7.232. $\log _{x^{2}} 16+\log _{2 x} 64=3$.
Solution.

Domain of definition: $\left\{\begin{array}{l}0<x \neq \frac{1}{2}, \\ x \neq 1 .\end{array}\right.$

Transition to base 2. Then

$$
\begin{aligned}
& \frac{\log _{2} 16}{\log _{2} x^{2}}+\frac{\log _{2} 64}{\log _{2} 2 x}=3 \Leftrightarrow \frac{4}{2 \log _{2} x}+\frac{6}{1+\log _{2} x}-3=0 \Leftrightarrow \\
& \Leftrightarrow 3 \log _{2}^{2} x-5 \log _{2} x-2=0, \text { where } \log _{2} x \neq 0 \text { and } \log _{2} x \neq-1
\end{aligned}
$$

Solving this equation as a quadratic in terms of $\log _{2} x$, we get $\log _{2} x=-\frac{1}{3}, x_{1}=\frac{1}{\sqrt[3]{2}}=0.5 \sqrt[3]{4} ; \log _{2} x=2, x_{2}=4$.

Answer: $0.5 \sqrt[3]{4} ; 4$.