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    If I have -ln(2x-6) - 2ln(x-6) do I combine the terms to get -ln((2x-6)/(x-6)^2) or have I made a mistake somewhere?
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    (Original post by Luketheace)
    If I have -ln(2x-6) - 2ln(x-6) do I combine the terms to get -ln((2x-6)/(x-6)^2) or have I made a mistake somewhere?
    Your answer is wrong. If you take out the minus it should become clearer:

    \displaystyle- \ln(2x-6) - 2\ln(x-6) = -\left[\ln(2x-6) + 2\ln(x-6)\right]

    Now join together the logs inside the bracket.
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    Thanks for the response, I've seen the mistake now.
    I'm trying to integrate 1/((2x-6)(x-6)) using partial fractions and it just seems impossible at the moment
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    (Original post by Luketheace)
    Thanks for the response, I've seen the mistake now.
    I'm trying to integrate 1/((2x-6)(x-6)) using partial fractions and it just seems impossible at the moment
    If you're still having problems then please post your working up to the point you got stuck.
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