7.097. $\log _{3}(x-3)^{2}+\log _{3}|x-3|=3$.
## Solution.

Domain of definition: $x-3 \neq 0, x \neq 3$.

From the condition

$2 \log _{3}|x-3|+\log _{3}|x-3|=3, 3 \log _{3}|x-3|=3, \log _{3}|x-3|=1$,

hence $|x-3|=3$. Then $(x-3)_{1}=-3$ or $(x-3)_{2}=3$. From this, $x_{1}=0$, $x_{2}=6$.

Answer: $0 ; 6$.