Mathematics Advanced • Year 11 • Module 4 • Lesson 7

Laws of Logarithms

Build procedural fluency in the product, quotient and power laws — expand, condense and evaluate logarithmic expressions.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the three laws of logarithms:

Product law: log_a(xy) = ____________________

Quotient law: log_a(x/y) = ____________________

Power law: log_a(xⁿ) = ____________________

Q1.2 Each law requires x > ____________ and y > ____________, with base a > ____________ and a ≠ ____________.

Q1.3 Write "true" or "false":

A. log_a(x + y) = log_ax + log_ay ⇒ ____________

B. log_ax · log_ay = log_a(xy) ⇒ ____________

C. (log_ax)ⁿ = n log_ax ⇒ ____________

Stuck? Revisit lesson § Key Terms and § Common errors.

2. Worked example — expand log_a(x² √y / z³)

Follow each step. Each line applies exactly one log law.

Problem. Express log_a(x² √y / z³) as a sum/difference of log_ax, log_ay, and log_az.

Step 1 — Quotient law (top − bottom).

= log_a(x² √y) − log_a(z³)

Reason: log_a(P/Q) = log_aP − log_aQ.

Step 2 — Product law on the first term.

= log_a(x²) + log_a(√y) − log_a(z³)

Reason: log_a(MN) = log_aM + log_aN.

Step 3 — Rewrite √y as y^1/2.

= log_a(x²) + log_a(y^1/2) − log_a(z³)

Reason: prepare each argument for the power law.

Step 4 — Power law on each term.

= 2 log_ax + ½ log_ay − 3 log_az

Reason: log_a(xⁿ) = n log_ax.

Conclusion. log_a(x² √y / z³) = 2 log_ax + ½ log_ay − 3 log_az.

3. Faded example — fill in the missing steps

Condense 3 log_ax − log_ay + ½ log_az to a single log. 4 marks

Step 1 — Power law (move coefficients to exponents):

= log_a(x^____) − log_a(____) + log_a(z^____)

Step 2 — Combine the addition (product law):

= log_a(x³ · ____________) − log_a(y)

Step 3 — Combine the subtraction (quotient law):

= log_a( ____________ / ____________ )

Conclusion. 3 log_ax − log_ay + ½ log_az = log_a( ________________ ).

Stuck? Apply power law first, then group the positive terms (sum ⇒ product), then handle the subtraction (difference ⇒ quotient).

4. Graduated practice — evaluate, expand, condense

Show the law used at each step. Where a numerical answer is asked, give an integer or a simple fraction.

Foundation — direct application of one law (4 questions)

Q	Expression	Simplified form
4.1 1	log₅2 + log₅3
4.2 1	log₂24 − log₂3
4.3 1	log₃(9⁴)
4.4 1	log₂8 + log₂4 − log₂2

Standard — multiple laws combined (6 questions)

Show at least one line of working applying each law.

4.5 Evaluate log₃6 + log₃(3/2). 2 marks

4.6 Expand log_a(x³ y²) in terms of log_ax and log_ay. 2 marks

4.7 Expand log_a(x² y / √z) in terms of log_ax, log_ay, log_az. 2 marks

4.8 Condense 2 log_ax + 3 log_ay − log_az to a single log. 2 marks

4.9 Evaluate log₂80 − log₂5. 2 marks

4.10 Given log₁₀2 = p, express log₁₀50 in terms of p. 2 marks

Extension — proof-style and substitution (2 questions)

4.11 Given log₁₀2 = p and log₁₀3 = q, express log₁₀(60/√5) in terms of p and q only. (Use the fact 5 = 10/2.) 3 marks

4.12 Using the product law and the fact a^m · aⁿ = a^m+n, prove that log_a(xy) = log_ax + log_ay, where x, y > 0. 3 marks

Stuck on 4.12? Let log_ax = m and log_ay = n, then convert each to exponential form.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I combined the two sums using the product law to get log₅(2·3) = log₅6.

For 4.2 I combined the difference using the quotient law to get log₂(24/3) = log₂8 = 3.

For 4.3 I used the power law plus the fact 9 = 3², so log₃(9⁴) = 4 log₃9 = 4 · 2 = 8.

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Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — The three laws

Product: log_a(xy) = log_ax + log_ay. Quotient: log_a(x/y) = log_ax − log_ay. Power: log_a(xⁿ) = n log_ax.

Q1.2 — Domain conditions

x > 0 and y > 0; base a > 0 and a ≠ 1. Logs are only defined for positive arguments and positive bases ≠ 1.

Q1.3 — Pitfall tests

A. False — there is no law for log of a sum. B. False — product of two logs is not the log of a product. C. False — the power must be on the argument, not on the log.

Q3 — Faded condensing

Step 1: x³, y, z^1/2. Step 2: = log_a(x³ · √z) − log_a(y). Step 3: = log_a(x³√z / y). Conclusion: log_a(x³ √z / y).

Q4.1 — log₅2 + log₅3

= log₅(2 × 3) = log₅6. (Product law.)

Q4.2 — log₂24 − log₂3

= log₂(24/3) = log₂8 = log₂(2³) = 3.

Q4.3 — log₃(9⁴)

= 4 log₃9 = 4 × 2 = 8. (Power law; log₃9 = 2 since 3² = 9.)

Q4.4 — log₂8 + log₂4 − log₂2

= 3 + 2 − 1 = 4. (Or: = log₂(8 × 4 / 2) = log₂16 = 4.)

Q4.5 — log₃6 + log₃(3/2)

= log₃(6 × 3/2) = log₃9 = 2.

Q4.6 — log_a(x³ y²)

= log_a(x³) + log_a(y²) = 3 log_ax + 2 log_ay.

Q4.7 — log_a(x² y / √z)

= log_a(x²y) − log_a(z^1/2) = log_a(x²) + log_ay − ½ log_az = 2 log_ax + log_ay − ½ log_az.

Q4.8 — Condense 2 log_ax + 3 log_ay − log_az

= log_a(x²) + log_a(y³) − log_az = log_a(x²y³) − log_az = log_a(x² y³ / z).

Q4.9 — log₂80 − log₂5

= log₂(80/5) = log₂16 = 4.

Q4.10 — log₁₀50 in terms of p = log₁₀2

50 = 100/2, so log₁₀50 = log₁₀100 − log₁₀2 = 2 − p, i.e. 2 − p.

Q4.11 — log₁₀(60/√5) in terms of p = log₁₀2, q = log₁₀3

log₁₀60 = log₁₀(2² × 3 × 5) = 2p + q + log₁₀5. Now log₁₀5 = log₁₀(10/2) = 1 − p. So log₁₀60 = 2p + q + 1 − p = p + q + 1.
log₁₀(√5) = ½ log₁₀5 = ½(1 − p).
log₁₀(60/√5) = (p + q + 1) − ½(1 − p) = p + q + 1 − ½ + p/2 = (3p/2) + q + ½.

Q4.12 — Proof of product law

Let log_ax = m and log_ay = n. By definition, a^m = x and aⁿ = y.
Then xy = a^m · aⁿ = a^m+n (index law).
Taking log_a of both sides: log_a(xy) = m + n = log_ax + log_ay. ▮

1. Quick recall

2. Worked example — expand loga(x² √y / z³)

3. Faded example — fill in the missing steps

4. Graduated practice — evaluate, expand, condense

Foundation — direct application of one law (4 questions)

Standard — multiple laws combined (6 questions)

Extension — proof-style and substitution (2 questions)

5. Self-check the easy 3

Q1.1 — The three laws

Q1.2 — Domain conditions

Q1.3 — Pitfall tests

Q3 — Faded condensing

Q4.1 — log52 + log53

Q4.2 — log224 − log23

Q4.3 — log3(94)

Q4.4 — log28 + log24 − log22

Q4.5 — log36 + log3(3/2)

Q4.6 — loga(x³ y²)

Q4.7 — loga(x² y / √z)

Q4.8 — Condense 2 logax + 3 logay − logaz

Q4.9 — log280 − log25

Q4.10 — log1050 in terms of p = log102

Q4.11 — log10(60/√5) in terms of p = log102, q = log103