Not Quite Fibonacci

Let $L_n = \log_2(M_n)$

Since $M_n = M_{n-1} \times M_{n-2}$ for $n>2$ , this implies that:

$L_n = L_{n-1} + L_{n-2}$ for $n>2$

So, the $L_n$ are simply the Fibonacci numbers, but they are off by 1. (Since Fibonacci numbers are defined $F_0 = 0, F_1 = 1$ , and $F_2 = 1$ )

Or, $L_n = F_{n-1}$ where $F_n$ are the Fibonacci numbers!

So, $\log_2(M_{17}) = L_{17} = F_{16} = \boxed{987}$