Chapter 9 - Section 9.3 - Logarithmic Functions - 9.3 Exercises - Page 605: 14

$log_{16}\frac{1}{8}=-\frac{3}{4}$

For all positive numbers $a$, where $a\ne1$, and all positive numbers $x$, we know that $y=log_{a}x$ means the same as $x=a^{y}$. Therefore, $16^{-\frac{3}{4}}=\frac{1}{8}$ means the same as $log_{16}\frac{1}{8}=-\frac{3}{4}$.