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

Answer 1

To find \( \log_2(16 \cdot 64) \) using the property of logarithms that states \( \log_b(xy) = \log_b x + \log_b y \), we can break down the expression as follows:

1. Identify the logarithm expression: \[ \log_2(16 \cdot 64) = \log_2(16) + \log_2(64) \]

2. Calculate \( \log_2(16) \): Since \( 16 = 2^4 \), \[ \log_2(16) = \log_2(2^4) = 4 \]

3. Calculate \( \log_2(64) \): Since \( 64 = 2^6 \), \[ \log_2(64) = \log_2(2^6) = 6 \]

4. Now, combine the results: \[ \log_2(16 \cdot 64) = \log_2(16) + \log_2(64) = 4 + 6 = 10 \]

Therefore, \( \log_2(16 \cdot 64) = 10 \).