Doubling Time and Half-Life in a Cooling and Growth Model
A biology lab tracks two populations in separate containers. Population P grows according to P(t) = 250 · 2^(t/6), where t is in hours. Population Q decays according to Q(t) = 1800 · (1/2)^(t/9). A technician records that after 12 hours, P is about 1000 and Q is 450. A supervisor challenges the team to find when the two populations are equal, to express both models in base e, and to explain why the equation P(t) = Q(t) has exactly one solution. The lab's protocol requires reporting any crossover time to the nearest 0.1 hour.
1. Solve P(t) = Q(t) exactly in terms of logarithms, then give the crossover time to the nearest 0.1 hour.
Read the worked answer
Set 250 · 2^(t/6) = 1800 · (1/2)^(t/9).
Rewrite (1/2)^(t/9) = 2^(-t/9). Then: 250 · 2^(t/6) = 1800 · 2^(-t/9). Divide by 250: 2^(t/6) = 7.2 · 2^(-t/9). Multiply both sides by 2^(t/9): 2^(t/6 + t/9) = 7.2.
t/6 + t/9 = (3t + 2t)/18 = 5t/18, so 2^(5t/18) = 7.2.
Take log base 2: 5t/18 = log₂ 7.2, so t = (18/5) log₂ 7.2.
Numerically, log₂ 7.2 = ln 7.2 / ln 2 ≈ 1.97408 / 0.69315 ≈ 2.8480. Then t ≈ 3.6 · 2.8480 ≈ 10.25 hours, which rounds to 10.3 hours.
Check: P(10.25) = 250 · 2^(10.25/6) = 250 · 2^1.7083 ≈ 250 · 3.268 ≈ 817. Q(10.25) = 1800 · 2^(-10.25/9) = 1800 · 2^(-1.1389) ≈ 1800 · 0.4543 ≈ 818. The values agree, confirming the crossover time.
2. Rewrite both models in the form y = A e^(kt), state the growth/decay constants, and justify why there is exactly one crossover.
Read the worked answer
For P: 2^(t/6) = e^((t/6) ln 2), so P(t) = 250 e^((ln 2/6) t). The constant is k_P = ln 2 / 6 ≈ 0.1155 per hour (growth).
For Q: (1/2)^(t/9) = e^((t/9) ln(1/2)) = e^(-(ln 2/9) t), so Q(t) = 1800 e^(-(ln 2/9) t). The constant is k_Q = -ln 2 / 9 ≈ -0.0770 per hour (decay).
Crossover: 250 e^(0.1155t) = 1800 e^(-0.0770t). Divide: e^(0.1155t + 0.0770t) = 7.2, so e^(0.1925t) = 7.2, giving t = ln 7.2 / 0.1925 ≈ 1.9741 / 0.1925 ≈ 10.25 hours, matching the earlier result.
Uniqueness: define f(t) = P(t) - Q(t) = 250 e^(0.1155t) - 1800 e^(-0.0770t). Both terms are continuous and monotonic in t on [0, ∞): the first is strictly increasing and the second is strictly decreasing (so -Q is strictly increasing). Therefore f is strictly increasing. Since f(0) = 250 - 1800 = -1550 < 0 and f(t) → +∞ as t → ∞, the intermediate value theorem guarantees exactly one positive root. Hence there is exactly one crossover time.