Answer
$2-\frac{1}{2}log_{8}(x+1)$
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, $log_{8}(\frac{64}{\sqrt (x+1)})=log_{8}(\frac{64}{(x+1)^{\frac{1}{2}}})=log_{8}64-log_{8}(x+1)^{\frac{1}{2}}$.
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, $log_{8}64-log_{8}(x+1)^{\frac{1}{2}}=log_{8}64-\frac{1}{2}log_{8}(x+1)$.
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, $log_{8}64=2$, because $8^{2}=64$. So, $log_{8}64-\frac{1}{2}log_{8}(x+1)=2-\frac{1}{2}log_{8}(x+1)$.