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How do you write $b^{- (\frac{4}{7})}$ $b^- (4/7)$ in radical form?

1 Answer

Aug 28, 2017

$b^{- (\frac{4}{7})} = \frac{1}{\sqrt[7]{b^{4}}}$ $b^(-(4/7))=color(red)(1/(root(7)(b^4))$

Explanation:

General observations on exponents:
$XXX p^{- q} = \frac{1}{p^{q}}$ $color(white)("XXX")p^(-q)=1/(p^q)$

$XXX p^{\frac{r}{s}} = {(p^{r})}^{\frac{1}{s}} = \sqrt[s]{p^{r}}$ $color(white)("XXX")p^(r/s)=(p^r)^(1/s)=root(s)(p^r)$

Base on the above, we have
$XXX b (- \frac{4}{7}) = \frac{1}{b^{\frac{4}{7}}} = \frac{1}{\sqrt[7]{b^{4}}}$ $color(white)("XXX")b(-4/7)=1/(b^(4/7))=1/(root(7)(b^4))$

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