\(\log_2 \left(\dfrac{2}1\right) + \log_2\left(\dfrac32\right)+\cdots + \log_2 \left(\dfrac{64}{63}\right) \\ = \log_2 \left(\dfrac21 \times \dfrac32 \times \cdots \times \dfrac{64}{63}\right)\\ = \log_2 64\\ = 6\)
The answer is obvious at this point.
