Log_2 64 + log_2 16 - log_2 9

We need to find the power to which 2 must be raised to get 64.

$$2^{x} = 64$$

Since $2^{6} = 64$, we have $\log_2 64 = 6$.

We need to find the power to which 2 must be raised to get 16.

$$2^{y} = 16$$

Since $2^{4} = 16$, we have $\log_2 16 = 4$.

The term $\log_2 9$ cannot be simplified to an integer, as 9 is not an integer power of 2.
Substitute the values from Step 1 and Step 2 into the expression:

$$\log_2 64 + \log_2 16 - \log_2 9 = 6 + 4 - \log_2 9$$ $$6 + 4 - \log_2 9 = 10 - \log_2 9$$

The final answer is $10 - \log_2 9$.