Solve $x=e^\frac{a}{b+x}$ when $a,b,x>0$ 
If $a,b,x>0$, then solve 
  $$x=e^\frac{a}{b+x}$$

I just can solve this numerically with given values of $a,b$. Any idea how I can solve this analytically? 
 A: There seems to be no closed-form solution in general.  You could write a solution as a formal series in powers of $a$:
$$  
x = 1+\frac{a}{b+1}+{\frac {b-1}{2\, \left( b+1 \right) ^{3}
}}{a}^{2}+{\frac {{b}^{2}-7\,b+4}{6\, \left( b+1 \right) ^{5}}}{a}^{3}
+{\frac {{b}^{3}-25\,{b}^{2}+67\,b-27}{24\, \left( b+1 \right) ^{7}}}{
a}^{4}+{\frac {{b}^{4}-71\,{b}^{3}+531\,{b}^{2}-821\,b+256}{120\,
 \left( b+1 \right) ^{9}}}{a}^{5}+\ldots $$
A: There is no way to solve analytically such equations. The way to approach such problems is by numerical methods.
$$x=e^{a/(b+x)} \Leftrightarrow x- e^{a/(b+x)} = 0$$
Set a function :
$$f(x) := x- e^{a/(b+x)} $$
And procceed on finding the roots of the equation $f(x) = 0$ using a numerical method, such as for example the Newton Raphson Method, since this is a differentiable function.
For Newton Raphson Method specifically, you make a sequence, which converges to the solution of your problem, given a starting $x_0$ which can be found by sketching the graph with a computer program.
The sequence : 
$$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k )}$$
You start with : 
$$x_{1} = x_0 - \frac{f(x_0)}{f'(x_0 )}$$
and then proceed on, until you converge to the root.
You can calculate the error of the method with : 
$$x_k = x^* + e_k$$
$$\dots$$
$$e_{k+1} \approx \frac{f''(x^*)}{2f'(x^*)}e_{k}^2$$
