Question

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

Answer 1

Answer: 1

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