So the domain of ln(x) is (0, Infinity) and the domain of ln(x^2) would be (-infinity, 0)U(0, Infinity)?
But couldn't you rewrite ln(x^2) to be 2ln(x) and that would have a domain of (0, Infinity)?
A problem like this came up in class (more complex, involved calculus and relative extrema but that's besides the point) I said you would have to rewrite it as 2ln(abs(x)) to get the same domain as the original function but my teacher said no.
So who is right?
In going from the square of x (which is never negative) to just x (which is, "half" the time, negative), you would indeed need to use the absolute value. (This is one of those situations in which the more-careful textbooks will tell you to "assume all variables are positive", or something similar.)
If your teacher uses graphing calculators (which almost-certainly means the Texas Instruments TI-83 or -84), you might try showing your teacher the side-by-side "TABLE" values for Y1=ln(X^2) and Y2=2*ln(X). The domain problem is obvious as soon as you scroll up past X=0. For thoroughness (and to prove your case), you could also show Y3=2*ln(abs(X)), and see which of Y1 and Y2 it matches.