TITLE: Solving $x\log x=1$ without numerical methods
QUESTION [6 upvotes]: I tried solving $x\log x=1$ where $x \in (1,e) \subset \mathbb{R} $ but couldn't find a nice way to solve it. 

Given that $x \in (1,e)$, I tried $x:=e^\alpha$ for $\alpha \in [0,1]$ so that I obtain $\alpha e^\alpha=1$. This resembles the familiar exponential function $f(t)=e^{\alpha t}$.
I tried to recover some useful relations and I found the following:

a) $\forall t \in [0,1]\quad f^n(t)f^{n+1}(t)=\alpha^{2n}$
b) Using the mean-value theorem, I can show that:
$\int_{0}^{1} f^n(t)dt=\alpha^{n-1}(\frac{1}{\alpha}-1)$ which isn't too difficult to show given that $\exists c_n \in (0,1) \quad f^{n+1}(c_n)=f^n(1)-f^n(0)=\alpha^n f'(c_0)$
I found that $c_0 = \frac{1}{\alpha}\log(\frac{1}{\alpha^2}-\frac{1}{\alpha})$
But, after all this work I still couldn't figure out how to find $\alpha$. Perhaps there's an additional relation which I simply failed to discover?

Anyway, at this point I tried Newton's method:

Using $\large t_{n+1}=t_n-\frac{f(t_n)}{f'(t_{n+1)}}=t_n-\frac{t_n\ln(t_n)+t_n^2}{1+t_n}$
I found that $\alpha \approx 0.567143$. But, I don't consider this a nice solution as I haven't found the exact value of $\alpha$. Might there be a better way that doesn't use numerical methods?

REPLY [2 votes]: The exact solution requires the use of the Lambert W function.
$$x\log x=1$$
$$(\log x)e^{\log x}=1$$
Now use the relation of the Lambert W function, $W(z)e^{W(z)}=z$:
$$\log x=W(1)$$
$$x=e^{W(1)}=\frac1{W(1)}$$
$W(1)$ is sometimes called the omega constant, denoted as $\Omega$. This happens to be equivalent to the $\alpha$ you found with Newton's method.