Chapter 8.3, Problem 6CYU

To determine

To Evaluate: The expression.

Answer to Problem 6CYU

  ${\log }_2\frac{1}{128}=-7$

Explanation of Solution

Given: ${\log }_2\frac{1}{128}$

Formula used:

  $\begin{array}{l}{\log }_{b}x=yifandonlyif\\ b^y=x\end{array}$

Given: ${\log }_2\frac{1}{128}$

Let the expression is equals to $y$

Then

Using the logarithm form

  ${\log }_{b}x=yif{b}^y=x$

  $\begin{array}{l}{\log }_2\frac{1}{128}=y\\ b=2,x=\frac{1}{128}\\ then\\ 2^y=\frac{1}{128}\\ 2^y=\frac{1}{2^7}\\ 2^y=2^{-7}\\ y=-7\\ i.e.\\ {\log }_2\frac{1}{128}=-7\end{array}$

Hence, ${\log }_2\frac{1}{128}=-7$