Answer
$0$
Work Step by Step
Using $\log_b b=1$ and $\log_bx^y=y\log_bx$, the given expression, $
\log_7(\log_4 (\log_216))
,$ simplifies to
\begin{array}{l}\require{cancel}
\log_7(\log_4 (\log_22^4))
\\\\=
\log_7(\log_4 (4\log_22))
\\\\=
\log_7(\log_4 (4\cdot1))
\\\\=
\log_7(\log_4 4)
\\\\=
\log_7(1)
\\\\=
\log_71
\\\\=
0
.\end{array}
