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5X = (1+X)^15 ; solve for X

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  1. Feb 9, 2016 #1
    5X = (1+X)^15

    What to do with this kinds of problems? How do I solve for X? I encounter this problems on engineering economy(Annuities)... I tried doing logarithms but no success.
     
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  3. Feb 9, 2016 #2

    BvU

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    Hi Jove,

    What did you do to try logarithms ? Please post so we can see where things go wrong ...
    However, perhaps a numerical approach is in order for this kind of exercise (e.g. if x should be a natural number).
     
  4. Feb 9, 2016 #3

    PeroK

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    It would be difficult to get. get a closed form solution. You will need to do a numerical approximation.
    You have a polynomial of degree 15.
     
    Last edited: Feb 9, 2016
  5. Feb 9, 2016 #4
    I tried changing the equations into logarithm form:
    log(5X) = 15log(1+X)

    and afterwards i still can't solve for X



    Yeah the book also did the same (numerical approximation). I thought there's a way(s) to do it other than that :/
     
  6. Feb 9, 2016 #5

    BvU

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    one more step gets ##\log 5 + \log x = 15 \log (1+x)## and then it is indeed hopeless. So either a numerical approach or a graphical solution is the only way out.
     
  7. Feb 9, 2016 #6

    PeroK

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    ## u = x +1## is an obvious substitution. There must be an algebraic solution but I don't know any advanced techniques for so high a power.
     
  8. Feb 9, 2016 #7

    BvU

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    Is there a solution at all ? If I look at the numbers I don't think so !
    Or at the graph
     
    Last edited: Feb 9, 2016
  9. Feb 9, 2016 #8

    HallsofIvy

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    No "must" about it. There exist "algebraic solutions" for polynomial equations of degree 4 or less but there is no general "algebraic solution" for such equations of degree 5 or higher.

    And the substitution u= x+ 1 just "shifts" the problem. You have [itex]u^{15}= 5(u- 1)[/itex], still a polynomial equation of degree 15.
     
  10. Feb 9, 2016 #9
    By taking the log you assume that x+1>0 and eliminate solutions less than -1.

    The same Wolfram Alpha will give a real solution of about -2.17, for the initial problem.
     
  11. Feb 9, 2016 #10

    BvU

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    I wonder how this answer fits in the annuity context, though :smile:
     
  12. Feb 9, 2016 #11

    BvU

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    So jove, is your exercise really 5X = (1+X)^15 or does it ask what interest rate you need to have five times your initial deposit after 15 years ?
     
  13. Feb 9, 2016 #12

    PeroK

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    The solutions to this equation must be algebraic, by definition.
     
  14. Feb 9, 2016 #13
  15. Feb 9, 2016 #14

    jbriggs444

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    An "algebraic number" is one that is a solution of a non-zero polynomial equation in one variable with integer coefficients. Real numbers that are not "algebraic" are, instead, "transcendental".

    However, these "algebraic numbers" may not be expressible in terms of elementary operations -- addition, subtraction, multiplication, division and root extraction, hence the Abel Ruffini theorem.
     
  16. Feb 9, 2016 #15
  17. Feb 10, 2016 #16

    PeroK

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    I meant algebraic numbers. And I was wondering about techniques to solve a particular equation. You don't necessarily need a general solution to every polynomial of order 15 in order to solve this one.
     
  18. Feb 10, 2016 #17
    It does not of course; there appears to be a typo in the OP. A problem which does makes sense in the context of annuities (although it is a simple compound interest problem) is ## 5 = (1+x)^{15} ## which is easily solved for its unique real solution.
     
  19. Feb 10, 2016 #18

    BvU

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    I'd like to hear that from Jove himself (herself?) ...
    (as in: post #11)
     
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