Solve (-1)^x=1

  1. Feb 11, 2014 #1
    Ok. So the answer to finding the solution of
    [itex](-1)^x=1[/itex]
    is clear.

    But say we didnt know it and wanted to solve it. One approach is to take the log of both sides

    [itex]x\cdot log(-1)=log(1)=0[/itex]

    But now the right hand side is defined where as the left is not!

    What am I missing?
     
  2. jcsd
  3. Feb 11, 2014 #2

    Stephen Tashi

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    You're missing the fact that there is no mathematical principle that says you can always solve an equation by "doing the same thing to both sides".

    For example, you can't solve the equation [itex] \frac{x}{(x-1)} = \frac{1}{(x-1)} [/itex] by multiplying both sides by [itex] x-1 [/itex].

    Mathematical manipulations are intended as a way to abbreviate thinking, not as a way of eliminating it.

    When we have an equation of the form [itex] f(x) = g(x) [/itex] and do some manipulation on it to produce another equation [itex] h(x) = r(x) [/itex] then we are suppose to ask if the manipulation may have created an equation that has more or fewer solutions than the original equation.
     
  4. Feb 11, 2014 #3
    You need a logarithm that's defined for complex numbers. :wink:
     
  5. Feb 11, 2014 #4
    You're missing the point. It's not a matter of solving the equation, its the fact that by performing a completely legitimate operation we now have an equation where the RHS does not equal the LHS.
     
  6. Feb 11, 2014 #5
    Olivermsun, I believe in complex terms we have [itex]log(-1)=i\pi[/itex]. But again we face the same dilemma as
    [itex]i\pi\ne 0[/itex]
     
  7. Feb 11, 2014 #6
    Maybe let me ask this way: what are the complex logs of 1?
     
  8. Feb 11, 2014 #7

    Stephen Tashi

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    What do you mean by "a completely legitimate operation"?

    You aren't using the proper terminology for talking about equations. An equation in a variable x is a statement that two functions of x are equal. A solution to the equation is a value of x that makes the statement true. The set of all solutions to the equation is called "the solution set". The left hand side of an equation may not equal the right hand side of the equation for some , or for any values of the variable x.

    I think what you are trying to say is that taking the log of both sides of the equation [itex] (-1)^x = 1 [/itex] transforms it to another equation, which has a different solution set (namely the null set since the transformed equation has no solutions).

    There is no mathematical principle that says applying the natural logarithm function to both sides of an equation will transform it to a new equation that has the same solution set as the original equation.

    If you define a complex logarithm such that [itex] log(-1) = i \pi [/itex] and transform the new equation to [itex] x (i \pi) = 0 [/itex] then [itex] x = 0 [/itex] is a solution to the transformed equation.
     
  9. Feb 11, 2014 #8

    Vanadium 50

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    Best. Statement. Ever.
     
  10. Feb 11, 2014 #9
    For real numbers a and b, log (a^b) = a log(b) is not valid if a<0.
     
  11. Feb 11, 2014 #10
    Seconded. That's going on the board tomorrow.
     
  12. Feb 12, 2014 #11
    [tex]\\(-1)^x=1 \\ \\x=\log_{-1}(1) \\ \\x=\frac{\ln(1)}{\ln(-1)} \\ \\x=\frac{0}{i \pi} \\ \\x=0[/tex]
     
  13. Feb 12, 2014 #12
    you cannot add log because log(-1) is not defined.
     
  14. Feb 12, 2014 #13
    (-1)^2y = (-1)^ ( even integer) = 1

    (-2)^(2y+1) = (-1)^(odd integer) = -1

    since 2y and (2y+1) form all N ( natural number) then you can say the only solution will be if x is even, i.e x= 2y , and y belongs to N
     
  15. Feb 15, 2014 #14

    lurflurf

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    consider the equivalent equation
    $$e^{x \, \log(-1)}=e^0
    \\
    e^{x \, \pi \imath}=e^0$$

    In general we cannot conclude
    x=y
    from
    f(x)=f(y)

    In this case exp is a periodic function with period 2π i

    so from
    $$e^{x \, \pi \imath}=e^0
    \\ \text{we conclude}\\
    x \, \pi \imath=2n \, \pi \imath
    \\ \text{for some $n \in \mathbb{Z}$} $$
     
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