Answer
$\frac{1}{2}log_{4}x-3$
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, $log_{4}(\frac{\sqrt x}{64})=log_{4}(\frac{x^{\frac{1}{2}}}{64})=log_{4}x^{\frac{1}{2}}-log_{4}64$.
According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number).
Therefore, $log_{4}x^{\frac{1}{2}}-log_{4}64=\frac{1}{2}log_{4}x-log_{4}64$.
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, $\frac{1}{2}log_{4}x-log_{4}64=\frac{1}{2}log_{4}x-3$.
