x = log 2 a where, a ∈ (0, 1]

Sol. Let y = 2 x . Then, = 1 – 2 x

⇒ = 1 – y ⇒ a (y – 2) + 1 = (1 – y) 2

⇒ y 2 – 2y = a (y – 2) ⇒ (y – 2) (y – a) = 0

⇒ y = 2 or y = a

Now, y = 2 ⇒ 2 x = 2 ⇒ x = 1

But, x = 1 is not a solution of the given equation, because for x = 1, LHS = 1 and RHS = – 1

y = a

⇒ 2 x = a ⇒ a > 0 and x = log 2 a [  2 x > 0 for all x ∈ R]

Putting 2 x = a in the given equation, we get = 1 – a

⇒ = 1 – a ⇒ |a – 1| = 1 – a ⇒ a – 1 ≤ 0

⇒ a ≤ 1

Also, a > 0. Therefore, a ∈ (0, 1]

Hence, x = log 2 a is the solution of the given equation for all a ∈ (0, 1]. For a ≤ 0 and for a > 1, the equation has no solution

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