Using the log properties- math

This is a discussion on Using the log properties- math within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Kind of an offshoot of my previous post but this has to do with logarithmic properties Change of base: logbX ...

  1. #1
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    Using the log properties- math

    Kind of an offshoot of my previous post but this has to do with logarithmic properties

    Change of base:

    logbX = logaX/logaB

    and

    logb(X/Y ) = logbX - logbY

    and

    X/Y = X(1/Y) exponents


    here is the formula I have:
    logc (b / a) where c = 3, b = 88 , a = 3

    how can I write b/a: Is it actually

    logc (logcb - logxa) or

    logc (X(1/Y)

    confused on how to proceed with the b/c- do I simpify that first then convert from base 5 to base 10

    Insight would be appreciated...

  2. #2
    HelpingYouHelpUsHelpUsAll
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    How I learnt change of base: log of the number / log of the base. So to use a common base, to save me writing an extra 2 letters, and because my calculator can do it, I use natural log (ln). E.g.
    with b = 88 , a = 3:
    log3 (88) = ln (88) / ln (3) ~= 4.08

    Or you could get a calculator to do it all for you...
    long time no C; //seige
    You miss 100% of the people you don't C;
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    if (language != LANG_C && language != LANG_CPP)
        drown(language);

  3. #3
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by ninety3gd View Post
    Kind of an offshoot of my previous post but this has to do with logarithmic properties

    Change of base:

    logbX = logaX/logaB

    and

    logb(X/Y ) = logbX - logbY

    and

    X/Y = X(1/Y) exponents


    here is the formula I have:
    logc (b / a) where c = 3, b = 88 , a = 3

    how can I write b/a: Is it actually

    logc (logcb - logxa) or

    logc (X(1/Y)

    confused on how to proceed with the b/c- do I simpify that first then convert from base 5 to base 10

    Insight would be appreciated...
    You don't have 5, or 10 in the problem. Do you want to do change of base, or the quotient rule? The quotient rule would say logc(b/a) = logc b - logc a. The change of base rule just allows you to find a numerical value with a calculator, should it be necessary in some way.

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    I wrote a program to convert a number in one base into another, it revolved around the modulus operator, I'll do some Googling and see if I can learn anything that can help you
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  5. #5
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    Quote Originally Posted by ninety3gd View Post
    how can I write b/a: Is it actually

    logc (logcb - logxa) or

    logc (X(1/Y)
    logc(b/a) = logcb - logca, not logc(logcb - logca)
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    figured out my problem- thanks for all of the insight

  7. #7
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    what was the solution?
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