We hypothesize that: \[T(2^n) = n(5 \cdot 2^{2(n-1)})\]

Proof by induction:
This holds for the base cases \(n \in \{0, 1, 2, 3, 4\}\) (verifying this is left as an exercise for the reader).

Now we show that the above expression is true for all \(n \geq 0\). Given that \(T(2^n) = n(5 \cdot 2^{2(n-1)})\) is true for some \(n\), we need to show that \(T(2^{n+1}) = (n + 1)(5 \cdot 2^{2n})\). Using the above definition of \(T(n)\), we get: \[\begin{split} T(2^{n+1}) &= 5 \cdot (2^n)^2 + (2^2)T(2^n)\\ &= 5 \cdot 2^{2n} + (2^2)(n(5 \cdot 2^{2(n-1)}))\\ &= 5 \cdot 2^{2n} + n(5 \cdot 2^{2 + 2(n-1)})\\ &= 5 \cdot 2^{2n} + n(5 \cdot 2^{2n})\\ &= (n + 1)(5 \cdot 2^{2n}) \end{split}\] Q.E.D

This gives us that: \[\begin{split} T(n) &= (\log{n})(5 \cdot 2^{2(\log(n) - 1)})\\ &= (\log{n})(5 \cdot 2^{2(\log(n) - 1)})\\ &= (\log{n})(5 \cdot 2^{2\log(n) - 2})\\ &= (\log{n})(5 \cdot 2^{\log(n^2)} \cdot 2^{-2})\\ &= (\log{n})(\frac{5}{4}n^2)\\ &= \frac{5}{4}n^2\log{n}\\ T(n) &\in \mathcal{O}(n^2\log{n}) \end{split}\]