Simple Algebra problem - Solving for 2 of the same unknown

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The discussion focuses on solving the equation ln(0.0693/0.0475) = (70435/8.314)((T3-298)/(T3*298)) for T3, where the presence of two T3 variables complicates the process. A participant expresses confusion over the dual T3s, suggesting that algebraic manipulation might be flawed. Another contributor clarifies that by rearranging the equation, it can be simplified to eliminate one T3, making it easier to solve. They recommend multiplying both sides by the denominator to facilitate the solution. Ultimately, the key to solving the equation lies in reducing the complexity by managing the fractions effectively.
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Homework Statement



ln(0.0693/0.0475) = (70435/8.314)((T3-298)/(T3*298))

Solve for T3



Homework Equations





The Attempt at a Solution



The fact that there are 2 T3's is really throwing me off, I feel that I am doing something algebraically wrong.

0.377 = 8471.9((T3-298)/(T3*298))

How would I solve for T3?
 
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smaan said:

Homework Statement



ln(0.0693/0.0475) = (70435/8.314)((T3-298)/(T3*298))

Solve for T3

Homework Equations



The Attempt at a Solution



The fact that there are 2 T3's is really throwing me off, I feel that I am doing something algebraically wrong.

0.377 = 8471.9((T3-298)/(T3*298))

How would I solve for T3?
\displaystyle \frac{T_3-298}{298T_3}=\frac{T_3}{298T_3}-\frac{298}{298T_3}
\displaystyle =\frac{1}{298}-\frac{1}{T_3}​
Now there's only one T3 .
 
Sometimes I find "getting rid" of the fraction helpful for seeing the solution. Basically multiply both sides by the denominator (of the RHS) and expand. Then you collect like terms and everything else is easy.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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