This problem is from Problem in Analytic Number theory by M.Ram Murty (Page no. 128).\
$\pi(x,z) = \# (n \leq x: n \ \text{is not divisible by any prime} \ p < z)$
We have, $\pi(x,z) \ll \frac{x}{\log z} +O(2^z).$
One can write $\pi(x) \leq \pi(x,z) + z$, where $\pi(x)=$ number of prime upto $x$.
Choose $z= \log x$, prove that $\pi(x)= O\bigg(\frac{x}{\log \log x}\bigg)$.
After plugging $z= \log x$, I got $\pi(x)= \frac{x}{\log \log x} + x^{\log 2} + \log x$
I used the fact that $ x^{\log y}= y^ {\log x}$ in intermediate step.
Also, consider base of the logrithmic is e.