Question

log₃ $\frac{1}{9}$ + log₉ $\frac{1}{9}$ - $\frac{1}{3}$ log₂₇ $\frac{1}{9}$

Answer 1

$log_3\frac{1}{9}+log_9\frac{1}{9}-\frac{1}{3}\cdot log_{27}\frac{1}{9}=log_33^{-2}+log_99^{-1}-\frac{1}{3}\cdot log_{3^3}3^{-2}=\\\\=-2log_33-1\cdot log_99-\frac{1}{3}\cdot \frac{-2}{3}\cdot log_33=-2-1+\frac{2}{9}=-3+\frac{2}{9}=-2\frac{7}{9}$