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Find the exact logarithmic form of y at A where y= -x + tanh4x is

  1. May 23, 2009 #1
    [tex]\frac{2\sqrt{3} - ln(2+\sqrt{3})}{4}[/tex]

    Given that the x co-ordinate is [tex]ln(2+ \sqrt{3})[/tex] which I worked out in the first part of this qusetion given that at A there is a maximum turning point (so maximum turning point means dy/dx = 0), but I'm not sure that helps in this part of the question

    y= -x + tanh4x


    i tried [tex]y = -ln(2+ \sqrt{3}) + tanh4.ln(2+ \sqrt{3})[/tex] but cant really see how that simplifies :\

    Thanks
     
    Last edited: May 23, 2009
  2. jcsd
  3. May 23, 2009 #2

    HallsofIvy

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    [tex]tanh(4x)= \frac{e^{4x}- e^{-4x}}{e^{4x}+ e^{-4x}}[/tex]
    so
    [tex]tanh(4(2+ \sqrt{3}))= \frac{e^{8+ 4\sqrt{3}}- e^{-8-4\sqrt{3}}}{e^{-2-\sqrt{3}}+e^{-2-\sqrt{3}}}[/tex]
    Is that better?
     
  4. May 23, 2009 #3
    sorry hall of ivy I missed out a ln in my original latex post. However, I will try that method, but putting a ln in front of it (to cancel out e's). Hopefully that will get me somewhere

    Thanks :)
     
  5. May 23, 2009 #4
    nope that gives me (2(2+sqrt(3))^4)/ 0 which cannot be :\
     
  6. May 23, 2009 #5
    also did you make a mistake on the second line of your second fraction... shouldn't it be
    [tex]
    tanh(4(2+ \sqrt{3}))= \frac{e^{8+ 4\sqrt{3}}- e^{-8-4\sqrt{3}}}{e^{8+ 4\sqrt{3}}+ e^{-8-4\sqrt{3}}}[/tex]
    :P
     
  7. May 23, 2009 #6

    Cyosis

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    What is [itex]e^{\ln a}, e^{-\ln a}[/itex]?
     
  8. May 24, 2009 #7

    Mentallic

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    e and ln are inverses of each other, just like squaring and square-rooting.

    Therefore,

    [tex]e^{lna}=a[/tex]

    and

    [tex]e^{-lna}=(e^{lna})^{-1}=a^{-1}=\frac{1}{a}[/tex]

    Note that if a is positive, [tex]e^{ln(-a)}\neq -a[/tex] simply because the log of a negative doesn't exist.
     
  9. May 24, 2009 #8

    Cyosis

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    Yes.

    Then I don't understand how you got a 0 in the denominator, because [itex]\exp(\ln[(2+\sqrt{3})^4])=(2+\sqrt{3})^4, \; \exp(-\ln[(2+\sqrt{3})]^4)=1/(2+\sqrt{3})^4[/itex]
     
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