22nd Problem 2016

$2^{3020}=4^{1510}$ and now the answer is obvious.

Here's a not so technical solution: $\frac{3020}{2016}\approx1.5$ . We could then say $3=1.5\cdot2$ and $2^{3}<7^{2}$ . In the same way, we could assume $2^{3020}<7^{2016}$ .