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  4. The number of solutions of the equation log (x+1)(2 x2+7 x+5)+ log (2 x+5)(x+1)2-4=0 x > 0 is

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

JEE MainJEE Main 2021Complex Numbers and Quadratic Equations

Solution:

$\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0$
$\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 $
$t+\frac{2}{t}=3 \Rightarrow t^{2}-3 t+2=0 $
$t=1,2 $
$\log _{(x+1)}(2 x+5)=1\, \& \,\, \log _{(x+1)}(2 x+5)=2 $
$x+1=2 x+3\,\, \& \,\, 2 x+5=(x+1)^{2} $
$x=-4$ ( rejected) $\,\,\, x^{2}=4 \Rightarrow x=2,-2$ ( rejected)
So, $x=2$
No. of solution $=1$