I know how to solve equations using Lambert's W function like
$xe^x=k$
or
$e^x+x=k$
But how can I solve this two kinds of equations involving natural log ?
$e^x \ln(x)=k$
and
$e^x+\ln(x)=k$
I can't figure how to rewrite them in one way where I can apply $W$
If you place $z=\ln(x)$ the equation can be rewritten as:
$$z=k-e^{e^z}$$
Applying lagrange inversion you will find:
$$z=k+\sum_{n=1} \frac{ (-1)^n}{n!} \left[\left(\frac{d}{du}\right)^{n -1}e^{ne^u}\right]_{u=k}$$
now remembering the Rodrigues representation of Touchard polynomials: ( see http://en.wikipedia.org/wiki/Touchard_polynomials )
$$T_n(u)=e^{-e^{u}} \left(\frac{d}{du}\right)^n e^{e^{u}} $$
you can find a formal solution:
$$z(k)=k+\sum_{n=1} \frac{ (-1)^n}{n!} e^{ne^k} T_{n-1}(k+\ln(n))$$
Therefore replacing $z=\ln(x)$ we find:
$$\ln(x(k))=k+\sum_{n=1} \frac{ (-1)^n}{n!} e^{ne^k} T_{n-1}(k+\ln(n))$$
If a link with Lambert W function or its one parameter generalization exists I don't know.
References For details about transcendental equations which can be solved by Lagrange series of mixed exponential/hypergeometric polynomials (that is to say transseries) obtained by suitable Rodrigues formulas see:
"Generalization of Lambert W function, Bessel polynomials and transcendental equations" Giorgio Mugnaini
http://arxiv.org/abs/1501.00138 (pdf: http://arxiv.org/pdf/1501.00138v3 )
For simplicity, I give an answer for solutions in the reals.
a)
$$e^x\ln(x)=k$$ $x\to e^t$: $$te^{e^t}=k$$
This equation is an equation of elementary functions. It's an equation in dependence of $t$ and $e^{e^t}$, algebraic over $\mathbb{C}$. Because the terms $t,e^{e^t}$ are algebraically independent, we don't know how to solve the equation for $t$ by rearranging by only elementary operations (means elementary functions).
I don't know if the equation has solutions in the elementary numbers.
Your equation cannot be solved in terms of Lambert W but in terms of Hyper Lambert W:
$$G(1;t)=k$$ $$t=HW(1;k)$$ $$x=e^{HW(1;k)}$$
see also: https://mathoverflow.net/questions/315204/expressions-for-the-inverse-function-of-fx-lnxex
b)
$$e^x+\ln(x)=k$$ $$e^x=k-\ln(x)$$ $$e^{e^x}=e^{k-\ln(x)}$$ $$e^{e^x}=\frac{e^k}{x}$$ $$xe^{e^x}=e^k$$
This equation is an equation of elementary functions. It's an equation in dependence of $x$ and $e^{e^x}$, algebraic over $\mathbb{C}$. Because the terms $x,e^{e^x}$ are algebraically independent, we don't know how to rearrange the equation for $x$ by only elementary operations (means elementary functions).
I don't know if the equation has solutions in the elementary numbers.
Your equation cannot be solved in terms of Lambert W but in terms of Hyper Lambert W:
$$G(1;x)=e^k$$ $$x=HW(1;e^k)$$ $\ $
So we have closed forms for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.
