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B Solve for n

  1. Apr 29, 2017 #1

    vcsharp2003

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    How do I go about solving the following equation for n? From inspection, it seems that this equation is not possible since 2n is always positive so positive + 2= 0 is impossible.

    2n+ 2 = 0
     
  2. jcsd
  3. Apr 29, 2017 #2
    If ##n## is supposed to be a real number (or integer) the equation has no solution. But, if ##n## can be complex the equation has solutions.
     
  4. Apr 29, 2017 #3

    vcsharp2003

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    Ok that makes sense. It is not mentioned that n is a real number. For complex solutions how would I start solving this?
     
  5. Apr 29, 2017 #4
    Rearranging the equation, you have
    ##2^n=-2## or
    ##2^{n-1}=-1##
    Use then the polar form ##z=re^{i\theta}## of a complex number.
     
  6. Apr 29, 2017 #5

    vcsharp2003

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    So, I would express 2 in polar form and also -1 in polar form.
    ∴ 2 = 2 e2kπi where k is any integer
    and -1 = eπi.

    I am not sure what value of k would I take when expressing 2 in polar form since k could be any integer?
     
  7. Apr 30, 2017 #6

    vcsharp2003

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    I think it might be easier to take original equation so we have 2n= -2, then take ln of both sides and express only -2 in polar form.

    Another important idea relating to solving for an unknown in an equation is that when the unknown variable appears in a power, then taking log of both sides can help as a first step to the solution.
    In this problem, the unknown n appears in the power, therefore it would be a good idea to take log of both sides as a first step, which is what is done below.

    ln(2n) = ln(-2)
    n ln2 = ln (2eπi)
    n ln2 = ln2 + ln(eπi)
    n ln2 = ln2 + iπ lne
    n ln2 = ln2 + iπ
    n = 1 + i (π/ln2)
     
    Last edited: Apr 30, 2017
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