The sum of the roots of the equation
x + 1 - 2log ₂(3 + 2ˣ) + 2log ₄(10 - 2⁻ˣ) = 0, is :

Solution

Maths sir · 4 steps

Put everything in base 2 and set t = 2ˣ. The sum of the x-roots comes from the product of the t-roots.

1. Bring to base 2. Since log₄ y = 1/2 log₂ y, the term 2 log₄(10 - 2⁻ˣ) becomes log₂(10 - 2⁻ˣ). Also write x + 1 = log₂ 2ˣ⁺¹ and 2 log₂(3 + 2ˣ) = log₂(3 + 2ˣ)².
2. Remove the logs. Combine the logs on each side, then drop log₂. This gives 2ˣ⁺¹(10 - 2⁻ˣ) = (3 + 2ˣ)², so 20 · 2ˣ - 2 = (3 + 2ˣ)².

   t = 2ˣ: 20t - 2 = t² + 6t + 9 ⇒ t² - 14t + 11 = 0

3. Check the roots. The roots are t = 7 ± √38, about 0.84 and 13.16. Both are positive, and both keep 10 - 1/t > 0, so each gives a real x.
4. So. 2^(x₁) · 2^(x₂) = t₁ t₂ = 11, so 2^(x₁ + x₂) = 11. That gives x₁ + x₂ = log₂ 11, option (b).

The idea to keep. With the substitution t = aˣ, a sum of x-roots turns into a product of t-roots: a^(x₁ + x₂) = t₁ t₂. Use the product of roots, not the sum.

Answer from the NTA final answer key. Solution by PrepWiser.