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

Question

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?

Answer 1

There seems to be no closed-form solution in general. You could write a solution as a formal series in powers of $a$:

$$ x = 1+\frac{a}{b+1}+{\frac {b-1}{2\, \left( b+1 \right) ^{3} }}{a}^{2}+{\frac {{b}^{2}-7\,b+4}{6\, \left( b+1 \right) ^{5}}}{a}^{3} +{\frac {{b}^{3}-25\,{b}^{2}+67\,b-27}{24\, \left( b+1 \right) ^{7}}}{ a}^{4}+{\frac {{b}^{4}-71\,{b}^{3}+531\,{b}^{2}-821\,b+256}{120\, \left( b+1 \right) ^{9}}}{a}^{5}+\ldots $$

Answer 2

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 :

$$x_k = x^* + e_k$$ $$\dots$$ $$e_{k+1} \approx \frac{f''(x^*)}{2f'(x^*)}e_{k}^2$$