Algebra Help: Exponential Equations, Exponential Functions, and Exponential Word Problems

One of the harder of the math word problems to solve in Algebra is the exponential word problem. To solve these you first have to make sure you understand how to solve basic exponential equations. (This material is covered in Saxon Algebra 2, Lesson 115A.)

Basic Exponential Equations in Algebra
An important math rule is if two powers of the same base are equal the exponents must be equal.

So, if 10³ = 10^x, then x = 3.

2^6.7 = 2^x+4. Solving for x, we get 6.7 = x+4, x = 2.7.

We can use this same rule with less obvious exponential equations. How would we solve for x here?

23 = 10^x-2

The key is that we need to rewrite 23 as a power of 10. To do that, we need to use our logarithm function. Unless otherwise noted, we can assume that "log" is the same as "log₁₀."

log 23 = 1.36 (this is an approximation, rounded to two decimal places)

If you are not familiar with how logarithms work, the key is to remember that "logarithm" is just another name for "exponent." log 23 = 1.36, therefore, can be represented as:

10^1.36 = 23

So, back to our original problem, we have:

10^1.36 = 10^x-2
1.36 = x-2
x = 3.36

Introduction to Exponential Functions
There are many processes that are shown to grow or decrease exponentially. Examples include the growth of bacteria, the growth of money in finance, and the decay of radioactive material. These kinds of processes can be represented using what are called "exponential functions." These can all be represented using a general form of an exponential equation:

A_t = A₀*e^k*t

Where A_t represents the amount (of bacteria, decay, money, etc.) at time "t" and A₀ represents the amount at time "0" (the starting amount). "k" is some constant that is specific to our exponential function. When k is greater than zero (that is, is a positive number) the graph of the exponential function goes up; if k is less than zero (that is, is a negative number), the graph of the exponential function goes down. "e" is known as the natural logarithm and is an irrational number. The decimal approximation of e to three places is 2.718. e is very important in calculus and other advanced math courses. (This material is covered in more detail in Saxon Algebra 2, Lesson 115B.)

How to Solve Exponential Word Problems
The best way to learn how to solve exponential word problems are to work through examples. In Saxon these kind of problems follow a set pattern: the initial amount (A₀) is given, as well as some amount set in the future. The student is asked to find the amount for a different value for t. Using the given information the student calculates the constant k, then inputs the value of t requested in the problem.

Sample Exponential Word Problem #1:
In the beginning there were 150 bacteria in the dish. The number of bacteria increased exponentially. 5 days later there were 1800 bacteria in the dish. How many bacteria will there be in 15 days?

The statement "the number of bacteria increased exponentially" means we are working with our general exponential equation:

A_t = A₀*e^k*t

From our math word problem description we know:

A₀ = 150
A₅ = 150*e^k*5 = 1800

Our goal is to solve for k. First we divide both sides by 150 and rearrange our equation a little.

e^5*k = 12

We need to represent 12 with a base of e. For that we need to use our natural logarithm function, ln. (If you are unfamiliar with how the natural logarith works, ln 12 = 2.48 can be represented as e^2.48 = 12.)

e^5*k = e^2.48
5*k = 2.48
k = 0.496

We now know that our specific exponential function (exponential equation) for this Algebra math word problem is:

A_t = 150*e^0.496*t

To finish the problem we need to plug 15 in for t.

A₁₅ = 150*e^0.496*15 = 2.55 x 10⁵ bacteria

Sample Exponential Word Problem #2:
The number of weeds increased exponentially. At first there were only 10 weeds, but in two weeks there were 70. Assuming no one does any weeding, how many weeds will there be in six weeks?

From our math word problem we are given:

A_t = A₀*e^k*t
A₀ = 10
A₂ = 10*e^k*2 = 70

Do a little algebraic rearranging:

e^2*k = 7

We now have to take the natural log (ln) of 7 to represent it as base e.

e^2*k = e^1.946
2*k = 1.946
k = 0.973

Our specific exponential function for this problem, therefore, is:

A_t = 10*e^0.973*t

Our final answer is: A₆ = 10*e^0.973*6 = 3431 weeds.

Sample Exponential Word Problem #3:
In the beginning there were 10.67 ounces of radioactive material. it decayed exponentially. 5 years later there was only 8.46 ounces of radioactive material remaining. How much radioactive material should remain in 50 years?

From our math word problem we are given:

A_t = A₀*e^k*t
A₀ = 10.67
A₅ = 10.67*e^k*5 = 8.46

We then divide, rearrange our equation, and use our natural log to solve for k:

e^5*k = 0.793
e^5*k = e^-0.232
5*k = -0.232
k = -0.0464

Our specific equation for this Algebra word problem, therefore, is:

A_t = 10.67*e^-0.0464*t

And our final answer for the problem (how much radioactive decay is left in 50 years) is:

A₅₀ = 10.67*e^-0.0464*50 = 1.040 ounces.

Blessings!

Source
Saxon, John H., Jr. Algebra 2: An Incremental Development. Third Edition