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Math Help - Difficult A2 logarithm

  1. #1
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    Difficult A2 logarithm

    Given the simultaneous equations
    2^x=3^y
    x+y=1
    show that x=log3/log6
    Last edited by kingkaisai2; May 21st 2006 at 01:36 AM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by kingkaisai2
    Given the simultaneous equations
    2^x=3^y
    x+y=1
    show that x=log3/log6
    2^x = 3^y ----(1)
    x +y = 1 ---------(2)

    Take the logs of both sides of (1),
    x*log(2) = y*log(3) ----(1a)

    From (2),
    y = 1-x
    Substitute that into (1a),
    x*log(2) = (1-x)log(3)
    x*log(2) = log(3) -x*log(3)
    x*log(2) +x*log(3) = log(3)
    x[log(2) +log(3)] = log(3)
    x[log(2*3)] = log(3)
    x*log(6) = log(3)
    x = log(3) / log(6) ---------answer.
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by kingkaisai2
    Given the simultaneous equations
    2^x=3^y
    x+y=1
    show that x=log3/log6
    eliminate y
    taking log on both sides of the first equation
    xlog2=ylog3
    substituting y from second equation
    xlog2=(1-x)log3
    rearranging
    x(log2+log3)=log3
    therefore,x=log3/log6
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