How do you simplify $16^{- \frac{3}{4}}$ $16^(-3/4)$ ?

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Answer 1 · 2015-04-09T05:23:14

$\frac{1}{8}$ $1/8$

Firstly, you should change the 16 into the simplest base-power form.

$16 = 2^{4}$ $16=2^4$

$16^{- \frac{3}{4}} = {(2^{4})}^{- \frac{3}{4}}$ $16^(-3/4) = (2^4)^(-3/4)$

As we remember, the formula of multiplying powers, where
${(a^{m})}^{n} = a^{m n}$ $(a^m)^n = a^(mn)$

Therefore,
${(2^{4})}^{- \frac{3}{4}} = 2^{- \frac{12}{4}} = 2^{- 3}$ $(2^4)^(-3/4) = 2^(-12/4)=2^-3$

Here we can apply the Indew law of $a^{- n} = \frac{1}{a^{n}}$ $a^-n=1/a^n$
Hence, $2^{- 3} = \frac{1}{2^{3}} = \frac{1}{8}$ $2^-3=1/2^3=1/8$

$16^{- \frac{3}{4}} = \frac{1}{8}$ $16^(-3/4)=1/8$

How do you simplify 16−3416^(-3/4)?