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Math Help - Equation help

  1. #1
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    Question Equation help

    if log x / a-b = log y / b-c = log z / c-a then xyz = ?????

    answer - a)2 , b)-1 ,c) 1 ,d) 0

    please tell the steps.
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  2. #2
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    Have you read the "Sticky: List of rules used to moderate MHF - please read carefully before posting."?
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Also, making use of brackets to clear any confusion is important. For example, is your question:

    \dfrac{\log\ x}{a-b} = \dfrac{\log\ y}{b-c} = \dfrac{\log\ z}{c-a}

    or

    \log\ \dfrac{x}{a-b} = \log\ \dfrac{y}{b-c} = \log\ \dfrac{z}{c-a}

    ?
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  4. #4
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    Quote Originally Posted by Unknown008 View Post
    Also, making use of brackets to clear any confusion is important. For example, is your question:

    \dfrac{\log\ x}{a-b} = \dfrac{\log\ y}{b-c} = \dfrac{\log\ z}{c-a}

    or

    \log\ \dfrac{x}{a-b} = \log\ \dfrac{y}{b-c} = \log\ \dfrac{z}{c-a}

    ?
    the top one
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Have you read the "Sticky: List of rules used to moderate MHF - please read carefully before posting."?
    srry whats the problem? i thik this is fro algebra right? plzz guide me.
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  6. #6
    MHF Contributor harish21's Avatar
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    you have to show the steps where you are stuck instead of asking for the entire solution... Thats what HallsofIvy meant..
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by saha.subham View Post
    srry whats the problem? i thik this is fro algebra right? plzz guide me.
    The problem is that you have to show what you have tried so far, to solve this problem.
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    The problem is that you have to show what you have tried so far, to solve this problem.
    (x)/a-b = (y )/ b-c and (y)/b-c = (z)/c - a (after cancelling out log.)
    now what?
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  9. #9
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    Now, you have (b- c)x= (a- b)y and (c- a)y= (b- c)z which you can rewrite as bx- cx= ay- by and cy- ay= bz- cz. Play with those.
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Hm... actually, you cannot cancel log like this. If it were, because you have a denominator outside the log function. If you want to cancel log, you will have to do it like this:

    \dfrac{1}{a-b} \log (x) = \log x^{\left(\frac{1}{a-b}\right)}

    Then you get:

    x^{\left(\frac{1}{a-b}\right)} = y^{\left(\frac{1}{b-c}\right)} = z^{\left(\frac{1}{c-a}\right)}

    But I don't know if that will work...

    I tried using this:

    \log(x) = \dfrac{(a-b) \log (y)}{(b-c)}

    \log(z) = \dfrac{(c-a) \log (y)}{(b-c)}

    Then, we get:

    \log (x) + \log(y) + \log(z) = \dfrac{(a-b) \log (y)}{(b-c)} + \log(y) + \dfrac{(c-a) \log (y)}{(b-c)}

    Hence,

    \log(xyz) = \dfrac{(a-b) \log (y) + (b-c)\log(y)  + (c-a)\log(y)}{(b-c)}

    \log(xyz) = \dfrac{[(a-b) + (b-c) + (c-a)]\log (y)}{(b-c)}

    \log(xyz) = \dfrac{\log (y)}{(b-c)}

    Though I'm not too sure where to go from here.

    EDIT: From the before last line:::

    \log(xyz) = \dfrac{[(a-b) + (b-c) + (c-a)]\log (y)}{(b-c)}

    \log(xyz) = \dfrac{[0]\log (y)}{(b-c)}

    \log(xyz) = \dfrac{0}{(b-c)} = 0
    Last edited by Unknown008; January 30th 2011 at 04:45 AM.
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    Now, you have (b- c)x= (a- b)y and (c- a)y= (b- c)z which you can rewrite as bx- cx= ay- by and cy- ay= bz- cz. Play with those.
    i am not being able to find the value of xyz plzz help
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  12. #12
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    If you want to find xyz

    then you can obtain it from logx+logy+logz=log(xyz)

    using the properties of logs.

    You need to get an expression for the sum of logs.
    Hence

    \frac{logx}{a-b}=\frac{logy}{b-c}\Rightarrow\ logy=\left(\frac{b-c}{a-b}\right)logx

    Also

    \frac{logx}{a-b}=\frac{logz}{c-a}\Rightarrow\ logz=\left(\frac{c-a}{a-b}\right)logx

    Therefore

    logx+logy+logz=logx\left(1+\frac{b-c}{a-b}+\frac{c-a}{a-b}\right)=logx\left(\frac{a-b+b-c+c-a}{a-b}\right)=logx\left(\frac{0}{a-b}\right)=0

    logx+logy+logz=log(xyz)=0\Rightarrow\ xyz=\;?
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  13. #13
    MHF Contributor Unknown008's Avatar
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    Oh, I was close but didn't see it... my bad
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