7.231. $\log _{4} \log _{2} x+\log _{2} \log _{4} x=2$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\log _{2} x>0, \\ \log _{4} x>0,\end{array} \Leftrightarrow x>1\right.$.

Switch to base 2. We have

$$
\begin{aligned}
& \frac{1}{2} \log _{2} \log _{2} x+\log _{2}\left(\frac{1}{2} \log _{2} x\right)=2 \Leftrightarrow \\
& \Leftrightarrow \log _{2} \log _{2} x+2 \log _{2}\left(\frac{1}{2} \log _{2} x\right)=4 \Leftrightarrow \\
& \Leftrightarrow \log _{2} \log _{2} x+\log _{2}\left(\frac{1}{4} \log _{2}^{2} x\right)=4 \Leftrightarrow \log _{2}\left(\log _{2} x \cdot \frac{1}{4} \log _{2}^{2} x\right)=4 \Leftrightarrow \\
& \Leftrightarrow \frac{1}{4} \log _{2}^{3} x=16, \log _{2}^{3} x=64
\end{aligned}
$$

Then $\log _{2} x=4, x=2^{4}=16$.

Answer: 16.