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

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Steve M Share
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=:

$A = \ln a \iff a = {e}^{A}$,

$B = \ln b \iff b = {e}^{B}$

$C = \ln \left(\frac{a}{b}\right) \iff \frac{a}{b} = {e}^{C}$

From the last definition we have:

$\frac{a}{b} = {e}^{C} \implies {e}^{C} = \frac{{e}^{A}}{{e}^{B}}$

And using the law of indices:

${e}^{C} = \left({e}^{A}\right) \left({e}^{-} B\right) = {e}^{A - B}$

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

$C = A - B$

And so:

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

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