Power Rule of Exponents

The formalized definition of the power rule of exponents is...

\((x^{a})^{b} = x^{ab}\)

Basically, it tells us that if we have a power raised to another exponent, we can multiply the two exponents to simplify the expression. You can also use this rule in reverse.

The power rule of exponents is also known as the "power property of exponents" or the "powers of powers" rule.

- Identify the exponent that is being raised to another exponent.
- Multiply the exponents.
- Write the simplified power with the result of Step 2 as the exponent of the original base.

\((2^{3})^{4}=2^{12}\)

\((7^{5})^{3}=7^{15}\)

\((y^{9})^{2}=y^{18}\)

To understand why the power rule of exponents works, let's calculate \((2^{3})^{4}\) with order of operations instead of using the power rule.

\((2^{3})^{4}\)

Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (3) tells me that I need to multiply 2 by itself 3 times.

\((2 \times 2 \times 2)^{4}\)

The exponent outside the parentheses (4) tells me that I need to multiply all the stuff inside the parentheses by itself 4 times.

\((2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times2 \times 2) \times (2 \times2 \times 2) \)

So, when \((2^{3})^{4}\) is fully expanded, we have twelve 2's multiplied by each other.

We can rewrite that as \(2^{12}\).

So, that is why...

\((2^{3})^{4}= 2^{12}\)

The reason you multiply the exponents is because the outside exponent multiplies the effect of the inside exponent. This is also true when variables are involved.

\((y^{9})^{2}\)

Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (9) tells me that I need to multiply y by itself 9 times.

\((yyyyyyyyy)^{2}\)

The exponent outside the parentheses (2) tells me that I need to multiply the stuff inside the parentheses by itself 2 times.

\((yyyyyyyyy)(yyyyyyyyy)\)

There are 18 y's multiplied together, which is \(y^{18}\).

So, that is why...

\((y^{9})^{2}= y^{18}\)

Sometimes, it can be helpful to use the reverse power rule of exponents to write a power with one exponent as a power of powers.

\(x^{ab}=(x^{a})^{b}\)

For example, if you were trying to factor perfect squares or perfect cubes, you may want to write \(x^{30}\) like this...

\(x^{30}=(x^{15})^{2}\)

OR

\(x^{30}=(x^{10})^{3}\)

Or if you were trying to rewrite rational exponents as roots, you may want to write \(x^{\frac{3}{2}}\) like this...

\(x^{\frac{3}{2}}=\)\((x^{\frac{1}{2}})^{3}\)

OR

\(x^{\frac{3}{2}}=\)\((x^{3})^{\frac{1}{2}}\)

This would be helpful because an exponent of \(\frac{1}{2}\) is the same thing as a square root.

If you have multiple exponents (as powers within powers), you should multiply all of the exponents.

\([(3^{4})^{7}]^{2}=3^{56}\)

\([(x^{2})^{5}]^{6}=x^{60}\)

\([(y^{3})^{7}]^{4}=y^{84}\)

I'll simplify an example using order of operations, so you can see WHY we multiply all of the exponents...

\([(8^{2})^{4}]^{3}\)

I'll start by simplifying the stuff inside the inner-most set of parentheses. The exponent (2) tells me to multiply 8 by itself two times.

\([(8\times8)^{4}]^{3}\)

The next exponent (4) tells me to multiply the stuff inside the inner parentheses by itself 4 times.

\([(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]^{3}\)

The last exponent tells me to multiply the stuff inside the outermost parentheses by itself 3 times.

\([(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\times\)

\([(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\times\)

\([(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\)

There are twenty-four 8's multiplied by each other, so…

\([(8^{2})^{4}]^{3}=8^{24}\)

If you have multiple numbers inside the parentheses, you can use the distributive property of exponents to distribute the outer exponent to each factor within the parentheses.

If any of the bases don't have an exponent written, remember that they have an "invisible" exponent of 1.

\((a^{2}b^{5})^{3}=a^{6}b^{15}\)

\((2x^{7})^{3}=8x^{21}\)

\((x^{2}y^{3}z^{4})^{5}=x^{10}y^{15}z^{20}\)

I'll simplify an example with order of operations, so you can see WHY we distribute the exponent to each factor inside the parentheses...

\((x^{3}y^{4}z)^{2}\)

I'll simplify the inside of the parentheses first. 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. The z has an "invisible" exponent of 1, so it is multiplied once.

\((xxxyyyyz)^{2}\)

The outer exponent tells me that I need to multiply the stuff inside parentheses by itself 2 times.

\((xxxyyyyz)(xxxyyyyz)\)

There are 6 x’s, 8 y’s, and 2 z's multiplied together, so…

\((x^{3}y^{4}z)^{2}=x^{6}y^{8}z^{2}\)

If you have a fraction inside the parentheses, you can use the division version of the distributive property of exponents to distribute the outside exponent to the numerator and denominator of the fraction.

\((\frac{7^2}{5^3})^4 = \frac{7^{8}}{5^{12}}\)

\((\frac{a^6}{9^4})^2 = \frac{a^{12}}{9^{8}}\)

\((\frac{x^{12}}{y^3})^5 = \frac{x^{60}}{y^{15}}\)

In this example, you can see WHY we distribute the exponent to the numerator and denominator...

\((\frac{x^2}{y^5})^4\)

I'll simplify the inside of the parentheses first. The exponent (2) tells me that I need to multiply x by itself 2 times. The exponent (5) tells me that I need to multiply y by itself 5 times.

\((\frac{xx}{yyyyy})^{4}\)

The next exponent tells me I need to multiply the stuff inside the parentheses by itself 4 times.

\((\frac{xx}{yyyyy})(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})\)

There are 8 x’s in the numerator and 20 y’s in the denominator, so...

\((\frac{x^2}{y^5})^4 = \frac{x^{8}}{y^{20}}\)

If one (or more) of the exponents is negative, multiply the exponents together using the rules for multiplying negative numbers.

\((2^{-4})^{3}= 2^{-12}\)

\((x^{4})^{-7}= x^{-28}\)

\((y^{-3})^{-6}= y^{18}\)

Read this page if you need a refresher on how to work with negative exponents.

In this example, you can see WHY we can multiply the exponents as negative numbers...

\((5^{-3})^{2}\)

I'll start by simplifying the inside of the parentheses first. The (-3), tells me that I need to multiply 5 by itself 3 times in the denominator of a fraction.

\((\frac{1}{5\times5\times5})^{2}\)

The next exponent (2) tells me that I need to multiply the stuff inside the parentheses by itself two times.

\((\frac{1}{5\times5\times5})(\frac{1}{5\times5\times5})\)

There are six 5's multiplied in the denominator of a fraction, so...

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

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Khan Academy - Powers of Products & Quotients

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