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

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Answer 1 · 2017-09-12T08:16:04

#((3x^-1)/(9y^2))^-2=9x^2y^4#

You will need the following index laws

(1) #(x/y)^a=x^a/y^a#

(2) #(kx^a)^b=k^bx^(ab)#

(3) #x^-a=1/x^a#

(4) #1/x^-a=x^a#

First apply (1)

#((3x^-1)/(9y^2))^-2=(3x^-1)^-2/(9y^2)^-2#

Then apply (2)

#(3x^-1)^-2/(9y^2)^-2=(3^-2x^2)/(9^-2y^-4)#

Finally, apply (3) and (4)

#(3^-2x^2)/(9^-2y^-4)=(9^2x^2y^4)/3^2=9x^2y^4#