How to solve equations where the power of $x$ is a function of $x$?











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I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?



Any help will be greatly appreciated.










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    I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?



    Any help will be greatly appreciated.










    share|cite|improve this question









    New contributor




    user3430220 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?



      Any help will be greatly appreciated.










      share|cite|improve this question









      New contributor




      user3430220 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I have been trying to find a solution for equations of the type $x^{px-c} = a$. I know how to use Lambert W. function to find solutions for $x^x = a$, but the function of $x$ at the exponent is making things much difficult for me. I do not how to proceed in this scenario. Should I try series expansion or some approximation method?



      Any help will be greatly appreciated.







      polynomials exponential-function lambert-w






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      edited Dec 2 at 5:32









      Rócherz

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      asked Dec 2 at 5:29









      user3430220

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          I do not think that explicit solutions could be obtained for the zero's of
          $$f(x)=x^{p,x-c} - a$$
          and you will, more than likely, use numerical methods.



          In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
          $$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
          $x^x=a^{frac1p}$.



          For $cneq 0$, consider instead the function
          $$g(x)=(px-c)log(x)-log(a)$$
          $$g'(x)=-frac{c}{x}+p log (x)+p$$
          $$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
          $$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
          $$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
          c}{p}right)right)$$
          Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
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            up vote
            0
            down vote













            I do not think that explicit solutions could be obtained for the zero's of
            $$f(x)=x^{p,x-c} - a$$
            and you will, more than likely, use numerical methods.



            In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
            $$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
            $x^x=a^{frac1p}$.



            For $cneq 0$, consider instead the function
            $$g(x)=(px-c)log(x)-log(a)$$
            $$g'(x)=-frac{c}{x}+p log (x)+p$$
            $$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
            $$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
            $$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
            c}{p}right)right)$$
            Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.






            share|cite|improve this answer

























              up vote
              0
              down vote













              I do not think that explicit solutions could be obtained for the zero's of
              $$f(x)=x^{p,x-c} - a$$
              and you will, more than likely, use numerical methods.



              In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
              $$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
              $x^x=a^{frac1p}$.



              For $cneq 0$, consider instead the function
              $$g(x)=(px-c)log(x)-log(a)$$
              $$g'(x)=-frac{c}{x}+p log (x)+p$$
              $$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
              $$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
              $$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
              c}{p}right)right)$$
              Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                I do not think that explicit solutions could be obtained for the zero's of
                $$f(x)=x^{p,x-c} - a$$
                and you will, more than likely, use numerical methods.



                In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
                $$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
                $x^x=a^{frac1p}$.



                For $cneq 0$, consider instead the function
                $$g(x)=(px-c)log(x)-log(a)$$
                $$g'(x)=-frac{c}{x}+p log (x)+p$$
                $$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
                $$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
                $$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
                c}{p}right)right)$$
                Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.






                share|cite|improve this answer












                I do not think that explicit solutions could be obtained for the zero's of
                $$f(x)=x^{p,x-c} - a$$
                and you will, more than likely, use numerical methods.



                In the case where $c=0$, there is a solution, still expressed in terms of Lambert function. It would be
                $$x=frac{log (a)}{p, Wleft(frac{log (a)}{p}right)}$$ obtained writing
                $x^x=a^{frac1p}$.



                For $cneq 0$, consider instead the function
                $$g(x)=(px-c)log(x)-log(a)$$
                $$g'(x)=-frac{c}{x}+p log (x)+p$$
                $$g''(x)=frac{c+p x}{x^2}$$ The first derivative cancels at
                $$x_*=frac{c}{p,Wleft(frac{e c}{p}right)}$$ At this point
                $$g''(x_*)=frac{p^2}{c} Wleft(frac{e c}{p}right) left(1+Wleft(frac{e
                c}{p}right)right)$$
                Considering the value of $g(x_*)$ and the above would give you indications about the existence of the roots, their number and where they are located.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 at 7:29









                Claude Leibovici

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                117k1156131






















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