# How to simplify ((3x^-1)/(9y^2))^-2 and have the answer in positive exponential notation?

Sep 12, 2017

${\left(\frac{3 {x}^{-} 1}{9 {y}^{2}}\right)}^{-} 2 = 9 {x}^{2} {y}^{4}$

#### Explanation:

You will need the following index laws

(1) ${\left(\frac{x}{y}\right)}^{a} = {x}^{a} / {y}^{a}$

(2) ${\left(k {x}^{a}\right)}^{b} = {k}^{b} {x}^{a b}$

(3) ${x}^{-} a = \frac{1}{x} ^ a$

(4) $\frac{1}{x} ^ - a = {x}^{a}$

First apply (1)

${\left(\frac{3 {x}^{-} 1}{9 {y}^{2}}\right)}^{-} 2 = {\left(3 {x}^{-} 1\right)}^{-} \frac{2}{9 {y}^{2}} ^ - 2$

Then apply (2)

${\left(3 {x}^{-} 1\right)}^{-} \frac{2}{9 {y}^{2}} ^ - 2 = \frac{{3}^{-} 2 {x}^{2}}{{9}^{-} 2 {y}^{-} 4}$

Finally, apply (3) and (4)

$\frac{{3}^{-} 2 {x}^{2}}{{9}^{-} 2 {y}^{-} 4} = \frac{{9}^{2} {x}^{2} {y}^{4}}{3} ^ 2 = 9 {x}^{2} {y}^{4}$