$ This constant $r$ can then be interpreted as the

For example, $r=0.06$ means that the quantity increases by $6%$ over every unit of time.

Example

Magdalena, excited for biology lab, is exploring about bacterial growth.

Bacteria are known to duplicate themselves within a certain amount of time. This means that after some time, there will be
Magdalena begins her experiment with only 1 bacteria of E. coli in her petri dish. She watches how it duplicates as time goes by.

a Write an equation that models this exponential growth in terms of the time elapsed since the start of the experiment, in minutes.

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b How much bacteria will there be after two hours?

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a The initial amount of bacteria is $1$ and they *duplicate* each $20$ minutes.

b Write $2$ hours as minutes. Then substitute that into the equation obtained in Part A.

a Exponential growths are modeled using exponential functions of the following form.

$y=a⋅b_{t} $

In the above equation, $a$ represents the $initial$ $number$ of bacteria. Since the experiment starts with $1$ bacteria, $a$ equals $1.$ The bacteria $duplicate$ themselves every $20$ minutes. That means the $base$ should be equal to $2.$
$y=1(2)_{t}⇒y=2_{t} $

Notice that this function does $y=2_{t}⇒y=2_{20t} $

Therefore, $y=2_{20t}$ correctly models the situation for Time Elapsed, $t$ | $y=2_{20t}$ | Number of Bacteria, $y$ |
---|---|---|

$0$ | $2_{200}=1$ | $1$ |

$20$ | $2_{2020}=2_{1}$ | $2$ |

$40$ | $2_{2040}=2_{2}$ | $4$ |

$60$ | $2_{2060}=2_{3}$ | $8$ |

$80$ | $2_{2080}=2_{4}$ | $16$ |

b Since the exponential growth function is a function of time in minutes, $2$ hours must expressed in minutes as well. That can be done using a conversion factor.

$2h⋅1h60min =120min $

The converted time can now be substituted into the equation.
$y=2_{20t}$

Substitute

$t=120$

$y=2_{20120}$

CalcQuot

Calculate quotient

$y=2_{6}$

CalcPow

Calculate power

$y=64$

Discussion

Applications of exponential growth can also be encountered in the world of finance. Some people use the power of *compound interest* to grow their wealth exponentially.

Compound interest is the interest earned depending on both the initial investment and previously earned interest. To find the balance $A$ of an account that earns compound interest, an exponential growth function can be used.

In this function, $P$ stands for the principal, or the initial amount of money, $r$ is the interest rate in decimal form, and $n$ is the number of times the interest is compounded per year. For an account with the principal $$100$ and an annual interest of $18%$ compounded twice a year, the balance in the account after $t$ years is shown in the graph.

Notice that the function grows continuously, whereas, in reality, the account balance only increases at the times of compound. When calculating compound interest, the number of compounding periods $n$ creates a difference. That is, the higher the number of compounding periods, the greater the amount of compound interest.

Example

Kriz, determined and focused, won an online video game competition. The first place prize was $$1000!$

Kriz decides to not spend the prize money. Instead, their parent suggests placing all of it into a Certificate of Deposit. This is a type of savings account with compound interest. The catch is that the money cannot be taken out for a certain period of time. Ngân Hàng, a local bank, offers a Certificate of Deposit with the interest rate at $3%$ compounded monthly.

a Write an equation that models the local bank's compound interest.

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b How much money will Kriz have $2$ years after opening the account? Round the amount to $2$ decimal places.

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a The meaning of

compounded monthlyis that the interest is compounded each month of the year — meaning twelve times per year.

b Substitute $2$ for $t$ in the equation obtained in Part A.

a The local bank offers Kriz an interest rate of $3%$ that is compounded $monthly.$ Using the exponential growth function that models compound interest, first write the percentage as a decimal.

$3%⇒0.03 $

It is given that the interest is compounded monthly. Therefore it will compound $12$ times a year. Knowing that Kriz will store all of the $$1000$ in prize money, there is now enough information to write the equation that models this compound interest.
Using this equation, Kriz can calculate their earning after $t$ years.
b In order to find how much money Kriz would have in their savings account after $2$ years, $2$ will be substituted for $t$ into the equation solved for in Part A.

Discussion

When the base of an exponential function is a number greater than $0$ and less than $1,$ the function is said to be decreasing. In such cases, the function represents what is known as *exponential decay*.

When a quantity *decreases* by the same factor over equal intervals of time it is said that such quantity is in exponential decay. Exponential decay is modeled using exponential functions with $a>0$ and a base $b$ that is between $0$ and $1.$ *rate of decay*, in decimal form.

$y=a(b)_{t} $

In this form, $a$ is the initial amount, the base $b$ is the decay factor, and $t$ usually represents time. Like any other exponential function, $a$ also represents the $y$-intercept.
As seen, the closer the base gets to $0,$ the faster the exponential function decays. Since the base $b$ is less than $1,$ it can be written as $1$ minus a positive number $r$ between $0$ and $1.$ This constant $r$ can be interpreted as the

A value of $0.12,$ for instance, would mean that the quantity decreases by $12%$ over every unit of time.

Example

Diego has saved for the past few years dreaming of buying a car with a drop top so he can cruise the streets looking fly. Diego runs to the nearest car dealer and is met by Mr. Peterson, a car salesmen. They come to an agreement where Diego trades in his old car to help pay for the new car.

Diego bought his car five years ago at the same dealer for $$20000.$ Mr. Peterson states that the car depreciates at a rate of $15%$ annually.

How much is Mr. Peterson going to give Diego for his old car? Round Mr. Peterson's offer to two decimal places.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mord\">$<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["8874.11"]}}

Since the value of the car depreciates, the situation can be modeled using an exponential decay function.

Previously, Diego bought a car with a value that has depreciated since the time of purchase. Therefore, this scenario can be modeled using an exponential decay function. The car was bought for $$20000,$ which is the initial amount.
Considering the depreciation rate of $15%,$ Mr.Peterson offers $$8874.11$ for Diego's $5-$year-old car. If Diego accepts the offer, he will have a capital of $$8874.11$ to put toward the purchase of a new car from the dealership.

$y=a(1−r)_{t}⇓y=20000(1−r)_{t} $

The car depreciates at a rate of $15%$ annually. That value needs to be written as a decimal.
$15%⇒0.15 $

There is now enough information to write an equation for the value of the old car.
It has been $five$ years since Diego bought his car. Therefore, to find the current value of the car, $5$ should be substituted for $t$ into the function.
$y=20000(0.85)_{t}$

Substitute

$t=5$

$y=20000(0.85)_{5}$

UseCalc

Use a calculator

$y=8874.10625$

RoundDec

Round to $2$ decimal place(s)

$y=8874.11$

Pop Quiz

Select the option that best describes the table of values given below.

Pop Quiz

Exponential functions $y=ab_{t}$ can model exponential decay as well as exponential growth. Identify the rate of decay or growth $r$ for the indicated function. Write the corresponding rate in decimal form.

Closure

This lesson introduced the interesting concepts of compound interest, exponential growth, and exponential decay. Using the knowledge gained from this lesson, the introductory challenge can be modeled using an exponential decay function. Recall what the archaeologist had to say.

Since the half-life of carbon$-14$ is $5730$ years, an initial amount $A$ of carbon$-14$ will decay by half that amount in $5730$ years. Let $y$ be the final amount of carbon$-14$ and write an equation that models this exponential decay.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["A","t"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["A*0.5^{\\dfrac{t}{5730}}","\\dfrac{A}{2}^{\\dfrac{t}{5730}}"]}}

Find the decay factor of this situation.

Exponential decay is modeled using exponential functions with a base $b$ that is between $0$ and $1.$

$y=a(b)_{t} $

Since the given information is about the $half-$life of carbon$-14,$ the base is $21 ,$ or $0.5.$ It is also given that the $initial$ $amount$ of carbon is represented by $A.$
$y=a(b)_{t}⇓y=A(0.5)_{t} $

The half-life of carbon$-14$ is $5730$ years, which means that after $5730$ years, only half of the initial amount will remain. Therefore, $t$ should be divided by $5730$ so that for every $5730$ years half the previous amount remains.
$y=A(0.5)_{5730t} $

Wow! That is a great amount of time that carbon$-14$ takes in order to decay. That fact makes for a good reason to use it when determining the age of a fossil.