Find the order of magnitude of the equation solution

Question

Find the order of magnitude of the following equation solution: $$ x(\ln x)^{2001}=n $$

Answer 1

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, $$ x=e^{2001\mathrm{W}(n^{1/2001}/2001)} $$

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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 $$ x=O\left(n\log(n)^{-2001}\right)\tag{3} $$

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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 $\log(\log(n))$.

Answer 2

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.