Solve equation $\exp(ax)+\exp(bx)=1$

The equation is $$ \exp\left(ax\right)+\exp\left(bx\right)=1, $$ where $a$ and $b$ are known real constants, $x$ is unknown. I would like to have the solution in form of relatively known special function (something like Lambert $W$ function, or generalized hyper-geometric $F$).

Solution 1:

If we place $y=\exp(x)$ we can rewrite the problema as: $$y^a+y^b=1$$ Such trinomial equation is a particular case of more general: $$y^a+z y^b=1$$ where we have supposed $a>b$. The solution of the trinomial equation can be found by means of Lagrange inversion series or also by Mellin transform as: $$y(a,b,z)=\frac{1}{a}\sum_{r=0}\frac{\Gamma\left(\frac{1+rb}{a}\right)}{\Gamma\left(\frac{1+rb}{a}+1-r\right)r!}(-1)^rz^r$$ So $$x(a,b)=\log\left[\frac{1}{a}\sum_{r=0}\frac{\Gamma\left(\frac{1+rb}{a}\right)}{\Gamma\left(\frac{1+rb}{a}+1-r\right)r!}(-1)^r\right]$$ References

"The Functions of Mathematical Physics ", 1986, Harry Hochstadt

** EDITED to fix several typos **

Solution 2:

Assuming that "a" and "b" are not zero, this equation does not have any explicit solution except if, say, "b" is a multiple of "a".

So, for the general case, this equation would be solved using a root-finder method such as Newton. What is nice is that, knowing the values of "a" and "b", a reasonably good guess of the solution can be easily made.

If you want to see that working, just give me the "a" and "b" you want and I shall post the path to solution for you.