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Solve for x

  1. Feb 26, 2007 #1
    i have to solve for x:
    [tex]x^a - x = 1[/tex] where [tex]a = \frac{\ln 6}{\ln 5}[/tex]

    taking [itex]\ln[/itex] in both side, i get,
    [tex]a\ln x = \ln(x+1)[/tex]

    [tex]\Rightarrow\frac{\ln(x+1)}{\ln x} = a = \frac{\ln 6}{\ln 5}[/tex]

    here we can see that x = 5.

    but what i wanted to know is the general solution of the equation for any [itex]a \in \mathbb{R}[/itex] .
     
    Last edited: Feb 26, 2007
  2. jcsd
  3. Feb 26, 2007 #2
    I just did quick graph of a(x), and it looks like no real roots for 0<a<1 exist.
     
  4. Feb 26, 2007 #3

    dextercioby

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    There's a simple argument for the uniqueness of the solution to the problem.

    Consider the function

    [tex] f(x)=\ln 5 \ln (x+1) -\ln 6 \ln x [/tex]

    x=5 is clearly a solution. Since the derivative of the function is monotonic negative on [itex] \mathbb{R}_{+} [/itex], it follows that the function is everly decreasing on the positive semiaxis, therefore the x=5 solution (zero of the function) is unique.
     
  5. Feb 26, 2007 #4
    but what is the general solution for any [itex]a \in \mathbb{R}[/itex]?
     
  6. Feb 26, 2007 #5

    dextercioby

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    In the general case, the problem's more complicated, as, for example the case a=4 shows.

    [tex] x+1=x^{4} [/tex]

    Solve it...
     
  7. Feb 26, 2007 #6
    its approximately x = 0.72449
    i solved it by numerical methods. but how can i solve it algebraically?
     
    Last edited: Feb 26, 2007
  8. Feb 26, 2007 #7

    dextercioby

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    The point i was trying to make is that you generally can't (solve it algebraically). Your problem was simple, but a general one isn't...
     
  9. Feb 26, 2007 #8

    HallsofIvy

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    You do understand, don't you, that there exist polynomial equations (or degree 5 or higher) that have NO solutions in terms of radicals?
     
  10. Feb 26, 2007 #9
    well at least there's a limit for x....as a-->infinity....x-->1 :D
     
  11. Feb 27, 2007 #10

    Gib Z

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    There exist methods of solving polynomials in terms of radicals and co-effecients, but only if the degree is 5 or lower, As Halls said. But even those methods, to me at least, are very long and cumbersome. In fact i remember seeing the equations which give x for a general quartic equation. There were 4 different equations for the 4 values of x, and each equation went at least across your entire screen, taking 2 lines.
     
  12. Feb 27, 2007 #11
    can we solve the following algebraically:
    [tex]x^x = e[/tex]

    i got this:
    [tex]\ln(x^x) = 1[/tex]
    [tex]\Rightarrow x\ln x = 1[/tex]
     
    Last edited: Feb 27, 2007
  13. Feb 27, 2007 #12

    dextercioby

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    Nope, it can't be done algebraically. Graph intersection.
     
  14. Feb 28, 2007 #13

    uart

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    It cant be written in terms of elementry function but it can be written in terms of the lambert W function.

    [tex] x = e^{W(1)} [/tex]
     
    Last edited: Feb 28, 2007
  15. Feb 28, 2007 #14

    HallsofIvy

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    My understanding was "only if the degree is lower than 5"- there exist polynomials of degree 5 which cannot be solved "by radicals" (because S5[/sup] is not a solvable group).
     
  16. Mar 1, 2007 #15
    how can we prove that we can't solve [tex]x^x = e[/tex] algebraically? can anyone please help?
     
    Last edited: Mar 1, 2007
  17. Mar 1, 2007 #16

    That's a transcendental equation.:smile:
    Not of polynomial type.
     
  18. Mar 2, 2007 #17

    Gib Z

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    Halls: My bad sorry, you're correct of course. Abel-Ruffeni's Theorem, i think it is, just incase anyone is interested.

    murshid_islam: Prove [itex] x = e^{W(1)} [/itex] isn't the solution to any polynomial with rational co-effecients.
     
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