SOLUTION: log base 2 (y+6)+log base 2(y-1)=1+log base 2 (3y)

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Question 869410: log base 2 (y+6)+log base 2(y-1)=1+log base 2 (3y)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%28y%2B6%29%29%2Blog%282%2C%28y-1%29%29=1%2Blog%282%2C%283y%29%29
log%282%2C%28%28y%2B6%29%28y-1%29%29%2F%283y%29%29=1
%28%28y%2B6%29%28y-1%29%29%2F%283y%29=2
%28%28y%2B6%29%28y-1%29%29=6y
y%5E2-y%2B6y-6=6y
y%5E2-y-6=0
%28y-3%29%28y%2B2%29=0
Two solutions:
y-3=0
y=3
and
y%2B2=0
y=-2
The second solution would not work since log%282%2C%28-3%29%29 and log%282%2C%28-6%29%29
are not defined.