Solving 5 to the Negative 4th Power

Instructor: Stephanie Matalone

Stephanie taught high school science and math and has a Master's Degree in Secondary Education.

In this lesson, you will learn the steps of solving 5 to the negative 4th power. In doing so, you will learn the basics of exponents and negative exponents and how to solve them.

Steps to Solving the Problem

You are deciding whether or not to enter a raffle. Your smart friend thinks it will be funny to tell you that you have a 5 to the negative 4th power chance of winning the raffle if you buy a ticket. Should you spend the $5? Is it worth it?

Before you spend your money, you want to know what 5 to the negative 4th power even means! Well, the word power tells you that you are dealing with exponents. In our problem, -4 is the exponent and 5 is the base. When we write this out, the exponent -4 will be the small number written like a superscript after the base of 5, which is written in normal size.

This is how it looks when written:


An exponent can also be written out like this:


Exponents tell you how many times to multiply the base number by itself. Typically, this is pretty easy when you have a positive exponent. But, how do you multiply 5 by itself -4 times? The easy way to do this is by writing a fraction with the base and positive exponent in the denominator, or bottom of the fraction. You will then put a 1 in the numerator, or top of the fraction.

As a fraction, it will look like this:

Exponent as fraction

The reason why you can put negative exponents in the denominator is because of inverse operations. The expression x^0 is is equal to 1 and the expression x^1 is equal to x. To go from x^0 to x^1, we just multiply by x. The expression x^2 is equal to x x x and x^3 is equal to x x x x x. To go from x^2 to x^3, we again multiply by x.

To go backwards from x^3 to x^2, we will use the inverse operation of multiplication, which is division. Thus, we will divide by x which is the same as multiplying by 1/x. Let's say we have x^-5, we would be dividing x^0 or 1 by x^5, which is the same as multiplying it by 1/x^5. Thus, x^-5 is equal to to 1 times 1/x^5 which is 1/x^5.

Now that you have a positive exponent in the denominator, you will treat it like a normal exponent problem. In the denominator, you will expand the exponent into its true multiplication. Since the exponent is four, we know we need to multiply 5 by itself four times so we will write that out:


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