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## Main Question or Discussion Point

To express the ##\cosh^{-1}## function as a logarithm, we start by defining the variables ##x## and ##y## as follows:

$$y = \cosh^{-1}{x}$$

$$x = \cosh{y}$$

Where ##y ∈ [0, \infty)## and ##x ∈ [1, \infty)##.

Using the definition of the hyperbolic cosine function, rearranging, and multiplying through by ##e^y##, we get:

$$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 -1})}$$

How exactly do we get rid of the minus sign?

$$y = \cosh^{-1}{x}$$

$$x = \cosh{y}$$

Where ##y ∈ [0, \infty)## and ##x ∈ [1, \infty)##.

Using the definition of the hyperbolic cosine function, rearranging, and multiplying through by ##e^y##, we get:

$$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 -1})}$$

How exactly do we get rid of the minus sign?