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(D) Given equation: $\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}-4=0$.Factorizing the argument: $2x^2+7x+5 = (2x+5)(x+1)$.Using log properti

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(D) Given equation: $\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}-4=0$.Factorizing the argument: $2x^2+7x+5 = (2x+5)(x+1)$.Using log properties: $\log _{(x+1)}(2x+5)(x+1) + 2\log _{(2x+5)}(x+1) - 4 = 0$.$\log _{(x+1)}(2x+5) + \log _{(x+1)}(x+1) + 2\log _{(2x+5)}(x+1) - 4 = 0$.Since $\log _{(x+1)}(x+1) = 1$,we have $\log _{(x+1)}(2x+5) + 1 + 2\log _{(2x+5)}(x+1) - 4 = 0$.Let $t = \log _{(x+1)}(2x+5)$. Then $\log _{(2x+5)}(x+1) = \frac{1}{t}$.The equation becomes $t + 1 + \frac{2}{t} - 4 = 0$,which simplifies to $t + \frac{2}{t} - 3 = 0$.Multiplying by $t$: $t^2 - 3t + 2 = 0$,so $(t-1)(t-2) = 0$.Case $1$: $t=1$ $\Rightarrow \log _{(x+1)}(2x+5) = 1$ $\Rightarrow 2x+5 = x+1$ $\Rightarrow x = -4$. Since $x > 0$,this is rejected.Case $2$: $t=2$ $\Rightarrow \log _{(x+1)}(2x+5) = 2$ $\Rightarrow 2x+5 = (x+1)^2$ $\Rightarrow 2x+5 = x^2+2x+1$ $\Rightarrow x^2 = 4$.Since $x > 0$,$x = 2$.Thus,there is only $1$ solution.

The number of solutions of the equation $\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}-4=0$ for $x > 0$ is:

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