# Thread: Need Help solving/manipulating hard equation

1. 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°)*(10.5°C-Tc)%

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

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

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

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

2. Originally Posted by rwd5013
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 $T_c$

3. $RH = 9.5e^{(-17.29Tc/Tc+238.3)(10.5°C-Tc)}$

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

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

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

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

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

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

$A=17.29$

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

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

$Tc = \frac{-B \mp \sqrt{B^2 -4AC}}{2A}$

4. Just do ln of both sides and you will solve it. Nothing special and challenging.