Chapter 16.7, Problem 1MRE

To determine

To find: ${125}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Answer to Problem 1MRE

  $\frac{1}{25}$

Explanation of Solution

Given: ${125}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Formula used:

  $a^{-b}=\frac{1}{a^b}$

  $a^{bc}={(a^c)}^b$

Using above formula:

  $\begin{array}{l}{125}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{{125}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{1}{{\left({125}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^2}\\ \Rightarrow \frac{1}{{(5)}^2}=\frac{1}{25}\end{array}$