Originally Posted by
ChristopherDunn One of the principles of logs is $\displaystyle \log\frac{x}{y}=\log x-\log y$.
Since the $\displaystyle \log_{x}1=0$ for any $\displaystyle x$, you can simplify to $\displaystyle 0-\log_{16}8$.
There are probably a few ways of going on, but the easiest to me is splitting 8 into $\displaystyle \frac{16}{2}$. The $\displaystyle \log_{16}16=1$ and so we are left with $\displaystyle 0-1-\log_{16}2$.
As 2 is the fourth root of 16, $\displaystyle 16^\frac{1}{4}=2$, and so the $\displaystyle \log_{16}2=\frac{1}{4}$. We can now solve $\displaystyle 0-1-\frac{1}{4}=-\frac{3}{4}$