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What Are the Power Rules?

The rules of powers are shown below. It’s important that you learn these. We’ll take a closer look at the rules and see how they’re used.

Rule

Rules for Powers

\begin{array}{l} a^{n} \cdot a^{m} & = a^{n + m} & \frac{a^{n}}{a^{m}} & = a^{n - m} \\ {(a^{b})}^{c} & = a^{b \cdot c} & {(a \cdot b)}^{n} & = a^{n} \cdot b^{n} \\ {(\frac{a}{b})}^{n} & = \frac{a^{n}}{b^{n}} & \sqrt{a} & = a^{\frac{1}{2}} \\ a^{- n} & = \frac{1}{a^{n}} & a^{0} & = 1 \end{array}

Remember that the letters $a$ and $b$ can be any number or letter. These rules are valid when calculating with both numbers and letters.

Let’s have a look at some examples where we use the rules for powers.

Rule 1

First, I will show you how to use Rule $1$ .

Rule

a^{n} \cdot a^{m} = a^{n + m}

Example 1

Let’s see why this rule is valid:

\begin{array}{l} 3^{4} \cdot 3^{3} & = (3 \cdot 3 \cdot 3 \cdot 3) \cdot (3 \cdot 3 \cdot 3) \\ = 3^{4 + 3} \\ = 3^{7} \end{array}

This gives us that

3^{4} \cdot 3^{3} = 3^{4 + 3} = 3^{7}

Rule 2

Next, we will look at Rule $2$ .

Rule

{(a^{b})}^{c} = a^{b \cdot c}

Example 2

Let’s see why this rule is valid:

\begin{array}{l} {(4^{2})}^{4} & = (4^{2}) \cdot (4^{2}) \cdot (4^{2}) \cdot (4^{2}) \\ = (4 \cdot 4) \cdot (4 \cdot 4) \cdot (4 \cdot 4) \cdot (4 \cdot 4) \\ = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \\ = 4^{8} \end{array}

This means that

{(4^{2})}^{4} = 4^{2 \cdot 4} = 4^{8}

Rule 3

Now you will see how to use Rule $3$ .

Rule

{(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}

Example 3

Let’s see why this rule is valid:

\begin{array}{l} {(\frac{2}{3})}^{3} & = (\frac{2}{3}) \cdot (\frac{2}{3}) \cdot (\frac{2}{3}) \\ = \frac{2 \cdot 2 \cdot 2}{3 \cdot 3 \cdot 3} \\ = \frac{2^{3}}{3^{3}} \\ = \frac{8}{27} \end{array}

That gives us

{(\frac{2}{3})}^{3} = \frac{2^{3}}{3^{3}} = \frac{8}{27}

Rule 4

Here you will see how to use Rule $4$ .

Rule

\frac{a^{n}}{a^{m}} = a^{n - m}

Example 4

Let’s see why this rule is valid:

\begin{array}{l} \frac{3^{4}}{3^{2}} & = \frac{3 \cdot 3 \cdot 3 \cdot 3}{3 \cdot 3} \\ = 3^{2} \end{array}

This means that

\frac{3^{4}}{3^{2}} = 3^{4 - 2} = 3^{2}

Rule 5

Now, we will take a look at Rule $5$ .

Rule

a^{- n} = \frac{1}{a^{n}}

Example 5

Let’s see why this rule is valid:

\begin{array}{l} 7^{- 2} & = 7^{0 - 2} \\ = \frac{7^{0}}{7^{2}} \\ = \frac{1}{7^{2}}, \end{array}

showing us that

7^{- 2} = \frac{1}{7^{2}}

Rule 6

Now you will see how to use Rule $6$ .

Rule

{(a \cdot b)}^{n} = a^{n} \cdot b^{n}

Example 6

Let’s see why this rule is valid:

\begin{array}{l} {(5 \cdot 2)}^{4} & = (5 \cdot 2) \cdot (5 \cdot 2) \cdot (5 \cdot 2) \cdot (5 \cdot 2) \\ = (5 \cdot 5 \cdot 5 \cdot 5) \cdot (2 \cdot 2 \cdot 2 \cdot 2) \\ = 5^{4} \cdot 2^{4}, \end{array}

showing that

{(5 \cdot 2)}^{4} = 5^{4} \cdot 2^{4}

Rule 7

Now, let’s look at Rule $7$ .

Rule

\sqrt{a} = a^{\frac{1}{2}}

Example 7

Let’s see why this rule is valid. Here we need to be a bit more thorough. We know that ${\sqrt{3}}^{2} = 3$ . With this as a starting point, we get

\begin{array}{l} {\sqrt{3}}^{2} & = 3 \\ {({\sqrt{3}}^{2})}^{\frac{1}{2}} & = 3^{\frac{1}{2}} \\ {(\sqrt{3})}^{2 \cdot \frac{1}{2}} & = 3^{\frac{1}{2}} \\ {(\sqrt{3})}^{\frac{2}{2}} & = 3^{\frac{1}{2}} \\ {(\sqrt{3})}^{1} & = 3^{\frac{1}{2}} \\ \sqrt{3} & = 3^{\frac{1}{2}} \end{array}

Rule 8

Finally, I will show you how to use Rule $8$ .

Rule

a^{0} = 1

Example 8

Let’s see why this rule is valid. Here we will use a trick, namely that $0 = 1 - 1$ . It might seem a bit funny, but as long as it works, which it does, we can use it to help us out.

\begin{array}{l} 6 5^{0} & = 6 5^{1 - 1} \\ = \frac{6 5^{1}}{6 5^{1}} \\ = \frac{65}{65} \\ = 1 \end{array}

This shows us that

6 5^{0} = 1

Combined examples

Here are some examples that require you to use several rules at once.

Example 9

\begin{array}{l} {(2 a)}^{3} \cdot {(a \cdot b)}^{2} & = 2^{3} \cdot a^{3} \cdot a^{2} \cdot b^{2} \\ = 8 \cdot a^{3} \cdot a^{2} \cdot b^{2} \\ = 8 \cdot a^{3 + 2} \cdot b^{2} \\ = 8 a^{5} b^{2} \end{array}

Example 10

\begin{array}{l} \frac{a^{4} \cdot {(a^{2} \cdot b)}^{5}}{{(a \cdot b)}^{3}} & = \frac{a^{4} \cdot {(a^{2})}^{5} \cdot b^{5}}{a^{3} \cdot b^{3}} \\ = \frac{a^{4} \cdot a^{10} \cdot b^{5}}{} a^{3} \cdot b^{3} \\ = \frac{a^{14} \cdot b^{5}}{a^{3} \cdot b^{3}} \\ = a^{14 - 3} \cdot b^{5 - 3} \\ = a^{11} \cdot b^{2} \end{array}

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