# su6.6.82 evaluate cosh(3)

#### karush

##### Well-known member
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??

#### MarkFL

Staff member
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:

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

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

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

Both roots are positive, thus:

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

#### Country Boy

##### Well-known member
MHB Math Helper
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.