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Math Help - Problem with logarithm

  1. #1
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    Question Problem with logarithm

    If only log(2) and log(3) are given, then how can I calculate log(20) and log(30) ?

    (The base is 10)

    Please help,
    Thanks.
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  2. #2
    Super Member Bacterius's Avatar
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    I believe you need \log{(5)} too, since 20 = 2 \times 2 \times 5 and 30 = 2 \times 3 \times 5.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Bacterius View Post
    I believe you need \log{(5)} too, since 20 = 2 \times 2 \times 5 and 30 = 2 \times 3 \times 5.
    ... if the base is 10 , then is \log 5 = \log 10 - \log 2 = 1 - \log 2 ... but more simple is \log 20 = 1 + \log 2 and \log 30 = 1 + \log 3 ...

    Kind regards

    \chi  \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    ... if the base is 10 , then is \log 5 = \log 10 - \log 2 = 1 - \log 2 ... but more simple is \log 20 = 1 + \log 2 and \log 30 = 1 + \log 3 ...

    Kind regards

    \chi  \sigma

    Thanks a lot... that helps !
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  5. #5
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    Actually, you don't need log(5). log(30)= log(3(10))= log(3)+ log(10) and log(20)= log(2(10))= log(2)+ log(10)- and you know log(10).

    (Which I notice now is exactly what chi sigma said!)
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