Posted by Anonymous on Wednesday, July 15, 2009 at 11:57am.
log a, log ar, log ar2, ...log a (r)^n
now, log ar^n=log a + n log r
consider the next term
log ar^(n+1)= log a + (n+1)log r
then
log ar^(n+1)-log ar^n= log r which is a constant. Therefor, the progression is arithemetic.
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