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I only recently learned what the Lambert W function is and how to apply it to different problems. But this expression
$c=e^{-4cx}+c^2e^{-cx}$
is something that I was not able to solve using Lambert functions and such. Where I get caught up on is turning the above expression to the form $ae^a=z$ (where the $z$ term is a function of $x$) to apply the Lambert function afterwards.... Any help with this problem will be very much appreciated....

EDIT: I'm trying to solve for c in this problem

IV_
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alienare 4422
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2 Answers2

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As Claude Leibovici already wrote in his answer, the equation

$$c=e^b+c^2e^c:\ \ \ e^c=\frac{-e^b+c}{c^2}$$

can be solved for $c$ by Mezö's generalization of Lambert W because $e^c$ is a rational function of $c$.

But for your second equation

$$c=e^{-4cx}+c^2e^{-cx}\colon\ \ \ c(e^{xc})^4-c^2(e^{xc})^3-1=0,$$

$e^{xc}$ is an algebraic irrational function of $c$, not a rational function of $c$. Therefore your second equation cannot be solved by Mezö's generalization of Lambert W. Maybe it is solvable by Castle's generalization of Lambert W - see the reference below.

See the last part of my answer at Algebraic solution to natural logarithm equations like $1-x+x\ln(-x)=0$ for the general form of equations that are solvable by Lambert W.

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

IV_
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If you are trying to solve for $c$ the equation $$c=e^b+c^2\,e^c$$ you have $$e^{-c}=\frac{c^2}{c-e^b}$$ and the only explicit solution would be given in terms of the generalized Lambert function (have a look at equation $(4)$).

Claude Leibovici
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  • can this be generalized into the equation that I wrote in the question because even though it has the same 'format' I don't think it can be generalized that easily....can it? – alienare 4422 Jun 25 '22 at 06:48
  • what I want is for C in terms of x for the equation in the question – alienare 4422 Jun 25 '22 at 06:49
  • The same paper mentions a series expansion in “Theorem 1”, so maybe it can be included? Also, how does $e^{-x}=\frac{x^2}{x-e^b}$ go into the form of $e^{-c x}=a\frac{x-t}{x-s}$, from equation (4)? – Тyma Gaidash Jun 25 '22 at 13:35