## College Algebra (6th Edition)

$-\frac{1}{2}$
We know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (where $x\gt0$, $b\gt0$, and $b\ne1$). Therefore, $log_{3}\frac{1}{\sqrt 3}=-\frac{1}{2}$, because $3^{-\frac{1}{2}}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt 3}$.