 Solution

The Prime number theorem says $\pi (x)\sim x/\log x$ as $x\to \infty$. Bertrand's postulate says that for every integer $n\ge 2$ there is a prime number strictly between $n$ and $2n$.

To count S-smooth numbers, split each prime exponent into its parity and an even part. This gives the unique expression $n=a{b}^2$, where $a$ is a squarefree product of a subset of $S$. There are $2^{|S|}$ possible $a$, while $b\le \sqrt{x}$ whenever $n\le x$. The squarefree-part bound for smooth numbers therefore gives $f_S(x)\le 2^{|S|}\lfloor \sqrt{x}\rfloor \le 2^{|S|}\sqrt{x}$.

For an integer $x\ge 1$, take $S$ to be all prime numbers at most $x$. Every positive integer at most $x$ is then S-smooth, so $x\le 2^{\pi (x)}\sqrt{x}$. Taking logarithms yields