Question 666476
Solve for x: ln(x-2) + ln(2x-3) = 2ln(x). Are there any extraneous solutions?
ln(x-2) + ln(2x-3) = 2ln(x)
=ln[(x-2)(2x-3)]=ln(x^2) 
(x-2)(2x-3)=x^2
2x^2-7x+6=x^2
x^2-7x+6=0
(x-6)(x-1)=0
x=1(reject, (x-2)>0) (extraneous solution)
x=6