# Examples

Here are two examples where you need to use most of the rules you have learned in the previous lessons. Take the time to really understand these examples and you’ll be on your way to being a master of powers!

Example 1

Simplify $\frac{\sqrt[3]{{a}^{2}}\cdot \sqrt{a}\cdot {b}^{2}\cdot {b}^{3}}{\sqrt[6]{{a}^{7}}}\cdot {b}^{-4}$

You use the rules for roots, which gives you $\sqrt[3]{{a}^{2}}={a}^{\frac{2}{3}}$, $\sqrt{a}={a}^{\frac{1}{2}}$, and $\sqrt[6]{{a}^{7}}={a}^{\frac{7}{6}}$. You can also see that ${b}^{-4}=\frac{1}{{b}^{4}}$.

Put together, these become

 $\frac{{a}^{\frac{2}{3}}\cdot {a}^{\frac{1}{2}}\cdot {b}^{2}\cdot {b}^{3}}{{a}^{\frac{7}{6}}}\cdot \frac{1}{{b}^{4}}$

You use the rules for multiplication and get

 $\frac{{a}^{\frac{2}{3}+\frac{1}{2}}\cdot {b}^{2+3}}{{a}^{\frac{7}{6}}}\cdot \frac{1}{{b}^{4}}$

You write $\frac{2}{3}+\frac{1}{2}$ as a single fraction:

$\begin{array}{llll}\hfill \frac{2}{3}+\frac{1}{2}& =\frac{2\cdot 2}{3\cdot 2}+\frac{1\cdot 3}{2\cdot 3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{4}{6}+\frac{3}{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7}{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill \frac{2}{3}+\frac{1}{2}=\frac{2\cdot 2}{3\cdot 2}+\frac{1\cdot 3}{2\cdot 3}=\frac{4}{6}+\frac{3}{6}=\frac{7}{6}& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

That gives you
 $\frac{{a}^{\frac{7}{6}}\cdot {b}^{5}}{{a}^{\frac{7}{6}}\cdot {b}^{4}}=\frac{\text{a76}\cdot {b}^{5}}{\text{a76}\cdot {b}^{4}}={b}^{5-4}=b$

Note! ${b}^{1}=b$.

Example 2

Simplify $\frac{{2}^{7}\cdot \sqrt[3]{2}}{\phantom{\rule{-0.17em}{0ex}}{\left(\sqrt[3]{{2}^{5}}\right)}^{2}}$

Using the power rules you arrive at the following: $\begin{array}{llll}\hfill \frac{{2}^{7}\cdot \sqrt[3]{2}}{\phantom{\rule{-0.17em}{0ex}}{\left(\sqrt[3]{{2}^{5}}\right)}^{2}}& =\frac{{2}^{7+\frac{1}{3}}}{{2}^{\frac{5}{3}\cdot 2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{2}^{\frac{21}{3}+\frac{1}{3}}}{{2}^{\frac{10}{3}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{\frac{22}{3}-\frac{10}{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{\frac{12}{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$