$\sum_{p\le x} \frac{1}{pq}$ I was given that $\sum_{p\le x} \frac{1}{p}$ = $\log\log x$+O(1). 
I need to show that $\sum_{pq\le x} \frac{1}{pq} = (\log \log x)^2 + O(\log \log x)$.
Here we go:
Break the sum into two sums: $\sum_{p\le x} \frac{1}{p}\sum_{q\le \frac{x}{p}} \frac{1}{q}$
Using what I was given: $(\log \log x +O(1))(\log \log \frac{x}{p} +O(1))$
Log Rules: $(\log \log x +O(1))(\log( \log x - \log p) +O(1))$
Algebra: $\log \log x \cdot \log(\log x - \log p) + O(\log \log x)$ 
From here I am lost. Any ideas?
 A: I am assuming you mean that you are given that
$$
\sum_{p\le x}\frac1p=\log(\log(x))+O(1)\tag{1}
$$
and that you want to estimate
$$
\begin{align}
\sum_{pq\le x}\frac1{pq}
&=\sum_{p\le x}\sum_{q\le x/p}\frac1{pq}\tag{2}\\
&=\sum_{p\le x}\sum_{2\le q\le x/p}\frac1{pq}\tag{3}\\
&=\sum_{p\le x/2}\frac1p\left(\log(\log(x/p))+O(1)\right)\tag{4}\\
&=\sum_{p\le x/2}\frac1p\log(\log(x)-\log(p))+O(1)\sum_{p\le x}\frac1p\tag{5}\\
&=\sum_{p\le x/2}\frac1p\left[\log(\log(x))+\log\left(1-\frac{\log(p)}{\log(x)}\right)\right]+O(\log(\log(x)))\tag{6}\\
&=\log(\log(x))^2+O(\log(\log(x)))+\sum_{p\le x/2}\frac1p\left[\log\left(1-\frac{\log(p)}{\log(x)}\right)\right]\tag{7}\\
\end{align}
$$
Explanation:
$(2)$: write explicitly as a double sum
$(3)$: note there is no $q=1$ case
$(4)$: apply $(1)$ to the sum in $q$
$(5)$: incorporate the left sum into the error term from the right sum
$(6)$: apply $(1)$ to the error term
$(7)$: $\sum\limits_{x/2\lt p\le x}\frac1p\le\log(2)$ and $(1)$ imply $\sum\limits_{p\le x/2}\frac1p=\log(\log(x))+O(1)$
It can be shown that
$$
\sum_{p\le x/2}\frac1p\left[\log\left(1-\frac{\log(p)}{\log(x)}\right)\right]=O(1)\tag{8}
$$
but I don't see how it can be shown using just $(1)$.
