The sum of squares of all the real solutions of the equation log₍ₓ₊₁₎(2x²+5x+3)=4-log₍₂ₓ₊₃₎(x²+2x+1) is equal to ____.

Solution

Maths sir · 5 steps

Factor both arguments first, then one substitution turns this log equation into a quadratic.

1. Fix the domain. Each base must be positive and not 1. So x + 1 > 0 with x + 1 ≠ 1, and 2x + 3 > 0 with 2x + 3 ≠ 1. Together these give x > -1 and x ≠ 0.
2. Factor and expand. Use 2x² + 5x + 3 = (2x+3)(x+1) and x² + 2x + 1 = (x+1)². The left side becomes 1 + log₍ₓ₊₁₎(2x+3) and the right side becomes 4 - 2 log₍₂ₓ₊₃₎(x+1).
3. Substitute. Let t = log₍ₓ₊₁₎(2x+3). Then log₍₂ₓ₊₃₎(x+1) = 1/t, so the equation is 1 + t = 4 - 2/t.

   t² - 3t + 2 = 0 ⇒ t = 1 or t = 2

4. Back to x. For t = 1: 2x + 3 = x + 1 gives x = -2, which fails x > -1. For t = 2: 2x + 3 = (x+1)² gives x² = 2, so x = ±√2. Only x = √2 satisfies x > -1.
5. So. The only real solution is x = √2, and the sum of squares is (√2)² = 2.

The common slip: 4. Keeping x = -√2 as well gives 2 + 2 = 4. But then the base x + 1 is negative, so that root must be rejected.

The idea to keep. When two logs have each other's argument as base, set one equal to t and the other becomes 1/t. Always check every root against the base conditions at the end.

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