How to solve: $x+\ln(x^2-1)=0$ I'm trying to solve the following equation for a couple of hours with no success.
$x+\ln(x^2-1)=0$
I'm trying to find x.
I tried playing with logarithmic identities to transform it to something simpler, but it didn't lead me to an answer.
I would appreciate your help.
Thanks.

The above equation gives:      
$$e^x=\frac{1}{x^2-1}$$
I suppose you know the graph of $f(x) =e^x$,now if you know how to draw the graph of $g(x) =\frac{1}{x^2-1}$ using its derivatives then, the value of $x$ corresponding to the point of intersection of the two curves on both sides of equality is the required value of $x$. 
