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

RulesforPowers

$\begin{array}{llllllll}\hfill {a}^{n}\cdot {a}^{m}& ={a}^{n+m}\phantom{\rule{2em}{0ex}}& \hfill \frac{{a}^{n}}{{a}^{m}}& ={a}^{n-m}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left({a}^{b}\right)}^{c}& ={a}^{b\cdot c}\phantom{\rule{2em}{0ex}}& \hfill {\left(a\cdot b\right)}^{n}& ={a}^{n}\cdot {b}^{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left(\frac{a}{b}\right)}^{n}& =\frac{{a}^{n}}{{b}^{n}}\phantom{\rule{2em}{0ex}}& \hfill \sqrt{a}& ={a}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}^{-n}& =\frac{1}{{a}^{n}}\phantom{\rule{2em}{0ex}}& \hfill {a}^{0}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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: $\begin{array}{llll}\hfill {3}^{4}\cdot {3}^{3}& =\left(3\cdot 3\cdot 3\cdot 3\right)\cdot \left(3\cdot 3\cdot 3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={3}^{4+3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={3}^{7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This gives us that

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

Rule 2

Next, we will look at Rule $2$.

Rule

 $\phantom{\rule{-0.17em}{0ex}}{\left({a}^{b}\right)}^{c}={a}^{b\cdot c}$

Example 2

Let’s see why this rule is valid: $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left({4}^{2}\right)}^{4}& =\phantom{\rule{-0.17em}{0ex}}\left({4}^{2}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left({4}^{2}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left({4}^{2}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left({4}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(4\cdot 4\right)\cdot \left(4\cdot 4\right)\cdot \left(4\cdot 4\right)\cdot \left(4\cdot 4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={4}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This means that

 $\phantom{\rule{-0.17em}{0ex}}{\left({4}^{2}\right)}^{4}={4}^{2\cdot 4}={4}^{8}$

Rule 3

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

Rule

 $\phantom{\rule{-0.17em}{0ex}}{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}$

Example 3

Let’s see why this rule is valid: $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left(\frac{2}{3}\right)}^{3}& =\phantom{\rule{-0.17em}{0ex}}\left(\frac{2}{3}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(\frac{2}{3}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(\frac{2}{3}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\cdot 2\cdot 2}{3\cdot 3\cdot 3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{2}^{3}}{{3}^{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{8}{27}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

That gives us

 $\phantom{\rule{-0.17em}{0ex}}{\left(\frac{2}{3}\right)}^{3}=\frac{{2}^{3}}{{3}^{3}}=\frac{8}{27}$

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: $\begin{array}{llll}\hfill \frac{{3}^{4}}{{3}^{2}}& =\frac{\text{3}\cdot \text{3}\cdot 3\cdot 3}{\text{3}\cdot \text{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={3}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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: $\begin{array}{llll}\hfill {7}^{-2}& ={7}^{0-2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{7}^{0}}{{7}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{{7}^{2}},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

showing us that

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

Rule 6

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

Rule

 ${\left(a\cdot b\right)}^{n}={a}^{n}\cdot {b}^{n}$

Example 6

Let’s see why this rule is valid: $\begin{array}{llll}\hfill {\left(5\cdot 2\right)}^{4}& =\left(5\cdot 2\right)\cdot \left(5\cdot 2\right)\cdot \left(5\cdot 2\right)\cdot \left(5\cdot 2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(5\cdot 5\cdot 5\cdot 5\right)\cdot \left(2\cdot 2\cdot 2\cdot 2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={5}^{4}\cdot {2}^{4},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

showing that

 ${\left(5\cdot 2\right)}^{4}={5}^{4}\cdot {2}^{4}$

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 $\begin{array}{llll}\hfill {\sqrt{3}}^{2}& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left({\sqrt{3}}^{2}\right)}^{\frac{1}{2}}& ={3}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left(\sqrt{3}\right)}^{2\cdot \frac{1}{2}}& ={3}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left(\sqrt{3}\right)}^{\frac{2}{2}}& ={3}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{\left(\sqrt{3}\right)}^{1}& ={3}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{3}& ={3}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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. $\begin{array}{llll}\hfill 6{5}^{0}& =6{5}^{1-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6{5}^{1}}{6{5}^{1}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{65}{65}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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

$\begin{array}{llll}\hfill {\left(2a\right)}^{3}\cdot {\left(a\cdot b\right)}^{2}& ={2}^{3}\cdot {a}^{3}\cdot {a}^{2}\cdot {b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8\cdot {a}^{3}\cdot {a}^{2}\cdot {b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8\cdot {a}^{3+2}\cdot {b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8{a}^{5}{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 10

$\begin{array}{llll}\hfill \frac{{a}^{4}\cdot {\left({a}^{2}\cdot b\right)}^{5}}{{\left(a\cdot b\right)}^{3}}& =\frac{{a}^{4}\cdot {\left({a}^{2}\right)}^{5}\cdot {b}^{5}}{{a}^{3}\cdot {b}^{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{a}^{4}\cdot {a}^{10}\cdot {b}^{5}}{}{a}^{3}\cdot {b}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{a}^{14}\cdot {b}^{5}}{{a}^{3}\cdot {b}^{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{14-3}\cdot {b}^{5-3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{11}\cdot {b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$