Answer
Fill the blank entries with
$\left[\begin{array}{ll}
\log_{4}64=3 & \\
& 4^{1/2}=2\\
\log_{4}8=3/2 & \\
& 4^{-2}=16\\
& 4^{-1/2}=\frac{1}{2}\\
\log_{4}\frac{1}{32}=-\frac{5}{2} &
\end{array}\right]$
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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$\log_{4}64=3 \Leftrightarrow 4^{3}=64$
$\log_{4}2=1/2 \Leftrightarrow 4^{1/2}=2$
$\displaystyle \log_{4}\frac{1}{16}=-2 \Leftrightarrow 4^{-2}=16$
$\displaystyle \log_{4}\frac{1}{2}=-\frac{1}{2} \Leftrightarrow 4^{-1/2}=\displaystyle \frac{1}{2}$
$\displaystyle \log_{4}\frac{1}{32}=-\frac{5}{2} \Leftrightarrow 4^{-5/2}=\displaystyle \frac{1}{32}$
