Chapter 5.2, Problem 26WE

To determine

To find: Simplify the given expression.

Answer to Problem 26WE

  ${\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{h^4}{4k^6}$

Explanation of Solution

Given information:

  ${\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}$

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.

  ${\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}={\left(\frac{2}{h^2\cdot \left(\frac{1}{k^3}\right)}\right)}^{-2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}={\left(\frac{2}{h^2}\cdot \frac{k^3}{1}\right)}^{-2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}={\left(\frac{2\cdot k^3}{h^2\cdot 1}\right)}^{-2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}={\left(\frac{2k^3}{h^2}\right)}^{-2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{1}{{\left(\frac{2k^3}{h^2}\right)}^2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}={\left(\frac{h^2}{2k^3}\right)}^2$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{{\left(h^2\right)}^2}{2^2\cdot {\left(k^3\right)}^2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{{\left(h^2\right)}^2}{4{\left(k^3\right)}^2}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{h^{2\cdot 2}}{4k^{3\cdot 2}}$

  $\Rightarrow {\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{h^4}{4k^6}$

Therefore,

  ${\left(\frac{2}{h^2{k}^{-3}}\right)}^{-2}=\frac{h^4}{4k^6}$