When is possible to ignore an absolute value

Question

$e^{\frac{2}{x-1}\log\left|x-1\right|}+1\neq 0$

Since that this is an exponential function, this equation is verified $\forall x \in \mathbb{R}$? Or I have to consider the absolute value of the ligarithm?


Note: these are not homework.

Answer 1

Everywhere the LHS is defined, the equality holds, since $e^{y} \geq 0$ for all $y$ and for all $x \geq 0$, $x + 1 > 0$. The LHS is not defined for $x = 1$ (since neither $\log|x-1|$ nor $\frac{2}{x-1}$ is defined there).