MHB Natural Log Rule: $\frac{a}{b}=-\frac{b}{a}$?

AI Thread Summary
The discussion confirms that the natural logarithm of a fraction can be expressed as the negative logarithm of its reciprocal, specifically stating that $\ln\left(\frac{a}{b}\right)$ equals $-\ln\left(\frac{b}{a}\right)$. Participants agree that this relationship holds true, illustrating that $\ln(a) - \ln(b)$ simplifies to $\ln\left(\frac{a}{b}\right)$ and can also be represented as $-\left[\ln(b) - \ln(a)\right]$. The conversation emphasizes that inserting a negative sign in front of the logarithm effectively flips the fraction. Overall, the properties of logarithms regarding fractions and their reciprocals are clearly established.
tmt1
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If I have $\ln\left({a}\right) - \ln\left({b}\right)$ that would equal $\ln\left({\frac{a}{b}}\right)$ or $-(\ln\left({b}\right) - \ln\left({a}\right))$ which is also $- \ln\left({\frac{b}{a}}\right)$. So does this mean $\ln\left({\frac{a}{b}}\right)$ equals $- \ln\left({\frac{b}{a}}\right)$?
 
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Yes, you are correct...another way to think of it is:

$$\log_a\left(\frac{b}{c}\right)=\log_a\left(\left(\frac{c}{b}\right)^{-1}\right)=-\log_a\left(\frac{c}{b}\right)$$
 
tmt said:
If I have $\, \ln (a) - \ln (b)\,$ that would equal $\,\ln\left({\dfrac{a}{b}}\right)\,$ or $\,-\left[\ln\left({b}\right) - \ln\left({a}\right)\right]\;$ which is also $\,- \ln\left({\dfrac{b}{a}}\right)$
So does this mean $\,\ln\left({\dfrac{a}{b}}\right)\,$ equals $\,- \ln\left({\dfrac{b}{a}}\right)\:$?
If you discovered this while 'fooling around' with logs, good workl

Yes indeed!
If you have the log of a fraction, inserting a minus in front
will 'flip' the fraction.

That is: -\ln\left(\frac{a}{b}\right) \:=\:\ln\left(\frac{b}{a}\right)

 
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