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View Full Version : Log5 1=log5 x-log5 8


shadella
Mar 15, 2011, 09:25 PM

galactus
Mar 16, 2011, 04:20 AM
I assume that is base 5?

log_{5}(1)=log_{5}(x)-log_{5}(8)?

The left side of the equation is obviously 0. Because log(1)=0

Therefore, what must x be?