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  1. #1

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    Re: Logarithm help

    Yes yes much clearer now. So based on the definition of a logarithm
    would it be correct to say that we are looking for the approximation of the power?

    81 = 92 --> 9log(81) = 2 --> log(81) = 1.908......... -->
    92 = 81

  2. #2
    Fanatic Member twanvl's Avatar
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    Re: Logarithm help

    Not exactly, 9log(81) means base-9 logarithm of 81, not 9log(81) which means 9 raised to the base-e logarithm of 81.
    When a base is not specified it is usually assumed to be equal to e (2.71828....), so log(81) is acutally elog(81) = 1.908... this number is not related to 9log(81) = 2.
    The subscript notation for the base can be confusing sometimes, some people prefer to use superscript, as in 9log(81) = 2

  3. #3

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    Re: Logarithm help

    Ok say we have log(81). Since no based is specified we are using the natural log in the form of elog(x) or ln x where e = 2.718......... So we have 2.718log(81) = 1.908..... Now how is 1.908..... obtained?

  4. #4
    Fanatic Member twanvl's Avatar
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    Re: Logarithm help

    NOTE elog(81) is actually 4.3944... Your calculator's log function returns 10log(81)
    A logarithm without base can mean many different things, to mathematicians the base is e, to computer scienctists the base is 2, to economists the base is 10...

    To a mathetaician 10log(81)=1.908... is not obtained somehow, it is just the eternal truth.
    To a computer or a person who wants to see the actual number it is more difficult. There is no easy way to find a logarithm. There are algorithms that can calculate them, but that is not something you want to do yourself. A simple algorithm you could use yourself is simply trying:
    Code:
       10log(81) = x
       10^x = 81
    try 1:
       10^1 = 10  (too small)
    try 2:
       10^2 = 100 (too large)
    try 1.5:
       10^1.5 = 31.. (too small)
    try 1.75:
       10^1.75 = 56.2... (too small)
    etc.
    So basicly log(81)=1.908... because your calculator says so and because 10log(81)=81

    Your calculator probably does not have a logarithm key for any base, in that case you can use the formula:
    blog(x) = log(x)/log(b)
    This works regardless of the base of the logarithm used.

  5. #5

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    Re: Logarithm help

    Posted by twanvl

    So basicly log(81)=1.908... because your calculator says so and because 10log(81)=81
    Is that correct? 10log(81)=81 -->
    81 != 1081.

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