12. (1-logo18)(1 – log218)

(1 - log₉18)(1 - log₂18) = (log₉9 - log₉18)(log₂2 - log₂18) = (log₉\(\frac{9}{18}\)) (log₂\(\frac{2}{18}\)) = log₉\(\frac{1}{2}\) * log₂\(\frac{1}{9}\) = log₉2^-1 * log₂9^-1 = -log₉2 * -log₂9 = log₉2 * log₂9 = log₉2 * \(\frac{1}{log_{9}2}\) = 1

Ответ: 1