How to solve equation with exponents multiplied with a polynomial?











up vote
4
down vote

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I don't know how to tackle equation with exponents multiplied with a polynomial in general.



For example:



$100 =2^x(1+x)$



Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.






polynomials exponential-function







share|cite|improve this question




















  • 2




    You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
    – Anurag A
    Nov 23 at 18:07












  • What you have done is the best and only thing to do.
    – Ethan Bolker
    Nov 23 at 18:14










  • @AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
    – Arthur
    Nov 23 at 18:26

















up vote
4
down vote

favorite












I don't know how to tackle equation with exponents multiplied with a polynomial in general.



For example:



$100 =2^x(1+x)$



Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.






polynomials exponential-function







share|cite|improve this question




















  • 2




    You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
    – Anurag A
    Nov 23 at 18:07












  • What you have done is the best and only thing to do.
    – Ethan Bolker
    Nov 23 at 18:14










  • @AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
    – Arthur
    Nov 23 at 18:26















up vote
4
down vote

favorite









up vote
4
down vote

favorite











I don't know how to tackle equation with exponents multiplied with a polynomial in general.



For example:



$100 =2^x(1+x)$



Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.






polynomials exponential-function







share|cite|improve this question















I don't know how to tackle equation with exponents multiplied with a polynomial in general.



For example:



$100 =2^x(1+x)$



Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.






polynomials exponential-function




polynomials exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 18:08









Anurag A

25.5k12249




25.5k12249










asked Nov 23 at 18:03









Chuanyuan Liu

303




303








  • 2




    You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
    – Anurag A
    Nov 23 at 18:07












  • What you have done is the best and only thing to do.
    – Ethan Bolker
    Nov 23 at 18:14










  • @AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
    – Arthur
    Nov 23 at 18:26
















  • 2




    You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
    – Anurag A
    Nov 23 at 18:07












  • What you have done is the best and only thing to do.
    – Ethan Bolker
    Nov 23 at 18:14










  • @AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
    – Arthur
    Nov 23 at 18:26










2




2




You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07






You can possibly solve it numerically only, but not exactly (algebraically). If you just want existence of a solution, then continuity and intermediate value theorem will help. Fir example, we can say there will be at least one solution in the interval $(0,5)$.
– Anurag A
Nov 23 at 18:07














What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14




What you have done is the best and only thing to do.
– Ethan Bolker
Nov 23 at 18:14












@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26






@AnuragA I'm pretty certain that this is doable with the Lambert $W$ function. I would count that as an exact solution.
– Arthur
Nov 23 at 18:26












2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










You can solve this with the Lambert W function: Write you equation as
$$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
with $y=1+x$ you get
$$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
$$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
y = frac{W(200ln 2)}{ln2}$$
So finally




$$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$




As an alternative you can use a simple iteration scheme from the equation
$$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$






share|cite|improve this answer






























    up vote
    0
    down vote













    As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).



    In this particular case, you can reduce to the form



    $$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      You can solve this with the Lambert W function: Write you equation as
      $$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
      with $y=1+x$ you get
      $$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
      Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
      $$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
      y = frac{W(200ln 2)}{ln2}$$
      So finally




      $$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$




      As an alternative you can use a simple iteration scheme from the equation
      $$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
      With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$






      share|cite|improve this answer



























        up vote
        3
        down vote



        accepted










        You can solve this with the Lambert W function: Write you equation as
        $$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
        with $y=1+x$ you get
        $$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
        Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
        $$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
        y = frac{W(200ln 2)}{ln2}$$
        So finally




        $$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$




        As an alternative you can use a simple iteration scheme from the equation
        $$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
        With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$






        share|cite|improve this answer

























          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          You can solve this with the Lambert W function: Write you equation as
          $$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
          with $y=1+x$ you get
          $$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
          Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
          $$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
          y = frac{W(200ln 2)}{ln2}$$
          So finally




          $$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$




          As an alternative you can use a simple iteration scheme from the equation
          $$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
          With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$






          share|cite|improve this answer














          You can solve this with the Lambert W function: Write you equation as
          $$100 = 2^x(1+x)=frac{1}{2}2^{1+x}(1+x),$$
          with $y=1+x$ you get
          $$200 = y2^y=y e^{yln 2} = frac{1}{ln 2} (y ln 2) e^{y ln 2} .$$
          Now substitute $z=y ln 2$ and use the definition of $W$ to solve for $z$
          $$200 ln 2 = z e^z quad Rightarrow z = W(200ln 2) quad Rightarrow
          y = frac{W(200ln 2)}{ln2}$$
          So finally




          $$x = frac{W(200ln 2)}{ln2} -1 approx 4.2512070962222326$$




          As an alternative you can use a simple iteration scheme from the equation
          $$x_{n+1} = ln left(frac{100}{1+x_n}right) / ln 2 $$
          With your starting value $x_0=log_2(100) approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, dots$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 21:44

























          answered Nov 23 at 21:29









          gammatester

          16.6k21632




          16.6k21632






















              up vote
              0
              down vote













              As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).



              In this particular case, you can reduce to the form



              $$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.






              share|cite|improve this answer

























                up vote
                0
                down vote













                As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).



                In this particular case, you can reduce to the form



                $$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).



                  In this particular case, you can reduce to the form



                  $$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.






                  share|cite|improve this answer












                  As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).



                  In this particular case, you can reduce to the form



                  $$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 21:58









                  Yves Daoust

                  123k668219




                  123k668219






























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