SOLUTION: solve 2ln(x+1)=ln(x^2-1)+ln5
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Question 975203: solve 2ln(x+1)=ln(x^2-1)+ln5
Found 3 solutions by Alan3354, solver91311, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
solve 2ln(x+1)=ln(x^2-1)+ln5
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ln(x+1)^2 = ln(5*(x^2-1))
Can you finish?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ \ln(x^2\ -\ 1)\ +\ \ln(5))
^2\ -\ \ln\left((x\ -\ 1)(x\ +\ 1)\right)\ =\ \ln(5))
^2}{(x\ -\ 1)(x\ +\ 1)}\right)\ =\ \ln(5))
\ =\ \ln(5))
But
\ =\ \ln(b)\ \Leftrightarrow\ a\ =\ b\ \forall a,\, b\ \in\ \text{dom}\left{\ln(x)\right})
Hence
You can take it from here, I should imagine.
Don't forget to make sure that 
and 
are actually in the domain of the log function before you publish your solution.
John
My calculator said it, I believe it, that settles it
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
solve 2ln(x+1)=ln(x^2-1)+ln5
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ln(x+1)^2 = ln[5(x^2-1)]
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(x+1)^2 = 5x^2 - 5
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x^2 + 2x + 1 = 5x^2 - 5
4x^2 - 2x - 6 = 0
------
2x^2 - x - 3 = 0
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Factor:
(2x-3)(x+1) = 0
----
Positive solution:
x = 3/2
-------
Cheers,
Stan H.
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