Quick algebra question!!

simplify this expression using only positive exponents. (here, x>0 and y>0)
((9y^-2)/(x^2))^(-1/2)
the 9y^-2 is above the x^2 and the -1/2 is the exponent outside of the parenthesis of the expression. will vote best answer!!

first, this can be rewritten as (9/x^2*y^2)^-1/2 because the negative y exponent is the same as the inverted term (y^-2=1/y^2)
Second, the same inversion preformed on the y term can be done with the entire quantity because of the -1/2 exponent, leading to (x^2*y^2/9)^1/2
Third, the 1/2 exponent is the same as taking the square root of the quantity, and that can be done to each individual term in the quantity. This leads to a final answer of x*y/3

Well to change to positive exponents just flip them to the opposite side of the fraction bar.

((x^2)^1/2)/ ((9y^2))

For instance 5^-2 is the same as 1/ 5^2.

Two days of seeing this:
((9y)^(-2))/(x^2))^(-1/2) = ((9y)^((-2)*(-1/2))) / (x^(2*(-1/2))) = ((9y)*(2/2)) / (x*(-2/2)) = 9yx

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

First, 9y^(-2) = 9/y^2, so this can be rewritten as

9 / x^2 y^2) ^ (-1/2)

(3 / xy) ^ -1

xy / 3

(xy)/3