JEE Main 2021 — Basic of Mathematics Question with Solution

We have,

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

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

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

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

$\Rightarrow {\log }_{\left(x+1\right)}\left(2x+5\right)+2{\log }_{\left(2x+1\right)}\left(x+1\right)=3$

$\Rightarrow {\log }_{\left(x+1\right)}\left(2x+5\right)+\frac{2}{{\log }_{\left(x+1\right)}\left(2x+5\right)}=3$

Put ${\log }_{(x+1)}(2x+5)=t$, then

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

$\Rightarrow t^2-3t+2=0$

$\Rightarrow t=1, 2$

Then,

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

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

$x=-4$ (rejected) as $x>0$

And,

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

$\Rightarrow x^2=4$

 $\Rightarrow x=2,-2$ (rejected)

So, $x=2$

Number of solution is $1$.