How to prove $\theta(2^r) \leq 2^{r+1}\log2$ 
Show that $\theta(2^r) \leq 2^{r+1}\log2$ where $\theta(n)= \sum_{p\leq n}\log p$ where $p$ is prime.

I have approached by induction: let $\theta(2)=\log2$ and $\theta(2^r) \leq 2^{r+1}\log2$ then
$$\theta(2^{r+1})=\sum_{p\leq 2^r}\log p + \sum_{2^r \leq p\leq 2^{r+1}}\log p \leq 2^{r+1}\log2+\sum_{2^r \leq p\leq 2^{r+1}}\log p$$
Now how to deal with next part.
 A: The key is the observation that for $k \geqslant \frac{n}{2}$, every prime between $k$ (exclusive) and $n$ (inclusive) divides $\binom{n}{k}$. With the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ that is immediate, since such a prime divides the numerator but not the denominator. So we have
$$\prod_{k < p \leqslant n} p \mathrel{\Big\vert} \binom{n}{k}.\tag{1}$$
Also, since
$$\sum_{\substack{0 \leqslant m \leqslant n \\ m \text{ even}}} \binom{n}{m} = \sum_{\substack{ 0 \leqslant m \leqslant n \\ m \text{ odd}}} \binom{n}{m} = 2^{n-1}$$
it follows that $\binom{n}{k} \leqslant 2^{n-1}$ for $n \geqslant 2$. Taking logarithms in $(1)$ and inserting the bound for $\binom{n}{k}$, we obtain
$$\theta(n) - \theta(k) \leqslant \log \binom{n}{k} \leqslant (n-1)\cdot \log 2\tag{2}$$
for $n \geqslant 2$ and $k \geqslant n/2$. With $n = 2^{r+1}$ and $k = 2^r$, this completes your induction step.
More generally, the fact that
$$\biggl\lceil \frac{\lceil y\rceil}{2}\biggr\rceil = \biggl\lceil \frac{y}{2}\biggr\rceil$$
for all $y \in \mathbb{R}$ together with $(2)$ and $\theta(x) = 0$ for $x < 2$ yields
\begin{align}
\theta(x) &\leqslant \theta(\lceil x\rceil) \\
&= \sum_{k = 0}^{\lfloor\log_2 x\rfloor-1} \Biggl(\theta\biggl(\biggl\lceil\frac{x}{2^k}\biggr\rceil\biggr) - \theta\biggl(\biggl\lceil\frac{x}{2^{k+1}}\biggr\rceil\biggr)\Biggr) \\
&\leqslant \sum_{k = 0}^{\lfloor\log_2 x\rfloor -1} \biggl(\biggl\lceil\frac{x}{2^k}\biggr\rceil-1\biggr)\cdot \log 2 \tag{by $(2)$}\\
&< \log 2 \sum_{k = 0}^{\lfloor\log_2 x\rfloor-1} \frac{x}{2^k} \tag{$\lceil y\rceil - 1 < y$} \\
&< (2\log 2)\cdot x
\end{align}
for all $x > 0$.
