# math

posted by .

okay, this i cannot figure out anywhere.

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

evaluate into logarithm of a single quantity.

• math -

ln[x/(x^2-1)^2]

• math -

would you mind telling me how you got it?
i don't get this problem..

• math -

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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