The sum of squares of all the real solutions | vedclass

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(B) The given equation is $\log_{(x+1)}((x+1)(2x+3)) = 4 - \log_{(2x+3)}(x+1)^2$.Using the property $\log_a(bc) = \log_a b + \log_a c$, we get $1 + \l

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$2$

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(B) The given equation is $\log_{(x+1)}((x+1)(2x+3)) = 4 - \log_{(2x+3)}(x+1)^2$.Using the property $\log_a(bc) = \log_a b + \log_a c$, we get $1 + \log_{(x+1)}(2x+3) = 4 - 2 \log_{(2x+3)}(x+1)$.Let $y = \log_{(x+1)}(2x+3)$. Then $\log_{(2x+3)}(x+1) = 1/y$.The equation becomes $1 + y = 4 - 2/y$.Multiplying by $y$ (where $y 
eq 0$), we get $y + y^2 = 4y - 2$, which simplifies to $y^2 - 3y + 2 = 0$.Factoring gives $(y-1)(y-2) = 0$, so $y=1$ or $y=2$.Case $1$: $\log_{(x+1)}(2x+3) = 1 \implies x+1 = 2x+3 \implies x = -2$. Checking the base constraints $(x+1 > 0, x+1 
eq 1)$, $x = -2$ is invalid.Case $2$: $\log_{(x+1)}(2x+3) = 2 \implies (x+1)^2 = 2x+3 \implies x^2 + 2x + 1 = 2x+3 \implies x^2 = 2 \implies x = \pm \sqrt{2}$.Checking constraints: For $x = -\sqrt{2}$, the base $x+1 = 1-\sqrt{2}  0$ and $2x+3 = 2\sqrt{2}+3 > 0$ are valid.Thus, the only real solution is $x = \sqrt{2}$.The sum of squares of all real solutions is $(\sqrt{2})^2 = 2$.

The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2 + 5x + 3) = 4 - \log_{(2x+3)}(x^2 + 2x + 1)$ is equal to . . . . . . .

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