# Two kind of equations involving natural log and exponentiation

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 $W$

• I don't think you can get an explicit solution. Probably, only numerical methods would do the job (easily). – Claude Leibovici Mar 20 '15 at 14:32
• Yes, @ClaudeLeibovici but I'm interested in closed-forms involving only special functions, they are too easy to be solved with approximations. – Renato Faraone Mar 20 '15 at 14:34
• They would be very special functions ! – Claude Leibovici Mar 20 '15 at 14:36

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