Mathematics Advanced • Year 11 • Module 4 • Lesson 7

Laws of Logarithms

Past-paper-style problems on evaluating, expanding, condensing and proving identities with the three log laws.

Master · Past-Paper Style

1. Short-answer questions

1.1 Without a calculator, evaluate log₂48 − log₂3. Show each step and name the law used. 2 marks Band 3

1.2 Express log_a(x³ √y / z²) as a sum/difference of log_ax, log_ay, log_az. 3 marks Band 4

1.3 Given log₁₀2 = p and log₁₀3 = q, find log₁₀(15) in terms of p and q only. 3 marks Band 4

Stuck on 1.3? Write 15 in terms of 2, 3 and 10.

2. Extended response

2.1 (a) Prove the quotient law: for x, y > 0 and a > 0 with a ≠ 1, log_a(x/y) = log_ax − log_ay.
(b) Prove the power law: for x > 0 and any real n, log_a(xⁿ) = n log_ax.
(c) Hence simplify the expression E = log_a(a³ x²) − 2 log_a(ax) + log_a(x/a) as a single logarithm of a function of x and a, and show that E is independent of x.
7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — sets log_ax = m, log_ay = n, writes x = a^m, y = aⁿ, and forms x/y = a^m−n.

• 1 mark — takes log_a and concludes log_a(x/y) = m − n = log_ax − log_ay.

Part (b) — 2 marks

• 1 mark — sets log_ax = m, so x = a^m, then xⁿ = (a^m)ⁿ = a^mn.

• 1 mark — takes log_a and concludes log_a(xⁿ) = mn = n log_ax.

Part (c) — 3 marks

• 1 mark — applies the power law to each term: 3 + 2 log_ax − 2(1 + log_ax) + (log_ax − 1).

• 1 mark — collects the log_ax terms (2 − 2 + 1 = 1) and the constants (3 − 2 − 1 = 0).

• 1 mark — states E = log_ax explicitly. (The "independent of x" claim is false: a top response notices that E still depends on x — sees the trap and corrects course.)

Your response:

Stuck on (c)? Use log_a(a) = 1 to simplify each constant chunk.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — log₂48 − log₂3 (2 marks)

Sample response. By the quotient law, log₂48 − log₂3 = log₂(48/3) = log₂16. Since 16 = 2⁴, log₂16 = 4.

Marking notes. 1 mark — names/uses the quotient law and simplifies to log₂16. 1 mark — converts log₂16 to 4 by recognising 16 = 2⁴. A response that goes "log₂48 ≈ 5.585, log₂3 ≈ 1.585, difference = 4" scores 1/2 — no exact-arithmetic credit because no log law is named.

1.2 — Expand log_a(x³ √y / z²) (3 marks)

Sample response. Quotient law: = log_a(x³ √y) − log_a(z²). Product law on the first term: = log_a(x³) + log_a(y^1/2) − log_a(z²). Power law on each: = 3 log_ax + ½ log_ay − 2 log_az.

Marking notes. 1 mark each for the quotient, product, and power applications. Common error: students convert √y to y but forget the ½ coefficient — costs 0.5 on the power-law mark.

1.3 — log₁₀15 from log₁₀2 = p, log₁₀3 = q (3 marks)

Sample response. Write 15 = (3 × 10)/2. Then log₁₀15 = log₁₀3 + log₁₀10 − log₁₀2 = q + 1 − p, i.e. 1 + q − p.
(Alternative: 15 = 3 × 5 = 3 × 10/2 gives the same result.)

Marking notes. 1 mark — rewrites 15 in terms of 2, 3 and 10. 1 mark — applies product/quotient laws. 1 mark — substitutes log₁₀10 = 1 and arrives at 1 + q − p. Common error: students try 15 = 3 + 12 or attempt a sum-of-logs decomposition — log of a sum is not a sum of logs (Trap 01 from the lesson).

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

Sample Band-6 response.

Part (a) — quotient law. Let log_ax = m and log_ay = n. By the definition of logarithm, a^m = x and aⁿ = y. [1 mark.]
Form the quotient: x/y = a^m/aⁿ = a^m−n (index law). Taking log_a of both sides:

log_a(x/y) = m − n = log_ax − log_ay. ▮ [1 mark.]

Part (b) — power law. Let log_ax = m, so x = a^m. For any real n, xⁿ = (a^m)ⁿ = a^mn (index law). [1 mark.]
Taking log_a: log_a(xⁿ) = mn = n log_ax. ▮ [1 mark.]

Part (c) — simplify E. Use log_aa = 1 throughout.

log_a(a³ x²) = log_a(a³) + log_a(x²) = 3 + 2 log_ax [product law + power law]

2 log_a(ax) = 2 [log_aa + log_ax] = 2 [1 + log_ax] = 2 + 2 log_ax

log_a(x/a) = log_ax − log_aa = log_ax − 1 [1 mark for the three rewrites.]

Substitute:

E = (3 + 2 log_ax) − (2 + 2 log_ax) + (log_ax − 1)

= (3 − 2 − 1) + (2 − 2 + 1) log_ax

= 0 + 1 · log_ax = log_ax. [1 mark — collects coefficients correctly.]

Comment on the claim. The question asks to "show E is independent of x". The simplification gives E = log_ax, which does depend on x. The correct conclusion is therefore: E = log_ax, and the original claim of independence is not supported. A top-band response notices this and states it explicitly. [1 mark — identifies that the "independent of x" claim is wrong and states the correct relationship.]

Total: 7/7.

Band descriptors for marker.

Band 3: Attempts (a) and (b) with index-law statements but does not take log_a at the end; in (c) simplifies one or two pieces correctly but does not combine. ≈ 2-3 marks.

Band 4: Completes (a) and (b) correctly; in (c) reaches E = log_ax but accepts the false claim "E is independent of x". ≈ 4-5 marks.

Band 5: All algebra correct; (c) reaches E = log_ax and queries the claim but does not state explicitly that it is wrong. ≈ 5-6 marks.

Band 6: Proofs in (a) and (b) include the "x = a^m" setup and the closing log_a step; (c) correctly simplifies and explicitly rejects the "independent of x" claim with the corrected statement E = log_ax. 7/7.