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I am looking for a solution to the equation $x\, \exp(poly(x)) = c$ for some polynomial $poly(x)$ in $x$ and a real number $c$.

We know that the Lambert $W$-function provides an analytical solution to $x \, \exp (x) = c$ as $x=W(c)= \sum_n \frac{(-n)^{n-1}}{n!} c^n$ (I am leaving out the existence of branches to keep things simpler).

We can also go one step further and solve equations with non-matching exponents in $x.$ As an example, we pick $x \exp(x^2)= c$, and we first square both sides of the equation and the multiply by 2, namely obtaining \begin{equation} 2x^2 \exp (2 x^2) = 2 c^2 \ , \end{equation} and finally we substitute $y=2x^2$ to apply the $W$-function.

But what if we now have a polynomial in the exponential, e.g. \begin{equation} x \exp(x+x^2) = c \ , \end{equation} does this equation have a (non-numerical) solution? If yes, can completely generalize the solution to an expression of the form \begin{equation} x \exp(a_1x+a_2x^2+...+a_N x^N) = c \ ? \end{equation}

EmFed
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    The equation $\text{poly}(x)e^x=c$ has this arxiv solution, so maybe something similar can be done here. Were you able to find anything else on the bottom equation? – Тyma Gaidash Oct 14 '24 at 11:38
  • @ТymaGaidash Because I'm interested in such solutions: Is there a way to transform $xe^{P_1(x)}=c_1$ into $P_2(t)e^t=c_2$ or vice versa? – IV_ Oct 14 '24 at 17:30
  • @ТymaGaidash I found this similar answer, https://math.stackexchange.com/questions/4936798/solve-for-y-in-ye-frac12y-frac1y-x, mentioning the "Burniston-Siewert method to solve transcendental equations". However I am not really familiar with the technique, so I still have to figure out if it can be applied to my instance. – EmFed Oct 15 '24 at 09:13
  • @EmFed It looks like a complicated technique, but seems to be able to be applied to most equations. Maybe the user from there can answer. – Тyma Gaidash Oct 15 '24 at 13:00

1 Answers1

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We have to differentiate between closed-form solutions and analytic solutions.

a) Let $P(x)$ denote a polynomial in $x$.

$$xe^{P(x)}=c$$

It seems the equation is solvable by Lambert W only if $P(x)=a_0+a_1x^b\ $ ($a_0,a_1,b$ constant).

$$xe^{a_0+a_1x^b}=c\tag{1}$$

The solution for real $a_0,a_1,b,x$ is:

$$x=e^\frac{-a_0b+b\ln(c)-W_k\left(a_1be^{-a_0b+b\ln(c)}\right)}{b}\ \ (k\in\{-1,0\})\tag{2}$$

If we already have a solution, we can apply the operations performed on the right-hand side of equation (1) $(c)$ in reverse order to retrace the solution way:

Solving equation (2) for the $W$-term yields:

$$W_k\left(a_1be^{-a_0b+b\ln(c)}\right)=-a_0b+b\ln(c)-b\ln(x)$$ $$-a_0b+b\ln(c)-b\ln(x)=W_k\left(a_1be^{-a_0b+b\ln(c)}\right)\tag{3}$$

We know: $xe^x=c\implies x=W(c)$ and $x=W(c)\implies xe^x$.

Equation (3) implies therefore:

$$(-a_0b+b\ln(c)-b\ln(x))e^{-a_0b+b\ln(c)-b\ln(x)}=a_1be^{-a_0b+b\ln(c)}\tag{4}$$

See the operations at $c$ on the right-hand side of equation (4).

$$xe^{a_0+a_1x^b}=c$$ $$\ln(xe^{a_0+a_1x^b})=\ln(c)$$ $$\ln(x)+a_0+a_1x^b=\ln(c)$$ $$b\ln(x)+a_0b+a_1bx^b=b\ln(c)$$ $$a_1bx^b+b\ln(x)=-a_0b+b\ln(c)$$ $$x^be^{a_1bx^b}=e^{-a_0b+b\ln(c)}$$ $$a_1bx^be^{a_1bx^b}=a_1be^{-a_0b+b\ln(c)}$$ $$a_1bx^b=W_k\left(a_1be^{-a_0b+b\ln(c)}\right)$$ $$x^b=\frac{W_k\left(a_1be^{-a_0b+b\ln(c)}\right)}{a_1b}$$ $$e^{b\ln(x)}=\frac{W_k\left(a_1be^{-a_0b+b\ln(c)}\right)}{a_1b}$$ $$\ln(e^{b\ln(x)})=\ln\left(\frac{W_k\left(a_1be^{-a_0b+b\ln(c)}\right)}{a_1b}\right)$$ $\ln(W(y))=\ln(y)-W(y)$: $$b\ln(x)=-a_0b+b\ln(c)-W_k\left(a_1be^{-a_0b+b\ln(c)}\right)$$ $$\ln(x)=\frac{-a_0b+b\ln(c)-W_k\left(a_1be^{-a_0b+b\ln(c)}\right)}{b}$$ $$x=e^{\frac{-a_0b+b\ln(c)-W_k\left(a_1be^{-a_0b+b\ln(c)}\right)}{b}}$$ $\ $

b) For applying Lambert W, we need the equation in a form

$$f(g(x)e^{g(x)})=c,$$

where $f$ and $g$ are functions of one variable. But with the elementary operations (are elementary functions), it's only possible for relatively simple functions of $x$ in the coefficient and in the exponent to make the coefficient and the exponent equal.

IV_
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