Precalculus (6th Edition) Blitzer

$\log_{3} (-9)$ can not be evaluated.
$\log_{b}x=y$ is equivalent to $b^{y}=x$ --- $\log_{3} (-9)=y \quad$ is equivalent to $\quad 3^{y}=-9$ But, $\quad 3^{y}$ is never negative, so $\log_{3} (-9)$ can not be evaluated. Alternatively, arrive to the same conclusion like this: $\log_{3}x$ is defined only for positive x. $-9$ is not in the domain of $\log_{3}x$, which is why $\log_{3} (-9)$ can not be evaluated.