How to find x in this case $\frac{x}{2}-1=e^x$? $$\frac{x}{2}-1=e^x$$
Let say we can write above function as: 
$$\frac{x}{2(e^x-1)}=1$$
In Wolframalpha I get a solution $x\approx -1.59362$.
How can I find x?

There is no real solution since
$$e^x\geq 1+x$$
For $x> -4$ we have $e^x\geq 1+x>\frac{x}{2}-1$, so $e^x-\left (\frac{x}{2}-1\right)>0$. This means that they don't intersect for $x\in \left <-4,\infty\right >$. 
When $x\leq -4$ we have $e^x>0$ and so $e^x-\left (\frac{x}{2}-1\right )>0$ for $x\in \left <-\infty,-4\right]$. Therefore they don't intersect, and no real solution exists.
