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Natural logs (e) problem

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    e^x - e^-x all divided by 2 = -1
    express answer in natural logs

    2. Relevant equations

    No equations, just properties of logs and natural logs

    3. The attempt at a solution

    First, I multiplied everything by 2


    e^x - e^-x = -2

    then i took the natural log of both sides and brought down the exponent

    (x)Ln(e) + (x)Ln(e) = -2

    so you can simplify that i think to

    2xLn(e) = -2

    so x= -2/(2Lne)

    which Lne is 1 so

    x=-2/1

    x=-2?

    I don't think thats right because it says express answer in natural logs so it should simplify out that easily.
     
  2. jcsd
  3. Nov 11, 2007 #2
    I think you missed something.

    When you turn
    e^x - e^-x = -2

    into

    (x)Ln(e) + (x)Ln(e) = -2

    the -2 is part of it too. So it should be

    (x)Ln(e) + (x)Ln(e) = Ln(-2)
     
  4. Nov 11, 2007 #3

    HallsofIvy

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    Unfortunately you are both wrong. You can't just "turn e^x- e^-x= -2 into Ln(e^x)+ Ln(e^-x)".

    You can't "take the natural logarithm of both sides because, as sourlemon said, you must also take the natural logarithm of -2 and that is not defined: the domain of Ln(x)is "all positive x". Even if it were +2 on the right side, Ln(x+ y) is NOT Ln(x)+ Ln(y).

    Instead, go back to e^x- e^-x= -2 and multiply both sides by e^x: (e^x)^2- 1= 2(e^x) or y^2- 1= 2y (taking y= e^x) so y^2- 2y= 1. You can solve that by completing the square on the left: y^2- 2y+ 1= (y-1)^2= 2 so y= [itex]1\pm\sqrt{2}[/itex].
    Once you have [itex]e^x= 1+ \sqrt{2}[/itex], then take the Ln of both sides. (Do you see why [itex]e^x= 1- \sqrt{2}[/itex] is not possible?

    I beat you Compuchip!
     
    Last edited: Nov 11, 2007
  5. Nov 11, 2007 #4

    CompuChip

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    Also note that in general, log(a + b) is not equal to log(a) + log(b). For example, log(1 + 0) = log(1) = 0 =/= log(1) + log(0) = [itex]-\infty[/itex]. There is just a rule saying that log(a * b) = log(a) + log(b). Instead, multiply both sides by [itex]e^x[/itex] and first solve it for [itex]y = e^x[/itex] as a quadratic expression. Then you can take the logarithm.

    [edit]Wow, you are fast :smile:[/edit]
     
  6. Nov 11, 2007 #5
    I think you can't do [itex]e^x= 1- \sqrt{2}[/itex] because it would be a negative answer and logs have to be greater than zero.


    I understand how you solved the problem but I was never taught to do it that way, are there other way to solve the problem other than completing the square?

    Is it possible to factor it?
     
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