# How to solve equation $e^x \log x=2$ [closed]

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.

• $x=1.5372017025783550472\cdots$ --- 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. Jun 26 '20 at 8:19
• Is there a closed form solution not only the numerical solution? Jun 26 '20 at 8:51
• "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 $U(2)$. Jun 26 '20 at 9:01
• 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. Jun 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: $$x=Z(e,-1)$$
• 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. Jun 26 '20 at 9:34
• 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 $-1$. Jun 26 '20 at 9:57