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Homework Help: Equaton problem

  1. Apr 2, 2007 #1
    5n*[(2i)^(n-1) ]=11520


    the answer in the book is n=9

    notice i = (-1)^0.5
    how do i solve this equation

    i tried to solve it in many ways including logarithims

    but log and i (complex numbers)
    i dont know if its legal to do

    any way i didnt study this logarithims of i (complex numbers)
    so i dont think that this is the right way.

    plz help
    Last edited: Apr 2, 2007
  2. jcsd
  3. Apr 2, 2007 #2


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    Of course, 5n(2i)n-1 is the same as 5n(2n-1)(in-1). But i to any power is one of i, -i, -1, and 1. Since the right hand side of the equation is a positive real number, in-1 must be equal to 1. This equation is exactly the same as n (2n-1)= 11520 (with the provision that n-1 must be a multiple of 4 so that the power of i is, in fact, +1). You still are not going to be able to solve that equation using logarithms- you have n both inside and outside a logarithm.

    But since n must be a positive integer, there is nothing wrong with "trial and error"- it's just a little tedious. If n=1, then n-1= 0= 4(0), but obviously 5(1)(20)= 5 is NOT 11520. Okay, try n-1= 4, the next multiple of 4, so n= 5. Now 5(5)(24)= 25(16)= 400, not 11520. The next multiple of 4 is 8:if n-1= 8, then n= 9. 5(9)(28)= 45(256)= 11520. Got it!
  4. Apr 2, 2007 #3
    are you sure that there is no other way?
    because i dont think that my teacher would like this
    trial and error technic
  5. Apr 2, 2007 #4
    I agree with HOI, trial and error is the way to go. But, you can severely limit the possible values of n by simply looking at the original statement.

    [tex] 5n \cdot (2i)^{n-1} = 11520 [/tex]

    Now, if we assume n > 0 then 5n > 0, and therefore, so must the exponential. But, this will only occur for multiples of 4 plus 1. In other words:

    [tex] n = 4m + 1 [/tex] for some m = 0, 1, 2,...

    A quick check of these values will land you an answer at m = 2, and thus, n = 9. Since the absolute value of the expression on the left is increasing without bound for n > 0, this is the only positive solution.

    If, on the other hand, we assume n < 0, then 5n < 0, and therefore, so must the exponential. This will only occur for multiples of 4 minus 1. Thus:

    [tex] n = 4m - 1 [/tex] for some m = 0, -1, -2,...

    However, note that for m = 0, we get a value of 5/4, and that for larger values of m, this result must be smaller. Therefore, there are no negative solutions.
  6. Apr 2, 2007 #5
    thank alot
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