this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

3/4=(e^x - e^-x)/(e^x + e^-x)

any help is very appreciated

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- Nov 29th 2009, 12:52 PMtwostep08how can i solve this equation for x
this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

3/4=(e^x - e^-x)/(e^x + e^-x)

any help is very appreciated - Nov 29th 2009, 12:54 PMe^(i*pi)
- Nov 29th 2009, 12:57 PMtwostep08
im not really allowed to use my calculator...there must be some nice answer using logs?

- Nov 29th 2009, 01:15 PMe^(i*pi)
Yeah there is. You can use standard algebraic techniques and logs. Ignore the second solution if you only want a real answer

$\displaystyle 3e^x+3e^{-x} = 4e^x-4e^{-x}$

$\displaystyle e^x-7e^{-x} = 0

$

$\displaystyle e^x - \frac{7}{e^x} = 0$

$\displaystyle e^{2x} - 7 = 0$

$\displaystyle (e^x-\sqrt{7})(e^x+\sqrt{7}) = 0$

$\displaystyle x = \frac{1}{2}ln(7)$ or $\displaystyle x = \frac{1}{2}ln(-7) = \frac{1}{2}ln(7)+ i \left(\frac{\pi}{2}\right)$ - Nov 29th 2009, 01:31 PMtwostep08
ok im following you all the way to the last step when you say e^x - (sq.rt.)7 = 1/2ln(7)

what makes that so? - Nov 29th 2009, 01:36 PMe^(i*pi)
It copmes from the difference of two squares when factorising the quadratic and the term is equal to 0

$\displaystyle e^x-\sqrt7 = 0$

From here add $\displaystyle \sqrt7$ to both sides and take the natural log

I used the log power law to bring the 1/2 to the front - Nov 29th 2009, 01:37 PMDefunkt
$\displaystyle e^x-\sqrt{7} = 0 \Rightarrow log(e^x) = x = log(\sqrt{7}) = \frac{1}{2}log7$

- Nov 29th 2009, 01:43 PMe^(i*pi)