7.233. $\left(3 \log _{a} x-2\right) \log _{x}^{2} a=\log _{\sqrt{a}} x-3 \quad(a>0, a \neq 1)$.
Solution.

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

Let's transition to the base $a$. We get

$$
\frac{3 \log _{a} x-2}{\log _{a}^{2} x}=2 \log _{a} x-3 \Leftrightarrow 2 \log _{a}^{3} x-3 \log _{a}^{2} x-3 \log _{a} x+2=0
$$

since $\log _{a} x \neq 0$. Further, we have

$$
\begin{aligned}
& 2\left(\log _{a}^{3} x+1\right)-3 \log _{a} x\left(\log _{a} x+1\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2\left(\log _{a} x+1\right)\left(\log _{a}^{2} x-\log _{a} x+1\right)-3 \log _{a} x\left(\log _{a} x+1\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left(\log _{a} x+1\right)\left(2 \log _{a}^{2} x-5 \log _{a} x+2\right)=0
\end{aligned}
$$

from which $\log _{a} x+1=0$ or $2 \log _{a}^{2} x-5 \log _{a} x+2=0$. From the first equation $\log _{a} x=-1, x_{1}=\frac{1}{a}$. From the second equation $\log _{a} x=\frac{1}{2}$ or $\log _{a} x=2$, from which $x_{2}=\sqrt{a}, x_{3}=a^{2}$.

Answer: $\frac{1}{a} ; \sqrt{a} ; a^{2}$.