How to solve the equation $x \log \log x = n$ I would like to solve the equation $x \log\log x = n$. I've seen a lot of post about the equation $x \log x$ but here I have a composition of $\log$.
How can I solve it ?
Thank you very much.
 A: With the estimate $x=\dfrac n{\log(\log(n))}$, you have
$$\dfrac n{\log(\log(n))}\log\left(\log\left(\dfrac n{\log(\log(n))}\right)\right)=\dfrac n{\log(\log(n))}\log\left(\log(n)-\log\left(\log(\log(n))\right)\right)$$
which is asymptotically $n$.
For example, for $n=100$,
$$x=\frac{100}{\log(\log(100))}=65.4801821\cdots$$
and
$$x\ln(\ln(x))=93.684410\cdots.$$
A: Let $x_n$ denote the solution of $x \log \log x = n$. Here is an asymptotic expansion for $x_n$.
Since $x \log\log x$ is monotone increasing and goes to $\infty$, it is clear that $x_n$ is motonone increasing and $\lim_n x_n = \infty$.
Since $0=\lim_n\frac{1}{\log \log x_n}=\lim_n \frac{x_n}{n}$, we have the estimate $x_n = o(n)$
Sharpening this is easy: $\ln x_n + \ln\ln\ln x_n = \ln n$ hence $\ln x_n + o(\ln x_n) =\ln n$ and $\ln x_n = \ln n +o(\ln n)$
Plugging the last estimate in the equation $\ln x_n + \ln\ln\ln x_n = \ln n$, $$\ln x_n + \ln\ln(\ln n +o(\ln n))=\ln n\\\ln x_n + \ln\ln [\ln n(1+o(1))]= \ln n\\ \ln x_n + \ln[\ln\ln n +\ln(1+o(1))]=\ln n\\ \ln x_n + \ln[\ln\ln n +o(1)]=\ln n\\ \ln x_n + \ln\ln\ln n + o\left(\frac1{\ln\ln n} \right)=\ln n\\$$
Hence $\displaystyle x_n = \exp(\ln n)\exp(-\ln\ln\ln n))\exp\left(o\left(\frac1{\ln\ln n} \right)\right)=\frac{n}{\ln \ln n}+o\left( \frac{n}{\ln \ln n}\right)$
Plugging the estimate $\ln x_n + \ln\ln\ln n + o\left(\frac1{\ln\ln n} \right)=\ln n$ back into $\ln x_n + \ln\ln\ln x_n = \ln n$ yields an even more precise estimate, ad libitum:
$$ x_n = \frac{n}{\ln \ln n} + \frac{ \ln\ln\ln n}{\ln n (\ln\ln n)^2}+o\left( \frac{ \ln\ln\ln n}{\ln n (\ln\ln n)^2}\right)$$
