HMMT-Feb 2018 guts, Problem 8

Suppose a real number $x>1$ satisfies \[ \log _{2}\left(\log _{4} x\right)+\log _{4}\left(\log _{16} x\right)+\log _{16}\left(\log _{2} x\right)=0 \] Compute \[ \log _{2}\left(\log _{16} x\right)+\log _{16}\left(\log _{4} x\right)+\log _{4}\left(\log _{2} x\right) \]

Solution: 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 $B=-\frac{1}{4}$. $\fbox{-\frac{1}{4}}$.

Answer: -\frac{1}{4}