SOLUTION: Please help me solve this equation. Round to the nearest thousandth. 2 ln(y+1) = ln(y^2 - 1) + ln 5

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help me solve this equation. Round to the nearest thousandth. 2 ln(y+1) = ln(y^2 - 1) + ln 5      Log On


   

Question 890920: Please help me solve this equation.
Round to the nearest thousandth.
2 ln(y+1) = ln(y^2 - 1) + ln 5
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
2%2Aln%28y%2B1%29+=+ln%285%28y%5E2+-+1%29%29
ln%28%28y%2B1%29%5E2%29+=+ln%285%28y%5E2+-+1%29%29
%28y%2B1%29%5E2=5%28y%5E2-1%29
y%5E2%2B2y%2B1=5y%5E2-5
4y%5E2-2y-6=0
2y%5E2-y-3=0
%28y%2B1%29%282y-3%29=0
Two possible solutions:
y%2B1=0
y=-1
Not allowed since ln%280%29 is not defined.
2y-3=0
2y=3
highlight%28y=3%2F2%29