Question

okay, this i cannot figure out anywhere.

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

evaluate into logarithm of a single quantity.

Answer 1

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

Answer 2

would you mind telling me how you got it?
i don't get this problem..
thanks forthe answer though!

Answer 3

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)]