Chapter 11, Problem 5GR

To determine

Simplify: $\frac{m^2{p}^{-3}{q}^4}{m^2{p}^2{q}^{-2}}$

Answer to Problem 5GR

  $\frac{m^2{p}^{-3}{q}^4}{m^2{p}^2{q}^{-2}}=\frac{q^5}{p^5}$

Explanation of Solution

Given:

Simplify: $\frac{m^2{p}^{-3}{q}^4}{m^2{p}^2{q}^{-2}}$

Concept Used:

Use the exponent rules:

  $\text{Negative}\text{exponent}\text{rule}:a^{-n}=\frac{1}{a^n}or\frac{1}{a^{-n}}=a^n$

  $\text{Product}\text{rule}=a^m·a^n=a^{m+n}$

Calculation:

Use the exponent rules:

  $\begin{array}{l}\frac{m^2{p}^{-3}{q}^4}{m^2{p}^2{q}^{-2}}=\frac{m^2·q^4·q^2}{m^2·p^2·p^3}[a^{-n}=\frac{1}{a^n}or\frac{1}{a^{-n}}=a^n]\\ \frac{m^2·q^{4+2}}{m^2·p^{2+3}}=\frac{m^2·q^5}{m^2·p^5}=\frac{m^2}{m^2}·\frac{q^5}{p^5}=\frac{q^5}{p^5}[a^m·a^n=a^{m+n}]\end{array}$

Thus, $\frac{m^2{p}^{-3}{q}^4}{m^2{p}^2{q}^{-2}}=\frac{q^5}{p^5}$