A question regarding Euler's number, Riemann's zeta function and the prime-number theorem

Question

Consider now, that $\zeta(s)$ is Riemann's zeta function, such that: $$\zeta(s)=\sum_{n=1}^\infty {1\over n^s}$$ Now, consider that the prime-number theorem states that: $$\lim_{x\to\infty} {\pi(x)\over x / \ln{x}}=1$$ Where $\pi$ is the prime-counting function. Now, consider the fact that Euler's number, $e$, can be written as: $$\lim_{n\to\infty} (1+{1\over n})^n=e$$ And now, consider that: $$\pi(x)={x\over \log_{e_x}{x}}$$ Where $e_x$, is Euler's number at the $x$th order. Now, consider that if we were looking forward to calculating Euler's number based on the prime-counting function we can say that, (based on the prime-number theorem): $$e_x=(\sqrt[x]{x})^{\pi(x)}\sim e$$ Now, we now, that as $x\rightarrow\infty$, $\sqrt[x]{x}\rightarrow 1$, and now, I want you to consider that this will be similar for a lot of functions. Say we have a function, $f(x)\subset\Bbb R$, (and with some other restraints), that as $x\rightarrow\infty$, $f(x)\rightarrow 1$, there will be a function, $g(x)$, such that: $$\lim_{x\to\infty} f(x)^{g(x)}=e$$ Now, what if, we let $f=\zeta$, we'd do so, because: $$\lim_{s\to\infty}\zeta(s)=1$$ And now, we have that, for some $g$: $$\lim_{s\to\infty}\zeta(s)^{g(s)}=e$$ In other words, my question is: what function, $g(s)$, satisfies the following? $$g(s)\sim\log_{\zeta(s)}{e}$$

Answer 1

We can take $g(s)=1/\ln\zeta(s)\;$ ($s>1$).