Question

I need to use the power rule (if m and n are real numbers and x doesn't equal 0, then (x^m)^n = x^mn... ) to simplify this fraction: (2x^2 y^-2 / y^5)^-3

Answer 1

`= 2x^(-6)y ^(21)\\\\`

Explanation:

Given the indicinal expression (2x^2 y^-2 / y^5)^-3

In indices

`(x^m)^n = x^{mn`

Applying to solve the question

`(2x^2 y^(-2) / y^5)^(-3)\\\\= (2x^(-6)y^6)/(y^(-15))\\= 2x^(-6) * (y^6)/(y^(-15)) \\= 2x^(-6) * y ^(6-(-15))\\= 2x^(-6) * y ^(6+15)\\= 2x^(-6) * y ^(21)\\= 2x^(-6)y ^(21)\\\\`