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?
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Hint : $(x+y)(x-y)=x^2 - y^2$ and $\log(\frac{1}{a})=-\log a$ – Evan William Chandra Oct 24 '19 at 03:15
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$\ln(a-\sqrt{a^2-1})=-\ln(a+\sqrt{a^2-1})$
because $\ln(a-\sqrt{a^2-1})+\ln(a+\sqrt{a^2-1})$
$=\ln[(a-\sqrt{a^2-1})(a+\sqrt{a^2-1})]=\ln[a^2-(a^2-1)]=\ln1=0$.
J. W. Tanner
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It's because $$\left(a-\sqrt{a^2-1}\right)\left(a+\sqrt{a^2-1}\right)=1,$$ at least for $a>1$.
Angina Seng
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