1. ## Exponential Growth

So, a scientist is growing a colony of bacteria. After 2 hours, there are 40 bacteria. After 3 hours, there are 120 bacteria. If the population of bacteria grows exponentially, then how many bacteria will there be after 4 hours?

We have (2, 40), (3, 120), and we need to find (4, ?). I've done the following:
$\displaystyle 40=ae^2k$, and $\displaystyle 120=ae^3k$
2) $\displaystyle a=40/e^2k$
3) $\displaystyle 120/40=(40/e^2k)e^3k$
4) $\displaystyle 3=3^3k/e^2k$

That's where I am. In most of the examples I've seen, in step 3, the equation is usually being multiplied by something simple, like $\displaystyle e^4k/e^2k$, but for this one, I guess I have to divide $\displaystyle e^3k/e^2k$, which in step 4, I am confused.

2. Originally Posted by BeSweeet
So, a scientist is growing a colony of bacteria. After 2 hours, there are 40 bacteria. After 3 hours, there are 120 bacteria. If the population of bacteria grows exponentially, then how many bacteria will there be after 4 hours?
Exponential functions can have any base, not just e. In this case, the colony of bacteria is trippling every hour. So the base is 3.

Say the colony started with 'a' bacteria $\displaystyle a * 3^0$.
After one hour it had $\displaystyle a * 3^1$.
After two hours it had $\displaystyle a * 3^2$.
After three hours it had $\displaystyle a * 3^3$.
After four hours it will have $\displaystyle a * 3^4$ or the number after three house times 3.

Notice what is changing in this sequence - the exponent. That makes this an exponential (when the variable is in the exponent).

If you want to see this using the $\displaystyle P = ke^{rt}$ formula, then follow this:

$\displaystyle 40 = k e^{r*2}$
$\displaystyle 120 = k e^{r*3}$ or $\displaystyle 40 = \frac{k e^{r*3}}{3}$
Since $\displaystyle 40 = 40$ you can say $\displaystyle k e^{2r}=\frac{k e^{3r}}{3}$
Multiply both sides by $\displaystyle \frac{3}{k}$ to get $\displaystyle 3e^{2r}=e^{3r}$
Then $\displaystyle 3=e^r$.

Plugging this back in to the $\displaystyle P = ke^{rt}$ you get $\displaystyle P = k*3^t$, which is the same as what is in the series above.