Exponential Growth: bacetria?

Question

The count in a bacteria culture was 700 after 20 minutes and 1000 after 40 minutes. What was the initial size of the culture?

I don't know why but I am having trouble starting this one. I tried to find the k in the equation y = yoe^kt, and I got 0.316. However, when I plug that value in to find the initial amount, I get 1.256, which is not right. Any suggestions?

Thanks

Answer 1

You are on the right track.

y(t) = y(0)*exp(k*t)

We need to find the growth constant, k and y(0). There are a couple of ways to do this. The most direct is to write down the equations for the two times you have data for. This gives you two equations in two unknowns, which can then be solved for.

1000 = y(0)*exp(k*40 min)
700 = y(0)*exp(k*20 min)

Divide the second equation by the first equation:

700/1000 = exp(k*20 min)/exp(k*40 min)
0.7 = exp(k*(20-40)) = exp(-k * 20 min)
-ln(0.7)/20 min = k
k = 1.78337*10^-2 min^-1

Plug this value back into one of the equations we started with to find y(0):

1000 = y(0)*exp(1.78337*10^-2 min^-1 * 40 min)
1000 = y(0)*exp(0.71335)
1000/exp(0.71335) = y(0)
y(0)= 490

As a check, we can see if this gives the right result for t = 20 min:

y(20 min) = 490*exp(1.78337*10^-2 min^-1 * 20 min)
= 490*exp(0.356675)
= 700 Check!

Answer 2

This particular problem has an extremely simple solution. Note that the second sample time is exactly twice the first. This means that initial count/count1 = count1/count2. Thus initial count = count1^2/count2 = 490.

Answer 3

You better don’t! I suggest an equation:
y(x)=y0*(1+k)^x, or Ln(y)-Ln(y0)=x*Ln(1+k), then
[Ln(1000)-Ln(y0)]/[Ln(700)-Ln(y0)]=40/20, then y0=490.
Am I right?