1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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 :\

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


    User Avatar
    Science Advisor

    [tex]tanh(4x)= \frac{e^{4x}- e^{-4x}}{e^{4x}+ e^{-4x}}[/tex]
    [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
    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]
  7. May 23, 2009 #6


    User Avatar
    Homework Helper

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


    User Avatar
    Homework Helper

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





    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


    User Avatar
    Homework Helper


    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook