Question

How do you multiply `(4x ^ { - 2} y ^ { 4} ) ^ { 2} * 3x ^ { - 2} y ^ { 5}`?

Answer 1

`=48x^-6y^13 = (48y^13)/x^6`

Explanation:

We know that `(x^a)^b = x^(a*b)` and `x^a * x^b = x^(a+b)`
Let us evaluate the parentheses first:
`(4x^-2y^4)^2 = 4^2*(x^-2)^2 * (y^4)^2 = 16x^-4y^8`

`16x^-4y^8 * 3x^-2y^5 = (16*3) * (x^(-4+(-2))) * (y^(8+5))`
`=48x^-6y^13 = (48y^13)/x^6`, making the negative exponent positive by bringing it to the denominator.