Asymptotic growth of solutions to $x - C \log x = 0$

Question

For $C>3$ a constant, I'm interested in the roots of the function $$ f(x) = x - C \log x, $$ defined for $x > 0$. It's straightforward to say that this function has two roots by observing $f$ has a single local minimum at $x=C$ with $f(C) < 0$. As $C \rightarrow \infty$, one of these roots approaches $1$, however, the second (larger) root grows at least linearly with $C$. Could the growth of this second root be bounded in terms of $C$? In particular, I want to find an asymptotic upper bound for the larger root of $f$ as $C$ grows.

One easy bound is $e^C$, as $x > e^C$ implies $x > \log^2 x > C \log x$. Testing numerically, I feel this bound is very weak, but I have no idea how this could be improved or what the best possible bound could be.

Answer 1

For any $\alpha > 1$, we can check that

$$f(\alpha C\log C)=(\alpha-1+o(1))C\log C \qquad\text{as}\quad C\to\infty.$$

In particular, this tells that $x < \alpha C\log C$ for large $C$. On the other hand,

$$ x = C\log x = C\log(C\log x) \geq C\log C. $$

So it follows that $x \sim C\log C$ as $C\to\infty$. Throwing this to the identity $x = C\log x$, we easily check that

$$ x = C \left( \log C + \log\log C + \mathcal{O}\left( \frac{\log\log C}{\log C}\right) \right) \qquad\text{as}\quad C\to\infty. $$

Answer 2

This is a historically famous problem. Notice that LHS of $x/\log x = C$ is the approximation of the number of primes $\le x$ so essentially what you are asking is how large must $x$ be so that the number of primes $\le x$ is equal to $C$. So clearly, $x$ bust be at least as large as the $C$-th prime. This was first inverted by Marin Cipolla in 1902 to get the asymptotic expansion of the $C$-th prime in terms of $C$. We have

$$ x = C\Big[\log C + \log \log C - 1 + \frac{\log\log C - 2}{\log C} + O\Big(\frac{1}{\log^2 C}\Big) \Big] $$

as $C \to \infty$.For the full expansion, refer to Theorem 2.1 in this link and the introduction in the paper of this link.