Why does lna - lnb = ln(a/b)lnalnb=ln(ab)?

1 Answer
Oct 1, 2017

It does not matter what base we use providing the same base is used for all logarithms, here we are using bease ee.

Let us define A,B.CA,B.C as follows=:

A = ln a iff a = e^A A=lnaa=eA,

B = ln b iff b = e^B B=lnbb=eB

C = ln (a/b) iff a/b = e^C C=ln(ab)ab=eC

From the last definition we have:

a/b = e^C => e^C = (e^A)/(e^B) ab=eCeC=eAeB

And using the law of indices:

e^C = (e^A) (e^-B) = e^(A-B) eC=(eA)(eB)=eAB

And as as the exponential is a 1:11:1 monotonic continuous function, we have:

C = A-B C=AB

And so:

ln (a/b) = ln a - ln b \ \ \ QED