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E / natural logs

  1. Mar 26, 2004 #1
    I'm trying to figure out how to do this problem...

    Solve for x:

    [tex]
    \frac{e^x-5e^{-x}}{4}=1
    [/tex]

    I know the answer is ln(5) [given by both the book and TI-89 calculator], but I can't figure out why.
     
    Last edited: Mar 26, 2004
  2. jcsd
  3. Mar 26, 2004 #2

    matt grime

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    Let y=e^x then you are solving y-5/y=4
     
  4. Mar 26, 2004 #3
    Got it. Thanks.
     
  5. Mar 27, 2004 #4

    HallsofIvy

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    Did you see WHY y- 5/y= 4 has TWO solutions but [itex]\frac{e^x-5e^{-x}}{4}=1[/itex] has only one?
     
  6. Mar 27, 2004 #5
    Couldn't we say that [itex]\frac{e^x-5e^{-x}}{4}=1[/itex] also has two solutions, but only one in {R}?
     
  7. Mar 27, 2004 #6
    I'm pretty sure I do, but should probably have someone confirm this...

    If you sub in y for e^x, multiple by y to get it out of the denominator, set equal to 0, and factor, you get: (y - 5)(y + 1) = 0

    Solutions: 5, -1.

    Then when you have...

    y = 5
    e^x = 5
    x = ln(5)

    ...which is valid, but...

    y = -1
    e^x = -1
    x = ln(-1)

    ...isn't a valid solution because you can't take the natural log of a negative number.
     
  8. Mar 27, 2004 #7
    True, but e^(pi * i) = -1. Doesn't that count?
     
  9. Mar 27, 2004 #8

    Integral

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    This means that the other solution is imaginary, therefore not in the real numbers. When you say

    "x= ln(-1) does not have a solution" It is not a correct statement you must add a qualifier:

    "x=ln(-1) does not have a REAL solution". Is a correct statement.

    In most intro math courses it is assumed that you are working in the Real numbers so the qualifier is often dropped.
     
  10. Mar 29, 2004 #9

    HallsofIvy

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    (I thought I had responded to this before but it doesn't appear.)

    No, we would have to say that it has an infinite number of solutions, one of which is real.
     
  11. Mar 29, 2004 #10

    Zurtex

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    Oh? Please could you explain.
     
  12. Mar 29, 2004 #11

    matt grime

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    Consider z a complex number as [tex]re^{i\theta}[/tex] the naive way to define log of z is logr + i\theta (sorry for the pseudo tex) but theta isn't determined uniquely one can add on two pi's at will. Just as one has two square roots and chooses the principal branch, one can do the same for logs, and define log(z) as log(|z|) +iArg(z) by taking the principal argument.

    Imagine walking round the unit circle noting the value of log at each point. as you appoach where you started from you see a problem coming up - you're suddenly going to have to jump from being close to i2pi to zero again.

    Taking that idea one starts to think about Riemann Surfaces where you dont' jump back down again but keep going and switch to another copy of the complex plane. You can imagine doing this again and again and the surface you create is an infinite spiral of copies of C glued together. If you did this for square roots, you'd see that after the second loop you do you're back to where you started. Informally sqrt has two C's gleud together (and there are two choices of sqrt) and log has infinitely many copies glued together, and there are infinitely many possible values of log.
     
  13. Mar 29, 2004 #12

    Zurtex

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    I'm sorry matt grime but I don't understand. Are you saying that there are an infinte number of answers in the form [itex]re^{i\theta}[/itex]? If so I don't quite understand how that is an infinite number of solutions because it is still actually the same numbers in the form of [itex]x + iy[/itex]. Oh and please what is a "Riemann Surfaces"?
     
  14. Mar 29, 2004 #13

    matt grime

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    You agree that exp^{i\pi} = exp^{3i\pi}=exp^{5i\pi}....?

    Well, taking naive logs (ie log(exp(z)) = z) says that i\pi=3i\pi=5i\pi....

    So we have to pick one of these options, just as we pick sqrt(4) to be 2 and not -2.

    The Riemann Surface is a topological way of encoding this data.

    The way you think about that is the paths idea. Don't worry about it, it's not important to you at the moment.
     
  15. Mar 29, 2004 #14

    Zurtex

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    Oww right I've never thought about it like that thanks. Very intresting thanks :biggrin:
     
  16. Mar 29, 2004 #15

    matt grime

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    So, do you see how the path analogy to the square root works? for sqaure root, you take the positive square root of the modulus and half the argument. Now imagine walking round the unit circle, at 1, thte sqaure root is 1, wander to the argument pi/4 and the square root there has argument pi/8, and so on. Now as you get round to 1 again from the other side you've got a problem, as by halving arugments you should be getting closer to minus 1... so imagine there were two copies of the complex plane, both split long the positive part of the real axis. Now as you approach 1 from below, you climb from one plane to the other where we have the other answer to the root - sqrt(1)=-1 there, and if you carry on around the circle in that plane, as you get back to the 1 again you see that the sqrt here is approching 1 again, so yuo climb back down to the plane below. The 'surface' is what you get if you glue the two planes along the slit. for logs once you start going round and round you never get back to the start though.
     
  17. Mar 29, 2004 #16

    Zurtex

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    Sorry you have totally lost me now :confused:

    Could you perhaps provide a link or something? I have no objection to reading through a lot of text I find it all very fascinating.
     
  18. Mar 29, 2004 #17

    matt grime

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    Not sure I can think of a good link off the top of my head, try wolfram or similar. There ought to be some web animations of this somewhere. Note to self: make one if not.
     
  19. Mar 29, 2004 #18

    Zurtex

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    Well the diagrams are quite scary http://mathworld.wolfram.com/RiemannSurface.html put reading over it and what you have put I think I understand the principle of the whole thing thanks :smile:.
     
  20. Mar 29, 2004 #19

    matt grime

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    yeah, that's not what i had in mind. the bit it links to on branch cuts is more useful
     
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