Use the rule logb(xy)=logbx+logby to find log2(16⋅64) .

Answer 1

To find \( \log_2(16 \cdot 64) \) using the rule \( \log_b(xy) = \log_b x + \log_b y \), we can express it as follows:

\[ \log_2(16 \cdot 64) = \log_2(16) + \log_2(64) \]

Now, we can calculate \( \log_2(16) \) and \( \log_2(64) \):

1. \( 16 \) can be expressed as \( 2^4 \), so: \[ \log_2(16) = \log_2(2^4) = 4 \]

2. \( 64 \) can be expressed as \( 2^6 \), so: \[ \log_2(64) = \log_2(2^6) = 6 \]

Now, we can substitute these values back into our equation:

\[ \log_2(16 \cdot 64) = \log_2(16) + \log_2(64) = 4 + 6 = 10 \]

Thus, the final result is:

\[ \log_2(16 \cdot 64) = 10 \]