Understanding roots of complicated equations

Consider the function $$x-left(A-frac{B e^x}{e^x+1}right)$$ plugging $x=log left(frac{A}{B-A}right)$ we get $$log left(frac{A}{B-A}right)+frac{A B}{(B-A) left(frac{A}{B-A}+1right)}-A=log left(frac{A}{B-A}right)$$ thus when $Bsim 2A$ we have $log left(frac{A}{B-A}right)sim 0$ and $x^*=log left(frac{A}{B-A}right)$ is a good approximation of the solution.

If we take $A=1;;B=10000$ the approximation gives $-9.2$ while the root is $-7.1$

For $A=60000;;B=100000$ we get respectively $0.405465;;0.405448$ with an error less than $10^{-4}$.

Hope this helps

In fact, there is a formal solution to this equation. Rewrite it as $$e^{-x}=-frac{x-(A-B)}{x-A}$$ and the solution is given in terms of the generalized Lambert function (have a look at equation $(4)$). The bad news is that, from a practical point of view, it does not bring much.

What we could try is to use an high order iterative method which would give for example $$x=-frac{4 (2 A-B) left(-B left(-2 A^2+2 A B+(B+6) (B+12)right)-96right)}{(B+4) left(B left(-8 A^2+8 A B+B (B+36)+144right)+192right)}$$

Trying for $A=1234$ and $B=5678$, this would give as an estimate $x=-1.28244$ while the solution is $x=-1.27997$. Pure luck ?

In this case, your guess is impressive since $-log left(frac{2222}{617}right)sim -1.28129$ (close to mine - which is not !)

Trying two orders above (formula not typed - just too long), the guess is $x=-1.27994$.

Unfortunately, it doesn't exist a general method, because every equation can easily be slightly different from each other.

I show you what I'd do in this case,

First things first, I'd see the entire equation as a rational one. How do I do that? I simply found a common denominator. Then, solve a system of two equations (numerator and denominator). In order to obtain the right result, you must ask yourself: "which solution satisfies both equations at the same time?" Seeing this on a number line can help a lot.

Tell me if you have any doubt.