Mathematics Advanced • Year 11 • Module 4 • Lesson 8

Change of Base & Logarithmic Equations

Past-paper-style problems on change of base, solving log/exponential equations, with full domain analysis and an extended response.

Master · Past-Paper Style

1. Short-answer questions

1.1 Evaluate log₄15 to 3 decimal places, showing the change-of-base step. 2 marks Band 3

1.2 Solve log₃(x − 1) + log₃(x + 1) = 1, stating the domain and rejecting any extraneous solution. 3 marks Band 4

1.3 Solve 4^{x + 1} = 7^x, giving the answer in exact form (a single fraction of natural logs) and then to 2 decimal places. 4 marks Band 4-5

Stuck on 1.2? Use the product law, exponentiate, then check that x > 1.

2. Extended response

2.1 The radioactive decay of a sample is modelled by N(t) = N₀ e^−λt, where N₀ is the initial number of atoms, λ > 0 is the decay constant (per year), and t is in years.
(a) Define the half-life T_½ as the time required for N(t) to fall to ½ N₀. Derive a formula for T_½ in terms of λ using natural logarithms.
(b) A geological sample currently contains 12% of its original radioactive isotope. The isotope has half-life 1.25 × 10⁹ years. Find λ exactly, then find the age t of the sample to 3 significant figures.
(c) Show, using the change-of-base formula, that the equivalent base-½ model can be written N(t) = N₀ (½)^{t / T_½}, and explain in one sentence why this form makes the "fraction remaining" calculation in (b) more transparent.
7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — sets ½ N₀ = N₀ e^{−λ T_½}, cancels N₀, and takes ln.

• 1 mark — concludes T_½ = (ln 2)/λ (presented as positive).

Part (b) — 3 marks

• 1 mark — λ = (ln 2) / (1.25 × 10⁹) ≈ 5.545 × 10⁻¹⁰ per year.

• 1 mark — sets 0.12 = e^−λt, takes ln, solves t = −ln(0.12)/λ.

• 1 mark — evaluates to t ≈ 3.82 × 10⁹ years (3 s.f.).

Part (c) — 2 marks

• 1 mark — substitutes λ = (ln 2)/T_½ into e^−λt, shows e^{−(ln 2) t / T_½} = (½)^t/T_½.

• 1 mark — explains: the exponent (t / T_½) is the number of half-lives, so the fraction remaining is just (½)^{that many half-lives}.

Your response:

Stuck on (c)? Use e^{ln 2} = 2, so e^{−(ln 2)k} = 2^−k = (½)^k.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — log₄15 (2 marks)

Sample response. log₄15 = ln 15 / ln 4 ≈ 2.708 / 1.386 ≈ 1.953.

Marking notes. 1 mark — change-of-base step explicit (ln 15 / ln 4 or log 15 / log 4). 1 mark — correct numerical answer to 3 d.p. A response that gives "1.953" with no change-of-base line scores 1/2.

1.2 — log₃(x − 1) + log₃(x + 1) = 1 (3 marks)

Sample response. Domain: x − 1 > 0 and x + 1 > 0, so x > 1. By the product law, log₃[(x − 1)(x + 1)] = 1 ⇒ (x − 1)(x + 1) = 3¹ = 3 ⇒ x² − 1 = 3 ⇒ x² = 4 ⇒ x = ±2. Reject x = −2 (violates x > 1). x = 2.

Marking notes. 1 mark — domain stated as x > 1 before solving. 1 mark — algebra correct to x = ±2. 1 mark — extraneous root x = −2 explicitly rejected with reason. A response that gives "x = ±2" as the final answer scores 2/3 — no domain check.

1.3 — 4^{x + 1} = 7^x (4 marks)

Sample response. Take ln: (x + 1) ln 4 = x ln 7 ⇒ x ln 4 + ln 4 = x ln 7 ⇒ ln 4 = x(ln 7 − ln 4) ⇒ x = ln 4 / (ln 7 − ln 4) = ln 4 / ln(7/4).
Numerical: x ≈ 1.386 / (1.946 − 1.386) = 1.386 / 0.560 ≈ 2.48.

Marking notes. 1 mark — take logs of both sides and apply power law. 1 mark — expand and collect x terms. 1 mark — exact form ln 4 / (ln 7 − ln 4) (or equivalent). 1 mark — numerical 2.48 (2 d.p.). Common error: students drop ln 4 instead of putting it on the right side — costs 1 mark.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a) — half-life formula. By definition, N(T_½) = ½ N₀:

½ N₀ = N₀ e^{−λ T_½}.

Cancel N₀ and take ln of both sides:

ln(½) = −λ T_½ ⇒ −ln 2 = −λ T_½. [1 mark — substitution and ln step.]

T_½ = (ln 2) / λ. ▮ [1 mark — solved positive.]

Part (b) — age of the geological sample.

λ = (ln 2) / (1.25 × 10⁹) ≈ 0.6931 / (1.25 × 10⁹) ≈ 5.545 × 10⁻¹⁰ per year. [1 mark.]

Fraction remaining: N(t)/N₀ = 0.12 ⇒ 0.12 = e^−λt ⇒ ln(0.12) = −λ t ⇒ t = −ln(0.12) / λ = ln(1/0.12) / λ. [1 mark — sets up and solves for t.]

t ≈ ln(8.333) / (5.545 × 10⁻¹⁰) ≈ 2.120 / 5.545 × 10¹⁰ ≈ 3.82 × 10⁹ years (3 s.f.). [1 mark.]

Part (c) — equivalent base-½ form. Starting from N(t) = N₀ e^−λt, substitute λ = (ln 2)/T_½:

e^−λt = e^{−(ln 2) t / T_½} = (e^{ln 2})^{−t / T_½} = 2^{−t / T_½} = (½)^{t / T_½}.

So N(t) = N₀ (½)^{t / T_½}. ▮ [1 mark.]

The exponent t/T_½ counts the number of half-lives elapsed, so the fraction remaining is read off immediately as (½) raised to that count — far more transparent than computing e^−λt. [1 mark — explanation.]

Total: 7/7.

Band descriptors for marker.

Band 3: Derives T_½ = (ln 2)/λ in (a) but does not pursue (b) algebraically; substitutes numbers without setting up the equation. ≈ 2-3 marks.

Band 4: Parts (a) and (b) complete and numerically correct; (c) attempted but does not link the base change to the half-life interpretation. ≈ 4-5 marks.

Band 5: All algebra correct; (c) shows the equivalence and gives a partial explanation but does not name "number of half-lives". ≈ 5-6 marks.

Band 6: Half-life derivation cleanly handles the sign; (b) gives λ and t to 3 s.f. with explicit "fraction remaining = 0.12" step; (c) shows the e^{ln 2} = 2 identity and explicitly names the exponent as "number of half-lives". 7/7.