Answer
$3-log_{4}y$
Work Step by Step
Based on the quotient rule of logarithms, we know that $log_{b}(\frac{M}{N})=log_{b}M-log_{b}N$ (where $b$, $M$, and $N$ are positive real numbers and $b\ne1$).
Therefore, $log_{4}(\frac{64}{y})=log_{4}64-log_{4}y$.
Based on the definition of the logarithmic function, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$ and $b\gt0$, $b\ne1$).
Therefore, $log_{4}64=3$, because $4^{3}=64$. So, $log_{4}64-log_{4}y=3-log_{4}y$.
