# How do you write (-2x)^-4 with positive exponents?

Mar 25, 2018

This is a law of exponents
${a}^{- m} = \frac{1}{a} ^ m$

Then...
${\left(- 2 x\right)}^{-} 4 = \frac{1}{- 2 x} ^ 4$

This is also a law of exponents
${\left(a b\right)}^{m} = {a}^{m} {b}^{m}$

$\frac{1}{- 2 x} ^ 4 = \frac{1}{16 {x}^{4}}$

But.. as you said you wanted it with positive exponent... this step won't be necessary

Mar 25, 2018

It is equal to $\frac{1}{16 {x}^{4}}$.

#### Explanation:

Use these two exponents rules:

${\textcolor{b l u e}{x}}^{-} \textcolor{red}{m} = \frac{1}{\textcolor{b l u e}{x}} ^ \textcolor{red}{m}$

${\left(\textcolor{b l u e}{x} \textcolor{g r e e n}{y}\right)}^{\textcolor{red}{m}} = {\textcolor{b l u e}{x}}^{\textcolor{red}{m}} {\textcolor{g r e e n}{y}}^{\textcolor{red}{m}}$

Here are these two rules applied to our problem:

$\textcolor{w h i t e}{=} {\left(- 2 x\right)}^{-} 4$

$= \frac{1}{- 2 x} ^ 4$

$= \frac{1}{- 2 \cdot x} ^ 4$

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

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

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

$= \frac{1}{1 \cdot {2}^{4} \cdot {x}^{4}}$

$= \frac{1}{1 \cdot 16 \cdot {x}^{4}}$

$= \frac{1}{16 \cdot {x}^{4}}$

$= \frac{1}{16 {x}^{4}}$