# How do you evaluate ( \frac { 8} { 27} ) ^ { - \frac { 2} { 3} } without using a calculator?

Jun 19, 2017

${\left(\frac{8}{27}\right)}^{- \frac{2}{3}} = \frac{9}{4}$

#### Explanation:

As ${x}^{- 1} = \frac{1}{x}$, ${x}^{\frac{1}{n}} = \sqrt[n]{x}$ and ${x}^{m n} = {\left({x}^{m}\right)}^{n}$

${\left(\frac{8}{27}\right)}^{- \frac{2}{3}}$

= ${\left({\left(\frac{8}{27}\right)}^{- 1}\right)}^{\frac{2}{3}}$

= ${\left(\frac{1}{\frac{8}{27}}\right)}^{\frac{2}{3}}$

= ${\left({\left(\frac{27}{8}\right)}^{\frac{1}{3}}\right)}^{2}$

= ${\left(\frac{\sqrt[3]{27}}{\sqrt[3]{8}}\right)}^{2}$

= ${\left(\frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{2 \times 2 \times 2}}\right)}^{2}$

= ${\left(\frac{3}{2}\right)}^{2}$

= $\frac{9}{4}$

Jun 19, 2017

See a solution process below:

#### Explanation:

First, rewrite this expression as:

${\left(\frac{8}{27}\right)}^{- \frac{2}{3}} \implies {\left(\frac{8}{27}\right)}^{\frac{1}{3} \times - 2}$

We can rewrite this using this rule for exponents:

${x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}} = {\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}}$

${\left(\frac{8}{27}\right)}^{\textcolor{red}{\frac{1}{3}} \times \textcolor{b l u e}{- 2}} \implies {\left({\left(\frac{8}{27}\right)}^{\textcolor{red}{\frac{1}{3}}}\right)}^{\textcolor{b l u e}{- 2}}$

We can now write this in radical form using this rule:

${x}^{\frac{1}{\textcolor{red}{n}}} = \sqrt[\textcolor{red}{n}]{x}$

${\left({\left(\frac{8}{27}\right)}^{\frac{1}{\textcolor{red}{3}}}\right)}^{-} 2 = {\left(\sqrt[\textcolor{red}{3}]{\frac{8}{27}}\right)}^{-} 2$

We can now use this rule for dividing radicals to evaluate the radical:

$\sqrt[n]{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}} = \frac{\sqrt[n]{\textcolor{red}{a}}}{\sqrt[n]{\textcolor{b l u e}{b}}}$

${\left(\sqrt[3]{\frac{\textcolor{red}{8}}{\textcolor{b l u e}{27}}}\right)}^{-} 2 \implies {\left(\frac{\sqrt[3]{\textcolor{red}{8}}}{\sqrt[3]{\textcolor{b l u e}{27}}}\right)}^{-} 2 \implies {\left(\frac{\sqrt[3]{\textcolor{red}{2 \cdot 2 \cdot 2}}}{\sqrt[3]{\textcolor{b l u e}{3 \cdot 3 \cdot 3}}}\right)}^{-} 2 \implies {\left(\frac{2}{3}\right)}^{-} 2$

Yet again, we can rewrite this as:

${\left(\frac{2}{3}\right)}^{-} 2 \implies {\left(2 \times \frac{1}{3}\right)}^{-} 2 \implies {2}^{-} 2 \times \frac{1}{3} ^ - 2$

Now, using these rules of exponents, yes, we can rewrite this again and then evaluate:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$ and $\frac{1}{x} ^ \textcolor{b l u e}{a} = {x}^{\textcolor{b l u e}{- a}}$

${2}^{\textcolor{red}{- 2}} \times \frac{1}{3} ^ \textcolor{b l u e}{- 2} \implies \frac{1}{2} ^ \textcolor{red}{- - 2} \times {3}^{\textcolor{b l u e}{- - 2}} \implies \frac{1}{2} ^ \textcolor{red}{2} \times {3}^{\textcolor{b l u e}{2}} \implies \frac{1}{4} \times 9 \implies$

$\frac{9}{4}$