The sum of all the real solutions of the equation

$\log_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 \log_{(6 x + 10)} (x^{2} + 6 x + 9)$ is equal to. [2026]

Question

The sum of all the real solutions of the equation

$\log_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 \log_{(6 x + 10)} (x^{2} + 6 x + 9)$ is equal to. [2026]

Smart Achievers · Accepted Answer

(1)

$\log_{x + 3} [(x + 3) (6 x + 10)] = 5 - 2 \log_{6 x + 10} {(x + 3)}^{2}$

$1 + \log_{x + 3} (6 x + 10) = 5 - 4 \log_{6 x + 10} (x + 3)$

$Let \log_{x + 3} (6 x + 10) = A$

$\Rightarrow A + \frac{4}{A} = 4 or A = 2$

$\Rightarrow \log_{x + 3} (6 x + 10) = 2$

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

$\Rightarrow 6 x + 10 = x^{2} + 9 + 6 x$

$\Rightarrow x^{2} = 1, x = \pm 1$

$So sum of roots = 0$

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