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 $F$).
 A: 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 **
A: 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.
A: Suppose without loss of generality, that $a=1$ and $b<1$ and set $y=\mathrm e^{x}$, with $0<y<1$. 


*

*If $b<0$, the minimum of $x\mapsto x+x^b$ for $x\in]0,+\infty[$ is
$$(-b)^{\frac1{1-b}}+\left(-b\right)^{\frac b{1-b}}=(-b)^{\frac1{1-b}}+\left(\frac1{-b}\right)^{\frac{-b}{1-b}}>1,$$
because if $b\neq-1$ either $-b$ or $-1/b$ is larger than 1. If $b=-1$ this minimum obviously equals $2$.
There is no solution.

*If $b=0$, as $\mathrm e^x+1>1$, there is no solution either.

*If $0<b<1$, the function $y\mapsto 1-y^{b}$ is a decreasing function on $[0,1]$. The equation rewrites $y+y^{b}=1$ or
$$ y=1-y^{b}.\tag{1}$$
Thus the solution is of the form $$x(b)=\ln f(b),$$
where $f(b)$ is the fixed point of $y\mapsto 1-y^b$ in $]0,1[$.$$ $$
An algorithm to find the numerical value of $x(b)$: define $z_0\in]0,1[$ and compute the values of $z_n=1-z_{n-1}^b$ for $n\geq 1$. Then the sequence $(z_n)_n$ converges towards $f(b)$.
As it is said in the comments to your questions, even for several values of  $b\in\mathbf Q\cap]0,1[$, there is no analytic expression for $x(b)$ or $f(b)$, then there is no possibility for finding a function (hypergeometric series, Lambert, etc...) for all $b\in]0,1[$.
A: $$e^{ax}+e^{bx}=1\tag{1}$$
$x\to\ln(y)$:
$$y^a+y^b=1\tag{2}$$
For $a,b,x\in\mathbb{N}$, equation (2) is a Diophantine equation.
For rational $a,b$, equation (2) is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.
For real or complex $a,b\neq 0,1$, equation (2) is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore.
see also: How to isolate $x$ in $a^x + b^x = c$? (For use in medical statistics)
Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104
Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106
