<s>[INST] 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 $r=s$
 [/INST] Looks good.
But note that it's really just a matter of observing that $x \pm \sqrt{x^2-1}$ are reciprocals, since
$$\frac{1}{x + \sqrt{x^2-1}} = \frac{1}{x + \sqrt{x^2-1}}\cdot\frac{x - \sqrt{x^2-1}}{x - \sqrt{x^2-1}}$$
$$=\frac{x - \sqrt{x^2-1}}{x^2-(x^2-1)}$$
$$= \frac{x - \sqrt{x^2-1}}{1}$$
$$=x - \sqrt{x^2-1}$$
This means that their logarithms are opposites, since $\log u^{-1} = -\log u$.
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