Let m be the minimum possible value of log 3 3 y 1 + 3 y 2 + 3 y 3 , where y 1 , y 2 , y 3 are real numbers for which y 1 + y 2 + y 3 = 9 . Let M be the maximum possible value of log 3 x 1 + log 3 x 2 + log 3 x 3 , where x 1 , x 2 , x 3 are positive real numbers for which x 1 + x 2 + x 3 = 9 . Then the value of log 2 m 3 + log 3 M 2 is

Step-by-step solution

Applying A.M-G.M inequality, 3 y 1 + 3 y 2 + 3 y 3 3 ≥ 3 y 1 ⋅ 3 y 2 ⋅ 3 y 3 1 3 = 3 y 1 + y 2 + y 3 1 3 ⇒ 3 y 1 + 3 y 2 + 3 y 3 ≥ 81 So, m = log 3 81 = 4 Now, log 3 x 1 + log 3 x 2 + log 3 x 3 = log 3 x 1 . x 2 .   x 3 ∵ x 1 + x 2 + x 3 3 ≥ x 1 . x 2 . x 3 1 3 ⇒ x 1 . x 2 . x 3 ≤ 27 (applying A.M-G.M inequality) So, M = log 3 27 = 3 Thus, log 2 m 3 + log 3 M 2 = 8