A bacteria population doubles in 20hours. when was the population 1/8 of its present population?

(present population) = (original population)2^(t / (doubling time))





(1 / 8)(original population) = (original population)2^(t / 20)





1 / 8 = 2^(t / 20)





LN(1 / 8) = LN[2^(t / 20)]





LN(1 / 8) = (t / 20)(LN(2))





LN(1 / 8) / LN(2) = t / 20





20LN(1 / 8) / LN(2) = t





20LN(2^(-3)) / LN (2) = t





3(-20)LN(2) / LN(2) = t





-60 = t





Answer: The poopulation was 1 / 8 of the original size 60 hours ago

A bacteria population doubles in 20hours. when was the population 1/8 of its present population?
20 hours ago 1/2 of current population


40 h ago ago 1/4


60 h ago 1/8


so 60 hours ago it was 1/8 of the current population.
Reply:Three doublings ago--that is to say, 60 hours, or 2 1/2 days.
Reply:Growth equation: P = Pe^(rt)





2 = e^(r*20)


ln2 = 0.693 = 20r; r = (ln2)/20





P/8 = Pe^(ln2*t/20)


ln(1/8) = ln2*t/20


-ln8/ln2*20 = t


t = -60 hours, in other words 60 hours ago.





A shortcut to this is to use P.f/P.i = 2^(t/20)


1/8 = 2^(t/20)


-3 = t/20


t= -60 hours
Reply:60 hours ago.





1/2 20 hours ago


1/4 40 hours ago


1/8 60 hours ago
Reply:Present population = P





We're looking for 1/8th of X.





P = X


1/2P = X/2


1/2(1/2P) = X/4


1/2(1/2(1/2P)) = X/8


^ ....^......^





That's three doublings. Three doublings takes 60 hours.