Solving the equation $frac{ln (x)}{ln (1-x)} = frac{1}{x} - 1$












6














I'm trying to solve the following equation (which has solution $x = 1/2$)



$$frac{ln (x)}{ln (1-x)} = frac{1}{x} - 1 $$



I can't seem to do it analytically. Any ideas?






calculus algebra-precalculus logarithms







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  • 1




    Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=left(1-xright)^{left(1-xright)}$ but I'm not sure if that leads anywhere.
    – Jam
    Dec 7 at 13:58


















6














I'm trying to solve the following equation (which has solution $x = 1/2$)



$$frac{ln (x)}{ln (1-x)} = frac{1}{x} - 1 $$



I can't seem to do it analytically. Any ideas?






calculus algebra-precalculus logarithms







share|cite|improve this question




















  • 1




    Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=left(1-xright)^{left(1-xright)}$ but I'm not sure if that leads anywhere.
    – Jam
    Dec 7 at 13:58
















6












6








6


2





I'm trying to solve the following equation (which has solution $x = 1/2$)



$$frac{ln (x)}{ln (1-x)} = frac{1}{x} - 1 $$



I can't seem to do it analytically. Any ideas?






calculus algebra-precalculus logarithms







share|cite|improve this question















I'm trying to solve the following equation (which has solution $x = 1/2$)



$$frac{ln (x)}{ln (1-x)} = frac{1}{x} - 1 $$



I can't seem to do it analytically. Any ideas?






calculus algebra-precalculus logarithms




calculus algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 at 9:27









Martin Sleziak

44.6k7115270




44.6k7115270










asked Dec 7 at 8:09









user308485

49039




49039








  • 1




    Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=left(1-xright)^{left(1-xright)}$ but I'm not sure if that leads anywhere.
    – Jam
    Dec 7 at 13:58
















  • 1




    Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=left(1-xright)^{left(1-xright)}$ but I'm not sure if that leads anywhere.
    – Jam
    Dec 7 at 13:58










1




1




Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=left(1-xright)^{left(1-xright)}$ but I'm not sure if that leads anywhere.
– Jam
Dec 7 at 13:58






Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=left(1-xright)^{left(1-xright)}$ but I'm not sure if that leads anywhere.
– Jam
Dec 7 at 13:58












3 Answers
3






active

oldest

votes


















11














Let $f(x)=x ln x$. Then the given equation is
$$f(x)=f(1-x).$$
This is symmetric about $x=1/2$. Hence $x=1/2$ is a solution.






share|cite|improve this answer

















  • 4




    Very nice solution!
    – gimusi
    Dec 7 at 8:17






  • 5




    I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
    – Batominovski
    Dec 7 at 8:41








  • 2




    Yep, IMO too many upvotes as the existence of other solutions should be discussed.
    – Yves Daoust
    Dec 7 at 12:53



















6














The function $xlog x-(1-x)log(1-x)$ is only defined in $(0,1)$, is differentiable, has a root at $x=frac12$ (by inspection) and tends to zero at the interval endpoints.



The derivative,



$$log x+log(1-x)+2$$ has two roots in $(0,1)$, where $x(1-x)=e^{-2}$.



Hence the function is negative in $(0,frac12)$, with a single minimum, and positive in $(frac12,1)$, with a single maximum, and there are no other roots.



enter image description here






share|cite|improve this answer































    3














    Let $g:Itomathbb{R}$, where $I:=left[-dfrac12,+dfrac12right]$, be the function defined by $$g(t):=begin{cases}left(dfrac{1}{2}+tright),lnleft(dfrac12+tright)-left(dfrac{1}{2}-tright),lnleft(dfrac12-tright)&text{if }tinleft(-dfrac12,+dfrac12right),,\0&text{if }tinleft{-dfrac12,+dfrac12right},.end{cases}$$ (Observe that $g$ is continuous on the whole $I$, and is smooth on $left(-dfrac12,+dfrac12right)$.) Then, we are to solve for $xin(0,1)$ from $$gleft(x-dfrac12right)=0,,$$
    which is equivalent to solving for $yin left(-dfrac12,+dfrac12right)$ such that
    $$g(y)=0text{ by setting }y:=x-dfrac12,.$$



    We claim that $g$ has only three roots $-dfrac12,0,+dfrac12$, and this means the only $yinleft(-dfrac12,+dfrac12right)$ such that $g(y)=0$ is $y=0$, making $x=dfrac12$ the sole solution to the required equation. Suppose contrary that $g$ has more than three roots. Then, by symmetry, on the interval $left[0,dfrac12right]$, $g$ has at least three roots. Using Rolle's Theorem, $g'$ has at least two roots on $left(0,dfrac12right)$, and so $g''$ has at least one root on $left(0,dfrac12right)$. However, we have
    $$g'(t)=lnleft(frac12+tright)+lnleft(frac12-tright)+2,,$$
    $$g''(t)=frac{1}{frac{1}{2}+t}-frac{1}{frac{1}{2}-t},,$$
    and
    $$g'''(t)=-frac{1}{left(frac{1}{2}+tright)^2}-frac{1}{left(frac12-tright)^2}<0$$
    for all $tinleft(0,dfrac12right)$, which is a contradiction we seek.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11














      Let $f(x)=x ln x$. Then the given equation is
      $$f(x)=f(1-x).$$
      This is symmetric about $x=1/2$. Hence $x=1/2$ is a solution.






      share|cite|improve this answer

















      • 4




        Very nice solution!
        – gimusi
        Dec 7 at 8:17






      • 5




        I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
        – Batominovski
        Dec 7 at 8:41








      • 2




        Yep, IMO too many upvotes as the existence of other solutions should be discussed.
        – Yves Daoust
        Dec 7 at 12:53
















      11














      Let $f(x)=x ln x$. Then the given equation is
      $$f(x)=f(1-x).$$
      This is symmetric about $x=1/2$. Hence $x=1/2$ is a solution.






      share|cite|improve this answer

















      • 4




        Very nice solution!
        – gimusi
        Dec 7 at 8:17






      • 5




        I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
        – Batominovski
        Dec 7 at 8:41








      • 2




        Yep, IMO too many upvotes as the existence of other solutions should be discussed.
        – Yves Daoust
        Dec 7 at 12:53














      11












      11








      11






      Let $f(x)=x ln x$. Then the given equation is
      $$f(x)=f(1-x).$$
      This is symmetric about $x=1/2$. Hence $x=1/2$ is a solution.






      share|cite|improve this answer












      Let $f(x)=x ln x$. Then the given equation is
      $$f(x)=f(1-x).$$
      This is symmetric about $x=1/2$. Hence $x=1/2$ is a solution.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 7 at 8:11









      Anurag A

      25.5k12249




      25.5k12249








      • 4




        Very nice solution!
        – gimusi
        Dec 7 at 8:17






      • 5




        I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
        – Batominovski
        Dec 7 at 8:41








      • 2




        Yep, IMO too many upvotes as the existence of other solutions should be discussed.
        – Yves Daoust
        Dec 7 at 12:53














      • 4




        Very nice solution!
        – gimusi
        Dec 7 at 8:17






      • 5




        I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
        – Batominovski
        Dec 7 at 8:41








      • 2




        Yep, IMO too many upvotes as the existence of other solutions should be discussed.
        – Yves Daoust
        Dec 7 at 12:53








      4




      4




      Very nice solution!
      – gimusi
      Dec 7 at 8:17




      Very nice solution!
      – gimusi
      Dec 7 at 8:17




      5




      5




      I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
      – Batominovski
      Dec 7 at 8:41






      I think this is insufficient. Proving that $x=dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0text{ for all }xtext{ such that }0<x<frac12,,$$ and by symmetry $$f(x)-f(1-x)>0text{ for all }xtext{ such that }frac12<x<1,.$$
      – Batominovski
      Dec 7 at 8:41






      2




      2




      Yep, IMO too many upvotes as the existence of other solutions should be discussed.
      – Yves Daoust
      Dec 7 at 12:53




      Yep, IMO too many upvotes as the existence of other solutions should be discussed.
      – Yves Daoust
      Dec 7 at 12:53











      6














      The function $xlog x-(1-x)log(1-x)$ is only defined in $(0,1)$, is differentiable, has a root at $x=frac12$ (by inspection) and tends to zero at the interval endpoints.



      The derivative,



      $$log x+log(1-x)+2$$ has two roots in $(0,1)$, where $x(1-x)=e^{-2}$.



      Hence the function is negative in $(0,frac12)$, with a single minimum, and positive in $(frac12,1)$, with a single maximum, and there are no other roots.



      enter image description here






      share|cite|improve this answer




























        6














        The function $xlog x-(1-x)log(1-x)$ is only defined in $(0,1)$, is differentiable, has a root at $x=frac12$ (by inspection) and tends to zero at the interval endpoints.



        The derivative,



        $$log x+log(1-x)+2$$ has two roots in $(0,1)$, where $x(1-x)=e^{-2}$.



        Hence the function is negative in $(0,frac12)$, with a single minimum, and positive in $(frac12,1)$, with a single maximum, and there are no other roots.



        enter image description here






        share|cite|improve this answer


























          6












          6








          6






          The function $xlog x-(1-x)log(1-x)$ is only defined in $(0,1)$, is differentiable, has a root at $x=frac12$ (by inspection) and tends to zero at the interval endpoints.



          The derivative,



          $$log x+log(1-x)+2$$ has two roots in $(0,1)$, where $x(1-x)=e^{-2}$.



          Hence the function is negative in $(0,frac12)$, with a single minimum, and positive in $(frac12,1)$, with a single maximum, and there are no other roots.



          enter image description here






          share|cite|improve this answer














          The function $xlog x-(1-x)log(1-x)$ is only defined in $(0,1)$, is differentiable, has a root at $x=frac12$ (by inspection) and tends to zero at the interval endpoints.



          The derivative,



          $$log x+log(1-x)+2$$ has two roots in $(0,1)$, where $x(1-x)=e^{-2}$.



          Hence the function is negative in $(0,frac12)$, with a single minimum, and positive in $(frac12,1)$, with a single maximum, and there are no other roots.



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 at 13:17

























          answered Dec 7 at 13:10









          Yves Daoust

          124k671221




          124k671221























              3














              Let $g:Itomathbb{R}$, where $I:=left[-dfrac12,+dfrac12right]$, be the function defined by $$g(t):=begin{cases}left(dfrac{1}{2}+tright),lnleft(dfrac12+tright)-left(dfrac{1}{2}-tright),lnleft(dfrac12-tright)&text{if }tinleft(-dfrac12,+dfrac12right),,\0&text{if }tinleft{-dfrac12,+dfrac12right},.end{cases}$$ (Observe that $g$ is continuous on the whole $I$, and is smooth on $left(-dfrac12,+dfrac12right)$.) Then, we are to solve for $xin(0,1)$ from $$gleft(x-dfrac12right)=0,,$$
              which is equivalent to solving for $yin left(-dfrac12,+dfrac12right)$ such that
              $$g(y)=0text{ by setting }y:=x-dfrac12,.$$



              We claim that $g$ has only three roots $-dfrac12,0,+dfrac12$, and this means the only $yinleft(-dfrac12,+dfrac12right)$ such that $g(y)=0$ is $y=0$, making $x=dfrac12$ the sole solution to the required equation. Suppose contrary that $g$ has more than three roots. Then, by symmetry, on the interval $left[0,dfrac12right]$, $g$ has at least three roots. Using Rolle's Theorem, $g'$ has at least two roots on $left(0,dfrac12right)$, and so $g''$ has at least one root on $left(0,dfrac12right)$. However, we have
              $$g'(t)=lnleft(frac12+tright)+lnleft(frac12-tright)+2,,$$
              $$g''(t)=frac{1}{frac{1}{2}+t}-frac{1}{frac{1}{2}-t},,$$
              and
              $$g'''(t)=-frac{1}{left(frac{1}{2}+tright)^2}-frac{1}{left(frac12-tright)^2}<0$$
              for all $tinleft(0,dfrac12right)$, which is a contradiction we seek.






              share|cite|improve this answer


























                3














                Let $g:Itomathbb{R}$, where $I:=left[-dfrac12,+dfrac12right]$, be the function defined by $$g(t):=begin{cases}left(dfrac{1}{2}+tright),lnleft(dfrac12+tright)-left(dfrac{1}{2}-tright),lnleft(dfrac12-tright)&text{if }tinleft(-dfrac12,+dfrac12right),,\0&text{if }tinleft{-dfrac12,+dfrac12right},.end{cases}$$ (Observe that $g$ is continuous on the whole $I$, and is smooth on $left(-dfrac12,+dfrac12right)$.) Then, we are to solve for $xin(0,1)$ from $$gleft(x-dfrac12right)=0,,$$
                which is equivalent to solving for $yin left(-dfrac12,+dfrac12right)$ such that
                $$g(y)=0text{ by setting }y:=x-dfrac12,.$$



                We claim that $g$ has only three roots $-dfrac12,0,+dfrac12$, and this means the only $yinleft(-dfrac12,+dfrac12right)$ such that $g(y)=0$ is $y=0$, making $x=dfrac12$ the sole solution to the required equation. Suppose contrary that $g$ has more than three roots. Then, by symmetry, on the interval $left[0,dfrac12right]$, $g$ has at least three roots. Using Rolle's Theorem, $g'$ has at least two roots on $left(0,dfrac12right)$, and so $g''$ has at least one root on $left(0,dfrac12right)$. However, we have
                $$g'(t)=lnleft(frac12+tright)+lnleft(frac12-tright)+2,,$$
                $$g''(t)=frac{1}{frac{1}{2}+t}-frac{1}{frac{1}{2}-t},,$$
                and
                $$g'''(t)=-frac{1}{left(frac{1}{2}+tright)^2}-frac{1}{left(frac12-tright)^2}<0$$
                for all $tinleft(0,dfrac12right)$, which is a contradiction we seek.






                share|cite|improve this answer
























                  3












                  3








                  3






                  Let $g:Itomathbb{R}$, where $I:=left[-dfrac12,+dfrac12right]$, be the function defined by $$g(t):=begin{cases}left(dfrac{1}{2}+tright),lnleft(dfrac12+tright)-left(dfrac{1}{2}-tright),lnleft(dfrac12-tright)&text{if }tinleft(-dfrac12,+dfrac12right),,\0&text{if }tinleft{-dfrac12,+dfrac12right},.end{cases}$$ (Observe that $g$ is continuous on the whole $I$, and is smooth on $left(-dfrac12,+dfrac12right)$.) Then, we are to solve for $xin(0,1)$ from $$gleft(x-dfrac12right)=0,,$$
                  which is equivalent to solving for $yin left(-dfrac12,+dfrac12right)$ such that
                  $$g(y)=0text{ by setting }y:=x-dfrac12,.$$



                  We claim that $g$ has only three roots $-dfrac12,0,+dfrac12$, and this means the only $yinleft(-dfrac12,+dfrac12right)$ such that $g(y)=0$ is $y=0$, making $x=dfrac12$ the sole solution to the required equation. Suppose contrary that $g$ has more than three roots. Then, by symmetry, on the interval $left[0,dfrac12right]$, $g$ has at least three roots. Using Rolle's Theorem, $g'$ has at least two roots on $left(0,dfrac12right)$, and so $g''$ has at least one root on $left(0,dfrac12right)$. However, we have
                  $$g'(t)=lnleft(frac12+tright)+lnleft(frac12-tright)+2,,$$
                  $$g''(t)=frac{1}{frac{1}{2}+t}-frac{1}{frac{1}{2}-t},,$$
                  and
                  $$g'''(t)=-frac{1}{left(frac{1}{2}+tright)^2}-frac{1}{left(frac12-tright)^2}<0$$
                  for all $tinleft(0,dfrac12right)$, which is a contradiction we seek.






                  share|cite|improve this answer












                  Let $g:Itomathbb{R}$, where $I:=left[-dfrac12,+dfrac12right]$, be the function defined by $$g(t):=begin{cases}left(dfrac{1}{2}+tright),lnleft(dfrac12+tright)-left(dfrac{1}{2}-tright),lnleft(dfrac12-tright)&text{if }tinleft(-dfrac12,+dfrac12right),,\0&text{if }tinleft{-dfrac12,+dfrac12right},.end{cases}$$ (Observe that $g$ is continuous on the whole $I$, and is smooth on $left(-dfrac12,+dfrac12right)$.) Then, we are to solve for $xin(0,1)$ from $$gleft(x-dfrac12right)=0,,$$
                  which is equivalent to solving for $yin left(-dfrac12,+dfrac12right)$ such that
                  $$g(y)=0text{ by setting }y:=x-dfrac12,.$$



                  We claim that $g$ has only three roots $-dfrac12,0,+dfrac12$, and this means the only $yinleft(-dfrac12,+dfrac12right)$ such that $g(y)=0$ is $y=0$, making $x=dfrac12$ the sole solution to the required equation. Suppose contrary that $g$ has more than three roots. Then, by symmetry, on the interval $left[0,dfrac12right]$, $g$ has at least three roots. Using Rolle's Theorem, $g'$ has at least two roots on $left(0,dfrac12right)$, and so $g''$ has at least one root on $left(0,dfrac12right)$. However, we have
                  $$g'(t)=lnleft(frac12+tright)+lnleft(frac12-tright)+2,,$$
                  $$g''(t)=frac{1}{frac{1}{2}+t}-frac{1}{frac{1}{2}-t},,$$
                  and
                  $$g'''(t)=-frac{1}{left(frac{1}{2}+tright)^2}-frac{1}{left(frac12-tright)^2}<0$$
                  for all $tinleft(0,dfrac12right)$, which is a contradiction we seek.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 at 9:28









                  Batominovski

                  33.7k33292




                  33.7k33292






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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