7.248. $\log _{2} x \cdot \log _{3} x=\log _{3}\left(x^{3}\right)+\log _{2}\left(x^{2}\right)-6$.
## Solution.

Domain of definition: $x>0$.

Let's switch to base 2. We have

$$
\begin{aligned}
& \frac{\log _{2} x \cdot \log _{2} x}{\log _{2} 3}=\frac{3 \log _{2} x}{\log _{2} 3}+2 \log _{2} x-6 \Leftrightarrow \\
& \Leftrightarrow \log _{2}^{2} x-\left(3+2 \log _{2} 3\right) \log _{2} x+6 \log _{2} 3=0
\end{aligned}
$$

Solving this equation as a quadratic equation in terms of $\log _{2} x$, we get $\log _{2} x=\log _{2} 9$ or $\log _{2} x=3$, from which $x_{1}=9, x_{2}=8$.

Answer: $8 ; 9$.