How Do You Solve cosh(x) = 3 for x?

[karush]

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??

View attachment 8568​

 

[MarkFL]

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:

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

 

[HOI]

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.