Question

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

Answer 1

`(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`

Answer 2

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`