Chapter 7.3, Problem 44P

Complete each equation.

${\left(n^{-8}\right)}^{\square }=n$

To determine

To complete the equation ${\left(n^{-8}\right)}^{\left[\right]}=n$

Answer to Problem 44P

  ${\left(n^{-8}\right)}^{\left[-\frac{1}{8}\right]}=n$

Explanation of Solution

Given information :

Given equation ${\left(n^{-8}\right)}^{\left[\right]}=n$

As we know that $\left(a^m\right)\left(a^n\right)=a^{m+n}\mathrm{........}\text{(i)},{\left(a^m\right)}^n=a^{m\times n}and\left(a^m\right)=\left(a^n\right)\Rightarrow m=n\mathrm{........}(iii)$

Let $x$ be the unknown value .

Now consider ${\left(n^{-8}\right)}^{\left[x\right]}=n$

  $\begin{array}{l}\Rightarrow \left(n^{-8x}\right)=n^1\text{from}\text{rule}\text{(ii)},\text{sinc}\text{e}{n}^1=n\\ \Rightarrow -8x\text{=}1\text{from}\text{rule}\text{(iii)}\\ \therefore \text{x=}-\frac{1}{8}\end{array}$

Hence, ${\left(n^{-8}\right)}^{\left[-\frac{1}{8}\right]}=n$