# Exponential pop. growth when only given population at two instances of time.

I have a problem where I'm only given the population of a "bacteria culture" at two instances in time: 2 hours and 4 hours. The problem says the population of bacteria is 125 after 2 hours, and 350 after 4 hours. It specifically says the bacteria in the culture increases according to the law of exponential growth.

I know the formula $y(t) = y_{0}e^{kt}$ where $k$ is the growth constant (and $t$ is time, of course). I know how to find k if you're given the initial population, or time. But in this case, I'm not sure what to do. It asks to find the initial population, then write the exponential growth model. I know how to write the growth model once I find the initial population, because then I can solve for $k$. But I just don't know how to find $k$ when given only two populations at given points of time with exponential growth. Can anyone help?

• After thinking more, I think it might be possible to write a system of equations: $125 = y_{0}e^{k(2)}$ and $350 = y_{0}e^{k(4)}$ and then this would solve for both k and the initial population simultaneously. Is this the right approach? – Sabien Mar 17 '14 at 6:08
• Just as a "cultural" remark, it is an interesting characteristic that just as two points determine a line, two points also determine an exponential curve. – colormegone Mar 17 '14 at 6:58

## 2 Answers

Let $y_0$ be the initial population, and let $k$ be the growth constant. Then the population at time $t$ is $y_0e^{kt}$.

Putting $t=2$, we get $y_0e^{2k}=125$. Similarly, $y_0e^{4k}=350$.

Divide. We get $e^{2k}=\frac{350}{125}$.

Now that we know $e^{2k}$, we can find $y_0$. And now that we know $e^{2k}$, we can find $k$ by taking the logarithm.

Remark: I see that you saw how to approach the problem. If you wish to weite up your conclusions as an answer, I will delete mine.

• Thank you very much that helps. – Sabien Mar 17 '14 at 6:20
• No, I tried my method but it didn't work, well at least I must have done the Algebra wrong. Your explanation is more clear, no need to delete it. – Sabien Mar 17 '14 at 6:22
• You are welcome. If there any step that you have trouble completing, please leave a message. – André Nicolas Mar 17 '14 at 6:24

For the sake of variety, without using template formulas, you can think in the following way. Exponential growth is about multiplying by some common growth factor every unit of time. We could call that growth factor $b$. So here, after two hours, we went from $125$ to $350$. that means $$125\cdot b^2=350$$ from which you can solve for $b$ and find that the growth factor is $\sqrt{\frac{350}{125}}$.

Now what was the original population? Well moving backwards in time two hours: $$125\div\left(\sqrt{\frac{350}{125}}\right)^2=\frac{125^2}{350}$$ So the population at $t$ hours after the initial moment is $$P(t)=\frac{125^2}{350}\left(\sqrt{\frac{350}{125}}\right)^t$$

• Hey, this is a really cool way to think about it. Thank you for the insight. – Sabien Mar 17 '14 at 6:35