A High School Exponential Decay Question

Question

I've been teaching a student, when we encountered a question that I just couldn't work out. It goes like this:

A city has a population of 10 million, decreasing by 1% every year. In a hundred years, the population will be:

a) half

b)less than half, but more than a quarter

c)less than a quarter, but more than a fifth

d) less than a fifth

e) none of the above

It seems to me that I want to compute $\left(\frac{99}{100}\right)^{100}$, but that obviously doesn't help. By using the rule of 70(/72/69), the answer intuitively is b), but it doesn't seem likely that the exam would expect the student to know that (and if so, then what is the fifth doing in there?). Am I missing something?

Thank you for help.

Answer 1

$$y=ce^{-kt}$$

The above is the equation for exponential decay (derived of course form $\frac{dy}{dt}=-ky$). This is actually a pretty simple problem. Let's analyze the question - it says A city has a population of 10 million. What does that mean? Well, for one that we are dealing with population (so $y$ will be $P(t)$ to make it better), and that the initial population is $10$ million. As you probably know, the $c$ is really the $P(0)$ ($e^{-k(0)}=1$). Hence we have the equation:

$$P(t)=P_0e^{-kt}=10000000e^{-kt}$$

The next thing we want to do is find $k$. That's next part of the problem: decreasing by 1% every year. Double check that our units are consistent (everything is in years) and now know what this means. This means that, after 1 year, the population will be $99\%$ of the initial population. Hence: $$\require{cancel}P(1)=0.99\cancel{P_0}=\cancel{P_0}e^{-k(1)} \\ \ln0.99=-k \\ \therefore k \approx 0.01005$$

From this we can set up the full equation: $$P(t)=10000000e^{-0.01005t}$$

Now, we are looking for the population after $100$ years, $P(100)$. Simple plugging in gives us: $$P(100)\approx 3660323.413$$

Which, in terms of the initial population, is:

$$\frac{P(100)}{P(0)}\approx 36.6032\%$$

Ends up with the answer B, less than half but more than a quarter. This is how your students will solve it.

Answer 2

Use that $(\frac{99}{100})^{100} = (1 -\frac{1}{100})^{100} \approx \frac{1}{e} \approx 0.37$. This way you can do it in your head without any calculations.