solving for $-\frac{x}{\gamma} = \ln\left( \frac{1}{2}x + \lambda\right)$ I am trying to solve the equation: 
$-\frac{x}{\gamma} = \ln\left( \frac{1}{2}x + \lambda\right)$
for $x$. $\lambda$ and $\gamma$ are constants. By using,
$e^y = x \leftrightarrow \ln x = y$,
I get,
$e^{-\frac{x}{\gamma}} = \frac{1}{2}x + \lambda$
$e^{-\frac{x}{\gamma}}-\frac{1}{2}x= \lambda$,
where I am stuck on how to solve for $x$. Can this equation be solved with a closed form solution for $x$ ? Much help/suggestion is appreciated.
 A: 
$e^{-\frac{x}{\gamma}}-\frac{1}{2}x= \lambda$,
where I am stuck on how to solve for $x$. Can this equation be solved with a closed form solution for $x$ ? Much help/suggestion is appreciated.

It can't, unless you allow special functions such as the Lambert-W-function (see here).


Can you elaborate more on "allow special function" ? for the solution of the equation ?

Yes. You should look up the documentation of that special function (follow the link above) to see what it does. It allows to give an explicit form of the solution, but in a sense this is hiding (or 'moving') the problem in that special function. If this is useful or not, depends on what your goal is.

If we don't allow such functions, that means there is no solution for x ?

No, there's a difference between "no solution" and "no explicit, closed form". Certain equations can have one or more solutions, without the possibility of giving them in a closed form. That doesn't mean the solutions don't exist, you could e.g. find them numerically or graphically.
