Simultaneous Equations with exponents

[cvr]

Given the two equatiions:

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

solve for x ant T. (a,b,c, and d are constants)

I don't see a simplifying substitution beyond

x = e^c(d-1/T) and 1 = e^a(b-1/T) + e^c(d-1/T)

ln(1) = 0 but how do you simplify ln(RHS) ?


Thanks for any suggestions.

 

[Gene]

You can't split up ln(1-x). Logs don't know about subtraction.
ln(1-x) = a(b-1/T)
ln(1-x)/a = b-1/T
ln(1-x)/a - b = -1/T

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

ln(1-x)/a-b = ln(x)/c-d
ln((1-x)^(1/a))-b = ln(x^(1/c))-d
ln((1-x)^(1/a)) - ln(x^(1/c)) = b-d
ln((1-x)^(1/a)/x^(1/c)) = b-d
(1-x)^(1/a)/x^(1/c) = e^(b-d)
I have to admit I don't see how to solve for x with generalized constants, but I think this is the way to go. As a test, if a=b=c=d=1 then x=1/2 and T=.5906161091 works.
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Gene