The sum of all the real solutions of the equation \(\log _{(x+3)}\left(6 x^{2}+28 x+30\right)=5-2 \log _{(6 x+10)}\left(x^{2}+6 x+9\right)\) is equal to :

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Basic Mathematics

The sum of all the real solutions of the equation

lo g_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 lo g_{(6 x + 10)} (x^{2} + 6 x + 9)

is equal to :

Answer: (C)

Note

6 x^{2} + 28 x + 30 = (x + 3) (6 x + 10)

and

x^{2} + 6 x + 9 = (x + 3)^{2}

. The equation becomes

1 + lo g_{(x + 3)} (6 x + 10) = 5 - 4 lo g_{(6 x + 10)} (x + 3)

. Writing

A = lo g_{(x + 3)} (6 x + 10)

so that

lo g_{(6 x + 10)} (x + 3) = \frac{1}{A}

, this is

1 + A = 5 - \frac{4}{A}

, i.e.

A^{2} - 4 A + 4 = 0

, so

(A - 2)^{2} = 0

and

A = 2

. Then

(x + 3)^{2} = 6 x + 10

gives

x^{2} = 1

, so

x = \pm 1

; both satisfy the base and argument conditions. The sum of the real solutions is

1 + (- 1) = 0

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option