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

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.

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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).

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Yes! I set up $x > 1$ as existence condition of the logarithm in the system (that equation is a part of a domain study). So, the absolute value is not important in this case? –  Overflowh Feb 2 '12 at 17:42
    
If $x > 1$, $|x - 1| = x - 1$, so you wouldn't even need to use the absolute value in the LHS... for the same reason we don't write $|x^{2}|$. –  Patrick87 Feb 2 '12 at 18:13
    
Okok, understood. It is like saying: since I set up $x>1$ in the system, there aren't values of $x$ that could make negative the argument of the logarithm; so, this equation is verified $\forall x \in\mathbb{R}$. Thank you! –  Overflowh Feb 2 '12 at 18:20

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