@DanielFischer A ok, we haven't said anything about it in class.
We have the recurrence relation $T(n)=7T\left( \frac{n}{2}\right)+n^2$ and the solution is $T(n)=\Theta(n^{\log_{4}{49}})$.
We also have the recurrence relation $T'(n)=\alpha T'\left( \frac{n}{4} \right)+n^2$.
$a=\alpha, b=4, f(n)=n^2$
$n^{\log_b a}=n^{\log_4 \alpha}$
For $\alpha>16$, $f(n)=O(n^{\log_b{\alpha}}- \epsilon), \epsilon>0$
So for $\alpha>16$: $T'(n)=\Theta(n^{\log_4{\alpha}})$
So that $A'$ is asymptotically faster than $A$, it has to hold $n^{\log_4{\alpha}}< n^{\log_4{49}} \Rightarrow \log_4{\alpha}<\log_4{49…