NDA Mathematics Question (2026) — Solution

Step-by-step solution

Let t = _ x x. The given equation becomes t + 1 t = 2. t^2 - 2t + 1 = 0 (t - 1)^2 = 0 t = 1 _ x x = 1 x = x x = 1 The smallest positive value of x is 4 . For x = 4 , x = 1 2 and x = 1 2 , which satisfy the conditions for the base and argument of logarithms (positive and not equal to 1). Answer: 4