Recently saw a question that asked me to show that $\ln\left(\sqrt{2} - 1\right) = - \ln\left(\sqrt{2} + 1\right)$. How can I demonstrate that the LHS equals the RHS?
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$-\ln x=\ln(1/x)$. – Andrés E. Caicedo Oct 02 '13 at 23:23
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For natural log, it should be $\ln$ or $\log$. Some use Ln, but certainly not In – Ross Millikan Oct 02 '13 at 23:26
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\begin{align} \log(\sqrt{2}-1) &= \log((\sqrt{2}-1)\frac{\sqrt{2}+1}{\sqrt{2}+1}) \\ &= \log(\frac{2-1}{\sqrt{2}+1})) \\ &= \log((\sqrt{2}+1))^{-1}) \\ &= -\log(\sqrt{2}+1) \end{align}
snar
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How did your mind even bring you to multiply inside the parentheses by (√2 + 1)/(√2 + 1)??? – Vladimir Nabokov Oct 02 '13 at 23:37
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Not a problem, please accept the answer.
Math, like most things in life, is a question of experience. I have the same reaction you just had day-in, day out, when I see people so clever it's frightening. That's good though, because it teaches you to embrace the feeling of confusion, and that's a supremely useful life-lesson.
When driving you swerve out of the way of a cat, yet you barely notice a plastic bag. How did your mind even bring you to distinguish these two?
– snar Oct 03 '13 at 00:16