Modeling with Exponentials & Logs
- Apply half-life reasoning to a decaying quantity
- Choose an appropriate model for a real-world growth or decay scenario
- Solve an exponential equation using logarithms
Half-life: repeated halving
Many decaying quantities — radioactive isotopes, drug concentrations — lose a fixed fraction over each fixed period. A half-life is the time for a quantity to fall to half its value. After one half-life, 80 g becomes 40 g; after another, 20 g; after another, 10 g. The model is A = A₀·(1/2)^(t/T), where T is the half-life.
Solving for a variable in the exponent
When the unknown sits in the exponent, as in 3ˣ = 20, you cannot isolate it with ordinary algebra — you need a logarithm. Take the log of both sides and use the power rule to bring the exponent down: log(3ˣ) = log 20 becomes x·log 3 = log 20, so x = log 20 / log 3. This is the standard technique for any exponential equation.
Solve 3ˣ = 20 for x.
- 1.Take the logarithm of both sides: log(3ˣ) = log(20).
- 2.Apply the power rule to bring the exponent down: x·log 3 = log 20.
- 3.Divide both sides by log 3: x = log 20 / log 3.
- 4.Numerically, log 20 ≈ 1.301 and log 3 ≈ 0.477, so x ≈ 2.73.
Carbon-14 decays with a fixed half-life. A sample starts at 80 g; how much remains after exactly one half-life?
A half-life halves the amount each period — it does not subtract a fixed number of grams. After two half-lives an 80 g sample is 80 → 40 → 20 g, not 80 → 40 → 0 g.
To solve 3ˣ = 20 for x, which method is appropriate?
Pick the log base that simplifies the work: natural log (ln) pairs with base e, common log with base 10, but log₃ 20 = log 20 / log 3 works for any base via the change-of-base formula. State the exact answer before rounding.
Answer the 2 checkpoints as you read.
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