Answer
Fill the blank entries with
$\left[\begin{array}{ll}
& 8^{1}=8 \\
& 8^{2}=64\\
\log_{8}4=2/3 & \\
\log_{8}512=3 & \\
& 8^{-1}=\dfrac{1}{8}\\
\log_{8}\dfrac{1}{64}=-2 &
\end{array}\right]$
Work Step by Step
By definition, $\log_{a}x=y \Leftrightarrow a^{y}=x$
($\log_{a}x$ is the exponent to which the base $a$ must be raised to give $x$.)
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$\log_{8}8=1 \Leftrightarrow 8^{1}=8$
Since $64=8^{2},\qquad \log_{8}64=2 \Leftrightarrow 8^{2}=64$
$8^{2/3}=4\Leftrightarrow \log_{8}4=2/3$
$8^{3}=512\Leftrightarrow \log_{8}512=3$
Since $\displaystyle \frac{1}{8}=8^{-1},\qquad \log_{8}\frac{1}{8}=-1 \Leftrightarrow 8^{-1}=\displaystyle \frac{1}{8}$
$8^{-2}=\displaystyle \frac{1}{64}\Leftrightarrow \log_{8}\frac{1}{64}=-2$
