From Förberedande kurs i matematik 1

Because we are working with \displaystyle \log _{9}, we express 1/3 as a power of 9,

\displaystyle \frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}

Using the logarithm laws, we get

\displaystyle \log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}