How to Solve Exponential Equations

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  • 0:04 An Exponential Equation
  • 0:53 Using the Log
  • 2:43 The Natural Log
  • 4:00 Example
  • 5:21 Lesson Summary
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer

Amy has a master's degree in secondary education and has taught math at a public charter high school.

Read this lesson to learn the steps you need to take to solve exponential equations. You'll see that it's not too difficult; you just need to make use of one other mathematical operation in order to solve your problems.

An Exponential Equation

Solving for your variable, usually x, is pretty straightforward when your x can easily be isolated through addition, subtraction, multiplication, or division. But if your variable is an exponent, then it becomes a bit harder to solve for it. These types of problems are called exponential equations. Here are some examples of exponential equations:

ex = 40

10x + 2 = 30

e2x - 7ex + 10 = 0

At first look, it may seem that isolating the x is a difficult, if not impossible, task to do. That is, until you realize that you can use the logarithmic function to help you isolate the x. Let's see how you can do that.

Using the Log

The log function is the inverse of an exponential function. It has a very useful property that can help you solve for your exponential equations.

x = by with the function y = logbx

This means that if the variable you are trying to solve for is in the exponent, then you can use the logarithm function, like this:

(logb)(xy) = ylogbx

Let's see if you can use this to help you solve this exponential equation:

10x + 2 = 30

Even though the x, the variable you are trying to solve for, is in the exponent, you are still going to use algebra techniques to solve this problem. This means trying to isolate your variable to the best of your ability. So you first subtract the 2 from both sides. Remember, whatever you do on one side, you have to do to the other as well.

10x + 2 - 2 = 30 - 2

10x = 28

Now you can take the log of both sides to help you move your variable out of the exponent so you can solve for it.

log(10x) = 28

You can now use the useful property of the logarithm function to move your x out of the exponent area.

(logb)(xy) = ylogbx

Applying this property, you get this:

x log (10) = log (28)

Using the rule (logb)(x) = 1 when b=x, our log (10) term cancels out and we're left with...

x = log (28)

x = 1.447

And you are done! You have solved the problem!

The Natural Log

For some problems, you might see a little e. This e is called Euler's number and is approximately equal to 2.71828. If you see an exponential equation with this number in it, then instead of using the log function, you'll use the natural log function ln, since the natural log has a base of e. You can use the natural log when your bases match. So if you have an e, then that means you have a base of e. If you see a 10 or any other number, then your base is a 10 and the standard log function will work.

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