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Simple Calculus question

  1. Apr 30, 2005 #1
    Hello all. Really quick question here....What are the intervals of increase and decrease for y= e^x = e^-2x. I found the first derivative : y'= e^x-2e^-2x and set it equal to 0 but that's where I got stuck. How would you solve for x? I know that the answer is (ln2)/3 but how would you get there? Thanks.
  2. jcsd
  3. Apr 30, 2005 #2


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    Homework Helper

    Working out the equation, and use some logarithm properties.

    Apply natural logaritm

    [tex] (e^{x})^{3} = 2 [/tex]
  4. Apr 30, 2005 #3
    Stupid question...but why to the exponent of 3?
  5. Apr 30, 2005 #4
    [tex] 0= e^x-\frac{2}{e^{2x}} [/tex]

    Just swing that fraction to the other side and multiply out.
  6. Apr 30, 2005 #5
    I can't believe I didn't get that....thanks.
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