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    Hi,
    How do you do this question?:

    Given that a>0, b>0, prove that ln(ab) = lna + lnb
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    You have got to know your log laws to answer all logs questions.

    Log(a) + Log(b) is identical to Log(ab)
    Log(a) - Log9b) is identical to Log(a/b)
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    (Original post by Earl_Earl)
    You have got to know your log laws to answer all logs questions.

    Log(a) + Log(b) is identical to Log(ab)
    Log(a) - Log9b) is identical to Log(a/b)
    Thanks for your reply yes I know that, but do you know how you prove it?
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    (Original post by Phoebus Apollo)
    Hi,
    How do you do this question?:

    Given that a>0, b>0, prove that ln(ab) = lna + lnb
    urm might be wrong yolo

    if you let

     x = \ln a
    &
     y = \ln b

    In terms of e, this would give:

     e^x = a
     e^y = b

     ab = e^xe^y
     ab = e^x^+^y
     \ln ab = \ln e^x^+^y
     \ln ab = (x + y )\ln e
     \ln ab = x + y
     \ln ab = \ln a + \ln b
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    (Original post by Naruke)
    urm might be wrong yolo

    if you let

     x = \ln a
    &
     y = \ln b

    In terms of e, this would give:

     e^x = a
     e^y = b

     ab = e^xe^y
     ab = e^x^+^y
     \ln ab = \ln e^x^+^y
     \ln ab = (x + y )\ln e
     \ln ab = x + y
     \ln ab = \ln a + \ln b
    Thanks, Naruke
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    (Original post by Phoebus Apollo)
    Thanks, Naruke
    No problemo
 
 
 
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