JEE Main 2026 · 23 January · evening · Single correct · +4 / −1
The sum of all the real solutions of the equation log₍ₓ₊₃₎(6x²+28x+30)=5-2 log₍₆ₓ₊₁₀₎(x²+6x+9) is equal to :
- (a) 1
- (b) 4
- (c) 0
- (d) 2
Answer and solution
Answer (c) 0
Solution
Maths sir · 5 steps
Factor both arguments first: everything turns into one log, log₍ₓ₊₃₎(6x+10), and its reciprocal.
- Factor the arguments. 6x²+28x+30 = 2(3x+5)(x+3) = (6x+10)(x+3) and x²+6x+9 = (x+3)². Both sides now involve only x+3 and 6x+10.
- Substitute t. Let t = log₍ₓ₊₃₎(6x+10), so log₍₆ₓ₊₁₀₎(x+3) = 1/t. The left side is 1 + t, and the right side is 5 - 2 × 2 × 1/t = 5 - 4/t.
1 + t = 5 - 4/t ⇒ t² - 4t + 4 = 0 ⇒ (t-2)² = 0 ⇒ t = 2
- Back to x. t = 2 means 6x+10 = (x+3)², that is x² - 1 = 0, so x = 1 or x = -1.
- Check the domain. We need x+3>0, x+3 ≠ 1, 6x+10>0 and 6x+10 ≠ 1, which means x > -5/3 with x ≠ -3/2. For x=1 the bases are 4 and 16, and for x=-1 they are 2 and 4, so both roots are valid.
- So. The real solutions are 1 and -1, and their sum is 1 + (-1) = 0.
The idea to keep. When two logs swap base and argument, set one equal to t and the other becomes 1/t. Solve for t first, then go back to x and check every base and argument.
Answer from the NTA final answer key. Solution by PrepWiser.