$$n=\frac{e^x}{\log x}$$
I've been thinking for several hours about how to find the inverse function, but I always get to a nested function.
They can help me or recommend some literature that allows me to solve this problem. In theory there is its inverse clearly not with elementary functions, I am interested in knowing and learning about new techniques for this type of functions. I appreciate the help.
Concerning the largest root (when it exists), you can have a quite good approximation of it after one single iteration of Halley method starting with $x_0=\log(n)$ writing the function as $$f(x)=x-\log(\log(x))-\log(n)$$ Using $$L_1=\log(n) \qquad\qquad L_2=\log(L_1)\qquad\qquad L_3=\log(L_2)$$ we obtain $$x_1=L_1+\frac{2 L_1 L_2L_3(L_1 L_2-1) }{2 (L_1 L_2-1)^2+(L_2+1) L_3}$$ Using this estimate, one single iteration of Newton method would give almost the solution $$x_2=x_1+\frac{x_1 \log (x_1) (\log (n)-x_1+\log (\log (x_1)))}{x_1 \log (x_1)-1}$$
Using $n=10^k$, some results $$\left( \begin{array}{cccc} k & x_1 & x_2 & \text{solution} \\ 2 & 5.09162713306291 & 5.09236408502652 & 5.09236407321316 \\ 3 & 7.61551194849834 & 7.61590897000061 & 7.61590896893251 \\ 4 & 10.0461676672324 & 10.0463804200754 & 10.0463804199298 \\ 5 & 12.4373422964443 & 12.4374674557571 & 12.4374674557281 \\ 6 & 14.8068649302967 & 14.8069444919917 & 14.8069444919841 \\ 7 & 17.1628132076291 & 17.1628668931673 & 17.1628668931649 \\ 8 & 19.5095090608892 & 19.5095470059521 & 19.5095470059513 \\ 9 & 21.8495239633877 & 21.8495517890709 & 21.8495517890705 \\ 10 & 24.1845060486885 & 24.1845270719237 & 24.1845270719235 \\ 11 & 26.5155717024025 & 26.5155879824867 & 26.5155879824866 \\ 12 & 28.8435102122394 & 28.8435230826814 & 28.8435230826813 \\ 13 & 31.1688990269469 & 31.1689093823764 & 31.1689093823764 \\ 14 & 33.4921726027437 & 33.4921810617013 & 33.4921810617012 \\ 15 & 35.8136655305610 & 35.8136725319265 & 35.8136725319265 \end{array} \right)$$
We could even do better using the first iterate of Householder method for the generation of $x_1$ but the formula would become quite nasty.