Proving that $-\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})$ Show $-\ln(x-\sqrt{x^2-1})=ln(x+\sqrt{x^2-1})$
Can somebody verify that my solution is correct (or incorrect!) and also show me alternative methods to do this? Thanks!
Let $r=-\ln(x-\sqrt{x^2-1})$ and $s=ln(x+\sqrt{x^2-1})$
$\rightarrow e^r=(x-\sqrt{x^2-1})^{-1}$ and $e^s = x+\sqrt{x^2-1}$
$\rightarrow \frac{e^s}{e^r} = (x+\sqrt{x^2-1})(x-\sqrt{x^2-1}) = x^2 - (x^2 - 1) = 1$
Thus $e^r=e^s$ and so $r=s$

Your proof looks fine to me. You can also do this: $\ln(x-\sqrt{x^2-1})+\ln(x+\sqrt{x^2-1})=\ln\left((x-\sqrt{x^2-1})(x+\sqrt{x^2-1})\right)=\ln 1 =0$
