Вычислить. log(3;2)*log(4;3)*log(5;4)*…*log(11;10), log(a;b), где a-основание, b-число

   1. Воспользуемся формулой для перехода к другому основанию логарифма:

• log(a, b) = log(c, b)/log(c, a), отсюда:
• log(c, b) = log(c, a) * log(a, b), или:
• log(c, b) = log(a, b) * log(c, a).

   2. Преобразуем выражение:

• x = log(3, 2) * log(4, 3) * log(5, 4) * log(6, 5) * log(7, 6) * log(8, 7) * log(9, 8) * log(10, 9) * log(11, 10);
• x = log(4, 2) * log(5, 4) * log(6, 5) * log(7, 6) * log(8, 7) * log(9, 8) * log(10, 9) * log(11, 10);
• x = log(5, 2) * log(6, 5) * log(7, 6) * log(8, 7) * log(9, 8) * log(10, 9) * log(11, 10);
• x = log(6, 2) * log(7, 6) * log(8, 7) * log(9, 8) * log(10, 9) * log(11, 10);
• x = log(7, 2) * log(8, 7) * log(9, 8) * log(10, 9) * log(11, 10);
• x = log(8, 2) * log(9, 8) * log(10, 9) * log(11, 10);
• x = log(9, 2) * log(10, 9) * log(11, 10);
• x = log(10, 2) * log(11, 10);
• x = log(11, 2).

   Ответ: log(11, 2).