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

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

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

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 $b\in]0,1[$.


The following solution (one real root for trinomial equation of bellow form), First - Define a function $f(x)$ Where $x$ is positive real number and less than one, the limit of $f(x)$ as $n$ becomes large enough or tends to infinity

$$f(x) = \frac{3x}{2} + (2x-1)\left(\frac{4x}{3} - 1\right) + \left(\frac{5x}{2} - 1\right)\left(\frac{5x}{3} - 1\right)\left(\frac{5x}{4} - 1\right) + \cdots + \left(\frac{nx}{2} - 1\right)\left(\frac{nx}{3} -1 \right)...\left(\frac{nx}{n-1} -1 \right)$$

How then one can find at least one polynomial real root of the following trinomial equation form, just by using the above introduced function $f(x)$ & without using any approximation methods

$$x^n +x^m = 1$$

Where ($n$, $m$) are two distinct positive integers, and $n$ is greater than $m$ Then the symbolic solution in terms of ($n$ and $m$) would be

$$x = \left[\frac{m}{n}f\left(1 - \frac{m}{n}\right)\right]^{1/n} \enspace .$$ Which is always an algebraic irrational number

Actually Lucian, had just outlined the content issue to solve the following arbitrary numerical example

Example : x^31 +x^7 = 1

Solution: $x = [(7/31)\, f(24/31)] ^ {1/31} = 0.95764070348\cdots$, an algebraic irrational number.

Reference:(unpublished book-1994) at this link: http://opac.nl.gov.jo/uhtbin/cgisirsi.exe/7514o2xvPE/MAIN/60530003/123

In this reference the general equation ($ax^n + bx^m +c = 0$), had been solved

  • $\begingroup$ I know, it is going to be deleted without obvious reasons, but for those who keep deleting my content, a simple question?, Did you find it wrong to downvote it, if so, come & refute it please, I shall be grateful to you if you kindly mistaken me, you may notice this simple formula & much more had been DOCUMENTED many years back, ALSO on the internet 12 years back, so deleting here will not in any case suppress the formula or many other formulas - is mathematics becoming boring to the limit that it is margin too narrow to contain a formula, What about others who find it absolutely correct?. $\endgroup$ – bassam karzeddin Jun 6 '15 at 16:49
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    $\begingroup$ Really nice, could this formula be extended to the case of arbitrary real $m$ and $n$? $\endgroup$ – 0x2207 Jun 6 '15 at 19:56
  • $\begingroup$ Yes of course this can be extended to reals, and that what I have tried many times to explain to mathematicians where they always delete without obvious reason, the generalization would be as following, if ($x^p+y^p=z^p$, implies $(x/z)^p=[log(z/y) / log(z/x)]*f(log(y/x) / log(z/x))$, where generally $(x, y z & p)$ are nonzero complex numbers, then you can limit it to any fields of numbers to conclude whatever you like, so if you choose ($x, y, & z$) nonzero integers, & p is greater than two corresponding to acute angle theorem I'm intended to announce, you find then $p$ is irrational number. $\endgroup$ – bassam karzeddin Jun 7 '15 at 8:35
  • $\begingroup$ I can't reply your second comment (ox2207) due to insufficient credit, there must be something interesting there, I think Dr. Robert Israel knows this better, because once in 2004 he did my formula using Gamma, I will analyse it and see what comes out! $\endgroup$ – bassam karzeddin Jun 7 '15 at 16:29
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    $\begingroup$ If you can show they are the same results, then it is possible (rediscovering), but my way to those results are very elementary, like a student handwritten work (few pages only), it has history of correspondence from 1986 also, at any case all the correct ways goes to Roma, for little more information please see this link: quora.com/What-is-the-new-Pythagoras-Theorem $\endgroup$ – bassam karzeddin Jun 8 '15 at 16:47

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