Chapter 4.3, Problem 5E

To determine

To find: The table for the appropriate logarithm or the exponential form of the equation.

Answer to Problem 5E

The completed table is shown in Figure 1

Explanation of Solution

Consider the expression for the log with the same base is,

  ${\log }_{a}a=1$

Then,

  ${\log }_{8}8=1$

Consider the expression for,

  $\begin{array}{c}{\log }_{8}64={\log }_8{8}^2\\ =2{\log }_{8}8\end{array}$

Then,

  $\begin{array}{c}{\log }_{8}64=2{\log }_{8}8\\ =2\end{array}$

Consider the given exponential form is,

  $8^{\frac{2}{3}}=4$

Then, take log on both side of the equation.

  $\begin{array}{l}{\log }_8{8}^{\frac{2}{3}}={\log }_{8}4\\ \frac{2}{3}{\log }_{8}8={\log }_{8}4\\ \frac{2}{3}={\log }_{8}4\\ {\log }_{8}4=\frac{2}{3}\end{array}$

Consider the given exponential form is,

  $8^3=512$

Then, take log on both side of the equation.

  $\begin{array}{l}{\log }_8{8}^3={\log }_{8}512\\ 3{\log }_{8}8={\log }_{8}512\\ 3={\log }_{8}512\\ {\log }_{8}512=3\end{array}$

Consider the expression for the given log equation,

  $\begin{array}{c}{\log }_8\left(\frac{1}{8}\right)={\log }_8{8}^{-1}\\ =-{\log }_{8}8\\ =-1\end{array}$

Consider the given exponential form is,

  $8^{-2}=\frac{1}{64}$

Then, take log on both side of the equation.

  $\begin{array}{l}{\log }_8{8}^{-2}={\log }_8\frac{1}{64}\\ -2{\log }_{8}8={\log }_8\left(\frac{1}{64}\right)\\ {\log }_8\left(\frac{1}{64}\right)=-2\end{array}$

The complete log table is shown in Table 1

Table 1

${\log }_{8}8=1$	$8^{\prime }=8$
${\log }_{8}64=2$	$8^2=64$
${\log }_{8}4=\frac{2}{3}$	$8^{\frac{2}{3}}=4$
${\log }_{8}512=3$	$8^{-1}=\frac{1}{8}$
${\log }_8\frac{1}{8}=-1$	$8^{-1}=\frac{1}{8}$
${\log }_8\left(\frac{1}{64}\right)=-2$	$8^{-2}=\frac{1}{64}$

Thus, the required table is shown in Figure 1