From Pi Mu Epsilon Journal, November 1950:
(1/2)3 2.
Taking the logarithm to the base 1/2 of each member of the above inequality, we write
3 log1/2(1/2) 1/2(1/2).
But logbb = 1. Therefore
3
From Pi Mu Epsilon Journal, November 1950:
(1/2)3 2.
Taking the logarithm to the base 1/2 of each member of the above inequality, we write
3 log1/2(1/2) 1/2(1/2).
But logbb = 1. Therefore
3
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