How do you write (x^-2/y^-3)^-2(x2y3)2using only positive exponents?

2 Answers
Oct 20, 2016

(x^2/y^3)^2 (x2y3)2 or x^4/y^6 x4y6

Explanation:

Use A^-1 = 1/AA1=1A

So (x^-2/y^-3)^-2 = (y^-3/x^-2)^2 (x2y3)2=(y3x2)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

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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