log(x-1)+2log(y)=2log(3), log(x)+log(y)=6; (shouldn't this be log(6)?)
log(y2(x-1))=log(9), y2(x-1)=9, x=1+9/y2
log(xy)=6, xy=b6, where b is the logarithmic base, x=b6/y.
Therefore b6/y=1+9/y2,
b6y=y2+9, y2-b6y+9=0, y=(b6±√(b12-36))/2=b6(1±√(1-36/b6))/2, which seems an unlikely solution.
Let's assume 6 should be log(6) in the question.
Instead of b6 we would have blog(6)=6 and for b12, we would have b2log(6)=blog(36)=36;
y=3, x=6/y=2.
Or:
log(x-1)+2log(y)=2log(3), log(x)+log(y)=log(6) should have been the question.
y2(x-1)=9, xy=6; x=1+9/y2=6/y, y2-6y+9=0=(y-3)2, y=3, x=6/3=2. x=2, y=3 is the most likely solution.