What is the condition that $e^x=ax+a-1$ has non-negative solution?












1












$begingroup$


Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots







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$endgroup$












  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58


















1












$begingroup$


Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58
















1












1








1





$begingroup$


Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots







share|cite|improve this question









$endgroup$




Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots




calculus systems-of-equations roots






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 9:51









Math_YMath_Y

605




605












  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58




















  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58


















$begingroup$
Can you please tell what methods have you tried to solve this question?
$endgroup$
– Jaideep Khare
Dec 28 '18 at 9:53




$begingroup$
Can you please tell what methods have you tried to solve this question?
$endgroup$
– Jaideep Khare
Dec 28 '18 at 9:53




1




1




$begingroup$
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 9:58






$begingroup$
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 9:58












3 Answers
3






active

oldest

votes


















-6












$begingroup$

Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It is not related to amount of $a$?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:48










  • $begingroup$
    Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:51






  • 2




    $begingroup$
    IMO, this brings you farther from the solution than the original setting of the question.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 12:45



















3












$begingroup$

Hint :



Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$



The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    you must also consider the case if $$a<0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:08










  • $begingroup$
    @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
    $endgroup$
    – Rebellos
    Dec 28 '18 at 10:10










  • $begingroup$
    Thank you very much. There exists a closed form relation for the solutions?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:29










  • $begingroup$
    @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
    $endgroup$
    – Rory Daulton
    Dec 28 '18 at 15:34





















0












$begingroup$

Hint:



The equation can be written



$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



enter image description here



Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






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









    -6












    $begingroup$

    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45
















    -6












    $begingroup$

    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45














    -6












    -6








    -6





    $begingroup$

    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$






    share|cite|improve this answer











    $endgroup$



    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 10:52

























    answered Dec 28 '18 at 10:36









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    79.1k42867




    79.1k42867












    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45


















    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45
















    $begingroup$
    It is not related to amount of $a$?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:48




    $begingroup$
    It is not related to amount of $a$?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:48












    $begingroup$
    Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:51




    $begingroup$
    Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:51




    2




    2




    $begingroup$
    IMO, this brings you farther from the solution than the original setting of the question.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 12:45




    $begingroup$
    IMO, this brings you farther from the solution than the original setting of the question.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 12:45











    3












    $begingroup$

    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34


















    3












    $begingroup$

    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34
















    3












    3








    3





    $begingroup$

    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






    share|cite|improve this answer











    $endgroup$



    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 10:21

























    answered Dec 28 '18 at 10:03









    RebellosRebellos

    15.7k31250




    15.7k31250












    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34




















    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34


















    $begingroup$
    you must also consider the case if $$a<0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:08




    $begingroup$
    you must also consider the case if $$a<0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:08












    $begingroup$
    @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
    $endgroup$
    – Rebellos
    Dec 28 '18 at 10:10




    $begingroup$
    @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
    $endgroup$
    – Rebellos
    Dec 28 '18 at 10:10












    $begingroup$
    Thank you very much. There exists a closed form relation for the solutions?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:29




    $begingroup$
    Thank you very much. There exists a closed form relation for the solutions?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:29












    $begingroup$
    @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
    $endgroup$
    – Rory Daulton
    Dec 28 '18 at 15:34






    $begingroup$
    @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
    $endgroup$
    – Rory Daulton
    Dec 28 '18 at 15:34













    0












    $begingroup$

    Hint:



    The equation can be written



    $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



    enter image description here



    Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



    The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint:



      The equation can be written



      $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



      enter image description here



      Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



      The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:



        The equation can be written



        $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



        enter image description here



        Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



        The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






        share|cite|improve this answer











        $endgroup$



        Hint:



        The equation can be written



        $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



        enter image description here



        Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



        The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 28 '18 at 12:54

























        answered Dec 28 '18 at 12:42









        Yves DaoustYves Daoust

        133k676232




        133k676232






























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            Bundesstraße 106

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            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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            Verónica Boquete

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            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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