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

[thomas49th]

$$\frac{2\sqrt{3} - ln(2+\sqrt{3})}{4}$$

Given that the x co-ordinate is $$ln(2+ \sqrt{3})$$ 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 + tanh4xi tried $$y = -ln(2+ \sqrt{3}) + tanh4.ln(2+ \sqrt{3})$$ but can't really see how that simplifies :\

Thanks

 

[HallsofIvy]

$$tanh(4x)= \frac{e^{4x}- e^{-4x}}{e^{4x}+ e^{-4x}}$$
so
$$tanh(4(2+ \sqrt{3}))= \frac{e^{8+ 4\sqrt{3}}- e^{-8-4\sqrt{3}}}{e^{-2-\sqrt{3}}+e^{-2-\sqrt{3}}}$$
Is that better?

 

[thomas49th]

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 :)

 

[thomas49th]

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

 

[thomas49th]

also did you make a mistake on the second line of your second fraction... shouldn't it be
$$tanh(4(2+ \sqrt{3}))= \frac{e^{8+ 4\sqrt{3}}- e^{-8-4\sqrt{3}}}{e^{8+ 4\sqrt{3}}+ e^{-8-4\sqrt{3}}}$$
:P

 

[Cyosis]

What is $e^{\ln a}, e^{-\ln a}$?

 

[Mentallic]

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

Therefore,

$$e^{lna}=a$$

and

$$e^{-lna}=(e^{lna})^{-1}=a^{-1}=\frac{1}{a}$$

Note that if a is positive, $$e^{ln(-a)}\neq -a$$ simply because the log of a negative doesn't exist.

 

[Cyosis]

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 $\exp(\ln[(2+\sqrt{3})^4])=(2+\sqrt{3})^4, \; \exp(-\ln[(2+\sqrt{3})]^4)=1/(2+\sqrt{3})^4$