Is there an exact answer to this equation? $$\frac{1}{x} +\ln(x)\ln(\ln x)=1$$
The solution to this equation is approximately $x \sim 5.13425\ldots$ but is there an exact answer?
 A: Denote this number by $\xi$. Then the exact answer is $$\xi={\rm the}\left\{x\in{\mathbb R}_{>1}\biggm|\>{1\over x}+\log x\log(\log x))=1\right\}\ ;$$
after you have proven that the set on the right hand side is indeed a singleton.
A: For comfort we transform the equation
$\frac{1}{x} +\ln(x)\ln(\ln x)=1\iff e^{\frac{x-1}{x}}=(ln\space x)^{ln\space x}$
The equation to solve is $\phi(x)= e^{\frac{x-1}{x}}-(ln\space x)^{ln\space x}=0$.
There is a clear solution $x=1$ (the domain of $(ln\space x)^{ln\space x}$ is $x\gt 1$ and the limit exists at $x=1$). On the other hand we verify that $\phi(5)$ and $\phi(6)$ have distinct sign so  there is a root of the equation in the interval $[5,6]$ (in fact $\phi(5)\approx0.07459\gt 0$ and $\phi(6)\approx-0.54304$). 
For, say $x\gt 2$, both
functions, $ e^{\frac{x-1}{x}}$ and $ (ln\space x)^{ln\space x}$,  are increasing and it is not hard to verify the solution in $[5,6]$ is the only other than $x=1$.
There are several methods of approximation of roots of a trascendental equation. Here without further details, for example, a first step $$5+\frac{(6-5)\phi (5)}{\phi(5)-\phi(6)}=5+\frac{0.07459}{0.07459-(-0.54304)}\approx 5,12076$$  And it is possible a better approximation if the process reiterates.
A: If the $\dfrac1x$ term were missing, letting $x=\exp(e^t)$, we'd be left with $te^t=1$, whose solution is the $\Omega$ constant, which is expressible in terms of the Lambert W function. As it stands, however, even the use of such special functions proves futile.
