Ln

[BobbyJones]

I having trouble with this Ln question. Have a go:

2Ln(x+1) - Ln(x-1) = Ln(x-3)

Firstly I took the 2 up as a squared, and minused the LHS


Ln(x+1)^2
_______ = Ln(x-3)
(x-1)

Then took away the Ln's and took the (x-1) over to the over side.

(x+1)^2 = (x-3)(x-1)

(x+1)(x+1) = (x-3)(x-1)

x^2 + 2x +1 = x^2 - 4x +3

2x+1 = -4x+3

6x = 2

x= 1/3

But this dosent equate when I put it back into the origial equation?


HELP!

 

[pka]

I having trouble with this Ln question. Have a go: 2Ln(x+1) - Ln(x-1) = Ln(x-3)

Because of domain issues with \(\displaystyle \ln(x-3)\) we must have \(\displaystyle x>3\).
So there is no solution.

 

[lookagain]

Because of domain issues with \(\displaystyle \ln(x-3)\) we must have \(\displaystyle x>3\).
So there is no solution.

It also has domain issues with \(\displaystyle \ln(x - 1).\) All the equation needs to
have "no solution" is when (x - 3) or (x - 1) is less than or equal to zero.
If one of these is shown not to work, then a person need not take time to
check where any other logs of expressions in the original equation are undefined.