How do you simplify $32^{- \frac{2}{5}}$ $32^(-2/5)$ ?

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How do you simplify $32^{- \frac{2}{5}}$ $32^(-2/5)$ ?

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Answer 1 · 2017-01-26T19:46:13

$\frac{1}{4}$ $1/4$

Explanation:

Recall that:

$a^{- b} = \frac{1}{a^{b}}$ $a^-b = 1/a^b$

and

$a^{\frac{b}{c}} = \sqrt[c]{a^{b}}$ $a^(b/c) = root(c)(a^b)$

combined, it is $a^{- \frac{b}{c}} = \frac{1}{\sqrt[c]{a^{b}}}$ $a^(-b/c) = 1/(root(c)(a^b))$

The above apply where defined.

Therefore, $32^{- \frac{2}{5}} = \frac{1}{\sqrt[5]{32^{2}}}$ $32^(-2/5) = 1/(root(5)(32^2)$

$= \frac{1}{{(\sqrt[5]{32})}^{2}}$ $=1/((root(5)(32))^2$ $= \frac{1}{2^{2}} = \frac{1}{4}$ $=1/2^2 = 1/4$ , because $2^{5} = 32$ $color(red)2^5 = 32$ .

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How do you simplify $32^{- \frac{2}{5}}$ $32^(-2/5)$ ?

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