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

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Solution

${log}_{(x + 1)} (x + 1) (2 x + 5) + {log}_{(2 x + 5)} (x + 1)^{2} = 4$
$\Rightarrow 1 + {log}_{(x + 1)} (2 x + 5) + 2 {log}_{(2 x + 5)} (x + 1) = 4$
$Let {log}_{(x + 1)} (2 x + 5) = t$
$then t + \frac{2}{t} = 3 \Rightarrow t = 1, 2$
$\Rightarrow 2 x + 5 = x + 1 or 2 x + 5 = (x + 1)^{2}$
$\Rightarrow x = - 4, + 2, - 2$ out of which only $x = 2$ is acceptable.

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