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Math Help - determine an expression for log3 interms of x and y

  1. #1
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    determine an expression for log3 interms of x and y

    no cal multiple choice
    Given Log 6 = x Log 8 = y determine an expression for log3 interms of x and y.

    a: x/3 - y
    b: x-y/3
    c: y/3-x
    d y-x/3


    Don't understand how to turn log 3 into log 6 and log8.. help

    ThankYou
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  2. #2
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    remember log(xy) = log(x) + log(y). So
    log 6 = log (2*3) = log 2 + log 3.
    Also log 8 = log 2*2*2 = log 2 + log 2 + log 2 = 3log2.
    Can you proceed?
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  3. #3
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    Sorry.. i can't proceed.
    If log2 + log3
    log2 + x

    3log2 = log2^3

    ..ah..>.<
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  4. #4
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    Start with the fact that

    \log{6} = x\log{8}

    \log{(2\cdot 3)} = x\log{\left(2^3\right)}

    \log{2} + \log{3} = 3x\log{2}

    \log{3} = 3x\log{2} - \log{2}.

    Call this Equation 1.


    Also, remember that

    \log{6} = y

    \log{(2\cdot 3)} = y

    \log{2} + \log{3} = y

    \log{2} = y - \log{3}.

    Substitute this back into Equation 1.


    \log{3} = 3x(y - \log{3}) - (y - \log{3})

    \log{3} = 3xy - 3x\log{3} - y + \log{3}

    0 = 3xy - 3x\log{3} - y

    3x\log{3} = 3xy - y

    \log{3} = y - \frac{y}{3x}.



    So none of your possibilities are correct...
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  5. #5
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    Hello, hovermet!

    Given: . \log 6 = x,\;\log 8 = y
    Determine an expression for \log3 in terms of x and y

    . . (a)\;\;\frac{x}{3} - y \qquad (b)\;\;x-\frac{y}{3} \qquad (c)\;\;\frac{y}{3}-x \qquad (d)\;\;y-\frac{x}{3}

    Note that: . \log 8 \,=\,y \quad\Rightarrow\quad \log(2^3) \,=\,y \quad\Rightarrow\quad 3\log 2 \,=\,y \quad\Rightarrow\quad \log 2 \,=\,\frac{y}{3}


    So we have: . \log 3 \;\;=\;\;\log\left(\frac{6}{2}\right) \;\;=\;\;\log 6 - \log 2 \;\;=\;\;x - \frac{y}{3}



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  6. #6
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    Thanks guys.. it seems like i forgot an AND between log 6 = x Log 8 = y ^^;
    In fact, i done the log test pretty well today with the understanding of the mistaken question Thanks^^
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