For any integer $x \geqslant 2$, define $\theta(x)=\sum_{p \leqslant x} \log p$, where the sum is taken over all primes $p \leqslant x$. Put $\theta(1)=0$. By studying the integer

$$\left(\begin{array}{c} 2 n \\ n \end{array}\right)$$

where $n \geqslant 1$ is an integer, prove that

$$\theta(2 n)-\theta(n)<2 n \log 2$$

Deduce that

$$\theta(x)<(4 \log 2) x$$

for all $x \geqslant 1$.