7.052. $\log _{0.5}^{2} 4 x+\log _{2} \frac{x^{2}}{8}=8$.
## Solution.

Domain of definition: $x>0$.

We have

$$
\begin{aligned}
& \log _{2}^{2} 4 x+\log _{2} \frac{x^{2}}{8}-8=0, \quad\left(\log _{2} 4+\log _{2} x\right)^{2}+\log _{2} x^{2}-\log _{2} 8-8=0 \\
& \log _{2}^{2} x+6 \log _{2} x-7=0
\end{aligned}
$$

Solving this equation as a quadratic equation in terms of $\log _{2} x$, we find $\left(\log _{2} x\right)_{1}=-7$, from which $x_{1}=2^{-7}=\frac{1}{128}$, or $\left(\log _{2} x\right)_{2}=1$, from which $x_{2}=2$. Answer: $\frac{1}{128} ; 2$.