# How do you write (x^-2/y^-3)^-2using only positive exponents?

Oct 20, 2016

${\left({x}^{2} / {y}^{3}\right)}^{2}$ or ${x}^{4} / {y}^{6}$

#### Explanation:

Use ${A}^{-} 1 = \frac{1}{A}$

So ${\left({x}^{-} \frac{2}{y} ^ - 3\right)}^{-} 2 = {\left({y}^{-} \frac{3}{x} ^ - 2\right)}^{2}$
$\therefore {\left({x}^{-} \frac{2}{y} ^ - 3\right)}^{-} 2 = {\left(\frac{\frac{1}{y} ^ 3}{\frac{1}{x} ^ 2}\right)}^{2}$
$\therefore {\left({x}^{-} \frac{2}{y} ^ - 3\right)}^{-} 2 = {\left(\left(\frac{1}{y} ^ 3\right) {x}^{2}\right)}^{2}$
$\therefore {\left({x}^{-} \frac{2}{y} ^ - 3\right)}^{-} 2 = {\left({x}^{2} / {y}^{3}\right)}^{2}$ or ${x}^{4} / {y}^{6}$

Oct 20, 2016

There are several laws of indices.

The ones which I think will be the easiest to apply here are:

${\left(x y\right)}^{m} = {x}^{m} {y}^{m} \text{ and } {\left({x}^{2} {y}^{3}\right)}^{4} = {x}^{8} {y}^{12}$

${\left({x}^{-} \frac{2}{y} ^ - 3\right)}^{-} 2 \text{ } \leftarrow$ neg x neg = pos

=${x}^{4} / {y}^{6}$

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OR: Recall: two other laws of indices

${x}^{-} m = \frac{1}{x} ^ m \text{ and } {\left({x}^{m} / {y}^{n}\right)}^{\textcolor{red}{- p}} = {\left({y}^{n} / {x}^{m}\right)}^{\textcolor{red}{\left(+ p\right)}}$

${\left({x}^{-} \frac{2}{y} ^ - 3\right)}^{\textcolor{red}{- 2}} = {\left({y}^{-} \frac{3}{x} ^ - 2\right)}^{\textcolor{red}{+ 2}}$

=${\left({x}^{2} / {y}^{3}\right)}^{2}$

=${x}^{4} / {y}^{6}$