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Math Help - Relationship between the logs of a number with different bases

  1. #1
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    Relationship between the logs of a number with different bases

    Can anyone explain step by step the relation between the logs of a number to different bases?

    If we take a number x and do the following:

    log_{a}x

    then

    log_{b}x

    What is the relationship of the two?

    I believe the formula that sums the relationship is:

    log_{b}x log_{a}x=log_{b}x

    Can anyone explain how this is reached?

    P.S sorry for the formatting!
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  2. #2
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    There is no relationship between them because the bases are different

    If you said that \log_b(x) = \log_a(x) then a=b but as two expressions there is no relationship


    You can use the change of base rule should you need to: \log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}
    Last edited by e^(i*pi); January 7th 2011 at 02:22 PM. Reason: expanding answer
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  3. #3
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    Really? I have a book that states the following:

    "The log of 0.278 is not equal to 1n 0.278. i.e. logarithms with different bases have different values. The different values are, however, related to eachother...."

    :?
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  4. #4
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    Quote Originally Posted by BIOS View Post
    Really? I have a book that states the following:
    "The log of 0.278 is not equal to 1n 0.278. i.e. logarithms with different bases have different values. The different values are, however, related to eachother...."
    Of course that is the case.
    If a\ne b~\&~N\ne 1 then \log_a(N)\ne \log_b(N).
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  5. #5
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    I think it's referring to a different relationship then i posted (and to example the book uses.)

    E.g. the relationship between:

    \log_{b}a and the \log_{a}x

    defining the relationship as:

    \log_{b}a\log_{a}x=\log_{b}x

    "This is the change of base formula which relates logarithms of a number relative to two different bases."

    How is that the case if the logarithms are of two different numbers, namely a and x? or am i misunderstanding?
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  6. #6
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    Uning the definitions \log_b(a)=\dfrac{\ln(a)}{\ln(b)}~\&~ \log_a(x)=\dfrac{\ln(x)}{\ln(a)} it is clear the by multiplying we

    get \,~~\log_b(a)\log_a(x)= \dfrac{\ln(x)}{\ln(b)}=\log_b(x) ,

    What is your question now?
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  7. #7
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    I think i have it now. Thanks Plato. I was looking at the definition a different way.
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  8. #8
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    =A(say)

    then in exponetial form of this is x=a^A

    =B(say)

    then in exponetial form of this is x=b^B

    then relation is a^A=b^B

    i think 2 question is not valid always.
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  9. #9
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    Quote Originally Posted by saravananbs View Post
    =A(say)
    then in exponetial form of this is x=a^A
    =B(say)
    then in exponetial form of this is x=b^B then relation is a^A=b^B
    i think 2 question is not valid always.
    What a confusing posting.
    Can you tell us what point you are trying to make?

    We do know that \log_b(a)=x is defined as b^x=a.
    So what is your point?
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