For example,$$-\ln\left(\frac{\ln\sqrt{...\sqrt{\pi}}}{\ln\pi}\right)$$ becomes $$\ln\left(\frac{\ln\pi}{\ln\sqrt{...\sqrt{\pi}}}\right)$$ if you take the reciprocal of the inside function. Is this just a rule to memorise?
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2It's the same exponent rule $$\log(a^b) = b\log a$$ with $b=-1$. Nothing "extra" to memorize. – Ninad Munshi Aug 23 '20 at 02:00
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2Does this answer your question? Why is $\ln\frac{1}{x}=-\ln x$? – Toby Mak Aug 23 '20 at 02:48
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Because $$\ln\left(\frac1a\right) =-\ln(a)$$
As a check: $$\ln\left(\frac1a\right) +\ln(a) = \ln\left(\frac1a\times a\right) =\ln(1)=0$$
Henry
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$$\log\left(a^b\right)=b\log(a)$$ for $a\gt0$. So $$\log\left(\frac xy\right)=\log\left(\left(\frac yx\right)^{-1}\right)=-\log\left(\frac yx\right)$$ This is why taking the reciprocal makes the $\log$ negative.
Just_A_User
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Ah I see. I was just confused because in the examples I searched up they only have fractions like $\frac{1}{x}$ which removes the fraction on the inside (to just $x$) – user71207 Aug 23 '20 at 02:33
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Define $\ln t = \int_1^t \mathrm d x/x$. Substituting $u = 1/x, \mathrm d u = -1/x^2 \ \mathrm d x$ gives:
$$\ln \left( \frac{1}{t} \right) = \int_1^{1/t} \frac{1}{x} \frac{\mathrm du}{-1/x^2} = \int_1^{1/t} u \frac{\mathrm du}{-u^2} = -\int_1^{1/t} \frac{\mathrm du}{u} = -\ln t.$$
Toby Mak
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