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Why does taking the logarithm of both sides not work?

  1. Jun 3, 2012 #1
    The problem statement, all variables and given/known data
    Differentiate [itex]x=e^{-x}[/itex]

    The attempt at a solution
    [tex]\ln{x}=\ln{e^{-x}}[/tex]
    [tex]\ln{x}=-x[/tex]
    [tex]\frac{d}{dx}\left(\ln{x}\right)=\frac{d}{dx}\left(-x\right)[/tex]
    [tex]\frac{1}{x}=-1[/tex]

    The correct answer is [itex]1+e^{-x}[/itex]. I know how to solve it that way; however, why is the above method wrong? I know there are usually two variables, but I can't see any mathematical errors in my method.

    Thank you.
     
    Last edited: Jun 3, 2012
  2. jcsd
  3. Jun 3, 2012 #2

    tiny-tim

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    Hi Hypercubes! :smile:
    Obviously a misprint, for "Differentiate [itex]x-e^{-x}[/itex]" :wink:
     
  4. Jun 3, 2012 #3
    Well, the exact question was "explain why the equation [itex]e^{-x} = x[/itex] has exactly one solution", and the answer goes on to define [itex]f(x) = x-e^{-x}[/itex] and says "Since [itex]f'(x) = 1 + e^{-x}[/itex], and [itex]1 + e^{-x} > 0[/itex] for all x, the function f is increasing..."

    So they set it to 0 to get the derivative. However, why was the method I used not valid?
     
    Last edited: Jun 3, 2012
  5. Jun 3, 2012 #4

    D H

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    Your expression, [itex]x=e^{-x}[/itex] is just that -- an expression. You cannot treat it as a function. In particular, you can't differentiate both sides and expect something meaningful to pop out. What you can do is to rewrite it as [itex]x-e^{-x} = 0[/itex], and now the quantity on the right hand side [itex]f(x)=x-e^{-x}[/itex] can be treated as a function. You aren't differentiating both sides of [itex]f(x)=0[/itex]. Just one.
     
  6. Jun 3, 2012 #5
    Ah, I see. Thanks you both!
     
  7. Jun 3, 2012 #6

    Matterwave

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    As a simple example. Take the equation 3x-2=x. This equation has 1 solution x=1. If you differentiate both sides though, you get 3=1 which is obviously not true.

    This is because the undifferentiated equation is asking a fundamentally different question than the differentiated equation. The equation 3x-2=x is asking "When is the function f(x)=3x-2 equal to the function g(x)=x?" While the differentiated equation is asking "When is the slope of the function f(x) equal to the slope of the function g(x)?"
     
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