What's $x^n = n!$ approximately What is the approximate solution to:
$x ^{1000} = 1000 !$
How can you solve for x?
More generally, for some constant $k$, how can you solve:
$x^k = k!$
 A: You can use stirling's approximation when $n \rightarrow +\infty$:
$$\ln(n!) = n\ln(n)-n+O(\ln(n))$$
So,
\begin{align}
x^k = k! & \Leftrightarrow  k\ln(x) =\ln(k!) \\
& \Leftrightarrow  k\ln(x) =k\ln(k)-k+O(\ln(k)) \\
& \Leftrightarrow  \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\
& \Leftrightarrow  \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\
& \Leftrightarrow  x =\frac{k}{e} \times\exp{(O(\frac{\ln(k)}{k}))} \approx \frac{k}{e}\\
\end{align}
A: Notice that by the Hierarchy of limits, $\frac{x^n}{n!}\rightarrow 0$ as $n\rightarrow\infty$, so $n$ would have to be 'quite small'. In response to your first problem, $1000\log{x}=log{1000!}$ so $\log{x}=\frac{log{1000!}}{1000}$. Can you see a way to complete the problem? If not, here is a hint:
What can you deduce about the function $f(x)=\frac{log{x!}}{x}$? Can you sketch a graph of it? How does its derivative compare with $\log{x}$? You should be able to use these to find an approximate intersection point.
[To differentiate ${\log{x!}}$, consider the Gamma function definition of factorials, then apply the Fundamental Theorem of Calculus to differentiate.]
