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    Hi

    How do I get from

    [-5ln(2/3)-2ln(5/3)]-[-5ln(5/6)-2ln(4/3)]

    to 5ln(5/4)+2ln(4/5)

    and eventually ln 125/64


    This is part of a question in which I can do the first part but can't rearange these logs. Thanks
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    aLn(\frac{b}{c}) - aLn(\frac{d}{e}) = aLn(\frac{be}{cd})

    aLn(\frac{b}{c}) + aLn(\frac{d}{e}) = aLn(\frac{bd}{ce})

    aLn(\frac{b}{c}) = Ln(\frac{b^{a}}{c^{a}})

    That help at all?
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    (Original post by StephenP91)
    aLn(\frac{b}{c}) - aLn(\frac{d}{e}) = aLn(\frac{be}{cd})

    aLn(\frac{b}{c}) + aLn(\frac{d}{e}) = aLn(\frac{bd}{ce})

    aLn(\frac{b}{c}) = Ln(\frac{b^{a}}{c^{a}})

    That help at all?
    Oh thanks, I can now get to 5ln(5/4)+2ln(4/5) but am stuck now. The 'a' is different so you can't add them right?
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    (Original post by thethinker)
    Oh thanks, I can now get to 5ln(5/4)+2ln(4/5) but am stuck now. The 'a' is different so you can't add them right?
    Now turn to the third "rule" he gave you?
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    (Original post by StephenP91)

    That help at all?
    (Original post by Mr M)
    Now turn to the third "rule" he gave you?
    Ok I've got it now

    Thanks to both of you

    I really need to do more work
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    (Original post by StephenP91)
    aLn(\frac{b}{c}) - aLn(\frac{d}{e}) = aLn(\frac{be}{cd})

    aLn(\frac{b}{c}) + aLn(\frac{d}{e}) = aLn(\frac{bd}{ce})

    aLn(\frac{b}{c}) = Ln(\frac{b^{a}}{c^{a}})

    That help at all?
    Just out of interest, are the first two basic rules, or are they derived from the other log rules? I can't seem to find those in my textbook
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    (Original post by thethinker)
    Just out of interest, are the first two basic rules, or are they derived from the other log rules? I can't seem to find those in my textbook
    You derive them. When I sat C2 there were questions asking you to derive those laws from logs. It's simple enough to do.
 
 
 
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