Question 611962
ln(3-x)=ln(3x-x^2)
3-x = 3x-x^2
-x+3 = 3x-x^2
x^2-x+3 = 3x
x^2-4x+3 = 0
(x-1)(x-3) = 0
x = {1,3}
since you can't take ln(0), we determine that x=3 is an extraneous solution -- throw it out leaving:
x = 1