Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$

Question

Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots are possible. Thus, $y=\ln\left(x\pm\sqrt{x^2-1}\right)$. Why did we fail to eliminate the minus?

Answer 1

The reason why we can dispose of the minus sign is that, firstly, $$(x+\sqrt{x^2-1})^{-1}=(x-\sqrt{x^2-1})$$

Therefore $$\ln(x\pm \sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})^{\pm1}=\pm\ln(x+\sqrt{x^2-1})$$

But $\operatorname{arcosh}(x)$ is defined as non-negative, so the minus sign is not required

Answer 2

The following is a graph of $y=\cosh(x):$

But the next one is the graph of $y=\operatorname{arccosh}(x)$, considering both real branches$:$

Applying domain restrictions or branch restrictions, one can remove the $\pm$, but I think the graphs make it clear why one would have the $\pm$ in the first place.

Answer 3

since the definition of the $$\sqrt{x^2-1}$$ we get $$|x|>1$$ which is for $$x-\sqrt{x^2-1}$$ impossible

Answer 4

As $y\ge 0$, $\;\mathrm e^y\ge 1$. Now $\mathrm e^y$ is a root of the quadratic polynomial $\;P(t)=t^2-2xt+1$. Just observe that, as $P(1)=2(1-x) < 0$ for $x>1$, $1$ separates the roots. Hence $\mathrm e^y$ is the greatest root — with a + sign.