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Math Help - Need Help solving/manipulating hard equation

  1. #1
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    I am trying to solve the following equation for Tc (Temperature in celsius).
    So far I was able to generate this equation from an expression I found online.


    RH = 9.5exp(-17.29Tc/Tc+238.3)*(10.5C-Tc)%


    1). ln(RH/9.5) = ln(e(-17.27Tc/Tc+238.3)*(10.5C-Tc))

    2). ln(RH/9.5) = ((-17.27Tc/Tc+238.3)*(10.5C-Tc))lne ; lne = 1

    3). ln(RH/9.5) = (-17.27Tc)/((10.5C-Tc)*(Tc+100C)) ; 238.3 = 100C

    How to solve resulting quadratic for Tc??


    Snow at above freezing temperatures | ScienceBits

    scroll down and find the equation for RH. Can you solve for Tc?

    Thanks,

    Rick
    Attached Thumbnails Attached Thumbnails Need Help solving/manipulating hard equation-untitled.png  
    Last edited by mr fantastic; June 7th 2010 at 03:59 PM. Reason: Posts moved and merged.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by rwd5013 View Post
    Snow at above freezing temperatures | ScienceBits

    scroll down and find the equation for RH. Can you solve for Tc?

    Thanks,

    Rick
    With some standard algebra you can get it into the form

    17.27T_c^2 - \left(181.335+\ln \left(\frac{RH}{9.5}\right)\right)T_c - 283.3\ln \left(\frac{RH}{9.5}\right) = 0

    Use the quadratic formula to find T_c
    Last edited by mr fantastic; June 7th 2010 at 04:00 PM. Reason: Moved from another thread.
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  3. #3
    MHF Contributor Amer's Avatar
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    RH = 9.5e^{(-17.29Tc/Tc+238.3)(10.5C-Tc)}


    e^{(-17.29Tc/Tc+238.3)(10.5-Tc)} = \frac{RH}{9.5}

    \ln (e^{(-17.29Tc/Tc+238.3)(10.5-Tc)}) = \ln (\frac{RH}{9.5})

    \left(\frac{-17.29Tc}{Tc+238.3}\right)(10.5-Tc) = \ln \left(\frac{RH}{9.5}\right)

    \frac{-17.29(10.5)Tc+17.29(Tc)^2}{Tc+238.3} = \ln \left(\frac{RH}{9.5}\right)

    -17.29(10.5)Tc+17.29(Tc)^2 = (Tc+238.3)\ln \left(\frac{RH}{9.5}\right)

    17.29(Tc)^2 +Tc\left(-17.29(10.5)-\ln \left(\frac{RH}{9.5}\right)\right)-238.3\left(\frac{RH}{9.5}\right)

    A=17.29

    B=\left(-17.29(10.5)+\ln \left(\frac{RH}{9.5}\right)\right)

    C = -238.3\left(\frac{RH}{9.5}\right)

    Tc = \frac{-B \mp \sqrt{B^2 -4AC}}{2A}
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  4. #4
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    Just do ln of both sides and you will solve it. Nothing special and challenging.
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