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

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Answer 1 · 2016-09-19T06:37:49

$81^(-3/2)=1/729$

$81^(-3/2)$

= $(3^4)^(-3/2)$

= $3^((4xx-3/2))$

= $3^(2xx-3)$

= $3^(-6)$

= $1/3^6$

= $1/729$

Answer 2 · 2016-09-19T07:07:05

$1/729$

Recall: 2 laws of indices $" " rarr " "x^-m = 1/x^m$

$color(white)(xxxxxxxxxxxxxxxxx) rarr" " x^(p/q) = rootq (x^p) = (rootq x)^p$

$81^(-3/2) = 1/(sqrt81^3)" "larr$ the cube of the square root of 81

= $1/(9^3)$

= $1/729$

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.

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