# How do you simplify 81^(-3/2)?

Sep 19, 2016

${81}^{- \frac{3}{2}} = \frac{1}{729}$

#### Explanation:

${81}^{- \frac{3}{2}}$

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

= ${3}^{\left(4 \times - \frac{3}{2}\right)}$

= ${3}^{2 \times - 3}$

= ${3}^{- 6}$

= $\frac{1}{3} ^ 6$

= $\frac{1}{729}$

Sep 19, 2016

$\frac{1}{729}$

#### Explanation:

Recall: 2 laws of indices $\text{ " rarr " } {x}^{-} m = \frac{1}{x} ^ m$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} \rightarrow \text{ } {x}^{\frac{p}{q}} = \sqrt[q]{{x}^{p}} = {\left(\sqrt[q]{x}\right)}^{p}$

${81}^{- \frac{3}{2}} = \frac{1}{{\sqrt{81}}^{3}} \text{ } \leftarrow$ the cube of the square root of 81

=$\frac{1}{{9}^{3}}$

=$\frac{1}{729}$

Note that 81 is both a square number and a 4th power.
However, the square root can be obtained immediately.

In work on indices, usually only the principal (+) square root is considered.