TITLE: Show $x^{\pi(x)} < 3^x$ using the PNT.
QUESTION [5 upvotes]: Using the Prime Number Theorem show that:
$$x^{\pi(x)} < 3^x$$ for sufficiently large $x$.

I started off by taking the $\log$ of the inequality such that:
$$\log(x^{\pi(x)}) < \log(3^x)$$
so $$\pi(x) \log(x) < x \log(3)$$
by PNT we know that $$\pi(x) \sim \dfrac{x}{\log(x)}$$
so can we then just sub this in and obtain:
$$\dfrac{x}{\log(x)} \log(x)<x\log(3)$$
$$x<x\log(3)$$
$$1<\log(3)$$
Can I do this? This is for a Distribution of Primes course, I'm trying to get my head around using the PNT! Thanks for any help!

REPLY [3 votes]: You need to be a bit careful when handling the $\sim$ relation. You can't just substitute it in the inequality, since it is only an asymptotic relation. This is one way to do it:
The $\sim$ operator is usually defined as 
$$f(x)\sim g(x) \iff \lim_{x \to \infty} \frac{f(x)}{g(x)}=1$$
So, 
$$\pi(x) \sim \frac x{\log x} \iff \lim_{x \to \infty}\frac{\pi(x) \log x}{x}= 1 $$
Dividing the inequality $\pi(x)\log x < x \log 3$ by $x$ gives you
$$\frac{\pi(x)\log x}x < \log 3$$
Finally, by taking the limit to infinity of the left hand side you can conclude that the value will approach $1$ for a large enough $x$, and you are done, since $3 > e \implies \log 3 >1$.