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Math Help - Formula Rearrange

  1. #1
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    Question Formula Rearrange

    Hi,

    I am generally not too bad at rearranging formulas, but the result I have for the below differs from that which my colleague has calculated. Other opinion(s) would be very much appreciated.
    If possible a couple of intermediate steps would be useful for my understanding.

    The equation is below and I would like to solve for X

    Formula Rearrange-equation-1.jpg

    When I rearrange i get:

    Formula Rearrange-equation-2.jpg

    [that is a B by the root sign]


    I would very much appreciate another interpretation.


    Regards

    (I hope this is the correct forum I did think about it!)
    Last edited by UberTyson; July 12th 2012 at 02:58 AM. Reason: missing info
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  2. #2
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    Re: Formula Rearrange

    We have 1 + \left(\frac{x}{C}\right)^B=\frac{A-D}{y-D}, from where \left(\frac{x}{C}\right)^B=\frac{A-D}{y-D}-1, \frac{x}{C}=\sqrt[B]{\frac{A-D}{y-D}-1} and x=C\sqrt[B]{\frac{A-D}{y-D}-1}=C\sqrt[B]{\frac{A-y}{y-D}}.
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  3. #3
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    Re: Formula Rearrange

    Thanks very much for the response.

    Would it be possible to explain 2 of the steps?:

    First step: how does "(y - D)" become the denominator as opposed to just "y"

    Last step: How (A - D)/(y - D) becomes (A - y)/(y - D)


    i apologise if I am being a bit thick here.


    Regards
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  4. #4
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    Re: Formula Rearrange

    Quote Originally Posted by UberTyson View Post
    First step: how does "(y - D)" become the denominator as opposed to just "y"
    In your version, the denominator is also y - D. We start from y=\frac{A-D}{1+\left(\frac{x}{C}\right)^B}+D, subtract D from both sides to get y - D = \frac{A-D}{1+\left(\frac{x}{C}\right)^B}, from where 1+\left(\frac{x}{C}\right)^B = \frac{A-D}{y-D} (just like 5 = 10 / 2 is equivalent to 2 = 10 / 5).

    Quote Originally Posted by UberTyson View Post
    Last step: How (A - D)/(y - D) becomes (A - y)/(y - D)
    It's not just (A - D)/(y - D), but (A - D)/(y - D) - 1. Represent 1 as (y - D) / (y - D); now you have the same denominator, and A - D - (y - D) = A - y.
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  5. #5
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    Re: Formula Rearrange

    Quote Originally Posted by UberTyson View Post
    Thanks very much for the response.

    Would it be possible to explain 2 of the steps?:

    First step: how does "(y - D)" become the denominator as opposed to just "y"
    Let's start from the beginning:

    y=\frac{A-D}{1+\left(\frac xC \right)^B} + D~\implies~y-D=\frac{A-D}{1+\left(\frac xC \right)^B}~\implies~1+\left(\frac xC \right)^B = \frac{A-D}{y-D}

    Last step: How (A - D)/(y - D) becomes (A - y)/(y - D)

    ...
    I'll take only the radicand:

    \frac{A-D}{y-D}-1 = \frac{A-D}{y-D}-\frac{y-D}{y-D} = \frac{A-D-y+D}{y-D} = \frac{A-y}{y-D}


    EDIT: This is a wonderful example of an unnecessary reply. Sorry, emakarov, I didn't see that you were online.
    Last edited by earboth; July 12th 2012 at 05:09 AM.
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  6. #6
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    Re: Formula Rearrange

    Thank you both for your help it is now clear in my mind. Makes me feel stupid when you solve it so easily!

    Appreciate the help,

    Regards
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