1498. a) log1 4 + log1 2; 6) logs+logs

1498. Вычислите:

a) $$log_{\frac{1}{8}} 4 + log_{\frac{1}{8}} 2 = log_{\frac{1}{8}} (4 \cdot 2) = log_{\frac{1}{8}} 8 = -1$$

б) $$log_{\frac{1}{4}} \frac{1}{8} + log_{\frac{1}{4}} \frac{1}{2} = log_{\frac{1}{4}} (\frac{1}{8} \cdot \frac{1}{2}) = log_{\frac{1}{4}} \frac{1}{16} = log_{\frac{1}{4}} (\frac{1}{4})^2 = 2$$

Ответ: a) -1, б) 2