Question

The doubling period of a baterial population is 20 minutes. At time t = 80 minutes, the baterial population was 80000. With t representing minutes, the formula for the population is p(t)=A e^{kt}.

1. k = .
2. The initial population at time t = 0 is .
3. The size of the baterial population after 5 hours is .

Answer 1

why use this when you know the doubling time?

Answer 2

huh?

Answer 3

its the last problem i dont understand thank you

Answer 4

at \(t=80\) the population was \(8000\)
the doubling time is \(20\) minutes,
so at \(t=60\) it was \(4000\)
and at \(t=40\) it was \(2000\)
at \(t=20\) it was \(1000\)
and at \(t=0\) it was \(500\)

Answer 5

that answers problem #2

Answer 6

i got that already and its 5000 not 500

Answer 7

i just dont know 3 and 1

Answer 8

3. The size of the baterial population after 5 hours is .
you start with \(5000\) and 5 hours has \(5\times 3=15\) 20 minute periods (15 doublings )
so the population will be an astronomical\[5000\times 2^{15}\]

Answer 9

and how do you find K?

Answer 10

anyone?

Answer 11

zencvbnmjkl - what you actually need is the FORMULA right?

Answer 12

yea i just need to find K

Answer 13

What is 'A' supposed to be?

Answer 14

its 2=e^20k

Answer 15

at 80 minutes population = 80,000
p(t)=A e^(kt)
population = A*2.7181828^(k*t)
80,000 = A * 2.7181828^(k*80)
Seems we need to know 'A'
you said it is
its 2=e^20k
but that's not a value - that's an equation
What is te value of 'A'?

Answer 16

I'm guessing that 'A' is the beginning amount of bacteria right?

Answer 17

i got the answer k=.03...

Answer 18

i forgot the rest but thank you very much for your time

Answer 19

someone gave me the answer in my class

Answer 20

After 80 minutes, population = 80,000
It doubles every 20 minutes so we have 4 doublings occurring.
2^4 = 16
So, the starting population was 80,000 / 16 = 5,000

Answer 21

yea i got that but thanks for your help ma dude

Answer 22

I can show you how to solve for 'k' if you want

Answer 23

do you know how?

Answer 24

Sure do

Answer 25

i know it involves ln

Answer 26

its ln2/20=.0346

Answer 27

yes ln means natural log - you are dealing with natural logs once 'e' is brought into an equation

Answer 28

80,000 = A * 2.7181828^(k*80)
We know A = 5,000
80,000 = 5,000 * 2.7181828^(k*80)
80,000 / 5,000 = 2.7181828^(k*80)
16,000 = 2.718281828^(k*80)
taking natural logs of both sides
ln (16,000) = k*80
9.6803440012 / 80 = k
k = 0.1210043

Answer 29

thanks

Answer 30

okay that's about 4 times bigger than the answer you gave me
I can check the answer if you want

Answer 31

80,000 = 5,000 * 2.7181828^(k*80)
We'll assume k = 0.1210043
80,000 = 5,000 * 2.7181828^(0.1210043*80)

80,000 = 5,000 * 16,000
80,000 = 80,000
Successful !!!