Unit 3: Exponential and Logarithmic Functions · ~25 min
Objective
Model and solve real-world problems involving compound interest, half-life, and logistic growth.
Every model in this lesson is built from pieces already in place: the exponential form from 3.1, the constant e from 3.2, and the equation-solving moves from 3.5. What's new is reading a word problem and recognizing which model — and which version of it — applies.
Compound interest
Money compounding n times per year for t years follows:
A = P(1 + \(\frac{r}{n} )^{nt}\)
P — principal (initial amount invested).
r — annual interest rate, as a decimal.
n — number of compounding periods per year (12 for monthly, 4 for quarterly, 365 for daily).
As n → ∞ (compounding continuously), this formula becomes the continuous growth model from 3.2: A = P\(e^{rt}\).
Half-life
A quantity that decays by half every fixed period h follows:
A = \(A_{0}\)(\(\frac{1}{2} )^{\(\frac{t}{h}\)}\)
This is a decay curve from 3.1 with base \(\frac{1}{2}\), where the exponent counts how many half-lives have passed rather than raw time — that's why t is divided by h.
Logistic growth
Exponential growth alone assumes nothing ever slows it down, which is unrealistic for populations limited by food, space, or resources. Logistic growth adds a ceiling, called the carrying capacity c:
P(t) = \(\frac{c}{1 + a\cdot e^{-bt}}\)
Early on, when P is small relative to c, the curve behaves almost exactly like exponential growth. As P approaches c, growth slows and the curve levels off — producing the characteristic S-shape, with y = c as a horizontal asymptote instead of unbounded growth.
Logistic growth levels off at a carrying capacity c — unlike unrestricted exponential growth, which has no ceiling
Worked Example 1 · Compound interest
Problem$1000 is invested at 5% annual interest, compounded quarterly, for 6 years. Find the final amount.
1P = 1000, r = 0.05, n = 4, t = 6: A = 1000(1 + \(\frac{0.05}{4} )^{4(6)}\) = 1000(1.0125\()^{24}\).
2(1.0125\()^{24}\) ≈ 1.3474, so A ≈ 1000(1.3474) ≈ 1347.4.
A ≈ $1347.40
Worked Example 2 · Half-life
ProblemA 50 mg dose of a medication has a half-life of 8 hours. How much remains after 20 hours?
1\(A_{0}\) = 50, h = 8, t = 20: A = 50(\(\frac{1}{2} )^{\(\frac{20}{8}\)}\) = 50(0.5\()^{2.5}\).
2(0.5\()^{2.5}\) ≈ 0.1768, so A ≈ 50(0.1768) ≈ 8.84.
A ≈ 8.84 mg
Worked Example 3 · Logistic growth
ProblemA population follows P(t) = \(\frac{1000}{1 + 9e^{-0.5t}}\), t in years. Find P(0), P(6), and the carrying capacity.
1.$2500 is invested at 4% annual interest, compounded monthly, for 3 years. Find the final amount ((1 + 0.04/12\()^{36}\) ≈ 1.1273).
Show answer
A = 2500(1.1273) ≈ $2818.25.
2.The same $2500 is invested at 4% compounded continuously for 3 years (\(e^{0.12}\) ≈ 1.1275). How does it compare to Problem 1?
Show answer
A = 2500\(e^{0.04(3)}\) = 2500\(e^{0.12}\) ≈ 2500(1.1275) ≈ $2818.75 — slightly more than monthly compounding, since continuous compounding is the limiting case as n → ∞.
3.A 200 g sample has a half-life of 5 years. How much remains after 15 years?
Show answer
\(\frac{t}{h}\) = 15/5 = 3, so A = 200(0.5)³ = 200(0.125) = 25 g.
4.A radioactive isotope decays from 80 g to 20 g in 12 years. Find its half-life h.
Show answer
80(0.5\()^{\(\frac{12}{h}\)}\) = 20 ⇒ (0.5\()^{\(\frac{12}{h}\)}\) = 0.25 = (0.5)², so \(\frac{12}{h}\) = 2. h = 6 years.
5.For the logistic model P(t) = \(\frac{500}{1 + 4e^{-0.3t}}\), find P(0) and the carrying capacity.
Show answer
P(0) = \(\frac{500}{1+4}\) = \(\frac{500}{5}\) = 100. As t → ∞, the e-term vanishes and P(t) → 500, the carrying capacity.