Solve equation $x \log(x) = k$ for $k > 2$ I am trying to solve this equation $x \log(x) = k$,
It's easy to show that the function $f(x)=x \log(x) -k$ has a unique solution $f(x)=0$ for $k>2$. But i need to get results about $x$ with a good precision.
 A: Let $f(x)=x\log x-k$. By Newton-Raphson Method:
$$x_{n+1}=x_n-\dfrac{x_n\log x_n-k}{\log x_n+1}$$
You can compute the solution for required values of $k$. You cannot explicitly find a general solution for any $k$ using numerical methods, the reason specifically for this method is that you need to make an initial "guess" about the value of $x_0$ which you'd be able to do if you have an actual value of $k$ in hand.
As pointed by @PierreCarre choosing $x_0=k$ gives us an iteration sequence that converges very fast to the solution of the equation.

For finding the general solution and then approximating using Lambert's $W$-function. Let $x=\exp u$, so we have:
$$\begin{aligned}x\log x&=k\\\exp u\exp \ln u&=k\\ u\exp u &=k \\ u&=W(k)\\ x&=\exp W(k)\\ x&=\dfrac{k}{W(k)}\end{aligned}$$
A: For $k>0$ this equation has the solution
$$x = \exp(\text{W}(k)),$$
in which $\text{W}$ is the Lambert function.
There are different ways to represent this function. For example, you could use continued fractions, such that you can calculate $\text{W}(k)$ and then $\exp(\text{W}(k)).$
A: Unfortunately, this equation can't be solved analytically but can be well tackled using Newton's method as follows $$x_0=k$$$$x_{n+1}={x_n+{k\over \ln 2}\over 1+\ln x_n}$$We continue iterations as soon as we attain to some precision.
