Solve: 2 log 2 log 2 x + log 1 2 log 2 2 2 x = 1

Given that 2 log 2 log 2 x + log 1 2 log 2 2 2 x = 1 ⇒ 2 log 2 log 2 x + log 2 - 1 log 2 2 2 x = 1 ⇒ 2 log 2 log 2 x + 1 - 1 log 2 log 2 2 2 x = 1 ⇒ log 2 log 2 x 2 - log 2 log 2 2 2 x = 1 ⇒ log 2 log 2 x 2 log 2 2 2 x = 1 ⇒ log 2 x 2 log 2 2 2 x = 2 1 ⇒ log 2 x 2 = 2 log 2 2 2 x ⇒ log 2 x 2 = 2 log 2 2 2 + log 2 x ⇒ log 2 x 2 = 2 log 2 2 3 2 + log 2 x ⇒ log 2 x 2 = 2 3 2 log 2 2 + log 2 x ⇒ log 2 x 2 = 2 3 2 × 1 + log 2 x ⇒ log 2 x 2 = 3 + 2 log 2 x ⇒ log 2 x 2 - 2 log 2 x - 3 = 0 ...... 1 Let, log 2 x = a Then, the quadratic equation 1 can be written as a 2 - 2 a - 3 = 0 ⇒ a 2 + a - 3 a - 3 = 0 ⇒ a a + 1 - 3 a + 1 = 0 ⇒ a + 1 a - 3 = 0 ⇒ a + 1 = 0 or a - 3 = 0 ⇒ a = - 1 or a = 3 ⇒ log 2 x = - 1 or log 2 x = 3 ⇒ x = 2 - 1 or x = 2 3 ∴ x = 1 2 or x = 8 Hence, the required solution is x = 1 2 or 8 .

Find the domain of the function f x = 1 log 10 1 - x + x + 2

Given: f x = 1 log 10 1 - x + x + 2 For log function to be defined 1 - x > 0 ⇒ x < 1 ⇒ x ∈ - ∞ , 1 . . . . . ( i ) As log function is written in denominator, so it should not be equals to 0 log 10 1 - x ≠ 0 ⇒ 1 - x ≠ 1 ⇒ x ≠ 0 . . . . ii Let h ( x ) = x + 2 For square root to be defined x + 2 ≥ 0 ⇒ x ∈ [ - 2 , ∞ ) . . . . iii Taking intersection of i , ii & iii , we get x ∈ [ - 2 , 0 ) ∪ 0 , 1