Chapter 5.3, Problem 11MRE

To determine

To Find: Simply the given expression.

Answer to Problem 11MRE

  $\left(6m^2\right){\left(2m\right)}^{-2}=\frac{3}{2}$

Explanation of Solution

Given information:

  $\left(6m^2\right){\left(2m\right)}^{-2}$

Concept:

  $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

Using law of exponents

  $\begin{array}{l}{a}^{-m}=\frac{1}{a^m}\\ a^m\cdot a^n=a^{m+n}\\ \frac{a^m}{a^n}=a^{m-n}\\ a^0=1\end{array}$

  ${\left(a^n\right)}^m=a^{mn}$

Where $a$ be any value.

  $\left(6m^2\right){\left(2m\right)}^{-2}$

  $\begin{array}{l}\Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=6m^2\cdot 2^{-2}\cdot m^{-2}\\ \Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=6m^2\cdot \frac{1}{2^2}\cdot \frac{1}{m^2}\\ \Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=\frac{2\cdot 3m^2}{2^2{m}^2}\\ \Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=3\cdot 2^{1-2}\cdot m^{2-2}\\ \Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=3\cdot 2^{-1}\cdot m^0\\ \Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=3\cdot \frac{1}{2}\cdot 1\\ \Rightarrow \left(6m^2\right){\left(2m\right)}^{-2}=\frac{3}{2}\end{array}$

Therefore,

  $\left(6m^2\right){\left(2m\right)}^{-2}=\frac{3}{2}$