Question 1007354
the problem as you show it is:


(x^2 * y^3) * (x^-2 * y^-2)


since they're all multiplications, you can remove the parentheses to get:


x^2 * y^3 * x^-2 * y^-2


you can use the associative law to regroup them as shown below:


(x^2 * x^-2) * (y^3 * y^-2)


x^2 * x^-2 = x^(2-2) = x^0


y^3 * y^-2 = y^(3-2) = y^1 = y


your answer is y.


the exponent property that allows this is"


x^a * x^b = x^(a+b)


when a = 2 and b = -2, this becomes:


x^2 * x^-2 = x^(2 + (-2)) which becomes x^(2-2) which becomes x^0 which is equal to 1.


similarly, y^a + y^b = y^(a+b)


when a = 3 and b = -2, this becomes:


y^3 * y^-2) = y^(3 + (-2)) which becomes y^(3-2) which becomes y^1 which is equal to y.


another exponent property that you could have used is:


x^-a = 1/x^2.


also x^a / x^b = x^(a-b).


using this property, you get x^-2 = 1/x^2 and y^-2 = 1/y^2


your equation of (x^2 * y^3) * (x^-2 * y^-2) becomes:


(x^2 * y^3) * (1/x^2 * 1/y^2).


re-associate as before to get:


(x^2 * 1/x^2) * (y^3 * 1/y^2) which becomes:


x^(2-2) * y^(3-2) which becomes x^0 * y^1 which becomes y.


you get the same answer either way.