# How to Use the Product Rule for Exponents

## Summary

The formalized definition of the product rule for exponents is...

$$(x^{a})(x^{b}) = x^{a+b}$$

Basically, it tells us that if we have two powers multiplied together, we can simplify the expression by adding the exponents together (as long as the bases are the same).

The product rule for exponents is also known as the "product property of exponents".

## How to Use the Product Rule for Exponents

1. Make sure that the bases of the powers are the same.
2. Make sure that the powers are multiplied together.
3. Identify the exponents and add them together.
4. Write the simplified power with the result of Step 3 as the exponent of the original base.

You can NOT use the product rule for exponents if the bases are different. If the bases are different but the exponents are the same, you can use the reverse distributive property for exponents.

You can NOT use the product rule for exponents if the powers are added or subtracted from each other. If the powers are divided by each other, you can use the quotient rule for exponents

## Examples

$$(2^{3})(2^{4}) = 2^{7}$$

$$x^{5}x^{6} = x^{11}$$

$$a^{4}a^{5} = a^{9}$$

## Why Does It Work?

To understand why the product rule for exponents works, let's simplify $$(a^{3})(a^{4})$$ with order of operations instead of the product rule.

$$a^{3}a^{4}$$

I'll start by simplifying the powers. The exponent (3) tells me that I need to multiply the first a by itself 3 times. The exponent (4) tells me that I need to multiply the second a by itself 4 times.

$$(aaa)(aaaa)$$

There are a total of seven a's multiplied together and we can rewrite that as  $$a^{7}$$.

So, that is why...

$$a^{3}a^{4}= a^{7}$$

The product rule for exponents also works if you have numbers instead of variables

$$(2^{4})(2^{5})$$

Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (4) tells me that I need to multiply the first 2 by itself 4 times. The exponent (5) tells me that I need to multiply the second 2 by itself 5 times.

$$(2\times2\times2\times2)(2\times2\times2\times2\times2)$$

There are nine 2's multiplied together, which is $$2^{9}$$.

So, that is why...

$$(2^{4})(2^{5})=2^{9}$$

## Reverse Product Rule for Exponents

Sometimes, it can be helpful to use the reverse product rule to write a single power as a product of two powers.

$$x^{a+b}=(x^{a})(x^{b})$$

For example, if you're trying to simplify square roots, you may want to write $$x^{5}$$ like this...

$$x^{5}=(x^{2})(x^{3})$$

OR

$$x^{5}=(x^{4})(x)$$

The reverse product rule for exponents can also be helpful for factoring out the GCF from polynomials

## Frequently Asked Questions

### Why do the bases have to be the same?

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The product rule for exponents is kind of like the exponent version of combining like terms. When you're adding products, you can't simplify $$3x+4y$$ because the terms don't have the same factors.

Similarly, when you're multiplying powers, you can't simplify $$x^{3}y^{4}$$ because they don't have the same base. This is really easy to see when you use order of operations to expand the powers.

$$x^{3}y^{4}$$

The exponent (3) tells me that I need to multiply x by itself 3 times. The exponent (4) tells me that I need to multiply y by itself 4 times.

$$xxxyyyy$$

There are 3 x’s and 4 y’s, which can be rewritten as $$x^{3}y^{4}$$.

However, this is the exact same expression that we were given. It can't be simplified any further because the variables are not the same.

### What if I have more than two powers?

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If you have more than two powers, you can still add all of the exponents. You just have to make sure that ALL of the powers are multiplied together and share the same base.

$$(7^{2})(7^{6})(7^{3})=7^{11}$$

$$(x^{5})(x^{8})(x^{2})=x^{15}$$

I'll simplify an example using order of operations, to show WHY the exponents can be added even if you have more than two powers...

$$(y^{2})(y^{3})(y^{4})$$

I'll start by simplifying each of the powers. The exponent (2) tells me to multiply the first y by itself two times. The exponent (3) tells me to multiply the second y by itself three times. The exponent (4) tells me to multiply the third y by itself four times.

$$(yy)(yyy)(yyyy)$$

There are nine y's multiplied together and we can rewrite that as $$y^{9}$$. So...

$$y^{2}y^{3}y^{4}=y^9$$

### What if one of the bases doesn't have an exponent?

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Remember that any number written without an exponent technically has an "invisible" exponent of 1. So, you can just add 1 to the other exponent (assuming that the bases are the same).

$$a^{2}a=a^{3}$$

$$xx^{6}=x^{7}$$

$$yy=y^{2}$$

This example shows WHY we can add an invisible exponent of 1 to the other exponent...

$$zz^8$$

The exponent (8) tells me that I need to multiply the second z by itself 8 times.

$$(z)(zzzzzzzz)$$

There are a total of 9 z's multiplied together, so…

$$zz^8=z^9$$

### What if there's a negative exponent?

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If one (or both) of the powers has a negative exponent, you can still add the exponents.

You just need to make sure to follow the rules for adding negative numbers and remember what a negative exponent means.

$$(6^{3})(6^{-5})=6^{-2}$$

$$(x^{-3})(x^{-2})=x^{-5}$$

$$(y^{-1})(y^{4})=y^{3}$$

This example shows WHY we can add negative exponents...

$$(y^{-1})(y^{4})$$

The exponent (-1) tells me that I need to multiply y by itself once and move it to the denominator. The exponent (4) tells me that I need to multiply y by itself 4 times.

$$\frac{yyyy}{y}$$

I can reduce the fraction by canceling out a y from the numerator and denominator.

$$\frac{yyy}{1}$$

There are 3 y’s in the simplified numerator, so...

$$(y^{-1})(y^{4})=y^{3}$$

If it's easier for you, you could also use the negative exponent rule to make the exponent positive and then use the quotient rule for exponents.

### What if there's a fractional exponent?

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If there's a fractional exponent, you can still add the exponents. You just need to make sure you follow the rules for adding fractions (or adding fractions and whole numbers).

$$x^\frac{1}{2}x^{3}= x^\frac{7}{2}$$

$$x^\frac{1}{4}x^\frac{1}{2}= x^\frac{1}{8}$$

To understand WHY this works, you first need to understand how to multiply square roots and the relationship between roots and fractional exponents.

After you've reviewed those pages, come back to this example :)

$$x^\frac{1}{2}x^{3}$$

I'll start by writing the exponents with common denominators. The common denominator is 2 and I don't have to change the first exponent because it already has a denominator of 2.

The exponent 3 can be rewritten as $$\frac{6}{2}$$

$$x^\frac{1}{2}x^\frac{6}{2}$$

Next, I'll simplify the powers. The exponent ($$\frac{1}{2}$$) tells me that I need to take the square root of x. The exponent ($$\frac{6}{2}$$) tells me that I need to take the square root of $$x^6$$.

$$\sqrt{x}\sqrt{xxxxxx}$$

When I multiply the square roots by each other, I get 7 x's under the square root symbol.

$$\sqrt{xxxxxxx}$$

That can be rewritten as $$x^\frac{7}{2}$$so...

$$x^\frac{1}{2}x^{3}= x^\frac{7}{2}$$

## Free Online Practice Problems

Khan Academy - Multiply Powers

Khan Academy - Multiply & Divide Powers (Integer Exponents)

## Free Printable Worksheets

To use the worksheet generator from Math Aids, select the options you want for your worksheet and then click the "Create It" button.