Logarithmic Model and Doubling Time
A bacterial culture grows so that population P after t hours is P(t) = 400e^(0.35t) for t ≥ 0. A lab protocol says the culture must be subcultured before the population reaches 5000. A student computes the time to reach 5000 and separately the doubling time, and gets two numbers that seem inconsistent; the student also wonders whether a linear model through (0, 400) and (10, P(10)) would be acceptable for short-term planning. You must resolve the timing, the doubling time, and judge the linear approximation.
1. Find the time to reach 5000 and the doubling time, and show they are consistent.
Read the worked answer
Reach 5000: 400e^(0.35t) = 5000 → e^(0.35t) = 12.5 → 0.35t = ln 12.5 ≈ 2.5257 → t ≈ 7.22 hours. Doubling: e^(0.35T) = 2 → 0.35T = ln 2 ≈ 0.6931 → T ≈ 1.98 hours. Consistency check: 12.5 = 400→5000 is a factor of 12.5; 2^(n) = 12.5 gives n = ln12.5/ln2 ≈ 3.64 doublings; 3.64 × 1.98 ≈ 7.21 hours, matching. The two numbers are not inconsistent; they answer different questions (absolute threshold vs proportional growth).
2. Evaluate whether a linear model through (0,400) and (10,P(10)) is acceptable for planning over the first 10 hours.
Read the worked answer
P(10) = 400e^3.5 ≈ 400(33.115) ≈ 13246. Linear slope = (13246 − 400)/10 = 1284.6 per hour, so L(t) = 400 + 1284.6t. Compare at t = 5: exponential P(5) = 400e^1.75 ≈ 400(5.755) = 2302; linear L(5) = 400 + 6423 = 6823. The linear model overestimates by nearly a factor of 3 at the midpoint because exponential growth is convex. Judgement: the linear model is poor for planning thresholds - it would trigger subculture far too early (L(t)=5000 at t≈3.6 h vs true 7.2 h). A linear approximation is only defensible over very short intervals where the curve is nearly straight, and even then should be justified by comparing midpoint values.