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

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Steve M
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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

Let us define

# 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

# C = A-B #

And so:

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

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