A town's population is 5000 and grows to 8000 in 4 years. Find k, then predict the population after 10 years.

1. Model: N = 5000e^kt. At t = 4, N = 8000: 8000 = 5000e^4k
2. e^4k = 1.6, so 4k = ln 1.6 and k = ln 1.64 = 0.1175… (keep this in the calculator)
3. At t = 10: N = 5000e^10k = 5000 × e^1.175… = 16,190.8…
4. Population after 10 years ≈ 16,200 (about 16,190 — round sensibly for people).

At what rate is that population growing after 10 years?

1. dNdt = kN, so use the value of N at t = 10, not N₀.
2. dNdt = 0.1175… × 16,190.8… = 1902.4…
3. The population is growing at about 1900 people per year at t = 10.

A radioactive sample of 100 g has a half-life of 30 years. Find k, the mass after 50 years, and the time until 10 g remains.

1. Half-life: 50 = 100e^30k, so e^30k = 12 and k = ln (1/2)30 = −ln 230 = −0.02310…
2. After 50 years: M = 100e^{−0.02310… × 50} = 100e^−1.155… = 31.5 g (to 1 decimal place)
3. Time to 10 g: 10 = 100e^kt, so e^kt = 0.1 and t = ln 0.1k = −2.3026…−0.02310… = 99.7 years
4. Sense check: 10 g is between 3 half-lives (12.5 g at 90 years) and 4 half-lives (6.25 g at 120 years). Yes.

Cooling: coffee at 80°C is left in a 22°C room. Its temperature follows T = 22 + 58e^−kt (t in minutes). After 5 minutes it is 60°C. When will it reach 40°C?

1. At t = 5: 60 = 22 + 58e^−5k, so 58e^−5k = 38 and e^−5k = 3858
2. −5k = ln (3858) = −0.4229…, so k = 0.08457…
3. For T = 40: 40 = 22 + 58e^−kt, so e^−kt = 1858
4. −kt = ln (1858) = −1.1701…, so t = 1.1701…0.08457… = 13.8 minutes

Write N = 3 × 2^t in the form N = N₀e^kt

1. 2 = e^{ln 2}, so 2^t = (e^{ln 2})^t = e^{(ln 2)t}
2. N = 3e^kt with k = ln 2 ≈ 0.693. The quantity doubles every unit of time, which matches t = ln 2k = 1.