Известно, что log₁₂ 27 = а. Найдите log₁₂ 16.

Дано:

• \[ log_{12} 27 = a \]

Найти:

• \[ log_{12} 16 \]

Решение:

1. \[ log_{12} 27 = log_{12} 3^3 = 3 log_{12} 3 = a \]
2. \[ log_{12} 3 = \frac{a}{3} \]
3. \[ log_{12} 12 = log_{12} (3 \times 4) = log_{12} 3 + log_{12} 4 = 1 \]
4. \[ log_{12} 4 = 1 - log_{12} 3 = 1 - \frac{a}{3} \]
5. \[ log_{12} 16 = log_{12} 4^2 = 2 log_{12} 4 = 2 \left( 1 - \frac{a}{3} \right) = 2 - \frac{2a}{3} \]

Ответ: \( 2 - \frac{2a}{3} \)