Isolating x analytically in this equation : $(xe^{ax})^x=f(y)$

Question

I have come upon the equation $(xe^{ax})^x=c$, where a and c can be regarded as real constants, and am looking for an analytical expression for x. The value is constrained to $x>0, \in \mathbb{R}$. I have delayed obtaining the exact range of possible values for $x$ as a function of $a$ and $c$, but it is likely that at the very least a maximum exists.

I don't think this can easily be solved using the Lambert W function. Here is what I have tried so far.

First, I have tried the change of variables $z\equiv axe^{ax}$, obtaining $(z/a)^{W(z)/a}=c$, which doesn't seem anyhow useful.

Second, I have tried taking the logarithm and finding the roots of the resulting quadratic expression in x, obtaining $(2ax+\ln(x)+\sqrt{(\ln(x))^2+4ac})(2ax+\ln(x)-\sqrt{(\ln(x))^2+4ac})=0$, from which I deduced that only the second root existed, since $x$ is strictly positive. As I still couldn't find a change of variables to isolate x using the Lambert function, I tried setting $z\equiv \ln(x)$ for clarity then using a binomial expansion on the square root, which gave

\begin{align} 2ae^z+z(1-sign(z))-sign(z)\sum_{k=1}^\infty \bigl( \begin{smallmatrix} 0.5 \\ k\end{smallmatrix} \bigr)\frac{(4ac)^k}{z^{2k-1}}&=0 \\ \Rightarrow 2ae^z+z(1-sign(z))-sign(z)\sum_{k=1}^\infty \bigl( \begin{smallmatrix} 0.5 \\ k\end{smallmatrix} \bigr)\frac{1}{k+1}\frac{(4ac)^{k+1}}{z^{2k+1}}&=0 \\ \Rightarrow 2ae^z+z(1-sign(z))-sign(z)\frac{4ac}{z}\sum_{k=1}^\infty \bigl( \begin{smallmatrix} 0.5 \\ k\end{smallmatrix} \bigr)\frac{1}{k+1}\bigl(\frac{4ac}{z^2}\bigr)^k&=0 \end{align}

If I could prove $x\leq1$, I could then obtain a simple relation between the exponential and a power series in z. I hoped this would allow me to convert this power series to the form used in "Taylor series for generalized Lambert W functions", by Paul Castle (2018), but did not figure out an easy way to get there as it would involve factoring an infinite series. I have also thought about using the power series form for $e^x$, but my series knowledge is rusty and I don't believe these series can cancel one another for any given value of k or power of z.

Finally, I have tried taking a step back and formulating the initial equation slightly differently, which gave

\begin{equation} e^{2cx}=a(x/2+\sqrt{x^2/4+b})(x/2-\sqrt{x^2/4+b}) \end{equation}

This approach did not take me any further.

I hope you have suggestions or clarifications, or would gladly accept any indication that one of my approaches is mistaken for whatever reason.

Thanks a lot!

EDIT :

By some transformations this equation can be brought to $x^2e^{k(x-1/x)}=a^2$. Posing $a\gamma\equiv kxe^{kx}\Rightarrow x=W(a\gamma)/k$ we also get $(x/k)e^{-k/x}=a/\gamma\Rightarrow x=k/W(\gamma/a)$. Dividing the two expressions, we get another implicit equation :

\begin{equation} W(a\gamma)W(\gamma/a)=k^2 \end{equation}

For $a=1$, this gives $\gamma=ke^k \Rightarrow x=W(ake^k)$. It is also interesting that $\gamma$ is symmetrical with respect to $log(a)$, but I could not go further with this form.

Another transformation gives some closure as to the existence of a closed form. Letting $z\equiv \log(x)$ and taking the logarithm, we can find an equation of the form $z+k(e^z-e^{-z})=\log(a)$, which can be transformed to

\begin{equation} z+2k\sinh{z}=\log(a) \end{equation}

I believe this last equation can be treated as a case of Kepler's equation in the hyperbolic case and has some explicit forms, although it is best computed numerically.

I don't really need an answer anymore, but this question has not been answered and any insight would still be interesting.

Answer 1

$\def\L{\operatorname L}\def\A{\operatorname A}$

Although there is a Kepler equation Fourier series solution, we apply Laplace inversion. It does not appear this solution is online:

$$f(x)=\sinh(x)+ax\implies f^{-1}(x)=h(x)=\L^{-1}_x(\L_s(h(t))$$

with the (inverse) Laplace transforms. We use integration by parts:

$$\L_s(h(t))=\int_0^\infty e^{-st} h(t)dt=\int_0^\infty t e^{-s f(t)}df(t)=\frac1s\int_0^\infty e^{-a s-s\sinh(t)}dt$$

and the associated Anger Weber function $\A_v(z)$:

$$\frac1s\int_0^\infty e^{-a s-s\sinh(t)}dt=\frac\pi s\A_{sa}(s)$$

Therefore the solution to the hyperbolic Kepler equation is:

$$\bbox[4px,border: 3px groove lightgreen]{f(x)=\sinh(x)+ax\implies f^{-1}(x)=\pi\L^{-1}_x\left(\frac{\A_{sa}(s)}s\right)=\int_{c-i\infty}^{c+i\infty}\frac{e^{sx}}{2is}\A_{sa}(s)ds}$$

Typing in {FindRoot[Sinh[x]+a*x-b,{x,b}],InverseLaplaceTransform[\[Pi]/s ResourceFunction["AngerWeberA"][s a,s],s,b]} for certain $a,b$ into Mathematica verifies the formula:

Answer 2

1)

$$(xe^{ax})^x=c$$ $$e^{\ln(xe^{ax})x}=c$$ $$ax^2+x\ln(x)=\ln(c)$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x,\ln(x)$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

$x\to e^t$: $$a(e^t)^2+te^t=\ln(c)\tag{1}$$

We see, equation (1) is an irreducible algebraic equation of $t$ and $e^t$ simultaneously if $a$ and $\ln(c)$ are algebraic numbers.
According to the theorems in [Lin 1983] and [Chow 1999], such kind of equations cannot have solutions except $0$ that are elementary numbers or explicit elementary numbers respectively in this case. Therefore $t$ cannot be an elementary number except $0$ in this case and equation (1) cannot have partial inverses that are elementary functions over non-discrete domains.

We see, because equation (1) is a polynomial equation of $t$ and different powers of $e^t$ greater than $1$, the equation is not in a form for applying Lambert W, Generalized Lambert W or Hyper Lambert W.

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

2)

$$z+2k\sinh(z)=\ln(a)$$ $z\to it$: $$it+2k\sinh(it)=\ln(a)$$ $$t+2k\sin(t)=-\ln(a)i$$

We see, this is Kepler's equation.