7.240. $\lg ^{4}(x-1)^{2}+\lg ^{2}(x-1)^{3}=25$.
## Solution.

Domain of definition: $x>1$.

From the condition, we have $16 \lg ^{4}(x-1)+9 \lg ^{2}(x-1)-25=0$. Solving this equation as a biquadratic equation in terms of $\lg (x-1)$, we get $\lg ^{2}(x-1)=1 \Rightarrow$ $\Rightarrow \lg (x-1)=-1$ or $\lg (x-1)=1$, from which $x_{1}=1.1, x_{2}=11$.

Answer: 1,$1 ; 11$.