# How do you evaluate and simplify 64^(-2/3)?

Jun 30, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

${64}^{\left(\frac{1}{3} \times - 2\right)}$

Using this rule of exponents we can rewrite this as:

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

${64}^{\textcolor{red}{\frac{1}{3}} \times \textcolor{b l u e}{- 2}} = {\left({64}^{\textcolor{red}{\frac{1}{3}}}\right)}^{\textcolor{b l u e}{- 2}}$

Next, use this rule to again rewrite and evaluate the term within the parenthesis:

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

${\left({64}^{\frac{1}{\textcolor{red}{n}}}\right)}^{-} 2 = {\left(\sqrt[\textcolor{red}{3}]{64}\right)}^{-} 2 = {4}^{-} 2$

Now, use this rule to complete the evaluation:

${4}^{\textcolor{red}{- 2}} = \frac{1}{4} ^ \textcolor{red}{- - 2} = \frac{1}{4} ^ 2 = \frac{1}{16}$

Jul 9, 2017

$= \frac{1}{16}$

#### Explanation:

The laws of indices state.

${x}^{-} m = \frac{1}{x} ^ m$

${x}^{\frac{p}{q}} = {\left(\sqrt[q]{x}\right)}^{p}$

Using these laws gives the following:

${64}^{- \frac{2}{3}} = \frac{1}{64} ^ \left(\frac{2}{3}\right) \text{ } \leftarrow$ make the index positive

$\frac{1}{64} ^ \left(\frac{2}{3}\right) = \frac{1}{\sqrt[3]{64}} ^ 2 \text{ } \leftarrow$ find the cube root

$= \frac{1}{{4}^{2}} \text{ } \leftarrow$ find the square

$= \frac{1}{16}$