Thank you for the opportunity to help you with your question!
1. log ((Y^2-3y+2)/(y^2-1)) We divide because with "logs" a + is like multiplying and a - is like dividing.
2. log((y-2)(y-1)/(y-1)(y+1)) Break down the top and the bottom. Y^2 - 3y +2 can break down into (y-2)(y-1) and y^2-1 can break into (y-1)(y+1)... now you can eliminate y-1 because it is on both sides and you get the answer:
ANS: log((y-2)/(y+1)) so log[(y-2)/(y+1)]
Please let me know if you need any clarification. I'm always happy to answer your questions.