Simultaneous Eqns with exponents

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The discussion revolves around solving the simultaneous equations involving logarithmic functions and exponents: ln(1-x) = a(b-1/T) and ln(x) = c(d-1/T). A user expresses confusion about the next steps after deriving x = e^c(d-1/T) and a related expression for T. Another participant suggests that the problem may not have a solution for arbitrary values of a and c, indicating that these parameters complicate the equations. They propose that substituting T back into the equation for x leads to a polynomial form, x + wx^(a/c) - 1 = 0, where w is a constant. The conversation emphasizes the need for a systematic approach to tackle the equations effectively.
cvr
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How do you solve:

ln(1-x) = a(b-1/T)
ln(x) = c(d-1/T)

for x and T ?

I see that x = e^c(d-1/T) and 1 = e^a(b-1/T) + e^c(d -1)/T but it is unclear to me what substitution to try next.

Thanks

cvr
 
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cvr,

1.) Homework goes in the Science Education Zone, not the Math section. I'm moving this thread to the right place.

2.) You must show an attempt at the problem in order to receive help.

I've soft deleted mathman's post, and will restore it once you have shown a reasonable attempt.
 
I've update the post to include what I've tried so far

Thanks,

cvr
 
Well I'm not sure if you can solve this particular set for any a,c. I say a,c, because when I play with them, its the letters a & c that get in the way.

If you solve the first equation for x and the second equation for T, you can then put this equation for T into your equation for x and you'll get something like:

x+wx^(a/c)-1=0

where w is just a number
 
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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