Properties of Chebyshev’s $\psi$ Function

Question

Given the definition $\psi(x)=\sum_{n\le x}\Lambda(n)$

How can one arrive to the conclusion below:

$$\psi(x)=\sum_{p\le x}\left\lfloor{\frac{\log x}{\log p}}\right\rfloor\log p $$

Answer 1

$$ \begin{aligned} \left\lfloor\log x\over\log p\right\rfloor &=\sum_{k\le\log x/\log p}1=\sum_{k\log p\le\log x}1 \\ &=\sum_{p^k\le x}1 \end{aligned} $$

Thus we have

$$ \begin{aligned} \psi(x) &=\sum_{n\le x}\Lambda(n)=\sum_k\sum_{p^k\le x}\log p \\ &=\sum_{p\le x}\left\lfloor\log x\over\log p\right\rfloor\log p \end{aligned} $$