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.

4$\begingroup$ $x=1.5372017025783550472\cdots $  I'm not sure what you mean by an "exact" solution  like in terms of some known special function? that is not likely to be forthcoming. $\endgroup$– Carlo BeenakkerJun 26 '20 at 8:19

$\begingroup$ Is there a closed form solution not only the numerical solution? $\endgroup$– ArcherJun 26 '20 at 8:51

5$\begingroup$ "closed form" in the sense of a known special function: unlikely; however, you could define a new special function $U(a)$ by $e^x \log x=a$, and then the number you seek is $U(2)$. $\endgroup$– Carlo BeenakkerJun 26 '20 at 9:01

5$\begingroup$ Maple does not find a closed form, and Maple is pretty good with the Lambert W, so it seems that W will not help with closed form for this. $\endgroup$– Gerald EdgarJun 26 '20 at 9:42
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: $$ x=Z(e,1) $$

3$\begingroup$ Of course $U$ could be boiled down to a function of $2$ variables; you can just put $x = e^{2U(e, 2)}$, in the obvious notation. $\endgroup$– LSpiceJun 26 '20 at 9:34

$\begingroup$ Yes: this is even better, because you simplify the study of convergence of such new function $\endgroup$ Jun 26 '20 at 9:35


4$\begingroup$ It seems to me, from $x=\exp(2\exp(x))$ we get not what you said, but instead $$ x = \exp\Bigg(2\exp\bigg(\exp\Big(2\exp(\dots)\big)\Big)\bigg)\Bigg)\\ $$ where we alternate coefficnents $2$ and $1$. $\endgroup$ Jun 26 '20 at 9:57

$\begingroup$ Thanks for noting this; I have now corrected the answer. $\endgroup$ Jun 26 '20 at 10:06