Finding the inverse of the function $f(k, x) = k^{x}x.$ Recently, I have been looking at the function $f(x) = e^{x}x,$ where its inverse is the Lambert W function.

I was intrigued by the fact that it is rather hard to calculate its solution, in comparison with other functions.

This intrigue encouraged me to develop my own function that would trace the inverse functions of functions of the form $f(x) = k^{x}x.$

$$g(k, 0, x) = k,$$

$$g(k, n, x) = \dfrac{1}{2} \log_{k}\left(\dfrac{k^{g(k, n - 1, x)} x}{g(k, n - 1, x)}\right),$$ and

$$\lim_{n \to \infty}{g(k, n, x)} = f^{-1}(k, x),$$ where

$$f(k, x) = k^{x}x.$$

Please, feel free to try it for yourself.

My question is, has this function ever been found before, and, if not, is it helpful in any way?

I will include a 'proof' of this, if you would like.
The inverse of $f(k,x)$ can be found in terms of the Lambert W. \begin{align*} y&amp;=k^xx \\ 
y &amp;= e^{x \ln k}x \\
y\ln k &amp;= e^{x \ln k}x\ln k \\
W(y\ln k) &amp;= x\ln k \\
x &amp;= \frac{W(y\ln k)}{\ln k} \end{align*}
You have that

$$f(k,x) = \frac{1}{\ln(k)} e^{x\ln(k)} (x\ln(k)) = y$$

Hence, the inverse of your function is 

$$\frac{1}{\ln(k)} W(y\ln(k))$$

Indeed,

$$f\left(k,\frac{1}{\ln(k)} W(y\ln(k)) \right) = \frac{1}{\ln(k)}\exp \left( \ln(k)\frac{1}{\ln(k)} W(y\ln(k)) \right)\ln(k)\frac{1}{\ln(k)} W(y\ln(k)) $$

$$= \frac{1}{\ln(k)}\exp \left( W(y\ln(k)) \right) W(y\ln(k))$$

$$= \frac{1}{\ln(k)}y\ln(k) = y$$

So your function can easily be expressed with Lambert function (and vice-versa)

$$W(y) = \ln(k) f^{-1}(\frac{x}{\ln(k)}, k )$$