# How do you calculate log_9(1/7)?

$- l o h \frac{7}{\log} 9 = - 0.8856$, nearly.
Use ${\log}_{b} {a}^{n} = n {\log}_{b} a \mathmr{and} {\log}_{b} a = {\log}_{c} \frac{a}{\log} _ c b$
Here, ${\log}_{9} \left(\frac{1}{7}\right) = {\log}_{9} \left({7}^{- 1}\right) = - {\log}_{9} 7 = - \log \frac{7}{\log} 9$