Question

log 2,75 16-log 2,75 121. Решите, пожалуйста.

Answer 1

$log_{2,75}16-log_{2,75}121=log_{2,75} \frac{16}{121}=log_{ \frac{11}{4}}( \frac{4}{11})^2=\\\\=log_{ \frac{11}{4}}( \frac{11}{4})^{-2}= -2* log_{ \frac{11}{4}} \frac{11}{4}=-2$