# Inverse hyperbolic functions (logarithmic form)

## 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?

SteamKing
Staff Emeritus
Homework Helper
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?
You haven't shown all the steps in your derivation of the inverse cosh function, especially how you came to have a "±" in your logarithmic expression.

You haven't shown all the steps in your derivation of the inverse cosh function, especially how you came to have a "±" in your logarithmic expression.
$$x = \cosh{y} = \frac{e^y + e^{-y}}{2}$$
$$e^y - 2x + e^{-y} = 0$$
$$e^{2y} - 2x e^y + 1 = 0$$
$$e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2}$$
$$e^y = x \pm \sqrt{x^2 - 1}$$
$$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 - 1})}$$

SteamKing
Staff Emeritus
Homework Helper
• If $x > \sqrt{x^2 - 1}$ for all $x \geq 1$, why should we ignore the second solution ($e^y = x - \sqrt{x^2 - 1}$)? I think I get it now. The requirement is that $x - sqrt{x^2 - 1}$ must be greater than 1, not just 0.