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Math Help - Logarithm

  1. #1
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    Logarithm

    If log4 N=p and log12 N=q, show that

    log
    3 48= Logarithm-msp310219e962cfa4gc2d1700001a0d133gdh9c3fh8.gif

    I tried so many times but it seems to be so tough for me... can anyone help?
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  2. #2
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    This might be sending you the long way home but write out \displaystyle \frac{p+q}{p-q} in terms of the logs given then change the base of all logs to base 3.
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  3. #3
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    Quote Originally Posted by MichaelLight View Post
    If log4 N=p and log12 N=q, show that

    log
    3 48= Click image for larger version. 

Name:	MSP310219e962cfa4gc2d1700001a0d133gdh9c3fh8.gif 
Views:	46 
Size:	695 Bytes 
ID:	20812

    I tried so many times but it seems to be so tough for me... can anyone help?
    How did "N" disappear? And where did "48" come from?
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  4. #4
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    Quote Originally Posted by pickslides View Post
    This might be sending you the long way home but write out \displaystyle \frac{p+q}{p-q} in terms of the logs given then change the base of all logs to base 3.
    I tried but i couldn't get the required answer, have you tried it before?

    Quote Originally Posted by HallsofIvy View Post
    How did "N" disappear? And where did "48" come from?
    I don't know...
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  5. #5
    Senior Member BAdhi's Avatar
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    try using \log_aN=\frac{1}{\log_Na} for \frac{p+q}{p-q}

    EDIT:
    forgot to mention

    you need \log_ab=\frac{\log_Nb}{\log_Na} as well. That is where "N" disappears.
    Last edited by BAdhi; February 15th 2011 at 06:31 AM. Reason: latex edit
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  6. #6
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    Hello, MichaelLight!

    My approach is worse that Pickslides' suggestion.

    I used the Base-Change formula.


    \text{If }\,\log_4 N=p\,\text{ and }\,\log_{12}N=q,\,\text{ show that: }\:\log_3 48\:=\: \dfrac{p+q}{p-q}

    \displaystyle\text{We have: }\:\log_348 \;\;=\;\;\frac{\ln48}{\ln3} \;\;=\;\;\frac{\ln(12\cdot4)}{\ln(\frac{12}{4})} \;\;=\;\;\frac{\ln12 + \ln4}{\ln12 - \ln4}


    Divide numerator and denominator by (\ln12\!\cdot\!\ln4)\!:

    . . \displaystyle =\;\;\frac{\;\dfrac{\ln12 + \ln4}{\ln12\cdot\ln4}\;} {\dfrac{\ln12 - \ln4}{\ln12\cdot\ln4}} \;\;=\;\;\frac{\;\dfrac{1}{\ln4} + \dfrac{1}{\ln12}\;}{\dfrac{1}{\ln4} - \dfrac{1}{\ln12}}


    Multiply by \dfrac{\ln N}{\ln N}\!:

    . . \displaystyle =\;\;\frac{\ln N}{\ln N}\cdot\frac{\dfrac{1}{\ln4} + \dfrac{1}{\ln12}}{\dfrac{1}{\ln4} - \dfrac{1}{\ln12}} \;\;=\;\;\frac{\;\dfrac{\ln N}{\ln 4} + \dfrac{\ln N}{\ln12}\;}{\dfrac{\ln N}{\ln 4} - \dfrac{\ln N}{\ln 12}}


    . . \displaystyle =\;\;\frac{\log_4\!N + \log_{12}\!N}{\log_4\!N - \log_{12}\!N}} \;\;=\;\;\frac{p + q}{p - q}

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