**Learning Objectives**

After completing this tutorial, you should be able to:

- Solve exponential growth problems.
- Solve exponential decay problems.

**Introduction**

In this tutorial I will step you through how to solve
problems that deal in exponential growth and decay. These
problems
will require you to know how to evaluate exponential expressions and
solve exponential equations.
If you need a review on these topics, feel free to go
to **Tutorial
42: Exponential Functions** and **Tutorial
45: Exponential Equations**. Ready, set, GO!!!!!

** Tutorial**

**Exponential Growth**

, (*k* > 0)

You can use this formula to find any of its variables, depending on the information given and what is being asked in a problem. For example, you may be given the values for

Some examples of exponential growth are population growth and financial growth. The information found, can help predict what a population for a city or colony would be in the future or what the value of your house is in ten years. Or you can use it to find out how long it would take to get to a certain population or value on your house.

The diagram below shows exponential growth:

A) What was the population of the city in 1994?

B) By what % is the population of the city increasing each year?

C) What will the population of the city be in 2005?

D) When will the city’s population be 60 thousand?

Since we are looking for the population, what variable
are we seeking? If you said *A* you are right on!!!!

The way the problem is worded, 1994 is what we call our initial year. This is when*t* = 0.

Plugging in 0 for*t* and solving for *A* we get:

The way the problem is worded, 1994 is what we call our initial year. This is when

Plugging in 0 for

*

When writing up the final answer, keep in mind that the
problem said that the population was in thousands.

**The population in 1994 would be 30,000.**

Another way that we could have approached this problem was noting that the year was 1994, which is our initial year, so basically it was asking us for the initial population, which is*A*o in the
formula. This happens to be the number in front of *e* which is 30 in this problem.

The reason I showed you using the formula was to get you use to it. Just note that when it is the initial year,*t* is 0,
so you will have *e* raised to the 0 power which means it will
simplify to be 1 and you are left with whatever *A*o is.

Another way that we could have approached this problem was noting that the year was 1994, which is our initial year, so basically it was asking us for the initial population, which is

The reason I showed you using the formula was to get you use to it. Just note that when it is the initial year,

As mentioned above, in the general growth formula, *k* is a constant that represents the growth rate. *k* is the
coefficient of *t* in *e*’s exponent.

So what would be our answer in terms of percent?

Well,*k* = .0198026, so converting that to percent we get **1.98026%
for our answer**.

So what would be our answer in terms of percent?

Well,

Since we are looking for the population, what variable
are we finding? If you said *A* give yourself a high five.

What are we going to plug in for*t* in this problem?

Our initial year is 1994, and since*t* represents years after
1994, we can get *t* from 2005 - 1994, which would be 11.

Plugging in 11 for*t* and solving for *A* we get:

What are we going to plug in for

Our initial year is 1994, and since

Plugging in 11 for

Again, when writing up the final answer, keep in mind
that the problem said that the population was in thousands.

**The population in 2005 would be approximately 37, 301.**

Looks like we have a little twist here. Now we
are given the population and we need to first find *t* to find
out how many years after 1994 we are talking about and then convert
that knowledge into the actual year.

We will still be using the same formula we did to answer the questions above, we will just be using it to find a different variable.

Plugging in 60 for*A* and solving for *t* we get:

We will still be using the same formula we did to answer the questions above, we will just be using it to find a different variable.

Plugging in 60 for

This means a little over 35 years after 1994, the
population will be 60 thousand.

**So our answer is during the year 2029.**

Since we are looking for when, what variable do we
need to find? If you said *t* give yourself a high
five.

What are we going to plug in for*A* in this problem? If
you said 200000, you are correct!

Plugging 200000 for*A* in the model we get:

What are we going to plug in for

Plugging 200000 for

Rounding 4.46 to the nearest whole number we get *t* = 4.

Since*t* is measured in years since 2002, the model indicates
that the population will reach 200000 in year 2002 + 4 = 2006.

Since

, (

You can use this formula to find any of its variables, depending on the information given and what is being asked in a problem. For example, you may be given the values for

Examples of exponential decay are radioactive decay and population decrease. The information found can help predict what the half-life of a radioactive material is or what the population will be for a city or colony in the future. The half-life of a given substance is the time required for half of that substance to decay or disintegrate.

The diagram below shows exponential decay:

If we are looking for the number of grams of carbon-14
present, what variable do we need to find? If you said *A* give
yourself a high five.

What are we going to plug in for*t* in this problem?

Since*t* represents the number of years, it looks like we will
be plugging in 10,000 for *t*.

Plugging in 10000 for*t* and solving for *A* we get:

What are we going to plug in for

Since

Plugging in 10000 for

*replace

If we are looking for the half-life of this isotope,
what variable are we seeking? If you said *t* you are
correct!!!!

It looks like we don’t have any values to plug into*A* or *A*o.
However, the problem did say that we were interested in the HALF-life,
which would mean ½ of the initial amount (*A*o) would be
present at the end (*A*) of that time. This means *A* can be
replaced with .5*A*o.

Replacing*A* with .5 *A*o and solving for *t* we get:

It looks like we don’t have any values to plug into

Replacing

*

Since we are looking for the age of the paintings, what
variable are we looking for? If you said *t* you are
correct!!!!

It looks like we don’t have any values to plug into*A* or *A*o.
However, the problem did say that the paintings that were found
contained 20% of the original carbon-14. This means *A* can
be replaced with .2*A*o (20% of the original).

Replacing*A* with .2 *A*o and solving for *t* we get:

It looks like we don’t have any values to plug into

Replacing

*

** Practice Problems**

These are practice problems to help bring you to the next level.
It will allow you to check and see if you have an understanding of these
types of problems. **Math works just like anything
else, if you want to get good at it, then you need to practice it.
Even the best athletes and musicians had help along the way and lots of
practice, practice, practice, to get good at their sport or instrument.**
In fact there is no such thing as too much practice.

To get the most out of these, **you should work the problem out on
your own and then check your answer by clicking on the link for the answer/discussion
for that problem**. At the link you will find the answer
as well as any steps that went into finding that answer.

Practice Problems 1a - 1b:Solve the given exponential growth or decay problem.

1a. The value of the property in a particular
block follows a pattern of exponential growth. In the year 2001,
your company purchased a piece of property in this block. The
value of the property in thousands of dollars, t years after 2001 is
given by the exponential growth model .

Use this model to solve the following:

A) What did your company pay for the property?

B) By what percentage is the price of the property in this block increasing per year?

C) What will the property be worth in the year 2010?

D) When will the property be worth 750 thousand dollars?

(answer/discussion to 1a)

Use this model to solve the following:

A) What did your company pay for the property?

B) By what percentage is the price of the property in this block increasing per year?

C) What will the property be worth in the year 2010?

D) When will the property be worth 750 thousand dollars?

(answer/discussion to 1a)

1b. An artifact originally had 10 grams of carbon-14
present. The decay model describes the
amount of carbon-14 present after t years.

Use this model to solve the following:

A) How many grams of carbon-14 will be present in this artifact after 25,000 years?

B) What is the half-life of carbon-14?

(answer/discussion to 1b)

Use this model to solve the following:

A) How many grams of carbon-14 will be present in this artifact after 25,000 years?

B) What is the half-life of carbon-14?

(answer/discussion to 1b)

** Need Extra Help on these Topics?**

**The following are webpages that can assist you in the topics that were covered on this page**.

**http://www.purplemath.com/modules/expoprob2.htm**

This webpage will help you with exponential growth problems.

**http://www.purplemath.com/modules/expoprob3.htm**

This webpage will help you with exponential decay problems.

**Go to Get
Help Outside the
Classroom found in Tutorial 1: How to Succeed in a Math Class for
some
more suggestions.**

Last revised on March 25, 2011 by Kim Seward.

All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.