We have,
log(x+1)(2x2+7x+5)+log(2x+5)(x+1)2−4=0,x>0
⇒log(x+1)(2x+5)(x+1)+2log(2x+1)(x+1)=4
⇒log(x+1)(2x+5)+log(x+1)(x+1)+2log(2x+1)(x+1)=4
⇒log(x+1)(2x+5)+1+2log(2x+1)(x+1)=4
⇒log(x+1)(2x+5)+2log(2x+1)(x+1)=3
⇒log(x+1)(2x+5)+log(x+1)(2x+5)2=3
Put log(x+1)(2x+5)=t, then
t+t2=3
⇒t2−3t+2=0
⇒t=1,2
Then,
log(x+1)(2x+5)=1 and log(x+1)(2x+5)=2
⇒2x+5=x+1 and 2x+5=(x+1)2
x=−4 (rejected) as x>0
And,
2x+5=(x+1)2
⇒x2=4
⇒x=2,−2 (rejected)
So, x=2
Number of solution is 1.