IB Maths AA SL Topic 1 — Number & Algebra Paper 1 & 2 🎯 Skill ~3 min practice

AA SL Logarithm Laws Skills

Three laws turn any messy log expression into something clean. Multiply becomes add, divide becomes subtract, power becomes a multiplier out front. Get fluent and “simplify” questions take 30 seconds flat.

The Method

log_a(x) = y ⇔ a^y = x log answers the question: “what power gives x?”

Law 1

Product

log(xy) = log x + log y

multiply → add

Law 2

Quotient

log(x/y) = log x − log y

divide → subtract

Law 3

Power

log(xⁿ) = n log x

power → coefficient

Ground rules to memorise

Same base only. All three laws assume the logs share the same base. log₂ + log₃ ≠ anything tidy.
log_a(1) = 0 for any base — because a⁰ = 1.
log_a(a) = 1 — because a¹ = a.
log_a(aⁿ) = n — base and inside cancel.
“ln” means log to base e. Same three laws apply.

Worked examples

WE 1 EASY

Write as a single logarithm: log 6 + log 4 − log 3

apply product law (add → multiply) log 6 + log 4 = log(6 × 4) = log 24apply quotient law (subtract → divide) log 24 − log 3 = log(24 / 3) = log 8log 8 work left to right — combine two terms at a time.

WE 2 MEDIUM

Expand fully: log₂(x³y / z²)

step 1 — split using quotient law log₂(x³y) − log₂(z²)step 2 — split top using product law log₂(x³) + log₂(y) − log₂(z²)step 3 — pull powers down 3 log₂(x) + log₂(y) − 2 log₂(z)3 log₂ x + log₂ y − 2 log₂ z “expand fully” = no products, quotients, or powers left inside any log!

WE 3 HARD

Write as a single logarithm: 2 log 5 + 3 log 2 − log 10

step 1 — push coefficients in (reverse power law) log 5² + log 2³ − log 10 = log 25 + log 8 − log 10step 2 — combine using product law log(25 × 8) − log 10 = log 200 − log 10step 3 — combine using quotient law log(200 / 10) = log 20log 20 when there are coefficients, push them inside as powers FIRST!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Aim to do all 5 in under 5 minutes — these should be quick once the laws click.

Q1 EASY Write as a single log: log 9 + log 4 Show answer ▼Hide answer ▲

product law log 9 + log 4 = log(9 × 4) log 36

Q2 EASY Write as a single log: log 50 − log 5 Show answer ▼Hide answer ▲

quotient law log 50 − log 5 = log(50 / 5) log 10

Q3 MEDIUM Expand fully: log₃(a²b) Show answer ▼Hide answer ▲

product law first log₃(a²) + log₃(b) power law 2 log₃ a + log₃ b

Q4 MEDIUM Write as a single log: 3 log 2 + log 5 Show answer ▼Hide answer ▲

push the 3 inside (power law) log 2³ + log 5 = log 8 + log 5 product law log(8 × 5) = log 40 log 40

Q5 HARD Expand fully: ln(x²√y / z) Show answer ▼Hide answer ▲

step 1 — quotient law ln(x²√y) − ln(z) step 2 — product law on top ln(x²) + ln(√y) − ln(z) step 3 — rewrite √y as y^(1/2), then power law 2 ln x + ½ ln y − ln z roots are powers — √y is y to the half. always rewrite!

⚠ Common mistakes

Trying to combine logs of different bases. log₂(8) + log₃(9) doesn’t simplify with these laws — bases must match.
Treating log(x + y) as log x + log y. The product law is for log(xy), not log(x + y). There’s no law for adding inside a log.
Forgetting to rewrite roots as powers before applying the power law. √x must become x^(1/2).
Power law dropped on the wrong term. 3 log x + log y ≠ 3(log x + log y) — the 3 only attaches to the first log.
Sign errors with the quotient law — log(x/y) is log x minus log y, not plus.
Confusing log x² with (log x)². Power law applies to the first; the second is the log squared and stays as it is.

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Want the theory?

Read the full Logarithms notes for the link to exponentials, the change-of-base formula, and how to use logs to solve a^x = b.

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