# How do you simplify ((5x^2)/x^-4)^-2?

Sep 6, 2016

$\frac{1}{25 {x}^{12}}$

#### Explanation:

We have: ${\left(\frac{5 {x}^{2}}{{x}^{- 4}}\right)}^{- 2}$

Using the laws of exponents:

$= \left(\frac{{\left(5 {x}^{2}\right)}^{- 2}}{{\left({x}^{- 4}\right)}^{- 2}}\right)$

$= \frac{{5}^{- 2} \cdot {x}^{- 4}}{{x}^{8}}$

$= {5}^{- 2} \cdot {x}^{- 4 - 8}$

$= {5}^{- 2} \cdot {x}^{- 12}$

$= \frac{1}{{5}^{2} \cdot {x}^{12}}$

$= \frac{1}{25 {x}^{12}}$

Sep 6, 2016

$\frac{1}{25 {x}^{12}}$

#### Explanation:

Recall: The law of indices : ${\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m}$

This shows that a fraction raised to a negative index can be inverted and the index becomes positive.

I prefer to sort out the negative indices first.

${\left(\frac{5 {x}^{2}}{x} ^ - 4\right)}^{\textcolor{red}{- 2}} = {\left({x}^{-} \frac{4}{5 {x}^{2}}\right)}^{\textcolor{red}{2}}$

=${\left(\frac{\textcolor{b l u e}{{x}^{-} 4}}{5 {x}^{2}}\right)}^{2} = {\left(\frac{1}{5 {x}^{2} \times \textcolor{b l u e}{{x}^{4}}}\right)}^{2}$

=${\left(\frac{1}{5 {x}^{6}}\right)}^{2}$

=$\frac{1}{25 {x}^{12}}$