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

  • Thread starter thomas49th
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  • #1
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[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
 
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Answers and Replies

  • #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?
 
  • #3
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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 :)
 
  • #4
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nope that gives me (2(2+sqrt(3))^4)/ 0 which cannot be :\
 
  • #5
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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
 
  • #6
Cyosis
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What is [itex]e^{\ln a}, e^{-\ln a}[/itex]?
 
  • #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.
 
  • #8
Cyosis
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Yes.

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 :)

nope that gives me (2(2+sqrt(3))^4)/ 0 which cannot be :\
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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