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su6.6.82 evaluate cosh(3)

karush

Well-known member
Jan 31, 2012
2,657
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??
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
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??
What I would do is multiply the original equation by \(2e^x\) so that we have:

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

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:

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

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
378
What you are "doing wrong" is that you are "going the wrong way"!

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