Use the rule logb(xy)=logbx+logby

Question

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, ${\log }_4(16\cdot 64)=5$.