To find \( \log_4(16 \cdot 64) \) using the rule \( \log_b(xy) = \log_b(x) + \log_b(y) \), we can break it down as follows:
-
Rewrite \( 16 \) and \( 64 \) in terms of base \( 4 \):
- \( 16 = 4^2 \)
- \( 64 = 4^{3} \) (since \( 64 = 4^{3/2} = (2^2)^{3} = 2^6 = 4^3 \))
-
Now apply the logarithm product rule:
\[ \log_4(16 \cdot 64) = \log_4(16) + \log_4(64) \]
-
Substitute the rewritten values from step 1:
\[ \log_4(16) = \log_4(4^2) = 2 \] \[ \log_4(64) = \log_4(4^3) = 3 \]
-
Combine the results:
\[ \log_4(16 \cdot 64) = \log_4(16) + \log_4(64) = 2 + 3 = 5 \]
Thus, \( \log_4(16 \cdot 64) = 5 \).