Why is $\ln(a-\sqrt{a^2-1})=-\ln(a+\sqrt{a^2-1})$?

Question

In a recent discussion thread, I came across an observation that $$\ln(a\pm\sqrt{a^2-1})=\pm\ln(a+\sqrt{a^2-1})$$ Since it's straightforward that $\ln(a+\sqrt{a^2-1})=+\ln(a+\sqrt{a^2-1})$, proving the above statement boils down to showing that $$\ln(a-\sqrt{a^2-1})=-\ln(a+\sqrt{a^2-1})$$ However, I'm having a difficult time solving this puzzle. What's the catch?

Answer 1

$\ln(a-\sqrt{a^2-1})=-\ln(a+\sqrt{a^2-1})$

because $\ln(a-\sqrt{a^2-1})+\ln(a+\sqrt{a^2-1})$

$=\ln[(a-\sqrt{a^2-1})(a+\sqrt{a^2-1})]=\ln[a^2-(a^2-1)]=\ln1=0$.

Answer 2

It's because $$\left(a-\sqrt{a^2-1}\right)\left(a+\sqrt{a^2-1}\right)=1,$$ at least for $a>1$.