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 $x&gt; -4$ we have $e^x\geq 1+x&gt;\frac{x}{2}-1$, so $e^x-\left (\frac{x}{2}-1\right)&gt;0$. This means that they don't intersect for $x\in \left &lt;-4,\infty\right &gt;$. 

When $x\leq -4$ we have $e^x&gt;0$ and so $e^x-\left (\frac{x}{2}-1\right )&gt;0$ for $x\in \left &lt;-\infty,-4\right]$. Therefore they don't intersect, and no real solution exists.
$$ \frac{x}{2} - 1 = e^x $$
$$ \frac{x}{2} - 2 = e^x - 1  $$
$$ \frac{x-4}{2} = e^x -1 $$
$$ \frac{x-4}{2(e^x -1)} = 1 $$


  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: