Answer
$3−\log_4{y}$
Work Step by Step
RECALL:
(1) $\log_b{\frac{M}{N}}=\log_b{M}-\log_b{N}$
(2) $\log_b{(b^x)}=x$
Use rule (1) above with $M=64$ and $N=y$ to obtain
$=\log_4{64}-\log_4{y}
\\=\log_4{(4^3)}-\log_4{y}$.
Simplify the first term by using rule (2) above to obtain
$=3−\log_4{y}$
