How do you multiply (4x ^ { - 2} y ^ { 4} ) ^ { 2} * 3x ^ { - 2} y ^ { 5}(4x−2y4)2⋅3x−2y5?

1 Answer
Aaditya Raghavan
Aug 30, 2017

=48x^-6y^13 = (48y^13)/x^6=48x−6y13=48y13x6

Explanation:

We know that (x^a)^b = x^(a*b)(xa)b=xa⋅b and x^a * x^b = x^(a+b)xa⋅xb=xa+b
Let us evaluate the parentheses first:
(4x^-2y^4)^2 = 4^2*(x^-2)^2 * (y^4)^2 = 16x^-4y^8(4x−2y4)2=42⋅(x−2)2⋅(y4)2=16x−4y8

16x^-4y^8 * 3x^-2y^5 = (16*3) * (x^(-4+(-2))) * (y^(8+5))16x−4y8⋅3x−2y5=(16⋅3)⋅(x−4+(−2))⋅(y8+5)
=48x^-6y^13 = (48y^13)/x^6=48x−6y13=48y13x6, making the negative exponent positive by bringing it to the denominator.

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