From Förberedande kurs i matematik 1

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

\displaystyle \frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}

Using the logarithm laws, we get

\displaystyle \log _{9}\frac{1}{3}=\log _{9}9^{-{1}/{2}\;}=-\frac{1}{2}\centerdot \log _{9}9=-\frac{1}{2}\centerdot 1=-\frac{1}{2}.