SOLUTION: Solve log2 (y+1) + log2 (3y-1) = 5

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Question 854923: Solve log2 (y+1) + log2 (3y-1) = 5
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
log2 (y+1) + log2 (3y-1) = 5
(y+1)(3y-1)) = 2^5
3y^2 + 2y - 1 = 32
3y^2 + 2y - 33 = 0
x = %28-2+%2B-+sqrt%28+400+%29%29%2F%286%29+ (Tossing out this negative root as Extraneous)
x =%28-2+%2B20%29%2F%286%29+=+3+