# Initial Population

• September 13th 2009, 10:10 PM
latavee
Initial Population
The question goes: The population of bacteria grows at a rate proportional to the number of bacteria at time t. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What is the initial number of bacteria?

y=ce^kt
y(0)=?
y(3)=400
y(10)=2000

I don't understand how I can solve for the two unknowns: k and c
ex. 400=ce^k(3)
• September 13th 2009, 10:22 PM
pickslides
Quote:

Originally Posted by latavee
The question goes: The population of bacteria grows at a rate proportional to the number of bacteria at time t. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What is the initial number of bacteria?

y=ce^kt
y(0)=?
y(3)=400
y(10)=2000

I don't understand how I can solve for the two unknowns: k and c
ex. 400=ce^k(3)

and $2000=ce^{k(10)} \Rightarrow c = \frac{2000}{e^{k(10)} }$

with your work $400=ce^{k(3)}\Rightarrow c = \frac{400}{e^{k(3)} }$

gives us $\frac{2000}{e^{k(10)} } = \frac{400}{e^{k(3)} }$

$\frac{2000}{400} = \frac{e^{k(10)} }{e^{k(3)} }$

$5 = e^{k(10)-k(3)}$

$5 = e^{10k-3k}$

$5 = e^{7k}$

You can now solve for k, which will then give you c.
• September 13th 2009, 10:40 PM
latavee
Thanks!!
Wow! I never thought about it that way, thanks! So when I plug my k back to find my c, I will actually have to find c1 and c2 correct?
• September 14th 2009, 01:48 AM
gustype
Hi all, I am a new member of forum
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• September 14th 2009, 01:56 PM
pickslides
Quote:

Originally Posted by latavee
Wow! I never thought about it that way, thanks! So when I plug my k back to find my c, I will actually have to find c1 and c2 correct?

Nope, just c, I don't see any others (Rofl)

After you have k and c you then sub t=0 into the equation for initial population.