# Hyperbolic Function (tanh & sech)

• May 8th 2008, 08:56 AM
Simplicity
Hyperbolic Function (tanh & sech)
Q: Solve the equation $4 \mathrm{tanh}x - \mathrm{sech} x = 1$ giving your solution in logarithmic form.
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My Method:

$4 \mathrm{tanh}x - \mathrm{sech} x = 1$

$\therefore4 \left( \frac{e^x - e^{-x}}{e^x + e^{-x}} \right) - \left( \frac{2}{e^x + e^{-x}} \right) = 1$

$\therefore \frac{4 (e^x - e^{-x}) -2}{e^x + e^{-x}} = 1$

$\therefore 4e^x - 4e^{-x} -2 = e^x + e^{-x}$

$\therefore 3e^x - 5e^{-x} - 2 = 0$

$\therefore 3e^{2x} - 2e^x - 5 = 0$

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However, this does not factorise and give an nice $\ln$ form answer so where have I gone wrong on my method? The answer is supposed to be $\ln \left( \frac 53 \right)$.
• May 8th 2008, 08:59 AM
Moo
Quote:

Originally Posted by Air
$\therefore 3e^{2x} - 2e^x - 5 = 0$

__________________
However, this does not factorise and give an nice $\ln$ form answer so where have I gone wrong on my method? The answer is supposed to be $\ln \left( \frac 53 \right)$.

Hello,

This factorises :)

$y=e^x$

$3e^{2x} - 2e^x - 5 = 0 \Longleftrightarrow 3y^2-2y-5=0$

$\Delta=4+4\cdot5\cdot3=8^2$

-> the solutions are...

(note that y>0 since y=e^x)
• May 8th 2008, 09:07 AM
Simplicity
Oh yes, It factorised to $(5e^x - 3)(e^x +1)$ and $e^x$ cannot be negative hence $(e^x +1)$ is ignored. Thanks.