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

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2 Answers
Steve M
Oct 20, 2016

# (x^2/y^3)^2 # or # x^4/y^6 #

Explanation:

Use #A^-1 = 1/A#

So # (x^-2/y^-3)^-2 = (y^-3/x^-2)^2 #
# :. (x^-2/y^-3)^-2 = ((1/y^3)/(1/x^2))^2 #
# :. (x^-2/y^-3)^-2 = ((1/y^3)x^2)^2 #
# :. (x^-2/y^-3)^-2 = (x^2/y^3)^2 # or # x^4/y^6 #

EZ as pi
Oct 20, 2016

There are several laws of indices.

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

#(xy)^m= x^my^m" and "(x^2 y^3)^4 = x^8y^12#

#(x^-2/y^-3)^-2" "larr# neg x neg = pos

=#x^4/y^6#

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OR: Recall: two other laws of indices

#x^-m = 1/x^m" and "(x^m/y^n)^color(red)(-p) = (y^n/x^m)^color(red)((+p))#

#(x^-2/y^-3)^color(red)(-2) = (y^-3/x^-2)^color(red)(+2)#

=#(x^2/y^3)^2#

=#x^4/y^6#

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