SOLUTION: I need help solving (x^2 * y^3) * (x^-2 * y^-2), with ^ representing to the power to (Exponent)

Question 1007354: I need help solving (x^2 * y^3) * (x^-2 * y^-2), with ^ representing to the power to (Exponent)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

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