75,000 live bacteria are present in a culture in a flask. when an antibiotic is added to the culture, the number of live bacteria is reduced by half every four hours. approximately how many hours have passed when there are 3000 bacteria left alive

N = 75,000*(1/2)^(h/4) . . . . N = number alive; h = hours

3000 = 75000*(1/2)^(h/4) . . . . . substitute given numbers
1/25 = (1/2)^(h/4) . . . . . . . . . . . . . .divide by 75000
log(1/25) = (h/4)*log(1/2) . . . . . . . take the log, then divide by the coefficient of x
h = 4*log(1/25)/log(1/2) ≈ 18.6 . . . . hours

___
Of course log(1/25) = -log(25) = -2log(5), so the expression for h can be written as h = 8log(5)/log(2)

Ugh h h t t u i i vick u