0
$\begingroup$

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.

$\endgroup$
4
  • 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. $\endgroup$ Jun 26 '20 at 8:19
  • $\begingroup$ Is there a closed form solution not only the numerical solution? $\endgroup$
    – Archer
    Jun 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 $U(2)$. $\endgroup$ Jun 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. $\endgroup$ Jun 26 '20 at 9:42
7
$\begingroup$

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: $$ x=Z(e,-1) $$

$\endgroup$
5
  • 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. $\endgroup$
    – LSpice
    Jun 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
  • $\begingroup$ Thanks for your discussion and help! $\endgroup$
    – Archer
    Jun 26 '20 at 9:46
  • 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.