How to simplify ${(\frac{3 x^{- 1}}{9 y^{2}})}^{- 2}$ $((3x^-1)/(9y^2))^-2$ and have the answer in positive exponential notation?

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Answer 1 · 2017-09-12T08:16:04

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

You will need the following index laws

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

(2) ${(k x^{a})}^{b} = k^{b} x^{a b}$ $(kx^a)^b=k^bx^(ab)$

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

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

First apply (1)

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

Then apply (2)

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

Finally, apply (3) and (4)

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

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