Engage your students with effective distance learning resources. ACCESS RESOURCES>>

Rumors


Alignments to Content Standards: F-LE.A.2

Task

Susanna heard some exciting news about a well-known celebrity.

Within a day she told 4 friends who hadn't heard the news yet.

By the next day each of those friends told 4 other people who also hadn't yet heard the news.

By the next day each of those people told four more, and so on.

  1. Assume the rumor continues to spread in this manner. Let $N$ be the function that assigns to $d$ the number of people who hear the rumor on the $d^{\rm th}$ day. Write an expression for $N(d)$.
  2. On which day will at least 100,000 people hear the rumor for the first time?
  3. How many people will hear the rumor for the first time on the 20th day?
  4. Is the answer to (c) realistic? Explain your reasoning.

IM Commentary

This problem is an exponential function example. The other tasks in this set illustrate F.BF.1a in the context of linear (Kimi and Jordan), quadratic (Skeleton Tower), and rational (Summer Intern) functions.

Solutions

Solution: Solution 1

  1. On day 1, Susanna told four friends, so $4\cdot 1$ people heard the rumor for the first time. On day 2, those four friends each told 4 people, so $4\cdot 4=4^2=16$ people heard the rumor for the first time. On day 3, each of those 16 people told 4 others, so $4\cdot 16=4\cdot 4^2=4^3=64$ people heard the rumor for the first time. In general, $N(d) = 4^d$ gives the number of people who learn of the rumor on day, $d$.
  2. The problem asks for $d$ given $N(d) = 100,\!000$. We let $$N(d) = 4^d = 100,\!000.$$ So $d = \log_4(100,000)$ is an exact answer. But most calculators do not compute logarithms with base 4. To obtain a decimal approximation, we take advantage of one of the properties of logarithms, namely that $\log_b(A^p)= p\log_b(A)$: \begin{align} \log 4^d &= \log 100,\!000\\ d\cdot\log 4 &= 5\\ d &=5/\log 4 \approx 8.3 \text{ days} \end{align} So, the $9$th day will be the first day that at least $100,\!000$ people will hear the rumor.
  3. The problem asks for $N(d)$ when $d=20$. We compute $$N(20) =4^{20} = 1,\!099,\!511,\!627,\!776.$$ So according to part (a), on the 20th day, $1,\!099,\!511,\!627,\!776$ people will hear the rumor for the first time.
  4. The answer in part (c) exceeds the number of people on Earth, so it is unrealistic. Eventually, the number of people hearing a rumor for the first time must cease to increase, because the number of people is finite.

Solution: Approximate Solution

  1. As in the previous solution, $N(d) = 4^d$ tells the number of people who learn of the rumor on day $d$, according to this model. Equivalently, since $4=2^2$, we write $N(d) = (2^2)^d = 2^{2d}$, so that we can think in powers of 2.
  2. For computer science and for thinking about exponential growth (based on doubling time) a very useful fact is that $2^{10} = 1,024.$ Let's approximate that as $2^{10} \approx 1000$. Because $4^5=2^{10}$, we can estimate $1,000$ people hear the rumor on the $5$th day, so that $64,000$ people hear it on the $8$th day, and $256,000$ people hear it on the $9$th day, the first day the number exceeds $100,000$ people.
  3. From the estimate that $2^{10} \approx 1000$, we have that $2^{20} = 2^{10} \times 2^{10} \approx 1000 \times 1000 = 1$million and similarly $2^{30} \approx 1$billion. Because $4^{15} = 2^{30}$, we can estimate that $1$ billion people would hear the rumor on the $15$th day. That means that 4 billion would hear it on the $16$th day, and on the $17$th day the number would exceed the population of the planet (which is estimated to reach $7$ billion on October $31$, $2011$). Realistically, there is no reason to proceed to $20$ days.
  4. The previous answer indicates that this model is not realistic for $20$ days. And none of these estimates consider the real-world difficulty of accounting for the people who have already heard the rumor.