# Bacteria Growth.

• October 13th 2007, 12:41 PM
Kane Nathaniel
Bacteria Growth.
A bacteria culture grows with constant relative growth rate. After 2 hours there are 800 bacteria and after 8 hours the count is 70,000.

All I need is the formula for the bacteria's growth, then I can answer the other questions on my own.

I don't even know where to start. I have worked with ones where it gives the initial value and how much hourly they grow. Our teacher hasn't explained this one.
Thanks for any help.
If you could explain how you got the formula, I would really appreciate it.

~Kane.
• October 13th 2007, 01:13 PM
TKHunny
I always find it hard to believe that you were given a problem with NO guidance.

Your book or teacher didn't translate this:

"constant relative growth rate"

Like this:

$\frac{dBacteria}{dt}\;=\;k*Bacteria(t)$

or like this:

$Bacteria(t)\;=\;A_{0}e^{k*t}$
• October 13th 2007, 01:58 PM
Kane Nathaniel
I have one of the worst Calculus professors. I don't even know why he is teaching...

He doesn't teach and we have a crappy book that doesn't help.
70% of the class is failing, so that's why I have to come to you guys and go to tutors...

It's pointless to go to class.
Thanks for the help.
• October 14th 2007, 09:46 PM
angel.white
So you know, the method to solve this will be to plug in your values, you will get

y(2)=800
y(2)=y(0)e^(kt)
800=y(0)e^(kt)

y(8)=70000
y(8)=y(0)e^(kt)
70,000=y(0)e^(kt)

Now you have 2 equations with 2 terms, you need to solve for each of them, and then you can plug them into your actual equation of
y(t)=y(0)e^kt

I think the best approach to solving for them would be to divide each by y(0), and then multiply by the numerator, so that each equation is 1/y(0)=... Then you can just set them equal to eachother, and use natural logarithms to solve for K

Tip for solving with natural logarithms: Remember ln(e^(8k)/70000) = ln(e^(8k)) - ln(70,000) = 8k - ln(70000)