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?
 A: 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.
A: There is no solution in terms of elementary functions. Furthemore, there is no real solutuon, as shown in the answer by @cansomeonehelpmeout. However, you can use the Lambert W-function to get complex solutions:
$${x \over 2} - 1 = e^x$$
$$x-2 = 2e^x = 2e^{x-2}e^2$$
$$(x-2)e^{2-x}=2e^2$$
$$(2-x)e^{2-x}=-2e^2$$
$$2-x=W(-2e^2)$$
$$x=2-W(-2e^2)$$
Actually, $W$ is multi-valued. Taking the principal branch, we get $W_0(-2e^2) \approx 1.6718+2.2169i$, leading to $x\approx 0.3281-2.2169i$.
A: $$ \frac{x}{2} - 1 = e^x $$
$$ \frac{x}{2} - 2 = e^x - 1  $$
$$ \frac{x-4}{2} = e^x -1 $$
$$ \frac{x-4}{2(e^x -1)} = 1 $$

Let say we can write above function as:

You can't do that. Because then $x-4 = x \implies -4 = 0$, which is nonsense.  
 
EDIT:

Assuming you meant:
$$ \frac{x}{2(e^x +1)} = 1 $$
Then you can just plot the function $f(x) = \frac{x}{2(e^x+1)}$ and solve $f(x) = 1$
EDIT 2:
I'm sure you're making a mistake:

