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Math 101: College Algebra13 chapters | 102 lessons | 11 flashcard sets

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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.

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:

*e**x* = 40

10*x* + 2 = 30

*e**2x* - 7*e**x* + 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.

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* = *b**y* with the function *y* = log*b**x*

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:

(log*b*)(*x**y*) = *y*log*b**x*

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

10*x* + 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.

10*x* + 2 - 2 = 30 - 2

10*x* = 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(10*x*) = 28

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

(log*b*)(*x**y*) = *y*log*b**x*

Applying this property, you get this:

*x* log (10) = log (28)

*x* = log (28)

*x* = 1.447

And you are done! You have solved the problem!

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.

The process of solving an exponential equation with an *e* in it instead of a base 10 number is the same process, except you're now using the natural log function. Let's try solving this problem:

*e**x* = 40

First you'll take the natural log of both sides.

ln *e**x* = ln (40)

Now you can move your *x* out of the exponent area to get this:

*x* ln *e* = ln (40)

*x* = ln (40)

*x* = 3.689

And you are done! Let's try solving another problem.

*e**2x* - 7*e**x* + 10 = 0

This one is a bit more interesting. Instead of having just one term with the *x* in the variable, you now have two of them. To solve this, you'll need to go about it in a different way. Looking closer at your numbers, you see that you can actually go ahead and factor your problem to help you solve it, like this:

*e**2x* - 7*e**x* +10 = 0

(*e**x* - 2)(*e**x* - 5) = 0

Once you've factored, now you can go ahead and solve each of your factors using the log method like you've done before.

*e**x* - 2 = 0

*e**x* = 2

ln *e**x* = ln (2)

*x* = ln (2)

*x* = 0.69

That's one solution. The other one is this:

*e**x* - 5 = 0

*e**x* = 5

ln *e**x* = ln (5)

*x* = ln (5)

*x* = 1.609

And you are done!

Let's review. An **exponential equation** is a problem with exponents that usually has a variable in the exponent area that you have to solve for. To solve these types of problems, you'll make use of the **logarithmic function**.

*x* = *b**y* with the function is *y* = log*b**x*, along with this property: log*b**x**y* = *y*log*b**x*

This property only applies when your bases are the same. Use the standard log with a base of 10 when working with base 10 numbers. If you see an *e*, then use the natural logarithm, the logarithm with a base of *e*. You'll use the same methods that you use to solve all your other algebra problems with the addition of the log function helping you to take the variable out of the exponent.

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Math 101: College Algebra13 chapters | 102 lessons | 11 flashcard sets

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