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

$\begin{array}{l}{\log }_{(\mathrm{x}+1)}\left(2{\mathrm{x}}^2+7\mathrm{x}+5\right)+{\log }_{(2\mathrm{x}+5)}(\mathrm{x}+1{)}^2-4=0\\ ={\log }_{(\mathrm{x}+1)}\left\{\right(2\mathrm{x}+5\left)\right(\mathrm{x}+1\left)\right\}+2{\log }_{(2\mathrm{x}+5)}(\mathrm{x}+1)-4=0\\ ={\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)+{\log }_{(\mathrm{x}+1)}(\mathrm{x}+1)+2{\log }_{(2\mathrm{x}+5)}(\mathrm{x}+1)-4=0\\ ={\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)+2{\log }_{(2\mathrm{x}+5)}(\mathrm{x}+1)-3=0\left[\because {\log }_{\mathrm{a}}\mathrm{a}=1\right]\\ ={\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)+2\frac{{\log }_{(\mathrm{x}+1)}(\mathrm{x}+1)}{{\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)}=3\end{array}$

Let   ${\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)=\mathrm{t}$

$\begin{array}{l}\mathrm{t}+\frac{2}{\mathrm{t}}=3\Rightarrow {\mathrm{t}}^2-3\mathrm{t}+2=0\\ (\mathrm{t}-1))\mathrm{t}-2)=0\end{array}$

$\Rightarrow \mathrm{t}=1, \mathrm{t}=2$

$\begin{array}{l}\Rightarrow {\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)=1\text{ and }{\log }_{(\mathrm{x}+1)}(2\mathrm{x}+5)=2\\ 2\mathrm{x}+5=(\mathrm{x}+1)\text{ and }2\mathrm{x}+5=(\mathrm{x}+1{)}^2\\ \mathrm{x}=-4\text{ and }2\mathrm{x}+5={\mathrm{x}}^2+1+2\mathrm{x}\\ \text{ i.e., }{\mathrm{x}}^2=4\\ \Rightarrow \mathrm{x}=+2,-2\end{array}$

Given, x > 0

$\mathrm{x}=-4, \mathrm{x}=-2$ are discarded.

$\therefore \mathrm{x}=2$ is only solution.