Is this pair of equations impossible to solve for x? $y_1 = x_2 - v^{pm 1}e^{-x_1}$, or equivalently $(x -...












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Original:



I'm trying to solve the following for $x_1$ and $x_2$,



$$ y_1 = x_2 - v, e^{-x_1} $$
$$ y_2 = x_1 - frac{1}{v}, e^{-x_2} $$



in terms of $y_1$, $y_2$, and $v$, which are known and real, and where $v$ is positive. This should be determined and the form seems familiar, but I can't find an inverse. Does anyone know how to go about solving them? Does a solution even exist?



Note 1:



These were simplified from the original relations,



$$ ln{gamma_1^∞} = 1 - lnleft( frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} right) - frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} $$



$$ ln{gamma_2^∞} = 1 - lnleft( frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} right) - frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} $$



where $gamma_1^∞$, $gamma_2^∞$, $V_1$, $V_2$, and $T$ are positive and $tau_{1,2}$ and $tau_{1,2}$ are real. Above,
$$v = frac{V_1}{V_2},, a=frac{tau_{1,2}}{T},, b=frac{tau_{2,1}}{T},, y_i = ln{gamma_i^∞} - (1 + ln frac{V_i}{V_{3-i}}) $$



Note 2:



Substituting and reducing yields the equivalent problem:



$$ (x - y)c^{exp(-x)} = z $$



(where $x = x_1, y=y_2, c=exp(v), z=e^{-y_1}/v$).



Can this equation be solved for x?










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    0












    $begingroup$


    Original:



    I'm trying to solve the following for $x_1$ and $x_2$,



    $$ y_1 = x_2 - v, e^{-x_1} $$
    $$ y_2 = x_1 - frac{1}{v}, e^{-x_2} $$



    in terms of $y_1$, $y_2$, and $v$, which are known and real, and where $v$ is positive. This should be determined and the form seems familiar, but I can't find an inverse. Does anyone know how to go about solving them? Does a solution even exist?



    Note 1:



    These were simplified from the original relations,



    $$ ln{gamma_1^∞} = 1 - lnleft( frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} right) - frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} $$



    $$ ln{gamma_2^∞} = 1 - lnleft( frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} right) - frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} $$



    where $gamma_1^∞$, $gamma_2^∞$, $V_1$, $V_2$, and $T$ are positive and $tau_{1,2}$ and $tau_{1,2}$ are real. Above,
    $$v = frac{V_1}{V_2},, a=frac{tau_{1,2}}{T},, b=frac{tau_{2,1}}{T},, y_i = ln{gamma_i^∞} - (1 + ln frac{V_i}{V_{3-i}}) $$



    Note 2:



    Substituting and reducing yields the equivalent problem:



    $$ (x - y)c^{exp(-x)} = z $$



    (where $x = x_1, y=y_2, c=exp(v), z=e^{-y_1}/v$).



    Can this equation be solved for x?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Original:



      I'm trying to solve the following for $x_1$ and $x_2$,



      $$ y_1 = x_2 - v, e^{-x_1} $$
      $$ y_2 = x_1 - frac{1}{v}, e^{-x_2} $$



      in terms of $y_1$, $y_2$, and $v$, which are known and real, and where $v$ is positive. This should be determined and the form seems familiar, but I can't find an inverse. Does anyone know how to go about solving them? Does a solution even exist?



      Note 1:



      These were simplified from the original relations,



      $$ ln{gamma_1^∞} = 1 - lnleft( frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} right) - frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} $$



      $$ ln{gamma_2^∞} = 1 - lnleft( frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} right) - frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} $$



      where $gamma_1^∞$, $gamma_2^∞$, $V_1$, $V_2$, and $T$ are positive and $tau_{1,2}$ and $tau_{1,2}$ are real. Above,
      $$v = frac{V_1}{V_2},, a=frac{tau_{1,2}}{T},, b=frac{tau_{2,1}}{T},, y_i = ln{gamma_i^∞} - (1 + ln frac{V_i}{V_{3-i}}) $$



      Note 2:



      Substituting and reducing yields the equivalent problem:



      $$ (x - y)c^{exp(-x)} = z $$



      (where $x = x_1, y=y_2, c=exp(v), z=e^{-y_1}/v$).



      Can this equation be solved for x?










      share|cite|improve this question











      $endgroup$




      Original:



      I'm trying to solve the following for $x_1$ and $x_2$,



      $$ y_1 = x_2 - v, e^{-x_1} $$
      $$ y_2 = x_1 - frac{1}{v}, e^{-x_2} $$



      in terms of $y_1$, $y_2$, and $v$, which are known and real, and where $v$ is positive. This should be determined and the form seems familiar, but I can't find an inverse. Does anyone know how to go about solving them? Does a solution even exist?



      Note 1:



      These were simplified from the original relations,



      $$ ln{gamma_1^∞} = 1 - lnleft( frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} right) - frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} $$



      $$ ln{gamma_2^∞} = 1 - lnleft( frac{V_1}{V_2} exp{-frac{tau_{1,2}}{T}} right) - frac{V_2}{V_1} exp{-frac{tau_{2,1}}{T}} $$



      where $gamma_1^∞$, $gamma_2^∞$, $V_1$, $V_2$, and $T$ are positive and $tau_{1,2}$ and $tau_{1,2}$ are real. Above,
      $$v = frac{V_1}{V_2},, a=frac{tau_{1,2}}{T},, b=frac{tau_{2,1}}{T},, y_i = ln{gamma_i^∞} - (1 + ln frac{V_i}{V_{3-i}}) $$



      Note 2:



      Substituting and reducing yields the equivalent problem:



      $$ (x - y)c^{exp(-x)} = z $$



      (where $x = x_1, y=y_2, c=exp(v), z=e^{-y_1}/v$).



      Can this equation be solved for x?







      systems-of-equations inverse-function elementary-functions






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      edited Dec 12 '18 at 5:54







      alexchandel

















      asked Dec 12 '18 at 1:34









      alexchandelalexchandel

      1013




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          $begingroup$

          This looks very much as the expressions of the binary activity coefficients at infinite dilution from Wilson or T-K-Wilson models.



          I have been facing the same problem as you long time ago and, unfortunately, I did not find any way to solve it (except, as you did, eliminating one of the $x$ and be left with one single equation in terms of the remaining variable). To tell the truth, I was hoping to get solutions in terms of generalized Lambert function. This is probably why were generated nomograms for this class of models.



          Then, I gave up and just used numerical methods.



          But, you must also know that there could be multiple solutions to the problem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
            $endgroup$
            – alexchandel
            Dec 12 '18 at 7:51











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          This looks very much as the expressions of the binary activity coefficients at infinite dilution from Wilson or T-K-Wilson models.



          I have been facing the same problem as you long time ago and, unfortunately, I did not find any way to solve it (except, as you did, eliminating one of the $x$ and be left with one single equation in terms of the remaining variable). To tell the truth, I was hoping to get solutions in terms of generalized Lambert function. This is probably why were generated nomograms for this class of models.



          Then, I gave up and just used numerical methods.



          But, you must also know that there could be multiple solutions to the problem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
            $endgroup$
            – alexchandel
            Dec 12 '18 at 7:51
















          0












          $begingroup$

          This looks very much as the expressions of the binary activity coefficients at infinite dilution from Wilson or T-K-Wilson models.



          I have been facing the same problem as you long time ago and, unfortunately, I did not find any way to solve it (except, as you did, eliminating one of the $x$ and be left with one single equation in terms of the remaining variable). To tell the truth, I was hoping to get solutions in terms of generalized Lambert function. This is probably why were generated nomograms for this class of models.



          Then, I gave up and just used numerical methods.



          But, you must also know that there could be multiple solutions to the problem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
            $endgroup$
            – alexchandel
            Dec 12 '18 at 7:51














          0












          0








          0





          $begingroup$

          This looks very much as the expressions of the binary activity coefficients at infinite dilution from Wilson or T-K-Wilson models.



          I have been facing the same problem as you long time ago and, unfortunately, I did not find any way to solve it (except, as you did, eliminating one of the $x$ and be left with one single equation in terms of the remaining variable). To tell the truth, I was hoping to get solutions in terms of generalized Lambert function. This is probably why were generated nomograms for this class of models.



          Then, I gave up and just used numerical methods.



          But, you must also know that there could be multiple solutions to the problem.






          share|cite|improve this answer











          $endgroup$



          This looks very much as the expressions of the binary activity coefficients at infinite dilution from Wilson or T-K-Wilson models.



          I have been facing the same problem as you long time ago and, unfortunately, I did not find any way to solve it (except, as you did, eliminating one of the $x$ and be left with one single equation in terms of the remaining variable). To tell the truth, I was hoping to get solutions in terms of generalized Lambert function. This is probably why were generated nomograms for this class of models.



          Then, I gave up and just used numerical methods.



          But, you must also know that there could be multiple solutions to the problem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 6:12

























          answered Dec 12 '18 at 6:04









          Claude LeiboviciClaude Leibovici

          122k1157134




          122k1157134












          • $begingroup$
            That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
            $endgroup$
            – alexchandel
            Dec 12 '18 at 7:51


















          • $begingroup$
            That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
            $endgroup$
            – alexchandel
            Dec 12 '18 at 7:51
















          $begingroup$
          That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
          $endgroup$
          – alexchandel
          Dec 12 '18 at 7:51




          $begingroup$
          That's right. I was hoping to obtain explicit interaction energy expressions from the infinite dilution activity coefficients of binary Wilson mixtures. But I've stuck on this equation. It does look vaguely Lambertian, except the tetration in the middle.
          $endgroup$
          – alexchandel
          Dec 12 '18 at 7:51


















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