I'm not sure if this is solvable. However, there appears to be one and only one solution (assuming you want real values for X
and T
), available by the following kludgy means:
First, solve the individual equations for T
:
s1 = Solve[-(8.314)*T*Log[X] == 8.3*(1400 - T), T]
s2 = Solve[-(8.314)*T*Log[1 - X] == 8.3*(1200 - T), T]
giving
{{T -> 11620./(8.3 - 8.314 Log[X])}}
{{T -> 9960./(8.3 - 8.314 Log[1. - 1. X])}}
Then equate the two answers:
Solve[s1[[1, 1, 2]] == s2[[1, 1, 2]], X]
giving a numerical method warning and the answer,
{{X -> 0.434986}}
from which T
can be obtained:
s1[[1, 1, 2]] /. X -> 0.4349859661954613`
s2[[1, 1, 2]] /. X -> 0.4349859661954613`
giving
763.423
763.423
Plotting the regions where the equations hold true in the $(X,T)$ plane suggests that this is the only solution. The red curve is the support of the first equations's validity, and the blue curve is the support of the second equation's validity:
ImageAdd @@ (ColorConvert[
ComplexPlotR2[
CCompileR2[10 #], {-2 + 10^-6 RandomReal[], 3,
0.005}, {-0 + 10^-6 RandomReal[], 6000, 10}], "RGB"] & /@ {1/
Abs[-8.3` (1400 - y) - 8.314` y Log[x]]^(1/2), -1/
Abs[-8.3` (1200 - y) - 8.314` y Log[1 - x]]^(1/2)})
producing the following colorful image:

where the following helper functions have been defined:
hue = Compile[{{z, _Complex}}, {(1.0 Arg[-z] + \[Pi])/(2 \[Pi]),
Exp[1 - Max[Abs[z], 1]], Min[Abs[z], 1]},
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
ComplexPlotR2[f_, {x0_, x1_, \[Delta]x_}, {y0_, y1_, \[Delta]y_}] :=
Image[hue[
f[#[[All, All, 1]], #[[All, All, 2]]] &@
Outer[List, Range[x0, x1, \[Delta]x],
Range[y0, y1, \[Delta]y]]]\[Transpose], ColorSpace -> Hue,
Magnification -> 1];
CCompileR2[expr_] :=
Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
There appears to only one intersection, which as mentioned before is at X -> 0.434986, T -> 763.423
.