7.192. $\sqrt{3 \log _{2}^{2} x-1-9 \log _{x}^{2} 2}=5$.
Solution.

Domain of definition: $\left\{\begin{array}{l}3 \log _{2}^{4} x-\log _{2}^{2} x-9 \geq 0, \\ 0<x \neq 1\end{array}\right.$

Square both sides of the equation. Then

$\frac{3 \log _{2}^{4} x-\log _{2}^{2} x-9}{\log _{2}^{2} x}=25 \Leftrightarrow 3 \log _{2}^{4} x-26 \log _{2}^{2} x-9=0$.

Solving this equation as a biquadratic equation in terms of $\log _{2} x$, we find $\left(\log _{2} x\right)_{1}=-3$ and $\left(\log _{2} x\right)_{2}=3$, from which $x_{1}=\frac{1}{8}, x_{2}=8$.

Answer: $\frac{1}{8} ; 8$.