Question

Use the rule logb(xy)=logbx+logby
to find log4(16⋅64)
.(1 point)

Answer 1

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:

1. Rewrite \( 16 \) and \( 64 \) in terms of base \( 4 \):

- \( 16 = 4^2 \)
- \( 64 = 4^{3} \) (since \( 64 = 4^{3/2} = (2^2)^{3} = 2^6 = 4^3 \))

2. Now apply the logarithm product rule:

\[
\log_4(16 \cdot 64) = \log_4(16) + \log_4(64)
\]

3. Substitute the rewritten values from step 1:

\[
\log_4(16) = \log_4(4^2) = 2
\]
\[
\log_4(64) = \log_4(4^3) = 3
\]

4. Combine the results:

\[
\log_4(16 \cdot 64) = \log_4(16) + \log_4(64) = 2 + 3 = 5
\]

Thus, \( \log_4(16 \cdot 64) = 5 \).