show that: logm to base a / log m to the base ab = 1+ log b to base a
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Let Log m base a = x.
Let log m base ab = y.
m = a^x = (ab)^y
So b^y = a^{x-y}
b = a^((x-y)/y)
Log b base a = (x-y)/y = x/y - 1
So x/y = 1 + log b base a.
Proved
Let log m base ab = y.
m = a^x = (ab)^y
So b^y = a^{x-y}
b = a^((x-y)/y)
Log b base a = (x-y)/y = x/y - 1
So x/y = 1 + log b base a.
Proved
kvnmurty:
Hope it's clear
very clear!
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