[Algebra]
Problem: $\quad$ Compute all real values of $x$ such that $\log _{2}\left(\log _{2} x\right)=\log _{4}\left(\log _{4} x\right)$.

Solution: If $y=\log _{a}\left(\log _{a} x\right)$, then $a^{a^{y}}=x$. Let $y=\log _{2}\left(\log _{2} x\right)=\log _{4}\left(\log _{4} x\right)$. Then $2^{2^{y}}=4^{4^{y}}=$ $\left(2^{2}\right)^{\left(2^{2}\right)^{y}}=2^{2^{2 y+1}}$, so $2 y+1=y, y=-1$, and $x=\sqrt{\mathbf{2}}$. (This problem is based on one submitted by ARML alum James Albrecht, 1986-2007.)
Raise 4 (or $2^{2}$ ) to the power of both sides to get $\left(\log _{2} x\right)^{2}=\log _{4} x$. By the change of base formula, $\frac{(\log x)^{2}}{(\log 2)^{2}}=\frac{\log x}{2 \log 2}$, so $\log x=\frac{\log 2}{2}$, thus $x=2^{1 / 2}=\sqrt{\mathbf{2}}$.
Let $x=4^{a}$. The equation then becomes $\log _{2}(2 a)=\log _{4} a$. Raising 4 to the power of each side, we get $4 a^{2}=a$. Since $a \neq 0$, we get $4 a=1$, thus $a=\frac{1}{4}$ and $x=\sqrt{2}$.

Final answer: $\sqrt{2}$