Question

At the start of an experiment, 2x10^4 bacteria are present in a colony. Eight hours later, the population is 3x10^4.
a.) Determine the growth constant k
b.) What was the population two hours after the start of the experiment?
c.)How long will it take for the population to triple?

Answer 1

@wolf1728 Could you help me with this?

Answer 2

Population \(P = A*e^{kt}\)
when t = 0, P = 2x10^4.
2x10^4 = A * e^0. A = 2x10^4
\(P = 2\times10^4 * e^{kt} \)
when t = 8, P = 3 x 10^4. Substitute and solve k which is the growth constant.

Answer 3

oh okay makes sense

Answer 4

@aum now for b.) i just useall that information into N(t)=Noe^kt

Answer 5

correct?

Answer 6

Yes. Once you find k from part a, you have the complete equation for P as a function of time t. For part b, all you have to do is set t = 2 and solve for P.

Answer 7

Okay the population goes from 20,000 to 30,000 in 8 hours
Therefore it increases by a factor of 1.5 every 8 hours
So the population increases by this equation:
20,000*1.5^(hours/8)

zero hours 20,000*1.5^(0/8) = 20,000
two hours 20,000*1.5^(2/8) = 22,134

Answer 8

ok

Answer 9

Time to triple is:
20,000*1.5^(X/8) = 60,000

1.5^(X/8) = 3

taking logs of both sides
(X/8) *log(1.5) = log(3)

(X/8) = log(3) / log(1.5)

(X/8) = 2.7095112914

X = 21.6760903308 hours

... and it's just that simple :-)
LOL

Answer 10

aum was quite helpful and used all that 'e' and natural log stuff
I prefer more traditional (common log) methods