Solving $\frac {e^x-3}{e^x-2}-|x|=0$ I'm trying to solve the following equation:
$$\frac {e^x-3}{e^x-2}-|x|=0$$
I have tried to solve it by algebric mode, but I got stuck at this point:
$$\left\{ 
\begin{array}{c}
x \ge0 \\  
x(2-e^x)=3-e^x
\end{array}
\right. \cup 
\left\{ 
\begin{array}{c}
x <0 \\  
x(e^x-2)=3-e^x
\end{array}
\right.;\ x\neq\log(2) $$
Someone could give me a hint for solve it?
If I would want to solve it by graphic mode, how should I proceed? 
 A: 
Let $$f(x) = \frac{\exp(x)-3}{\exp(x)-2}-|x|=1-\frac{1}{\exp(x)-2}-|x|$$
If $x <0$, $$ f(x) = 1-\frac{1}{\exp(x)-2}+x$$
$$f'(x)=\frac{\exp(x)}{(\exp(x)-2)^2}+1$$
Hence it is increasing.
You might like to examine $f(0)$ and $f(-2)$ and conclude that there is a unique solution in $(-2,0)$.
If $x \geq 0$,$$ f(x) = 1-\frac{1}{\exp(x)-2}-x$$
$$f'(x)=\frac{\exp(x)}{(\exp(x)-2)^2}-1$$
Find the asymptote and prove that it doesn't go close to $0$.
A: Let $$f(x) = \frac{e^x - 3}{e^x - 2} = \frac{e^x - 2 - 1}{e^x - 2} = 1 - \frac{1}{e^x-2}, \\ g(x) = |x|.$$  Then we seek the intersection of $f$ and $g$.  Note $f$ contains a singularity when $e^x = 2$, or $x = \log 2$.  We see also that $f < 0$ when $\log 2 < x < \log 3$, and since $\log 3 > 1$, it follows that $f < g$ when $x > \log 2$, and no solutions exist on this interval.
When $0 \le x < \log 2$, it is easy to see $f > g$, so again, no solutions here, and we can now restrict our attention to the case where $x < 0$, hence $g(x) = -x$, and now we seek the solution to $$(e^x - 2)(1+x) = 1.$$  This has no closed form solution.  Numeric approximation using Newton's method is possible; e.g., consider $$x_{n+1} = \frac{e^{x_n}(x_n^2+x_n-1)+3}{e^{x_n}(x_n+2)-2},$$ for some suitable initial guess $x_0 < 0$.  Convergence is rapid; we get $$x \approx -1.558781663973784196536245468626.$$
