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Math Help - Please help me 2

  1. #1
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    Please help me 2

    Does anyone know what the answer could be to this question as my dad is trying to work it out but has had no luck. We would be grateful for any help and advice.


    Express log9xy in terms of log 3x and log3y.
    Without using tables, solve for x and y the simultaneous equations
    Log 9xy = 5
    ---
    2

    Log 3x log3y = -6
    Expressing your answers as simply as possible
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Please help me 2

    \log_9 xy = \log_9 3 \times \log_3 xy = \frac 1 2 \log_3 xy = \frac 1  2 (\log_3 x + \log_3y)

    Sorry, but I'm not following what the second part of the question is asking. Do you mean this:

     \frac {log_9 xy} 2 = 5
     \log_3 x \log_3y = -6

    if so, then if you sub in the results of the first part you get two equations of the form:

     a+b= 20
     ab= -6

    Where  a= \log_3 x and b= \log_3 y

    hope this helps.
    Last edited by ebaines; July 20th 2014 at 10:30 AM.
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