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Logarithms

Section: Mathematics

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Question : 1 of 11

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The sum of squares of all real solution of the equation

\log_{(x+1)} (2x^{2}+5x+3) = 4 - \log_{(2x+3)} (x^{2}+2x+1)

is equal to ______ .

[JEE Main 8 apr 2026 shift 2]

Your Answer:

Solution:

\log_{(x+1)}(2x+3)(x+1) + \log_{(2x+3)}(x+1)^{2} = 4

1 + \log_{(x+1)}(2x+3) + 2 \log_{(2x+3)}(x+1) = 4

\log_{x+1}(2x+3) = t \;\; t + \frac{2}{t} = 3

t^{2} - 3t + 2 = 0 \Rightarrow t = 1, 2

for

t = 1 \Rightarrow \log_{(x+1)}(2x+3) = 1 \Rightarrow 2x+3 = x+1

\Rightarrow x = -2 \;\; (\text{rejected})

for

t = 2 \Rightarrow 2x+3 = x^{2} + 2x + 1 \Rightarrow x^{2} = 2

\Rightarrow x = \pm \sqrt{2}

rejecting

x = -\sqrt{2}

, we get

x = \sqrt{2}

∴ sum of squares of all the roots = 2

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