Simplify ${(16 x^{4})}^{- \frac{3}{4}}$ $(16x^4)^(-3/4)$ using only positive exponents?

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Answer 1 · 2017-10-28T01:08:17

$\frac{1}{8 x^{3}}$ $1/(8x^3)$

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

There are a few things going on here so let's do them one at a time.

The negative sign in the exponent

$x^{- 1} = \frac{1}{x}$ $x^-1=1/x$

So let's rewrite our expression:

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

Taking the 4th root

Next let's note that $16 = 2^{4}$ $16=2^4$ and so we can write:

$\frac{1}{{(16 x^{4})}^{\frac{3}{4}}} = \frac{1}{{(2^{4} x^{4})}^{\frac{3}{4}}}$ $1/(16x^4)^(3/4)=1/(2^4x^4)^(3/4)$

and from here we can take the 4th root:

$\frac{1}{{(2^{4} x^{4})}^{\frac{3}{4}}} = \frac{1}{{(2 x)}^{3}}$ $1/(2^4x^4)^(3/4)=1/(2x)^3$

Cube the denominator

$\frac{1}{{(2 x)}^{3}} = \frac{1}{8 x^{3}}$ $1/(2x)^3=1/(8x^3)$

Simplify (16x^4)^(-3/4)(16x4)−34(16x^4)^(-3/4) using only positive exponents?