math

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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..
    thanks forthe answer though!

  • 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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