Rumor Growth: Average Rates, Solving with Logarithms, and the Inverse Model
A high school has 1,200 students. A rumor begins with 12 students who hear it at 8:00 a.m. The number of students who have heard the rumor t hours later is modeled by N(t) = 12 * 2^(t/3). The principal checks the model at several times and wants to know how fast the rumor is spreading. A teacher calculates the average rate of change over the first 9 hours and gets about 9.33 students per hour; a student calculates the average rate over the interval from t = 3 to t = 6 and gets 8 students per hour. The teacher claims the student must have made an arithmetic error because both intervals are 3 hours long. The principal also asks when 500 students will have heard the rumor and whether the model can be inverted to estimate time from the number of students.
1. Compute both average rates of change and determine whether the student erred. Explain why two 3-hour intervals can have different average rates.
Read the worked answer
Evaluate the model:
N(0) = 12 * 2^0 = 12. N(9) = 12 * 2^(9/3) = 12 * 2^3 = 12 * 8 = 96.
Average rate on [0, 9]: (96 - 12)/9 = 84/9 = 28/3 ≈ 9.33 students per hour.
N(3) = 12 * 2^(3/3) = 12 * 2 = 24. N(6) = 12 * 2^(6/3) = 12 * 2^2 = 48.
Average rate on [3, 6]: (48 - 24)/3 = 24/3 = 8 students per hour.
The student did not err. For an exponential growth model, equal time intervals do not generally produce equal numbers of new students. The interval [0, 9] includes the later, steeper part of the curve near t = 9, which raises the overall average. The interval [3, 6] occurs while N is smaller, so it adds fewer students per hour.
2. Solve N(t) = 500 algebraically. Interpret the answer in context and check whether it is consistent with the school's population.
Read the worked answer
Set 12 * 2^(t/3) = 500. Divide by 12:
2^(t/3) = 500/12 = 125/3 ≈ 41.67.
Take log base 2 of both sides:
t/3 = log2(125/3) = ln(125/3)/ln(2) ≈ 3.7297/0.6931 ≈ 5.381.
So t ≈ 3 * 5.381 ≈ 16.14 hours. About 16.1 hours after 8:00 a.m. is roughly 12:08 a.m. the next day. Since 500 is less than the school population of 1,200, the numerical answer is at least possible, although the model may become less accurate as the rumor spreads.
3. Find the inverse function that gives t in terms of N, state its domain in context, and evaluate the proposed formula t = (1/3)log2(N/12) at N = 500.
Read the worked answer
Start with N = 12 * 2^(t/3). Divide by 12: N/12 = 2^(t/3). Take log base 2: log2(N/12) = t/3. Multiply by 3:
t = 3 log2(N/12).
In context, t >= 0 corresponds to N >= 12, so the relevant domain is N >= 12 (mathematically N > 0).
At N = 500, this gives t = 3 log2(125/3) ≈ 16.1 hours. The proposed formula t = (1/3)log2(N/12) gives about 1.79 hours, which is far too small. The error is dividing by 3 instead of multiplying by 3 after taking the logarithm.