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

Answer 1

(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

Answer 2

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.

Answer 3

Three doublings ago--that is to say, 60 hours, or 2 1/2 days.

Answer 4

60 hours ago.

1/2 20 hours ago
1/4 40 hours ago
1/8 60 hours ago

Answer 5

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.

Answer 6

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