# How do you evaluate log_3(1/9)?

$\setminus {\log}_{3} \left(\frac{1}{9}\right) = \setminus {\log}_{3} \left(1\right) - \setminus {\log}_{3} \left(9\right) = 0 - \setminus {\log}_{3} \left({3}^{2}\right) = - 2$