## EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS

In this section, we are going to see how to solve word problems on exponential growth and decay.

Before look at the problems, if you like to learn about exponential growth and decay,

please click here

Problem 1 :

David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007 ?

Number of years between 1999 and 2007 is

n = 2007 - 1999

n = 8

No. of stores in the year 2007 = P(1+r)ⁿ

Substitute P = 200, r = 8% or 0.08 and n = 8.

No. of stores in the year 2007 = 200(1 + 0.08) 8

No. of stores in the year 2007 = 200(1.08 ) 8

No. of stores in the year 2007 = 200(1.8509)

No. of stores in the year 2007 = 370.18

So, the number of stores in the year 2007 is about 370.

Problem 2 :

You invest $2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?

We have to use the formula given below to know the value of the investment after 3 years.

A = Pe rt

Substitute

P = 2500

r = 10% or 0.1

t = 10

e = 2.71828

Then, we have

A = 2500(2.71828) (0.1)10

A = 6795.70

So, the value of the investment after 10 years is $6795.70.

Problem 3 :

Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours ?

Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100.

We have to use the formula given below to find the percent of substance after 6 hours.

A = P(1 + r) n

P = 100

r = -3.5% or -0.035

t = 6

(Here, the value of "r" is taken in negative sign. because the substance decays)

A = 100(1-0.035) 6

A = 100(0.935 ) 6

A = 100(0.8075)

A = 80.75

Because the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75%

Problem 4 :

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture initially, how many bacteria will be present at the end of 8th hour?

Note that the number of bacteria present in the culture doubles at the end of successive hours.

Since it grows at the constant ratio "2", the growth is based is on geometric progression.

We have to use the formula given below to find the no. of bacteria present at the end of 8th hour.

A = ab x

a = 30

b = 2

x = 8

Then, we have

A = 30(2 8 )

A = 30(256 )

A = 7680

So, the number of bacteria at the end of 8th hour is 7680.

Problem 5 :

A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself ?

Let "P" be the amount invested initially. From the given information, P becomes 2P in 3 years.

Since the investment is in compound interest, for the 4th year, the principal will be 2P. And 2P becomes 4P (it doubles itself) in the next 3 years.

Therefore, at the end of 6 years accumulated value will be 4P. So, the amount deposited will amount to 4 times itself in 6 years.

## Related Topics

Doubling-Time Growth Formula

Half-Life Decay Formula

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## Exponential-Growth Word Problems

Log Probs Expo Growth Expo Decay

## What is the formula for exponential growth?

Exponential growth word problems work off the exponential-growth formula, A = Pe rt , where A is the ending amount of whatever you're dealing with (for example, money sitting in an investment, or bacteria growing in a petri dish), P is the beginning amount of that same whatever, r is the growth constant, and t is time.

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## MathHelp.com

Exponential Growth and Decay

The exponential-growth formula A = Pe rt is related to the compound-interest formula , and represents the case of the interest being compounded "continuously".

Note that the particular variables used in the equation may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:

A = P e r t A = P e k t Q = N e k t Q = Q 0 e k t

No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the growth constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

## How do you solve exponential growth problems?

To solve exponential growth word problems, you may be plugging one value into the exponential-growth equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the doubling-time information to find the growth constant (probably by solving the exponential equation by using logarithms), and then using that value to find whatever the exercise requested.

## What is an example of an exponential growth problem?

- A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant k for the bacteria? (Round k to two decimal places.)

For this exercise, the units on time t will be hours, because the growth is being measured in terms of hours.

The beginning amount P is the amount at time t = 0 , so, for this problem, P = 100 .

The ending amount is A = 450 at t = 6 hours.

The only variable I don't have a value for is the growth constant k , which also happens to be what I'm looking for. So I'll plug all the known values into the exponential-growth formula, and then solve for the growth constant:

450 = 100 e 6 k

4.5 = e 6 k

ln(4.5) = 6 k

ln(4.5) / 6 = k = 0.250679566129...

They want me to round my decimal value (found by punching keys on my calculator) to two decimal places. So my answer is:

Many math classes, math books, and math instructors leave off the units for the growth and decay rates. However, if you see this topic again in chemistry or physics, you will probably be expected to use proper units ("growth-decay constant ÷ time"). If I had included this information in my solution above, my answer would have been " k = 0.25 /hour". I doubt that your math class will even mention this, let alone require that you include this. Still, it's not a bad idea to get into the habit now of checking and reporting your units.

Note that the constant was positive, because it was a growth constant. If I had come up with a negative value for the growth constant, then I would have known to check my work to find my error(s).

- A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half?

In this problem, I know that time t will be in hours, because they gave me growth in terms of hours. As a result, I'll convert "a day and a half" to "thirty-six hours", so my units match.

I know that the starting population is P = 100 , and I need to find A at time t = 36 .

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But what is the growth constant k ? And why do they tell me what the doubling time is?

They gave me the doubling time because I can use this to find the growth constant k . Then, once I have this constant, I can go on to answer the actual question.

So this exercise actually has two unknowns, the growth constant k and the ending amount A . I can use the doubling time to find the growth constant, at which point the only remaining value will be the ending amount, which is what they actually asked for. So first I'll find the constant.

If the initial population were, say, 100 , then, in 6.5 hours (being the specified doubling time), the population would be 200 . (It doesn't matter what starting value I pick for this part of my solution, as long as my ending value is twice as much. It's the "twice as much" that matters here, more than the number that it's twice of.) I'll set this up and solve for k :

200 = 100 e 6.5 k

2 = e 6.5 k

At this point, I need to use logs to solve:

ln(2) = 6.5 k

ln(2) / 6.5 = k

I could simplify this to a decimal approximation, but I won't, because I don't want to introduce round-off error if I can avoid it. So, for now, the growth constant will remain this "exact" value. (I might want to check this value quickly in my calculator, to make sure that this growth constant is positive, as it should be. If I have a negative value at this stage, I need to go back and check my work.)

Now that I have the growth constant, I can answer the actual question, which was "How many bacteria will there be in thirty-six hours?" This means using 100 for P , 36 for t , and the above expression for k . I plug these values into the formula, and then I simplify to find A :

A = 100 e 36(ln(2)/6.5) = 4647.75313957...

I will take the luxury of assuming that they don't want fractional bacteria (being the 0.7313957... part), so I'll round to the nearest whole number.

about 4648 bacteria

You can do a rough check of this answer, using the fact that exponential processes involve doubling times. The doubling time in this case is 6.5 hours, or between 6 and 7 hours.

If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, 800 in eighteen hours, 1600 in twenty-four hours, 3200 in thirty hours, and 6400 in thirty-six hours.

If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twenty-one hours, 1600 in twenty-eight hours, and 3200 in thirty-five hours.

The answer I got above, 4678 bacteria in thirty-six hours, fits nicely between these two estimates.

Note: When you are given a nice, neat doubling time, another method for solving the exercise is to use a base of 2 . First, figure out how many doubling-times that you've been given. In the above case, this would start by noting that "a day and a half" is 36 hours, so we have:

36 ÷ 6.5 = 72/13

Use this as the power on 2 :

100 × 2 (72/13) = 4647.75314...

Not all algebra classes cover this method. If you're required to use the first method for every exercise of this type, then do so (in order to get the full points). Otherwise, this base- 2 trick can be a time-saver. And, yes, you'd use a base of 3 if you'd been given a tripling-time, a base of 4 for a quadrupling-time, etc.

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## Exponential Functions Word Problems Worksheet with Answers

Exponential functions are one of the most applicable mathematics topics to the real-world. Whether it’s calculating compound interest or exploring the exponential decay of a radioactive substance, there are just so many real-world applications that make excellent exponential functions word problems! But since word problems can be tricky, I have put together an exponential functions word problems worksheet with answers to help!

Let’s get to the bottom of how to solve exponential functions word problems by exploring some real-world applications!

## What is an Exponential Function Word Problem?

Exponential growth and decay happen everywhere in the real-world. From the bacteria growing on your toilet seat to the compound interest you earn in you bank account, there are so many applications of exponential growth and decay functions. So it makes sense that you will see a wide variety of applications when it comes to exponential function word problems.

Exponential functions are a specific type of equation that can be used to model certain relationships that exist in the real world. In particular, exponential functions can be used to model exponential relationships, or relationships that involve rapid growth or decay by some factor. This is much different than linear functions, for example, which have a constant rate of change.

Just like any other type of function, exponential functions are awesome because they allow us to make predictions about future amounts. And who doesn’t want to predict the future!?

An exponential function word problem will ask you to do exactly this! Your goal will be to use the information provided to make a prediction about how much of something there will be in the future.

For example, you could be tasked with finding anything ranging from the amount of a substance remaining, to the value of an investment, or the size of a population of bacteria after a certain amount of time.

There are also a wide variety of exponential function word problems types. But in general, you will be given some initial amount, a growth rate or a decay rate, and an amount of time that has passed. In some cases you will also be given an exponential expression that you can use. If not, you will have to create your own exponential growth function or decay function!

## How Do You Solve Exponential Function Word Problems?

The best way to solve exponential function word problems is by using a similar problem-solving process to what you would use in any other area of mathematics.

As a first step, I always recommend reading the question carefully. This sound obvious, but reading and re-reading the problem will help you to identify any key information that is given. This will help you determine exactly what values you will be using in your solution.

For example, you should ask yourself questions such as:

- Is the problem providing a growth rate or a decay rate?
- Do I know the initial value? Or is this what I am being tasked with determining?
- Is an exponential growth function or an exponential decay function being given? Do I have to write my own function?
- What does the problem not tell me?

Once you have identified all of the key information that you are given, it is time to identify the unknown quantity that you are being asked to determine. As a tip, this is usually mentioned in the final sentence of the problem. It is important to take note of the units provided so that you can include those in your final answer as well.

Finally, your last step should be to reflect to select the appropriate problem-solving strategy. In some problems, exponential equations are provided that you can use. In other cases, you have to write your own.

Most exponential function word problems will require you to make a substitution of some sort into an exponential function. After you perform this step, you will be required to perform a few basic operations in order to determine your answer.

I have covered exponential function word problems in the past on the Math By The Pixel YouTube channel , so I have linked a video walkthrough of a series of examples . Take a look at the following video as a primer before you download and complete the exponential functions word problems worksheet with answers linked below!

## Download Your Exponential Functions Word Problems Worksheet with Answers

Below you will find an exponential functions word problems worksheet with answers. I constructed this worksheet by using the best problems that I have used throughout my teaching career to help students develop an understanding of exponential functions in the real-world.

I have hand picked a good mix of exponential growth and exponential decay problems across a wide variety of applications. For example, you will see exponential growth function examples such as compound interest and population growth. You will also see exponential decay examples such as depreciation of an asset and radioactive decay of a substance.

I also made sure to include problems that require a wide range of problem solving strategies. For example, you will be asked to use a function that is provided, as well as write and use your own!

There are also a few problems that require you to identify the meaning of each parameter in a given exponential function. This will help deepen your understanding of how these important functions work!

Remember to check the answer keys provided to ensure that you fully understand each of these problems.

Download the worksheet by clicking below!

## Practicing Exponential Functions Word Problems

Remember that there are a wide variety of exponential functions word problems that you can encounter. Practice is always key when it comes to mastering any mathematics concept. And exponential functions word problems are no different!

Regular practice working with exponential expression word problems will help you become more comfortable with concepts like compound interest and radioactive decay.

When you complete these types of problems often you will start to notice patterns. And before long, you will start to develop a sense of mathematical intuition about how to solve specific real-world problems!

Did you find this exponential functions word problems worksheet with answers helpful? Share this post with a friend and subscribe to Math By The Pixel on YouTube for more helpful math content!

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## Exponential Growth and Decay

Exponential growth can be amazing.

The idea: something always grows in relation to its current value, such as always doubling.

## Example: If a population of rabbits doubles every month, we would have 2, then 4, then 8, 16, 32, 64, 128, 256, etc!

Amazing tree.

Let us say we have this special tree.

It grows exponentially , following this formula:

Height (in mm) = e x

e is Euler's number , about 2.718

- At 1 year old it is: e 1 = 2.7 mm high ... really tiny!
- At 5 years it is: e 5 = 148 mm high ... as high as a cup
- At 10 years: e 10 = 22 m high ... as tall as a building
- At 15 years: e 15 = 3.3 km high ... 10 times the height of the Eiffel Tower
- At 20 years: e 20 = 485 km high ... up into space!

No tree could ever grow that tall. So when people say "it grows exponentially" ... just think what that means.

## Growth and Decay

But sometimes things can grow (or the opposite: decay) exponentially, at least for a while .

So we have a generally useful formula:

y(t) = a × e kt

Where y(t) = value at time "t" a = value at the start k = rate of growth (when >0) or decay (when <0) t = time

## Example: 2 months ago you had 3 mice, you now have 18.

Start with the formula:

We know a=3 mice, t=2 months, and right now y(2)=18 mice:

18 = 3 × e 2k

Now some algebra to solve for k :

- The step where we used ln(e x )=x is explained at Exponents and Logarithms .
- we could calculate k ≈ 0.896 , but it is best to keep it as k = ln(6)/2 until we do our final calculations.

We can now put k = ln(6)/2 into our formula from before:

y(t) = 3 e (ln(6)/2)t

Now let's calculate the population in 2 more months (at t=4 months):

y( 4 ) = 3 e (ln(6)/2)× 4 = 108

And in 1 year from now ( t=14 months):

y( 14 ) = 3 e (ln(6)/2)× 14 = 839,808

That's a lot of mice! I hope you will be feeding them properly.

## Exponential Decay

Some things "decay" (get smaller) exponentially.

## Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher.

It decreases about 12% for every 1000 m: an exponential decay .

The pressure at sea level is about 1013 hPa (depending on weather).

- Write the formula (with its "k" value),
- Find the pressure on the roof of the Empire State Building (381 m),
- and at the top of Mount Everest (8848 m)
- a (the pressure at sea level) = 1013 hPa
- t is in meters (distance, not time, but the formula still works)
- y(1000) is a 12% reduction on 1013 hPa = 891.44 hPa

891.44 = 1013 e k×1000

Now we know "k" we can write :

y(t) = 1013 e (ln(0.88)/1000)×t

And finally we can calculate the pressure at 381 m , and at 8848 m :

y( 381 ) = 1013 e (ln(0.88)/1000)× 381 = 965 hPa

y( 8848 ) = 1013 e (ln(0.88)/1000)× 8848 = 327 hPa

(In fact pressures at Mount Everest are around 337 hPa ... good calculations!)

The "half life" is how long it takes for a value to halve with exponential decay.

Commonly used with radioactive decay, but it has many other applications!

## Example: The half-life of caffeine in your body is about 6 hours. If you had 1 cup of coffee 9 hours ago how much is left in your system?

- a (the starting dose) = 1 cup of coffee!
- t is in hours
- at y(6) we have a 50% reduction (because 6 is the half life)

0.5 = 1 cup × e 6 k

Now we can write :

y(t) = 1 e (ln(0.5)/6)×t

In 6 hours:

y( 6 ) = 1 e (ln(0.5)/6)× 6 = 0.5

Which is correct as 6 hours is the half life

And in 9 hours:

y( 9 ) = 1 e (ln(0.5)/6)× 9 = 0.35

After 9 hours the amount left in your system is about 0.35 of the original amount. Have a nice sleep :)

Have a play with the Half Life of Medicine Tool to get a good understanding of this.

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## 6.8: Exponential Growth and Decay

- Last updated
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- Page ID 2526

- Gilbert Strang & Edwin “Jed” Herman

## Learning Objectives

- Use the exponential growth model in applications, including population growth and compound interest.
- Explain the concept of doubling time.
- Use the exponential decay model in applications, including radioactive decay and Newton’s law of cooling.
- Explain the concept of half-life.

One of the most prevalent applications of exponential functions involves growth and decay models. Exponential growth and decay show up in a host of natural applications. From population growth and continuously compounded interest to radioactive decay and Newton’s law of cooling, exponential functions are ubiquitous in nature. In this section, we examine exponential growth and decay in the context of some of these applications.

## Exponential Growth Model

Many systems exhibit exponential growth. These systems follow a model of the form \(y=y_0e^{kt},\) where \(y_0\) represents the initial state of the system and \(k\) is a positive constant, called the growth constant. Notice that in an exponential growth model, we have

\[ y′=ky_0e^{kt}=ky. \label{eq1} \]

That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth. Equation \ref{eq1} involves derivatives and is called a differential equation .

## Exponential Growth

Systems that exhibit exponential growth increase according to the mathematical model

\[y=y_0e^{kt} \nonumber \]

where \(y_0\) represents the initial state of the system and \(k>0\) is a constant, called the growth constant.

Population growth is a common example of exponential growth. Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows. Figure \(\PageIndex{1}\) and Table \(\PageIndex{1}\) represent the growth of a population of bacteria with an initial population of 200 bacteria and a growth constant of 0.02. Notice that after only 2 hours (120 minutes), the population is 10 times its original size!

Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.

## Example \(\PageIndex{1}\): Population Growth

Consider the population of bacteria described earlier. This population grows according to the function \(f(t)=200e^{0.02t},\) where t is measured in minutes. How many bacteria are present in the population after \(5\) hours (\(300\) minutes)? When does the population reach \(100,000\) bacteria?

We have \(f(t)=200e^{0.02t}.\) Then

\[ f(300)=200e^{0.02(300)}≈80,686. \nonumber \]

There are \(80,686\) bacteria in the population after \(5\) hours.

To find when the population reaches \(100,000\) bacteria, we solve the equation

\[ \begin{align*} 100,000 &= 200e^{0.02t} \\[4pt] 500 &=e^{0.02t} \\[4pt] \ln 500 &=0.02 t \\[4pt] t &=\frac{\ln 500}{0.02}≈310.73. \end{align*} \nonumber \]

The population reaches \(100,000\) bacteria after \(310.73\) minutes.

## Exercise \(\PageIndex{1}\)

Consider a population of bacteria that grows according to the function \(f(t)=500e^{0.05t}\), where \(t\) is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach \(100\) million bacteria?

Use the process from the previous example.

There are \(81,377,396\) bacteria in the population after \(4\) hours. The population reaches \(100\) million bacteria after \(244.12\) minutes.

Let’s now turn our attention to a financial application: compound interest . Interest that is not compounded is called simple interest. Simple interest is paid once, at the end of the specified time period (usually \(1\) year). So, if we put \($1000\) in a savings account earning \(2%\) simple interest per year, then at the end of the year we have

\[ 1000(1+0.02)=$1020. \nonumber \]

Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every \(6\) months, it credits half of the year’s interest to the account after \(6\) months. During the second half of the year, the account earns interest not only on the initial \($1000\), but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have

\[ 1000 \left(1+\dfrac{0.02}{2}\right)^2=$1020.10. \nonumber \]

Similarly, if the interest is compounded every \(4\) months, we have

\[ 1000 \left(1+\dfrac{0.02}{3}\right)^3=$1020.13, \nonumber \]

and if the interest is compounded daily (\(365\) times per year), we have \($1020.20\). If we extend this concept, so that the interest is compounded continuously, after \(t\) years we have

\[ 1000\lim_{n→∞} \left(1+\dfrac{0.02}{n}\right)^{nt}. \nonumber \]

Now let’s manipulate this expression so that we have an exponential growth function. Recall that the number \(e\) can be expressed as a limit:

\[ e=\lim_{m→∞}\left(1+\dfrac{1}{m}\right)^m. \nonumber \]

Based on this, we want the expression inside the parentheses to have the form \((1+1/m)\). Let \(n=0.02m\). Note that as \(n→∞, m→∞\) as well. Then we get

\[ 1000\lim_{n→∞}\left(1+\dfrac{0.02}{n}\right)^{nt}=1000\lim_{m→∞}\left(1+\dfrac{0.02}{0.02m}\right)^{0.02mt}=1000\left[\lim_{m→∞}\left(1+\dfrac{1}{m}\right)^m\right]^{0.02t}. \nonumber \]

We recognize the limit inside the brackets as the number \(e\). So, the balance in our bank account after \(t\) years is given by \(1000 e^{0.02t}\). Generalizing this concept, we see that if a bank account with an initial balance of \($P\) earns interest at a rate of \(r%\), compounded continuously, then the balance of the account after \(t\) years is

\[ \text{Balance}\;=Pe^{rt}. \nonumber \]

## Example \(\PageIndex{2}\): Compound Interest

A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays \(5%\) annual interest compounded continuously. How much does the student need to invest today to have \($1\) million when she retires at age \(65\)? What if she could earn \(6%\) annual interest compounded continuously instead?

\[ 1,000,000=Pe^{0.05(40)} \nonumber \]

\[ P=135,335.28. \nonumber \]

She must invest \($135,335.28\) at \(5%\) interest.

If, instead, she is able to earn \(6%,\) then the equation becomes

\[ 1,000,000=Pe^{0.06(40)} \nonumber \]

\[ P=90,717.95. \nonumber \]

In this case, she needs to invest only \($90,717.95.\) This is roughly two-thirds the amount she needs to invest at \(5%\). The fact that the interest is compounded continuously greatly magnifies the effect of the \(1%\) increase in interest rate.

## Exercise \(\PageIndex{2}\)

Suppose instead of investing at age \(25\sqrt{b^2−4ac}\), the student waits until age \(35\). How much would she have to invest at \(5%\)? At \(6%\)?

At \(5%\) interest, she must invest \($223,130.16\). At \(6%\) interest, she must invest \($165,298.89.\)

If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from \(100\) to \(200\) bacteria as it does to grow from \(10,000\) to \(20,000\) bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have

\[ \begin{align*} 2y_0 &=y_0e^{kt} \\[4pt] 2 &=e^{kt} \\[4pt] \ln 2 &=kt \\[4pt] t &=\dfrac{\ln 2}{k}. \end{align*} \nonumber \]

## Definition: Doubling Time

If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by

\[\text{Doubling time}=\dfrac{\ln 2}{k}. \nonumber \]

## Example \(\PageIndex{3}\): Using the Doubling Time

Assume a population of fish grows exponentially. A pond is stocked initially with \(500\) fish. After \(6\) months, there are \(1000\) fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches \(10,000\). When will the owner’s friends be allowed to fish?

We know it takes the population of fish \(6\) months to double in size. So, if \(t\) represents time in months, by the doubling-time formula, we have \(6=(\ln 2)/k\). Then, \(k=(\ln 2)/6\). Thus, the population is given by \(y=500e^{((\ln 2)/6)t}\). To figure out when the population reaches \(10,000\) fish, we must solve the following equation:

\[ \begin{align*} 10,000 &=500e^{(\ln 2/6)t} \\[4pt] 20 &=e^{(\ln 2/6)t} \\[4pt] \ln 20 &=\left(\frac{\ln 2}{6}\right)t \\[4pt] t &=\frac{6(\ln 20)}{\ln 2} \\[4pt] &≈25.93. \end{align*} \nonumber \]

The owner’s friends have to wait \(25.93\) months (a little more than \(2\) years) to fish in the pond.

## Exercise \(\PageIndex{3}\)

Suppose it takes \(9\) months for the fish population in Example \(\PageIndex{3}\) to reach \(1000\) fish. Under these circumstances, how long do the owner’s friends have to wait?

\(38.90\) months

## Exponential Decay Model

Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant \(k\), we have

\[ y=y_0e^{−kt}. \nonumber \]

As with exponential growth, there is a differential equation associated with exponential decay. We have

\[ y′=−ky_0e^{−kt}=−ky. \nonumber \]

## Exponential Decay

Systems that exhibit exponential decay behave according to the model

\[y=y_0e^{−kt}, \nonumber \]

where \(y_0\) represents the initial state of the system and \(k>0\) is a constant, called the decay constant.

Figure \(\PageIndex{2}\) shows a graph of a representative exponential decay function.

Let’s look at a physical application of exponential decay. Newton’s law of cooling says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if \(T\) represents the temperature of the object and \(T_a\) represents the ambient temperature in a room, then

\[T′=−k(T−T_a). \nonumber \]

Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional \(T_a\) term. Fortunately, we can make a change of variables that resolves this issue. Let \(y(t)=T(t)−T_a\). Then \(y′(t)=T′(t)−0=T′(t)\), and our equation becomes

\[ y′=−ky. \nonumber \]

From our previous work, we know this relationship between \(y\) and its derivative leads to exponential decay. Thus,

\[ y=y_0e^{−kt}, \nonumber \]

and we see that

\[ T−T_a=(T_0−T_a)e^{−kt} \nonumber \]

\[ T=(T_0−T_a)e^{−kt}+T_a \nonumber \]

where \(T_0\) represents the initial temperature. Let’s apply this formula in the following example.

## Example \(\PageIndex{4}\): Newton’s Law of Cooling

According to experienced baristas, the optimal temperature to serve coffee is between \(155°F\) and \(175°F\). Suppose coffee is poured at a temperature of \(200°F\), and after \(2\) minutes in a \(70°F\) room it has cooled to \(180°F\). When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.

\[ \begin{align*} T &=(T_0−T_a)e^{−kt}+T_a \\[4pt] 180 &=(200−70)e^{−k(2)}+70 \\[4pt] 110 &=130e^{−2k} \\[4pt] \dfrac{11}{13} &=e^{−2k} \\[4pt] \ln \dfrac{11}{13} &=−2k \\[4pt] \ln 11−\ln 13 &=−2k \\[4pt] k &=\dfrac{\ln 13−\ln 11}{2} \end{align*}\]

Then, the model is

\[T=130e^{(\ln 11−\ln 13/2)t}+70. \nonumber \]

The coffee reaches \(175°F\) when

\[ \begin{align*} 175 &=130e^{(\ln 11−\ln 13/2)t}+70 \\[4pt]105 &=130e^{(\ln 11−\ln 13/2)t} \\[4pt] \dfrac{21}{26} &=e^{(\ln 11−\ln 13/2)t} \\[4pt] \ln \dfrac{21}{26} &=\dfrac{\ln 11−\ln 13}{2}t \\[4pt] \ln 21−\ln 26 &=\left(\dfrac{\ln 11−\ln 13}{2}\right)t \\[4pt] t &=\dfrac{2(\ln 21−\ln 26)}{\ln 11−\ln 13}\\[4pt] &≈2.56. \end{align*}\]

The coffee can be served about \(2.5\) minutes after it is poured. The coffee reaches \(155°F\) at

\[ \begin{align*} 155 &=130e^{(\ln 11−\ln 13/2)t}+70 \\[4pt] 85 &=130e^{(\ln 11−\ln 13)t} \\[4pt] \dfrac{17}{26} &=e^{(\ln 11−\ln 13)t} \\[4pt] \ln 17−\ln 26 &=\left(\dfrac{\ln 11−\ln 13}{2}\right)t \\[4pt] t &=\dfrac{2(\ln 17−\ln 26)}{\ln 11−\ln 13} \\[4pt] &≈5.09.\end{align*}\]

The coffee is too cold to be served about \(5\) minutes after it is poured.

## Exercise \(\PageIndex{4}\)

Suppose the room is warmer \((75°F)\) and, after \(2\) minutes, the coffee has cooled only to \(185°F.\) When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.

The coffee is first cool enough to serve about \(3.5\) minutes after it is poured. The coffee is too cold to serve about \(7\) minutes after it is poured.

Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant half-life. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have

\(\dfrac{y_0}{2}=y_0e^{−kt}\)

\(\dfrac{1}{2}=e^{−kt}\)

\(−\ln 2=−kt\)

\(t=\dfrac{\ln 2}{k}\).

Note: This is the same expression we came up with for doubling time.

## Definition: Half-Life

If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by

\[\text{Half-life}=\dfrac{\ln 2}{k}. \nonumber \]

## Example \(\PageIndex{5}\): Radiocarbon Dating

One of the most common applications of an exponential decay model is carbon dating. Carbon-14 decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon-14 was originally present in an object and how much carbon-14 remains, we can determine the age of the object. The half-life of carbon-14 is approximately 5730 years—meaning, after that many years, half the material has converted from the original carbon-14 to the new nonradioactive nitrogen-14. If we have 100 g carbon-14 today, how much is left in 50 years? If an artifact that originally contained 100 g of carbon-14 now contains 10 g of carbon-14, how old is it? Round the answer to the nearest hundred years.

\[ 5730=\dfrac{\ln 2}{k} \nonumber \]

\[ k=\dfrac{\ln 2}{5730}.\nonumber \]

So, the model says

\[ y=100e^{−(\ln 2/5730)t}.\nonumber \]

In \(50\) years, we have

\[y=100e^{−(\ln 2/5730)(50)}≈99.40\nonumber \]

Therefore, in \(50\) years, \(99.40\) g of carbon-14 remains.

To determine the age of the artifact, we must solve

\[ \begin{align*} 10 &=100e^{−(\ln 2/5730)t} \\[4pt] \dfrac{1}{10} &= e^{−(\ln 2/5730)t} \\ t &≈19035. \end{align*}\]

The artifact is about \(19,000\) years old.

## Exercise \(\PageIndex{5}\): Carbon-14 Decay

If we have 100 g of carbon-14 , how much is left after 500 years? If an artifact that originally contained 100 g of carbon-14 now contains 20 g of carbon-14, how old is it? Round the answer to the nearest hundred years.

A total of 94.13 g of carbon-14 remains after 500 years. The artifact is approximately 13,300 years old.

## Key Concepts

- Exponential growth and exponential decay are two of the most common applications of exponential functions.
- Systems that exhibit exponential growth follow a model of the form \(y=y_0e^{kt}\).
- In exponential growth, the rate of growth is proportional to the quantity present. In other words, \(y′=ky\).
- Systems that exhibit exponential growth have a constant doubling time, which is given by \((\ln 2)/k\).
- Systems that exhibit exponential decay follow a model of the form \(y=y_0e^{−kt}.\)
- Systems that exhibit exponential decay have a constant half-life, which is given by \((\ln 2)/k.\)

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## How to Write an Exponential Function: Word Problems

Any function in the form \(y=ab^x\) is called an exponential function. Unlike linear functions where the growth rate is constant, exponential functions are characterized by the fact that the growth rate of the function is proportional to the function's current value. In this article, the method of writing an exponential function is explained step by step.

## A step-by-step guide to Writing exponential function: word problems

Any function in the form \(y=ab^x\) is called an exponential function where \(a\) and \(b\) are fixed and \(x\) is an independent variable:

Variable \(y\): represents the output value

\(a\) represents the initial value

\(b\) represents the common ratio

\(x\) represents the number of times the original value has been multiplied by the common ratio

Unlike linear functions where the growth rate is constant, exponential functions are characterized by the fact that the growth rate of the function is proportional to the function’s current value and are used to model growth and decay processes, such as population growth, radioactive decay, and compound interest.

Follow the step-by-step procedure below to write an exponential function.

Step \(1\): Analyzing the problem and identifying the variables. At this stage, you should specify what the question is about and what the variables of the problem are.

Step \(2\): Using the variables, write the exponential function in the form of \(y=ab^x\).

Step \(3\): Determine the initial value (\(a\)) according to the information of the problem. \(a\) shows the value of the variable at the beginning of the time period.

Step \(4\): In this step, you must specify the common ratio (\(b\)). \(b\) is the coefficient by which the variable is multiplied in each time period.

Step \(5\): Specify the \(x\) variable that represents the number of time periods that have passed. For example, if in the problem it is said that a certain amount will be multiplied by \(4\) every month and after \(5\) months you will be asked for the amount, \(x=5\).

Step \(6\): Place the values of \(a\), \(b\) and \(x\) in the exponential function and simplify it.

Step \(7\): Check your work by inserting \(x\) values and seeing if they match the problem statement

Step \(8\): Write the final answer in one sentence

## Writing exponential function: word problems-Example 1:

If a bank’s profits start at \($60,000\) and increase by \(25%\) each year, how much will the profits be after \(4\) years?

This problem is solved by using the exponential function. In the said problem, the profit of the company starts from \(60,000\) dollars and increases by \(25%\) every year, the profit is multiplied by \(1.2\) every year. With this information, we write the exponential function:

\(y = 60000 * 1.25^x\)

In this function, \(x\) is the number of past years and \(y\) represents the profit after \(x\) years.

The profit amount after \(4\) years is obtained by inserting \(x = 4\) in the function: \(y = 40000 * 1.25^4 = 60000 * 1.25 * 1.25 * 1.25 * 1.25 = 146484\)

by: Effortless Math Team about 1 year ago (category: Articles )

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## Linear & Exponential Word Problems

These lessons, with videos, examples and step-by-step solutions, help High School students learn how to distinguish between situations that can be modeled with linear functions and with exponential functions.

Related Pages Linear and Non-Linear Functions Common Core (Functions) Common Core Mathematics

Constructing a linear function word problem

Example: Archimedes is draining his bathtub. Every 2 minutes that pass, 7 gallons of water are draines. Which of the following functions can represent the number of gallons of water in the tub, W, as a function of the minutes that have passed, T? W = 2/7 T W = 100 + 7/2 T W = 50 - 7/2 T W = 2/7 T + 20

Constructing and interpreting a linear function

Example: Entrance to the paintball court costs $5 and paint balls are paid separately. A ticket for entrance with 5 balls, for example, costs $8. Select all that apply. The relationship is proportional. Entrance with 10 balls costs $16. When the number of balls increases by 11, the price increases by $6.60. When the x-axis represents the number of paint balls, the slope of the graph of the relationship is 5/3.

Linear Function Word Problems This lesson demonstrates how linear functions can be applied to the real world. It discusses strategies for figuring out word problems.

Bathtub Problem: You pull the plug from the bathtub. After 40 seconds, there are 13 gallons of water left in the tub. One minute after you pull the plug, there are 10 gallons left. Assume that the number of gallons varies linearly with the time since the plug was pulled. a. Write the particular equation expressing the number of gallons (g) left in the tub in terms of the number of seconds (s) since you pulled the plug. b. How many gallons would ve left after 20 seconds? 30 seconds? c. At what time will there be 7 gallons left in the tub? d. Find the y-intercept (gallon-intercept). What does this number represent in the real world? e. Find the x-intercept (time-intercept). What does this number represent in the real world? f. Plot the graph of this linear function. Use a suitable domain. g. What is the slope? What does this number represent?

Thermal Expansion Problem: Bridges on highways often have expansion joints, which are small gaps between one bridge section and the next. The gaps are put there so the bridge will have room to expand when the weather gets hot. Suppose a bridge has a gap of 1.3 cm when the temperature is 22°C, and the gap narrows to 0.9 cm when the temperature warms to 30°C. Assume the gap varies linearly with the temperature. a. Write the particular equation for gap width (w) in terms of temperature (t). b. How wide would the gap be at 35°? At -10° C? c. AT what temperature would the gap close completely? What mathematical name is given to this temperature? d. What is the width-intercept? What does this tell you in the real world?

Word Problems with Exponential Functions How to solve word problems involving exponential functions?

- Write an exponential function to model the situation. Tell what each variable represents. A price of $125 increases 4% each year.
- Write an exponential function to model the situation. Then find the value of the function after 5 years to the nearest whole number. A population of 290 animals that increases at an annual rate of 9%.
- Write an exponential function to model the situation. Then find the value of the function after 5 years to the nearest whole number. A population of 300 animals that increases at an annual rate of -22%.

Word Problem Solving- Exponential Growth and Decay.

- Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours?
- Nadia owns a chain of fast food restaurants that operated 200 stores in 199. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007?

Solving an exponential word problem.

Example: The value of a mortgage brokerage company has decreased from 7.5 million dollars in 2010 to 6 million dollars in 2012. a) Assuming continuous exponential decay, find a formula of the form P 0 e kt for the worth of the company t years after 2010. b) Find when the worth of the company will be 4 million dollars.

Exponential growth and decay word problems Depreciation, Appreciation, Compounded, Compounded Continuously.

- My “Clown at your Party” purchases a van to drive to parties for $20,000. The value of the van depreciates at a rate of 9% each year. Write an exponential decay model for the value of the van and find the value of the van after 8 years.
- A house was purchased in 2012 for $150,000 If the value of the home increases by 3% each year, what will the house be worth in the year 2015?
- Your deposit $100.000 in an account earning 7.5% annual interest, compounded monthly. What will your balance be in 10 years?
- George Washington’s wooden teeth are decaying rapidly. P(in grams) is the weight of the tooth after t years and can be modeled by P = 3.2(0.98) t . How much did George’s teeth weigh after 225 years?
- You deposit $43,128 in an account that earns 2.5% annual interest compounded continuously. What will be the account balance after 12 years?

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## How To : Solve word problems involving exponential growth/decay

The first problem that is show is exponential decay. First you will need to create a table with hours and substance left. The next step is to find the trend by noting that we are left with a certain percentage of the substance. One you have the trend you will use this to calculate the amount of substance left in said hours, in this case 6 hours. The following example shows you how to solve exponential growth. It is essentially the same except for the fact that you will be increasing, so instead of multiplying by a number less than 1 you will have to multiply by a number greater than one. You still have to find the trend and then use that trend to solve for the given number of intervals, in this case 8 years.

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## How Do You Solve a Word Problem with Exponential Decay?

If something decreases in value at a constant rate, you may have exponential decay on your hands. In this tutorial, learn how to turn a word problem into an exponential decay function. Then, solve the function and get the answer!

- exponential
- exponential decay
- exponential function

## Background Tutorials

Introduction to algebraic expressions.

## How Do You Evaluate an Algebraic Expression?

Plugging variables into an expression is essential for solving many algebra problems. See how to plug in variable values by watching this tutorial.

## What is a Variable?

You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!

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## What's the Order of Operations?

Check out this tutorial where you'll see exactly what order you need to follow when you simplify expressions. You'll also see what happens when you don't follow these rules, and you'll find out why order of operations is so important!

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## Identifying and Evaluating Exponential Functions

## What's an Exponential Function?

Looking at an equation with a variable in the exponent? You have an exponential function! Learn about exponential functions in this tutorial.

## Further Exploration

Exponential growth.

## How Do You Solve a Word Problem with Exponential Growth?

If something increases at a constant rate, you may have exponential growth on your hands. In this tutorial, learn how to turn a word problem into an exponential growth function. Then, solve the function and get the answer!

- Terms of Use

## Exponential Growth Using Calculus

We can use Calculus to measure Exponential Growth and Decay by using Differential Equations and Separation of Variables. Note that we studied Exponential Functions here and Differential Equations here in earlier sections.

## Introduction to Exponential Growth and Decay

Remember that Exponential Growth or Decay means something is increasing or decreasing at an exponential rate (faster than if it were linear). We usually see Exponential Growth and Decay problems relating to populations, bacteria, temperature, and so on, usually as a function of time.

## Solving Exponential Growth Problems using Differential Equations

It turns out that if a function is exponential, as many applications are, the rate of change of a variable is proportional to the value of that variable. Thus, $ \displaystyle \frac{{dy}}{{dt}}=ky$ or $ {y}’=ky$. This is where the Calculus comes in: we can use a differential equation to get the following:

Exponential Growth and Decay Formula

For a function $ y>0$ that is differentiable function of $ t$, and $ {y}’=ky$:

$ y=C{{e}^{{kt}}}$

$ C$ is the initial value of $ y$, $ k$ is the proportionality constant. For $ k>0$, we have exponential growth, and for $ k<0$, we have exponential decay.

Here’s how we got to this equation (using a Differential Equation ), which is good to know for future problems. Note that since $ {{e}^{C}}$ is a constant, we can just turn this into another constant “$ C$”. (Note that $ y>0$.)

$ \displaystyle \frac{{dy}}{{dt}}=ky;\,\,\,dy=ky\cdot dt$

$ \displaystyle \frac{{dy}}{y}=k\,dt;\,\,\,\int{{\frac{1}{y}\,dy}}=\int{{kdt}}$

$ \displaystyle \begin{array}{c}\ln \left( y \right)=kt+{{C}_{1}}\\{{e}^{{\ln \left( y \right)}}}={{e}^{{kt+{{C}_{1}}}}}\\y={{e}^{{kt}}}\cdot {{e}^{{{{C}_{1}}}}}={{e}^{{kt}}}\cdot C=C{{e}^{{kt}}}\end{array}$

Before we get into the Exponential Growth problems, let’s do a few practice differential equation problems using Separation of Variables . Remember that we can cross-multiply to get started:

Use this same technique to solve an Exponential Growth problem:

## Exponential Growth Word Problems

Now let’s do some Exponential Growth and Decay Calculus problems:

Here are a few more Exponential Growth problems:

Learn these rules and practice, practice, practice!

On to Derivatives and Integrals of Inverse Trig Functions — you are ready!

## IMAGES

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## COMMENTS

How do you solve word problems involving exponential growth and decay? In this video, you will learn how to use a table and a formula to find the percentage of a radioactive substance that remains after a certain time.

This algebra and precalculus video tutorial explains how to solve exponential growth and decay word problems. It provides the formulas and equations / funct...

Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/algebra-home/alg-exp-and-log/al...

Solution : We have to use the formula given below to know the value of the investment after 3 years. A = Pert Substitute P = 2500 r = 10% or 0.1 t = 10 e = 2.71828 Then, we have A = 2500 (2.71828)(0.1)10 A = 6795.70 So, the value of the investment after 10 years is $6795.70. Problem 3 :

Exponential growth word problems work off the exponential-growth formula, A = Pert, where A is the ending amount of whatever you're dealing with (for example, money sitting in an investment, or bacteria growing in a petri dish), P is the beginning amount of that same whatever, r is the growth constant, and t is time. MathHelp.com

Exponential expressions word problems (algebraic) Ngozi earns $ 24,000 in salary in the first year she works as an interpreter. Each year, she earns a 3.5 % raise. Write a function that gives Ngozi's salary S ( t) , in dollars, t years after she starts to work as an interpreter. Do not enter commas in your answer.

6.3K views 3 years ago How do you solve exponential growth/decay problems? In this video, I use 3 examples to show you how solve exponential growth/decay problems using Natural...

How Do You Solve a Word Problem with Exponential Growth? Note: If something increases at a constant rate, you may have exponential growth on your hands. In this tutorial, learn how to turn a word problem into an exponential growth function. Then, solve the function and get the answer! Keywords: problem word growth exponential exponential growth

If you do 400 * 5%, the result you get is equal to just the interest earned for the 1st year. The problem is asking you to find the balance in the account after 3 years. -- At the end of year 1, the account balance = the original amount (400) + interest (400 * 0.05). This can be simplified into 400 * 1.05. -- At the end of year 2, the account ...

The best way to solve exponential function word problems is by using a similar problem-solving process to what you would use in any other area of mathematics. As a first step, I always recommend reading the question carefully. This sound obvious, but reading and re-reading the problem will help you to identify any key information that is given.

No tree could ever grow that tall. So when people say "it grows exponentially" ... just think what that means. Growth and Decay But sometimes things can grow (or the opposite: decay) exponentially, at least for a while. So we have a generally useful formula: y (t) = a × e kt Where y (t) = value at time "t" a = value at the start

Key Concepts. Exponential growth and exponential decay are two of the most common applications of exponential functions. Systems that exhibit exponential growth follow a model of the form y = y0ekt. In exponential growth, the rate of growth is proportional to the quantity present. In other words, y′ = ky.

955 74K views 3 years ago Algebra 1 In this video we learn how to solve exponential function word problems. We create the equation by converting the percentage growth/decay rate to a...

Word problems with variables as exponents. Estimated19 minsto complete. Progress. Practice Applications of Exponential Functions. Practice.

1.61051/1.4641 = 1.1. All the ratios are the same, so the sequence must be exponential. Basically, divide a term in a sequence by the previous term. If the quotient is always the same number, then you have an exponential sequence. (Note: the quotient could be fractional and even negative. 1, -0.5, 0.25. -0.125, 0.0625.

Solution: This problem is solved by using the exponential function. In the said problem, the profit of the company starts from \ (60,000\) dollars and increases by \ (25%\) every year, the profit is multiplied by \ (1.2\) every year. With this information, we write the exponential function: \ (y = 60000 * 1.25^x\)

Exponential functions often involve the rate of increase or decrease of something. When it's a rate of increase, you have an exponential growth function! Check out these kinds of exponential functions in this tutorial! Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and ...

Solving an exponential word problem. Example: The value of a mortgage brokerage company has decreased from 7.5 million dollars in 2010 to 6 million dollars in 2012. a) Assuming continuous exponential decay, find a formula of the form P 0 e kt for the worth of the company t years after 2010. b) Find when the worth of the company will be 4 ...

There are formulas that can be used to find solutions to most problems related to exponential growth. Here, we will look at a summary of exponential growth and the formulas that can be used to solve these types of problems.

498 40K views 7 years ago Logarithms - COMPOUND INTEREST Learn about compound interest. We will look at how to determine the final value, initial value, interest rate and years needed. We will...

Solve word problems involving exponential growth/decay. The first problem that is show is exponential decay. First you will need to create a table with hours and substance left. The next step is to find the trend by noting that we are left with a certain percentage of the substance. One you have the trend you will use this to calculate the ...

In this tutorial, learn how to turn a word problem into an exponential growth function. Then, solve the function and get the answer! Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In this non-linear system, users are ...

Solving Exponential Growth Problems using Differential Equations. It turns out that if a function is exponential, as many applications are, the Thus, $ \displaystyle \frac { {dy}} { {dt}}=ky$ or $ {y}'=ky$. This is where the Calculus comes in: we can use a differential equation to get the following: Here's how we got to this equation (using ...