Solving Bacterial Population Growth: A Math Problem

[Hussam Al-Tayeb]

If all I have given is that
1. Bacteria grows at a rate proportional to it's size.
2. It doubles in 2 days.
3. At 10 days, population is 1000.

I'm not given the initial bacteria count, I need help setting up the equation.

I did:
dy/dt = ky => dy/y = kdt => lny= kt + c => y=e^(kt) + c

y(10)= 1000 = e^(10k) + c

But I'm lost how to complete this. Any idea?

 

[Antineutron]

lny= kt + c => y=e^(kt) + c

Look on the right side, you neglicted the constant.

 

[Hussam Al-Tayeb]

so it should be y=ce^(kt)

y(o) = c e^0 = c
then y(2) = c e^(2k) = 2 * y (0)
=> 2k = ln(2) => k = ln(2) / 2

now y(10) = 1000 = c e^(10k)

=> c = 1000 / (e^(5ln2))

correct?

 

[Antineutron]

Yes! great job, keep up the good work! very good.

 

[HallsofIvy]

Of course, it would have been much simpler to argue that, since the population doubles every 2 days, we must have P(t)= C2^t/2 where t is in days. Then P(10)= C 2⁵= 1000 so C= 1000/32= 100/16= 10/8= 5/2.

P(t)= (5/2) 2^t/2.

(But Antineutron is right- great job!)