quicksort - Why is $\log_4 n$ used as an approximation instead of $\log_2 n$?
No, $\log_4 n$ is not the same as $\log_2 n$. $\log_4$ is exactly $\frac{\log_2 n}{2}$. At the end of the complexity computation you can obviously say that $O(\log_4 n) = O(\log_2 n)$. But at this point they aren't computing the complexity (yet). They are doing "exact" math. If you have an array with $n$ elements and you always divide it by $4$, it will take exactly $\lceil\log_4 n \rceil$ iterations until you reach the size $1$, and not $\lceil \log_2 n \rceil$. 

One more thing. They don't proof, that the average runtime is $\Theta(n \lg n)$. Because the array will obviously not always split with a ratio of 1:3. They only give you some intuition, why the complexity makes sence. Like they mention in the article:


  Showing that the average-case running time is also $\Theta(n \lg n)$ 
  takes some pretty involved mathematics, and so we won't
  go there.