HMMT-Feb 2020 alg, Problem 3

Let $a=256$. Find the unique real number $x>a^{2}$ such that \[ \log _{a} \log _{a} \log _{a} x=\log _{a^{2}} \log _{a^{2}} \log _{a^{2}} x \]

Solution: Solution: Let $y=\log _{a} x$ so $\log _{a} \log _{a} y=\log _{a^{2}} \log _{a^{2}} \frac{1}{2} y$. Setting $z=\log _{a} y$, we find $\log _{a} z=$ $\log _{a^{2}}\left(\frac{1}{2} z-\frac{1}{16}\right)$, or $z^{2}-\frac{1}{2} z+\frac{1}{16}=0$. Thus, we have $z=\frac{1}{4}$, so we can backsolve to get $y=4$ and $x=2^{32}$. $\fbox{2^{32}}$.

Answer: 2^{32}