Why does $ln a - ln b = ln (\frac{a}{b})$ $lna - lnb = ln(a/b)$ ?

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Why does $ln a - ln b = ln (\frac{a}{b})$ $lna - lnb = ln(a/b)$ ?

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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 $e$ $e$ .

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

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

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

$C = ln (\frac{a}{b}) \Leftrightarrow \frac{a}{b} = e^{C}$ $C = ln (a/b) iff a/b = e^C$

From the last definition we have:

$\frac{a}{b} = e^{C} \Rightarrow e^{C} = \frac{e^{A}}{e^{B}}$ $a/b = e^C => e^C = (e^A)/(e^B)$

And using the law of indices:

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

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

$C = A - B$ $C = A-B$

And so:

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

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Why does $ln a - ln b = ln (\frac{a}{b})$ $lna - lnb = ln(a/b)$ ?

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