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

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 e.

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

A = ln a iff a = e^A ,

B = ln b iff b = e^B

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

From the last definition we have:

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

And using the law of indices:

e^C = (e^A) (e^-B) = e^(A-B)

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

C = A-B

And so:

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