We know that any positive polynomial function grows faster than any polylogarithmic function.
Or, if $c > 0$ , $\log n = O(n^c)$
if we put upper bound of $\log n$ in the given recurrence relation it becomes :
$T(n) = {\color{Blue} 2T\left ( \frac{n}{4} \right ) } + O(n^c)$
Complexity of the first term in the right hand side is = $O(\sqrt n)$.
If we choose a small value for c such as 0.001 then, $\sqrt n > n^{0.001}$
=> $T(n) = O(\sqrt n)$
