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tuy
Answers (1)
so number = a e^(kt) 10,000 = a e^(3k) 40,000 = a e^(5k) divide the 2nd by the 1st: 4 = e^(2k) 2k = ln4 k = (ln4)/2 so N = a e^( (ln4)/2 t) so we need a a e^(3ln4/2) = 10,000 a = 10,000/8 = 1250 initial count is 1250 bacteria
