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

Question

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?

Answer 1

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.

Answer 2

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 $$P(t)=\frac{125^2}{350}\left(\sqrt{\frac{350}{125}}\right)^t$$