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    hi, i did maths A level 2 and a half years ago but now need my logs again and cant remember how to do them. can someone help me please.

    i know that the log of a concentration (logx) is 300. how do i find out what x is?
    i cant get the e to work
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    log(x) = a
    so x = 10^a (10 to the a)

    ln(x) = s
    so x = e^s

    log is base 10 unless otherwise specified.
    ln is base e. (I'll assume you know what I mean by base)

    in your example:

    log(x) = 300

    so x = 10^300.

    That's very concentrated.
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    (Original post by Melanie-v)
    log is base 10 unless otherwise specified.
    ...at A-level. If this is uni stuff, it's more likely that the log is to base e (or even base 2), so it's worth the OP checking what base is used as standard in her course.
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    (Original post by Dollaleigh)
    i cant get the e to work
    The mind boggles!

    As others have said. Whatever base you are using, raise it to the power of 300.
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    (Original post by nuodai)
    ...at A-level. If this is uni stuff, it's more likely that the log is to base e (or even base 2), so it's worth the OP checking what base is used as standard in her course.
    Why 2? Just a random thought half-lives etc.?
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    (Original post by Melanie-v)
    Why 2? Just a random thought half-lives etc.?
    I think it's usually computer scientists that use "log" to mean \log_2 because of its use with binary, but it's not out of the question for other types of scientist to use it... to be honest most of the time it doesn't really matter which base you use as long as you're consistent and you know which base you're working with.
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    (Original post by nuodai)
    I think it's usually computer scientists that use "log" to mean \log_2 because of its use with binary, but it's not out of the question for other types of scientist to use it... to be honest most of the time it doesn't really matter which base you use as long as you're consistent and you know which base you're working with.
    Ah, ok. Thanks
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    thanks for your help. in the end i just plotted a graph with log paper so i didnt have to worry about antilogging
 
 
 
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