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

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To solve the equation cosh(x) = 3, the identity cosh(x) = (e^x + e^(-x))/2 is used. By rearranging and multiplying the equation by 2e^x, it transforms into a quadratic form: e^(2x) - 6e^x + 1 = 0. Applying the quadratic formula yields e^x = 3 ± 2√2, leading to the solutions x = ln(3 ± 2√2). The discussion emphasizes the importance of correctly interpreting the problem as solving for x rather than evaluating cosh(3).
karush
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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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karush said:
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:

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

Arrange in standard quadratic form:

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

Apply quadratic formula:

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

Both roots are positive, thus:

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