Why is ln(5) the solution to e^x - 5e^{-x} = 4?

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Discussion Overview

The discussion revolves around solving the equation e^x - 5e^{-x} = 4, with a focus on understanding why ln(5) is considered a solution. Participants explore the implications of the equation in terms of real and complex solutions, as well as the nature of logarithmic functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the equation can be transformed into y - 5/y = 4 by letting y = e^x.
  • Another participant notes that the transformed equation has two solutions, but the original equation has only one valid solution in the real numbers.
  • There is a discussion about the nature of the solutions, with one participant asserting that while y = 5 leads to a valid solution x = ln(5), y = -1 leads to an invalid solution since e^x cannot equal a negative number.
  • Another participant introduces the idea that e^(pi * i) = -1, suggesting that this could imply an imaginary solution exists.
  • Participants discuss the concept of complex logarithms and the existence of infinitely many solutions due to the periodic nature of the exponential function in the complex plane.
  • One participant expresses confusion about the implications of having infinite solutions and the concept of Riemann surfaces in relation to logarithms.
  • There is a request for further resources to understand the concepts discussed, particularly regarding Riemann surfaces and branch cuts.

Areas of Agreement / Disagreement

Participants express differing views on the nature of solutions to the equation, particularly regarding the existence of real versus complex solutions. There is no consensus on the implications of complex solutions or the interpretation of logarithmic values.

Contextual Notes

The discussion includes assumptions about the domain of the logarithmic function and the nature of solutions in the real versus complex number systems. The implications of Riemann surfaces and the behavior of logarithmic functions in the complex plane are also explored but remain unresolved.

Inspector Gadget
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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.
 
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Let y=e^x then you are solving y-5/y=4
 
Got it. Thanks.
 
Did you see WHY y- 5/y= 4 has TWO solutions but [itex]\frac{e^x-5e^{-x}}{4}=1[/itex] has only one?
 
Couldn't we say that [itex]\frac{e^x-5e^{-x}}{4}=1[/itex] also has two solutions, but only one in {R}?
 
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.
 
True, but e^(pi * i) = -1. Doesn't that count?
 
Gunni said:
True, but e^(pi * i) = -1. Doesn't that count?
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.
 
Gunni said:
Couldn't we say that [itex]\frac{e^x-5e^{-x}}{4}=1[/itex] also has two solutions, but only one in {R}?

(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.
 
  • #10
HallsofIvy said:
(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.
Oh? Please could you explain.
 
  • #11
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.
 
  • #12
I'm sorry matt grime but I don't understand. Are you saying that there are an infinite 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"?
 
  • #13
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.
 
  • #14
Oww right I've never thought about it like that thanks. Very interesting thanks :biggrin:
 
  • #15
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.
 
  • #16
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.
 
  • #17
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.
 
  • #18
matt grime said:
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.
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:.
 
  • #19
yeah, that's not what i had in mind. the bit it links to on branch cuts is more useful
 

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