Posted by **strawberryfields** on Friday, January 23, 2009 at 10:18pm.

okay, this i cannot figure out anywhere.

2[ln(x)-ln(x+1)-ln(x-1)]

evaluate into logarithm of a single quantity.

- math -
**drwls**, Friday, January 23, 2009 at 10:21pm
ln[x/(x^2-1)^2]

- math -
**strawberryfields**, Friday, January 23, 2009 at 10:34pm
would you mind telling me how you got it?

i don't get this problem..

thanks forthe answer though!

- math -
**drwls**, Friday, January 23, 2009 at 11:38pm
Actually I should have written it as

ln{[x/(x^2-1]^2}

The -ln(x+1)-ln(x-1) in the denominator is ln {1/[(x+1)(x-1)] = ln [1/(x^2-1)]

Adding ln x puts the x in the numerator of [x/(x^2-1)]

The two in front of

2[ln(x)-ln(x+1)-ln(x-1)]

is the same as taking the log of the square of [ln(x)-ln(x+1)-ln(x-1)]

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