Chapter 7.3, Problem 14MRE

To determine

To Express:

In simplest form without negative exponents.

  ${\left(\frac{x^2}{y^3}\right)}^{-2}$

Answer to Problem 14MRE

  $\frac{y^6}{x^4}$

Explanation of Solution

Given:

The expression is :

  ${\left(\frac{x^2}{y^3}\right)}^{-2}$

Formula Used:

  $x^{-n}=\frac{1}{x^n}$

Calculation:

Given:

  $\begin{array}{l}{\left(\frac{x^2}{y^3}\right)}^{-2}\\ ={\left(\frac{y^3}{x^2}\right)}^2\mathrm{.................}[x^{-n}=\frac{1}{x^n}]\\ =\frac{y^6}{x^4}\mathrm{.....................}[{\left(x^m\right)}^n=x^{mn}]\end{array}$

So,

  $\frac{y^6}{x^4}$ is the solution.