Aspire Faculty ID #18017 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

The sum of all the real solutions of the equation

$\log_{x+3}(6x^2 + 28x + 30) - 5 = 2\log_{6x+10}(x^2 + 6x + 9)$

is equal to:

Solution

$\log_{x+3}((x+3)(6x+10)) = 5 - 2\log_{6x+10}(x+3)^2$ $1 + \log_{x+3}(6x+10) = 5 - 4\log_{6x+10}(x+3)$ Let $\log_{x+3}(6x+10) = A$ $\Rightarrow A + \frac{4}{A} = 4 \Rightarrow A = 2$ $\Rightarrow \log_{x+3}(6x+10) = 2$ $6x + 10 = (x+3)^2$ $x^2 - 1 = 0 \Rightarrow x = \pm 1$ Sum of roots $= 0$
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