If log 2 ⁡ 5 2 x + 1 , log 4 ⁡ 2 1 - x + 1 and 1 are in arithmetic progression, then x is equal to

Step-by-step solution

log 2 ⁡ 5 ⋅ 2 x + 1 , 1 2 log 2 ⁡ 2 1 - x + 1 , log 2 ⁡ 2 are in A.P. ⇒ 2 × 1 2 log 2 ⁡ 2 1 - x + 1 = log 2 ⁡ 5 ⋅ 2 x + 1 + l o g 2 2 ⇒ 2 1 - x + 1 = 5 ⋅ 2 x + 1 2 Let, 2 x = y ⇒ 2 y + 1 = 5 y + 1 2 ⇒ 2 + y = 10 y 2 + 2 y ⇒ 10 y 2 + y - 2 = 0 ⇒ y = 2 5 , - 1 2 ⇒ 2 x = 2 5 or - 1 2 (not possible) ⇒ 2 x = 2 5 ⇒ x = l o g 2 2 5 = 1 - log 2 ⁡ 5 = 1 - log ⁡ 5 log ⁡ 2