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

Answer:

$= 2x^{-6}y ^{21}$

Step-by-step explanation:

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

In indices

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

Applying to solve the question

$(2x^2 y^{-2} / y^5)^{-3}\\\\= \frac{2x^{-6}y^6}{y^{-15}}\\= 2x^{-6} * \frac{y^6}{y^{-15}} \\= 2x^{-6} * y ^{6-(-15)}\\= 2x^{-6} * y ^{6+15}\\= 2x^{-6} * y ^{21}\\= 2x^{-6}y ^{21}$