Population Growth of a Lab Culture
A biology class tracks a bacterial culture that doubles every 3 hours. At time $t = 0$ hours the population is 500 cells. The class wants to model the population with an exponential function and use logarithms to predict when the culture will reach certain thresholds. A student proposes the model $P(t) = 500 \cdot 2^{t/3}$ and another proposes $P(t) = 500 e^{kt}$. The teacher also asks them to consider a competing culture that starts at 800 cells but doubles every 5 hours.
1. Show the two models are equivalent and find k.
Read the worked answer
Set $500 \cdot 2^{t/3} = 500 e^{kt}$. Then $2^{t/3} = e^{kt}$. Taking natural logs: $(t/3)\ln 2 = kt \Rightarrow k = \frac{\ln 2}{3} \approx \frac{0.6931}{3} \approx 0.2310$ per hour. So $P(t) = 500 e^{0.231t}$ matches the doubling model.
2. When does the first culture reach 20,000 cells?
Read the worked answer
Solve $500 \cdot 2^{t/3} = 20000 \Rightarrow 2^{t/3} = 40$. Take $\log_2$: $t/3 = \log_2 40 = \log_2 8 + \log_2 5 = 3 + 2.322 = 5.322$. So $t \approx 15.97$ hours, about 16 hours. Check: $2^{16/3} \approx 2^{5.333} \approx 40.3$, times 500 ≈ 20,160, consistent.
3. At what time do the two cultures have equal populations?
Read the worked answer
Set $500 \cdot 2^{t/3} = 800 \cdot 2^{t/5}$. Divide: $2^{t/3 - t/5} = 800/500 = 1.6$. Exponent: $t(1/3 - 1/5) = t(2/15)$. So $2^{2t/15} = 1.6$. Take logs: $(2t/15)\ln 2 = \ln 1.6 \Rightarrow t = \frac{15 \ln 1.6}{2 \ln 2} = \frac{15(0.4700)}{2(0.6931)} \approx \frac{7.050}{1.386} \approx 5.09$ hours. At that time both are near $500 \cdot 2^{1.696} \approx 500(3.24) \approx 1620$ cells.