Two Exponential Bacteria Models and a Misleading Linear Approximation
A laboratory starts with 500 bacteria in a controlled culture. After 3 hours the count is 2000, and the growth is modeled as exponential, N(t) = 500·4^(t/3), where t is in hours. A second culture starts with 800 bacteria and doubles every 2 hours, so M(t) = 800·2^(t/2). The technician needs to know when the first culture reaches 10,000 bacteria, and when the two cultures have equal counts. A colleague proposes replacing the exponential model by a straight line through the two measured points (0, 500) and (3, 2000), arguing that the prediction will be close for the first 5 hours. The technician suspects that the linear model will be far too low near t = 6 and wants a quantitative comparison. The laboratory's protocol requires answers to the nearest 0.1 hour and a check that any logarithm used is defined. The two cultures are in separate rooms and do not interact, so the models can be compared directly.
1. Solve N(t) = 10,000. Give the exact logarithmic form and the answer to the nearest 0.1 hour.
Read the worked answer
Set 500·4^(t/3) = 10,000. Divide by 500: 4^(t/3) = 20. Since 4^(t/3) is positive for all real t, the logarithm is defined. Taking log base 4: t/3 = log_4(20), so t = 3 log_4(20).
Using natural logarithms, t = 3 ln(20)/ln(4) ≈ 3(2.9957)/(1.3863) ≈ 6.48 hours. To the nearest 0.1 hour, t ≈ 6.5 hours.
Check: N(6.48) = 500·4^(2.16) ≈ 500·19.99 ≈ 9995, consistent with the target.
2. Find the time when the two cultures have equal counts, to the nearest 0.1 hour.
Read the worked answer
Set 500·4^(t/3) = 800·2^(t/2). Write 4^(t/3) = 2^(2t/3). Dividing by 500 gives 2^(2t/3) = 1.6·2^(t/2). Divide both sides by 2^(t/2): 2^(2t/3 − t/2) = 1.6.
The exponent simplifies: 2t/3 − t/2 = 4t/6 − 3t/6 = t/6. Hence 2^(t/6) = 1.6. Taking log base 2: t/6 = log_2(1.6), so t = 6 log_2(1.6) = 6 ln(1.6)/ln(2) ≈ 6(0.4700)/(0.6931) ≈ 4.07 hours. To the nearest 0.1 hour, t ≈ 4.1 hours.
Check approximately: N(4.07) ≈ 500·4^(1.357) ≈ 3290 and M(4.07) ≈ 800·2^(2.035) ≈ 3290.
3. Compare the linear model through (0, 500) and (3, 2000) with the exponential model at t = 6. Is the colleague correct that the linear model is close for the first 5 hours?
Read the worked answer
The linear model has slope (2000 − 500)/3 = 500, so L(t) = 500 + 500t.
At t = 6, L(6) = 3500, while N(6) = 500·4^(6/3) = 500·4^2 = 8000. The linear model is low by 4500 bacteria, more than a factor of two.
At t = 5, L(5) = 3000. The exponential model gives N(5) = 500·4^(5/3) ≈ 500·10.08 ≈ 5040. The linear model is already low by about 2040 bacteria, roughly 40% below the exponential value. So the colleague's claim is not reliable even by t = 5; the two models agree only at the two measured points t = 0 and t = 3.