I want to know how to solve the equation
$$e^x\log x=2.$$
We can get a numerical solution but it seems difficult to get an exact solution. I know the Lambert W function but unable to use it for the above equation.



How to solve equation $e^x \log x=2$
I do not think that there is any known special function which solves your equation. However, a solution can be given in a sort of 'infinite exponent tower'. More precisely, rewrite the equation as:
$$ \ln x = 2e^{-x} \Rightarrow x = e^{2e^{-x}} $$
Then, the solution is:
$$ x=e^{2e^{-e^{2e^{...}}}} $$
Following a similar idea to the one given by @Carlo Beenakker in the comments, you could then define a function $U$ as follows:
$$ U(a,b,c):= a^{ba^{ ca^{ ba^{ ...}}}} $$
(alternating $b$ and $c$)
Then, you can study when this new function converges. It is clear that the solution to your equation is:
$$ x=U(e,2,-1) \approx 1.537201702578...$$
Notice that the function $U$ is a generalisation of the operation of infinite tetration. Indeed:
$$ U(a,1,1) = a^{a^{a^{ ...}}} = \frac{W(-\ln a)}{- \ln a} $$


EDIT: as noticed by @LSpice, we can even simplify this and consider the function (with only two variables):
$$ Z(a,b) := a^{-2b a^{ba^{ ...}}}$$
(again, alternating) and get:
$$ x=Z(e,-1) $$


