# Thread: Hyperbolic Function (tanh & sech)

1. ## Hyperbolic Function (tanh & sech)

Q: Solve the equation $\displaystyle 4 \mathrm{tanh}x - \mathrm{sech} x = 1$ giving your solution in logarithmic form.
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My Method:

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

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

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

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

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

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

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

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

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

This factorises

$\displaystyle y=e^x$

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

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

-> the solutions are...

(note that y>0 since y=e^x)

3. Oh yes, It factorised to $\displaystyle (5e^x - 3)(e^x +1)$ and $\displaystyle e^x$ cannot be negative hence $\displaystyle (e^x +1)$ is ignored. Thanks.