SOLUTION: Solve: ln(3-y)-ln(2y+1)=ln(4y)
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Question 35832: Solve: ln(3-y)-ln(2y+1)=ln(4y)
Answer by narayaba(40) (Show Source): You can put this solution on YOUR website!
ln(3-y)-ln(2y+1)=ln(4y)
ln(a) - ln(b) can be written as ln(a/b)
similarly
ln(3-y) - ln(2y+1) = ln((3-y)/(2y+1)) = ln(4y)
ln((3-y)/(2y+1)) = ln(4y)
logrrathims on both aides are to the same base
therefore
(3-y) / (2y+1) = 4y
3-y = (2y+1)*(4y)
3-y = 8y^2 + 4y
8y^2 + 5y -3 = 0 (you can write 5y as 8y - 3y, just makes it easy to solve it)
8y^2 +8y - 3y -3 = 0
(8y-3)(y+1) = 0
y = 3/8 or -1
to solve 8y^2 + 5y -3 = 0
you can also use the formula
and find y
b = 5, a = 8 and c = -3
ln of negative nunber is not possible so y = 3/8
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