The number of solutions of the equation

$\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0$

$\log _{(x+1)}(2 x+5)(x+1)+2 \log _{(2 x+5)}(x+1)=4$

$\log _{(x+1)}(2 x+5)+1+2 \log _{(2 x+5)}(x+1)=4$

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

$\mathrm{t}+\frac{2}{\mathrm{t}}=3 \Rightarrow \mathrm{t}^{2}-3 \mathrm{t}+2=0$

$t=1,2$

$\log _{(x+1)}(2 x+5)=1 \& \quad \log _{(x+1)}(2 x+5)=2$

$x+1=2 x+3 \quad \& \quad 2 x+5=(x+1)^{2}$

$x=-4$ (rejected) $\quad x^{2}=4 \Rightarrow x=2,-2$ (rejected)

So, $x=2$

No. of solution $=1$