I am trying to prove a relatively simple bound for the problem below. I want someone to check if my solution is good :)
For a $P=\{ p_1,p_2,...,p_k \}$ a set of prime factors, we define as $N_P= \{ n\in \mathbb{N}: p|n \Rightarrow p\in P \}$. We also define as $N_P(x)= \# \{n\in N_P : n \leq x \}$.
I want to prove that exist a constant $C$ and a $x_0$, which depend on $P$, such that: $$ N_P(x) \leq C (\log x)^k \; \text{for $x \geq x_0$} $$
My attempt:
My first idea was to use somehow use the Prime Number Theorem (PMT): $\pi(x) \sim x/\log x $, for a sufficiently large $x$. But since the $\log$ is on the denominator, I think this is hopeless.
So I made this argument: Let $n\in N_P$, then $n$ has prime factorization that contains only primes from the set $P$.
Thus $n=p_1^{n_1}p_2^{n_2}...p_k^{n_k}$. For $n \leq x$, we must have that: $ p_i^{n_i} \leq x \Rightarrow n_i \leq \frac{\log x}{\log p_i} $ for $\forall i$.
Thus for $x\geq p_k$ $$N_p \leq \frac{\log x}{\log p_1} \frac{\log x}{\log p_2} ... \frac{\log x}{\log p_k} = \frac{1}{\log p_1 \log p_2 ... \log p_k} (\log x)^k = C (\log x)^k $$ for $x\geq x_0=p_k $.
What do you think? Thanks in advance for your time!
There's a little problem in your proof:
After you get $n_i \le \frac{\log x}{\log p_i}$ you need to consider that $n_i$ can take $\left\lfloor \frac{\log x}{\log p_i}\right\rfloor \color{red}{+1}$ values, because $n_i$ can be $0$.