The number of solutions of the equation _ ( x + 1 ) (2 x ^ 2 + 7x +...

We have,

$\log_{( x + 1 )}\left(2{x}^{2} + 7x + 5\right) + \log_{( 2x + 5 )} ( x + 1{)}^{2} - 4 = 0, x > 0$

$⇒ \log_{\left(x + 1\right)}\left(2x + 5\right)\left(x + 1\right) + 2\log_{\left(2x + 1\right)}\left(x + 1\right) = 4$

$⇒ \log_{\left(x + 1\right)}\left(2x + 5\right) + \log_{\left(x + 1\right)}\left(x + 1\right) + 2\log_{\left(2x + 1\right)}\left(x + 1\right) = 4$

$⇒ \log_{\left(x + 1\right)}\left(2x + 5\right) + 1 + 2\log_{\left(2x + 1\right)}\left(x + 1\right) = 4$

$⇒ \log_{\left(x + 1\right)}\left(2x + 5\right) + 2\log_{\left(2x + 1\right)}\left(x + 1\right) = 3$

$⇒ \log_{\left(x + 1\right)}\left(2x + 5\right) + \dfrac{2}{\log_{\left(x + 1\right)}\left(2x + 5\right)} = 3$

Put $\log_{( x + 1 )} ( 2x + 5 ) = t$, then

$t + \dfrac{2}{t} = 3$

$⇒ {t}^{2} - 3t + 2 = 0$

$⇒ t = 1, 2$

Then,

$\log_{( x + 1 )} ( 2x + 5 ) = 1$ and $\log_{( x + 1 )} ( 2x + 5 ) = 2$

$⇒ 2x + 5 = x + 1$ and $2x + 5 = {\left(x + 1\right)}^{2}$

$x = - 4$ (rejected)$as x > 0$

And,

$2x + 5 = {\left(x + 1\right)}^{2}$

$⇒ {x}^{2} = 4$

$⇒ x = 2, - 2$ (rejected)

So, $x = 2$

Number of solution is $1$.