Solving $|x-1|^{\log^2(x)-\log(x^2)}=|x-1|^3$ 
Solve the equation:$$|x-1|^{\log^2(x)-\log(x^2)}=|x-1|^3.$$

There are three solutions of $x$: $10^{-1}$, $10^3$ and $2$. I obtained the first two solutions but I have been unsuccessful in getting $2$ as a solution. Please help.
 A: The domain gives $x>0$ and $x\neq1$.


*

*$|x-1|=1$.


With domain we get here $x=2$.


*$\log^2x-\log{x^2}=3$.


Let $\log{x}=t$.
We need to solve now $t^2-2t=3$, which gives $t=-1$ or $t=3$, which is 
$x=\frac{1}{10}$ or $x=1000$.
Finely we get the answer: $\{2,\frac{1}{10},1000\}$.
A: First, the presence of $\ln x$ in the equation, i.e. in $\ln^2 x - \ln(x^2)$, dictates that $x > 0$. Next, if $|x - 1| = 0$, then $x = 1$ and $\ln^2 x - \ln(x^2) = 0$, but $0^0$ is undefined and thus $x = 1$ is excluded.
Now the equation reduces to$$
|x - 1|^{\ln^2 x - \ln(x^2) - 3} = 1,
$$
which is equivalent to$$
|x - 1| = 1 \text{ or } \ln^2 x - \ln(x^2) - 3 = 0.
$$
Case 1: $|x - 1| = 1$.
In this case, $x = 0$ or $2$ and $x > 0$ implies that $x = 2$.
Case 2: $\ln^2 x - \ln(x^2) - 3 = 0$.
Note that $\ln(x^2) = 2\ln x$ for $x > 0$, so this equation becomes $(\ln x)^2 - 2 \ln x - 3 = 0$, or$$
(\ln x + 1)(\ln x - 3) = 0,
$$
which implies $\ln x = -1$ or $3$, i.e. $x = \dfrac{1}{10}$ or $1000$.
Therefore, the solutions are $\dfrac{1}{10}$, $2$, $1000$.
