Use the rule logb(xy)=logbx+logby

Question

to find log4(16⋅64)
.(1 point)

give me the answer in the simplest form

Answer 1

To find \(\log_4(16 \cdot 64)\), we can apply the rule \(\log_b(xy) = \log_b x + \log_b y\).

First, we can express the log as:

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

Now, we calculate \(\log_4(16)\) and \(\log_4(64)\).

1. For \(\log_4(16)\):

   • Since \(16 = 4^2\), \[ \log_4(16) = \log_4(4^2) = 2 \]

2. For \(\log_4(64)\):

   • Since \(64 = 4^3\) (as \(4^3 = 64\)), \[ \log_4(64) = \log_4(4^3) = 3 \]

Putting it all together:

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

Thus, the answer in its simplest form is:

\[ \boxed{5} \]