Estimate $n$ such that $\log(n^C)<n$

By a result of Comtet, if $x_n =n/ \log n$ then $$ n = x_n \left( {\log x_n + \log \log x_n + \mathcal{O}\left( {\frac{{\log \log x_n }}{{\log x_n }}} \right)} \right). $$ See https://gallica.bnf.fr/ark:/12148/bpt6k480298g/f1092.image p. 1087

The inequality $\log n^C < n$ is equivalent to $ \dfrac{\log n}{n} < \dfrac 1C$. Since $\dfrac{\log n}{n} \le \dfrac 1e$ for all $n \in \mathbf N$ we may as well assume that $C \ge e$.

This is a good place to apply the Lambert $W$-function, although not its usual branch. The function $y = xe^x$ is decreasing on $(-\infty,-1)$ and increasing on $(-1,\infty)$. Its minimum value at $x=-1$ is $y=-1/e$.

The $W$-function is usually defined as the inverse of $y = xe^x$ on the interval $[-1/e,\infty)$. Thus if $y \in [-1/e,\infty)$ then $W(y)$ is the unique $x \in [-1,\infty)$ satisfying $y=xe^x$.

On the other hand, if $y \in [-1/e,0)$ you can define $W_{-1}(y)$ as the unique $x \in (-\infty,-1)$ satisfying $y = xe^x$. This is the particular Lambert function that is useful here. Observe that $W_{-1}$ is decreasing.

With that out of the way you can do a simple calculation. With $x = \log n$ you get $$0 < xe^{-x} = \frac{x}{e^x} = \frac{\log n}{n} < \frac 1C \le \frac 1e$$ so that $$- \frac 1e \le - \frac 1C < -xe^{-x} < 0.$$ Thus $- \dfrac 1C$ and $-xe^{-x}$ both belong to the domain of $W_{-1}$ and since $W_{-1}$ is decreasing you get $$-x = W_{-1}(-xe^{-x}) < W_{-1} \left( - \frac 1C \right)$$ so that $$ n > e^{-W_{-1} \left( - \frac 1C \right)}.$$

For $y \in [-1/e,0)$ the value of $W_{-1}(y)$ can be computed in Wolfram using productlog$(-1,y)$. For instance, with $C = 100$ we find $$n > \mathrm{exp}(-\mathrm{productlog}(-1,-1/100)) \approx 647.2775.$$