$7.59 \quad \lg ^{4}(x-1)^{2}+\lg ^{2}(x-1)^{3}=25$.
7.59 Here $x-1>0$, hence $x>1$. Using the fact that
$\lg (x-1)^{2}=2 \lg (x-1), \lg (x-1)^{3}=3 \lg (x-1)$, we can rewrite the equation as

$$
16 \lg ^{4}(x-1)+9 \lg ^{2}(x-1)=25
$$

Let $\lg ^{2}(x-1)=y \geq 0$, then we get the equation $16 y^{2}+9 y-25=0$, which has roots $y_{1}=1, y_{2}=-\frac{25}{16}$ (not suitable). Therefore, $y=\lg ^{2}(x-1)=1$, hence $\lg (x-1)=1$ or $\lg (x-1)=-1$. From the first equation, we find $x_{1}=11$, from the second $x_{2}=1.1$. Both roots satisfy the condition $x>1$.

Answer: $x_{1}=11, x_{2}=1.1$.