How do you simplify $81^{- \frac{3}{2}}$ $81^(-3/2)$ ?

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How do you simplify $81^{- \frac{3}{2}}$ $81^(-3/2)$ ?

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Answer 1 · 2016-09-19T06:37:49

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

Explanation:

$81^{- \frac{3}{2}}$ $81^(-3/2)$

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

= $3^{(4 \times - \frac{3}{2})}$ $3^((4xx-3/2))$

= $3^{2 \times - 3}$ $3^(2xx-3)$

= $3^{- 6}$ $3^(-6)$

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

= $\frac{1}{729}$ $1/729$

Answer 2 · 2016-09-19T07:07:05

$\frac{1}{729}$ $1/729$

Explanation:

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

$\times \times \times \times \times \times \times \times x \to x^{\frac{p}{q}} = \sqrt[q]{x^{p}} = {(\sqrt[q]{x})}^{p}$ $color(white)(xxxxxxxxxxxxxxxxx) rarr" " x^(p/q) = rootq (x^p) = (rootq x)^p$

$81^{- \frac{3}{2}} = \frac{1}{{\sqrt{81}}^{3}} \leftarrow$ $81^(-3/2) = 1/(sqrt81^3)" "larr$ the cube of the square root of 81

= $\frac{1}{9^{3}}$ $1/(9^3)$

= $\frac{1}{729}$ $1/729$

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.

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