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

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Answer 1 · 2017-10-01T16:05:13

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

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