Answer
$-2$
Work Step by Step
Using $\log_bx^m=m\log_bx$ or the Power Rule of Logarithms, the given expression, $
\log_{8}\dfrac{1}{64}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_{8}64^{-1}
\\\\=
\log_{8}(8^2)^{-1}
\\\\=
\log_{8}8^{2(-1)}
\\\\=
\log_{8}8^{-2}
\\\\=
-2\log_{8}8
.\end{array}
Since the $\log_bb=1,$ the expression, $
-2\log_{8}8
,$ simplifies to
\begin{array}{l}\require{cancel}
-2(1)
\\\\=
-2
.\end{array}