Assuming base as $2$
$$\begin{align}\frac{\log_2{64}}{\log_2{16}} &= \frac{\log_2{2^6}}{\log_2{2^4}}\\[2em] &= \frac{6}{4} \qquad \Bigl \{ \text{using property: } \log_a{a^m} = m\\[1em] &= \frac{3}{2}\end{align}$$
Also, you don't even need to assume the base $2$ for $\log$. Remember the following property of logarithms:
$$\boxed{\frac{\log_x{\color{blue}{a}}}{\log_x{\color{red}{b}}} = \log_{\color{red}{b}}{\color{blue}{a}}}$$
So,
$$\begin{align}\frac{\log_x{\color{blue}{64}}}{\log_x{\color{red}{16}}} &= \log_{\color{red}{16}}{\color{blue}{64}}\\[1em] &= \frac{3}{2} \qquad \Bigl \{ \because \sqrt{16^3} = 16^{3/2} = 64 \end{align}$$
