Why $\sum_{p^k\leq x}\log(p)\lfloor\frac{x}{p^k} \rfloor= (\sum_{p\leq x}\log(p))\cdot (\sum_{p^k\leq x}\lfloor \frac{x}{p^k}\rfloor )$?

Question

Why $$\sum_{\substack{p^k\leq x\\ k\geq 1}}\log(p)\left\lfloor\frac{x}{p^k} \right\rfloor =\left(\sum_{p\leq x}\log(p)\right)\left(\sum_{\substack{p^k\leq x\\ k\geq 1}}\left\lfloor \frac{x}{p^k}\right\rfloor \right)\ \ ?$$

To me $$\sum_{\substack{p^k\leq x\\ k\geq 1}}\log(p)\left\lfloor\frac{x}{p^k} \right\rfloor=\sum_{p\leq 1}\log(p)\sum_{k\leq \frac{\log(x)}{\log(p)}}\left\lfloor \frac{x}{p^k}\right\rfloor .$$

Answer 1

This is a problem of notation, it meant for $x$ fixed : $$\sum_{p,k,p^k \le x}\log(p)\left\lfloor\frac{x}{p^k} \right\rfloor =\sum_{p\leq x}\log(p)\sum_{k, p^k \le x}\left\lfloor \frac{x}{p^k}\right\rfloor $$

Which is the same as your second formula