Find the order of magnitude of the equation solution Find the order of magnitude of the following equation solution:
$$
x(\ln x)^{2001}=n
$$
 A: Before computing the order of magnitude of the solution, we can get a solution using the Lambert W function:
$$
\begin{align}
n
&=x\log(x)^{2001}\\
&=\log(x)^{2001}\ e^{\log(x)}\\
n^{1/2001}/2001
&=\log(x)/2001\ e^{\log(x)/2001}\\
\mathrm{W}(n^{1/2001}/2001)
&=\log(x)/2001
\end{align}
$$
Therefore,
$$
x=e^{2001\mathrm{W}(n^{1/2001}/2001)}
$$

Note that
$$
\begin{align}
\lim_{n\to\infty}\frac{\log(n)}{\log(x)}
&=\lim_{n\to\infty}\frac{\log(x)+2001\log(\log(x))}{\log(x)}\\
&=1+2001\lim_{x\to\infty}\frac{\log(\log(x))}{\log(x)}\\[6pt]
&=1\tag{1}
\end{align}
$$
Therefore, as Antonio Vargas points out, $(1)$ implies that
$$
x=n\log(x)^{-2001}\sim n\log(n)^{-2001}\tag{2}
$$
Thus, in terms of orders of magnitude, $(2)$ is a more precise statement of
$$
x=O\left(n\log(n)^{-2001}\right)\tag{3}
$$

Here is a plot of $\log(\log(x))$ vs $\log(\log(n))$:
$\hspace{3.2cm}$
For $\log(\log(n))\le5$ (i.e. $n\le2.85112\times10^{64}$), $\log(\log(x))$ is very close to $0$ (i.e. $x\approx e$).
For $\log(\log(n))\ge12$ (i.e. $n\ge3.2197\times10^{70683}$, $\log(\log(x))$ is very close to $\log(\log(n))$.  
A: Taking into account robjohn's answer, the approximate value is $e$.  I give you a few solutions for $x(n)$. $$x(10^0)=2.71692$$ $$x(10^1)=2.72005$$ $$x(10^2)=2.72319$$ $$x(10^3)=2.72633$$ $$x(10^4)=2.72948$$  $$x(10^5)=2.73263$$  $$x(10^6)=2.73579$$ $$x(10^7)=2.73896$$ $$x(10^8)=2.74214$$ $$x(10^9)=2.74533$$ while $e=2.71828$.
You could check that we obtain a value $x=3$, which is only $10$% larger than $e$, it would be required that $n>1.61\times 10^{82}$ which is quite large.
Added after robjohn's last answers
As shown by robjohn, the solution of $$x \log
(x)^k=n$$ is given by $$x=e^{k\mathrm{W}(n^{1/k}/k)}$$ which corresponds to $$\log( x)=k\mathrm{W}(n^{1/k}/k)$$ For large values of its argument, Lambert function can be expanded as $$\mathrm{W}(y) \simeq \log (y)-\log (\log (y))+\frac{\log (\log (y))}{\log (y)}$$ Using $y=\frac{n^{\frac{1}{k}}}{k}$, $\log(y)=\frac{\log (n)}{k}-\log (k)$ and then $$\log(x) \simeq k \left(\frac{\log (n)}{k}-\log (k)-\log \left(\frac{\log (n)}{k}-\log
   (k)\right)+\frac{\log \left(\frac{\log (n)}{k}-\log
   (k)\right)}{\frac{\log (n)}{k}-\log (k)}\right)$$ which shows the respective roles of $n$ and $k$ and the fact that the asymptotic value is  $\log(x) \simeq \log(n)-k \log(k)$
For $k=2001$ and $n=10^{1000000}$, the previous expansion gives $x=2.08528\times 10^{987280}$ for an exact value $x=2.14287\times10^{987280}$ as reported by robjohn.
