# A bacterial culture starts with 2,000 bacteria and doubles in size every 6 hours. How do you find an exponential model for the size of the culture as a function of time t in hours?

Jul 2, 2016

$\ln \left(\frac{P}{2000}\right) = 0.115 t$

#### Explanation:

One thing we know about bacteria, if there's one, there'll be more.
So that means the rate of growth of the bacteria is dependent on how much of these lil' buggers there is in the start.

In mathematical terms, $\frac{d}{\mathrm{dt}} \left\{P\right\} = \setminus \lambda \cdot P$

Let's rearrange that equation, so we get $\frac{\mathrm{dP}}{P} = \setminus \lambda \cdot \mathrm{dt}$
Integrate, from ${P}_{o}$ to $P$ (where P is whatever the amount of bacteria while ${P}_{o}$ the initial, we'll substitute later).
so that means, time is from $t = 0$ to $t = t$

Integrating leaves us with $\ln \left(\frac{P}{P} _ o\right) = \setminus \lambda t$

We see that in the next six hours, $P = 2 {P}_{o}$ (it says that in the question). So that means $\frac{P}{P} _ o = 2$ at $t = 6 h r s$

So, rearranging the equation after substitution, $\ln \frac{2}{6} = \setminus \lambda$

Your calculator will help with the finalizing of what $\lambda$ is.
Nevertheless, now you see how the answer comes to as given above.