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

Question

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 $F$).

Answer 1

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.

Answer 2

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 $b\in]0,1[$.