# Simplify (16x^4)^(-3/4) using only positive exponents?

$\frac{1}{8 {x}^{3}}$

#### Explanation:

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

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

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

So let's rewrite our expression:

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

Taking the 4th root

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

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

and from here we can take the 4th root:

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

Cube the denominator

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