1. Вычислии: 1) log3 1/3; 2) (1/3) ^log3 7; 3) log2 56 + 2 log2 12 - log2 63

Решаем:

1. \(log_{3} \frac{1}{3} = log_{3} 3^{-1} = -1\)
2. \((\frac{1}{3})^{log_{3} 7} = (3^{-1})^{log_{3} 7} = 3^{-log_{3} 7} = 3^{log_{3} 7^{-1}} = 7^{-1} = \frac{1}{7}\)
3. \(log_{2} 56 + 2log_{2} 12 - log_{2} 63 = log_{2} 56 + log_{2} 12^2 - log_{2} 63 = log_{2} 56 + log_{2} 144 - log_{2} 63 = log_{2} \frac{56 \cdot 144}{63} = log_{2} \frac{8 \cdot 7 \cdot 16 \cdot 9}{7 \cdot 9} = log_{2} (8 \cdot 16) = log_{2} 2^3 \cdot 2^4 = log_{2} 2^7 = 7\)

Ответ: 1) -1; 2) 1/7; 3) 7