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