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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 $F$).

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    $\begingroup$ Since this is (up to logarightms) equivalent to solving $$y^a + y^b = 1,$$ I'm not sure if this is even doable. At least in a closed form. $\endgroup$ – 5xum Jan 20 '14 at 8:56
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    $\begingroup$ If $a=0$ or $b=0$, there is no solution. If $a<0$ and $b>0$, and $(-a/b)^{a/(b-a)}+(-a/b)^{b/(b-a)}>1$, there is no solution. Could you tell us more about the values of $a$ and $b$? $\endgroup$ – Tom-Tom Jan 20 '14 at 9:09
  • $\begingroup$ @V.Rossetto, it is assumed that there is solution, and $a$ and $b$ are appropriate. My question is about functions, or about the proof that there is no constructive way to explicitly define the function of a and b such that $F(a,b)$ is the solution to the equation. $\endgroup$ – 0x2207 Jan 20 '14 at 10:55
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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 **

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    $\begingroup$ Is it consistent @bassam-karzeddin formulae when $a$ and $b$ are integer? $\endgroup$ – 0x2207 Jun 7 '15 at 13:43
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    $\begingroup$ I have some reason to believe that @bassam-karzeddin rediscovered the series of Mellin to solve trinomial equations. $\endgroup$ – giorgiomugnaini Jun 8 '15 at 13:40
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    $\begingroup$ Adding some explanation of why you have that belief e.g. a solution for 0 would be very helpful... $\endgroup$ – Jay Nov 29 '18 at 18:12
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    $\begingroup$ Just saying such that then it would be slightly easier to adjust for testing to lessen any bias...Other spaces and theories (manifolds etc) aside I am just visualizing that it would be easier to apply for such given the starting point of 0... :) $\endgroup$ – Jay Nov 29 '18 at 19:32
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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.

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    $\begingroup$ I really wonder why a downvote ! $\endgroup$ – Claude Leibovici Jan 20 '14 at 9:04
  • $\begingroup$ You claimed something without proof: "this equation does not have any explicit solution" $\endgroup$ – T.J. Gaffney Jan 20 '14 at 9:05
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    $\begingroup$ It's enough for the ratio to be rational. $\endgroup$ – Lucian Jan 20 '14 at 9:11
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    $\begingroup$ @Gaffney: Regarding your complaint: Mr. Leibovici has discovered a truly marvellous proof of this, which this margin is too narrow to contain. - You happy now ? :-| $\endgroup$ – Lucian Jan 20 '14 at 9:17
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    $\begingroup$ Then our equation becomes $y^{31}+y^{7}-1=0$, where $y=e^{kx}$, for some unspecified k. Of course, the solution may not be expressibel in radicals, but we know that (at least some) algebraics can be expressed in terms of hypergeometric functions (see Bring radical, for instance, as well as this question), and our OP allowed for such. Then we further extract x as $\frac1k\ln y$, for some k. $\endgroup$ – Lucian Jan 20 '14 at 9:38
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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 $b\in]0,1[$.

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