Answer:
$$
-\frac{1}{4}
$$

Let $A$ and $B$ be these sums, respectively. Then
$$
\begin{aligned}
B-A & =\log _{2}\left(\frac{\log _{16} x}{\log _{4} x}\right)+\log _{4}\left(\frac{\log _{2} x}{\log _{16} x}\right)+\log _{16}\left(\frac{\log _{4} x}{\log _{2} x}\right) \\
& =\log _{2}\left(\log _{16} 4\right)+\log _{4}\left(\log _{2} 16\right)+\log _{16}\left(\log _{4} 2\right) \\
& =\log _{2}\left(\frac{1}{2}\right)+\log _{4} 4+\log _{16}\left(\frac{1}{2}\right) \\
& =(-1)+1+\left(-\frac{1}{4}\right) \\
& =-\frac{1}{4} .
\end{aligned}
$$

Since $A=0$, we have the answer $B=-\frac{1}{4}$.
8. [6] Suppose a real number $x>1$ satisfies
$$
\log _{2}\left(\log _{4} x\right)+\log _{4}\left(\log _{16} x\right)+\log _{16}\left(\log _{2} x\right)=0 \text {. }
$$

Compute
$$
\log _{2}\left(\log _{16} x\right)+\log _{16}\left(\log _{4} x\right)+\log _{4}\left(\log _{2} x\right) .
$$