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#### karush

##### Well-known member

- Jan 31, 2012

- 2,648

- Thread starter karush
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- Thread starter
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- Jan 31, 2012

- 2,648

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What I would do is multiply the original equation by \(2e^x\) so that we have:find x

$\displaystyle\frac{e^x+e^{-x}}{2}=3$

ok we have the indenty of

$$\displaystyle\cosh{x}=\frac{e^x+e^{-x}}{2}$$

presume then the x can be replaced by 3

$$\displaystyle\cosh{3}=\frac{e^3+e^{-3}}{2}$$

ok $W\vert A$ returns

$x = \ln(3 \pm 2 \sqrt 2)$

ok so how??

\(\displaystyle e^{2x}+1=6e^x\)

Arrange in standard quadratic form:

\(\displaystyle e^{2x}-6e^x+1=0\)

Apply quadratic formula:

\(\displaystyle e^x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(1)}}{2(1)}=\frac{6\pm\sqrt{32}}{2}=3\pm2\sqrt{2}\)

Both roots are positive, thus:

\(\displaystyle x=\ln\left(3\pm2\sqrt{2}\right)\)

- Jan 30, 2018

- 368

You titled this "Evaluate cosh(3)" and that is what you did. But the problem you state is to