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Math Help - Help with transposition

  1. #1
    Newbie
    Joined
    May 2009
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    13

    Help with transposition

    Hello,

    I'm having particular problems with re-arranging this formulae.
    If someone could help me with it i'd be most thankful. Here goes:

    X = y/z ln(1+r^3/1- r^3) Make r the subject.


    Thanks.
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  2. #2
    Super Member

    Joined
    May 2006
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    Lexington, MA (USA)
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    Hello, Andy!

    This is The Transposition Problem from Hell
    . . coming soon to a theater near you.


    Solve for r\!:\quad x \;=\;\frac{y}{z}\ln\left(\frac{1+r^3}{1-r^3}\right)

    We have: . \frac{y}{z}\ln\left(\frac{1+r^3}{1-r^3}\right) \;=\;x

    Multiply by \frac{z}{y}\!:\;\; \ln\left(\frac{1+r^3}{1-r^3}\right) \;=\;\frac{xz}{y}

    Exponentiate: . \frac{1+r^3}{1-r^3} \;=\;e^{\frac{xz}{y}}

    Multiply by (1-r^3)\!:\;\;1 + r^3 \;=\;e^{\frac{xz}{y}}(1 - r^3)


    Expand: . 1 + r^3 \;=\;e^{\frac{xz}{y}} - e^{\frac{xz}{y}}r^3


    Rearrange terms: . e^{\frac{xz}{y}}r^3 + r^3 \;=\;e^{\frac{xz}{y}} - 1


    Factor: . \left(e^{\frac{xz}{y}} + 1\right)r^3 \;=\;e^{\frac{xz}{y}} - 1

    Then: . r^3 \;=\;\frac{e^{\frac{xz}{y}} - 1}{e^{\frac{xz}{y}} + 1}

    Therefore: . r \;=\;\sqrt[3]{\frac{e^{\frac{xz}{y}}-1}{e^{\frac{xz}{t}} + 1}}

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  3. #3
    Newbie
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    May 2009
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    It's amazing how simple it looks once you have the answer!
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