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

[tmt1]

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 $- \ln\left({\frac{b}{a}}\right)$?

 

[MarkFL]

Yes, you are correct...another way to think of it is:

\(\displaystyle \log_a\left(\frac{b}{c}\right)=\log_a\left(\left(\frac{c}{b}\right)^{-1}\right)=-\log_a\left(\frac{c}{b}\right)\)

 

[soroban]

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