JEE Main 2026 — 23 January, Evening Shift — Question 7

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

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

Let $\log _{(x+3)}(6 \mathrm{x}+10)=\mathrm{A}$

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

$\Rightarrow \log _{(\mathrm{x}+3)}(6 \mathrm{x}+10)=2$

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

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

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

So sum of roots $=0$