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#sum _(a,b,c) (1/ (1+ log _a bc)) = 1 #Prove it?

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Ananda Dasgupta

Mar 21, 2018

Since #log_a b = log b/log a#, we have

#1/(1+log_a bc) = 1/(1+(log (bc))/log a) = log a/(log a+log (bc)) = log a/log(abc)#

Thus, the sum is

#log a/log(abc) + log b/log(abc) + log c/log(abc) = (log a +log b+ log c)/log(abc) = log(abc)/log(abc) = 1#

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