IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 ~9 min read

Laws of Logarithms

The log laws are three short rules that let you combine, split, and rearrange logarithms — exactly mirroring the index laws you already know. They’re the toolkit you’ll use to simplify expressions, solve equations, and squeeze cleaner answers out of the GDC. The three core rules are in the formula booklet; the useful corollaries are not, so those need memorising.

📘 What you need to know

The three core laws: product (logs add), quotient (logs subtract), and power (the exponent comes out as a coefficient).
Useful corollaries — log_a(1) = 0, log_a(a) = 1, log_a(a^k) = k, and a^log_a(x) = x.
The same laws apply to ln (natural log) — same rules, just base e.
Change of base: log_a(x) = log_b(x) / log_b(a) — useful for non-calculator questions.
The big trap: log_a(x + y) ≠ log_a(x) + log_a(y). The laws work on products and quotients, not sums.
When solving log equations, always check for invalid solutions at the end — anything that puts a negative number or zero inside any log gets rejected.

The three core laws

Each log law is the mirror image of an index law. That’s not a coincidence — they’re the same statement, written in different notation.

Log law

In plain English

Mirror of which index law?

log_a(xy) = log_a(x) + log_a(y)

Log of a product = sum of logs.

a^m × aⁿ = a^m+n

log_a(x/y) = log_a(x) − log_a(y)

Log of a quotient = difference of logs.

a^m ÷ aⁿ = a^m−n

log_a(x^m) = m · log_a(x)

A power inside the log comes out as a coefficient in front.

(aⁿ)^m = a^mn

All three core laws log_a(xy) = log_a(x) + log_a(y)
log_a(xy) = log_a(x) − log_a(y)
log_a(x^m) = m · log_a(x) ✓ in formula booklet (where a, x, y > 0)

🤔 Why does the product law work?

Say x = a^m and y = aⁿ. Then xy = a^m+n. Taking log_a of each: log_a(x) = m, log_a(y) = n, and log_a(xy) = m + n. So log_a(xy) = log_a(x) + log_a(y). The other two laws fall out the same way from the corresponding index laws.

Useful results that fall out of the laws

These come for free if you’ve understood the flip identity from the previous note. They’re worth memorising because they appear constantly in exam working.

Result

Why it’s true

Example

log_a(1) = 0

Because a⁰ = 1 — the only power that gives 1.

log₇(1) = 0

log_a(a) = 1

Because a¹ = a.

log₅(5) = 1

log_a(a^k) = k

By the power law: k · log_a(a) = k · 1 = k.

log₃(3⁵) = 5

a^log_a(x) = x

Logs and powers with the same base undo each other.

4^log₄(15) = 15

log_a(1x) = −log_a(x)

Quotient law: log(1) − log(x) = 0 − log(x).

log₂(½) = −log₂(2) = −1

For natural log (ln): ln(1) = 0 | ln(e) = 1 | ln(e^x) = x | e^ln(x) = x. The same patterns, with base e.

The three core laws are in the formula booklet. The useful results above are not — but they’re consequences, not new rules. If you forget one mid-exam, derive it from the laws you do have.

The big trap — sums don’t split

The single most common error on this topic is splitting a sum inside a log. It does not work.

⚠ NOT A LAW log_a(x + y) ≠ log_a(x) + log_a(y)

Quick test with numbers: log₁₀(10 + 90) = log₁₀(100) = 2. But log₁₀(10) + log₁₀(90) = 1 + 1.954… ≈ 2.954 — completely different. The laws apply to products and quotients, never sums.

Change of base

Sometimes a logarithm is awkward in its natural base — for instance log₄(8) doesn’t have a “nice” calculator button, and on a non-calculator paper you can’t enter it directly. Change of base lets you rewrite any log using whatever base is convenient.

Change of base log_a(x) = log_b(x)log_b(a) ✓ in formula booklet

🤔 Why does change of base work?

Let y = log_a(x), so a^y = x. Take log_b of both sides: log_b(a^y) = log_b(x). Apply the power law on the left: y · log_b(a) = log_b(x). Divide and you get y = log_b(x) / log_b(a).

Choose the base wisely: if both numbers are powers of the same prime (4 and 8 are both powers of 2; 27 and 81 are both powers of 3), pick that prime as the new base. Then both logs evaluate cleanly.

Worked examples

WE 1

Combine into a single logarithm

Write log₃(15) + log₃(6) − log₃(10) as a single logarithm and evaluate it.

Step 1: Apply the product and quotient laws = log₃(15 × 610) = log₃(9010) = log₃(9) Step 2: Read it as a question — “what power of 3 gives 9?” = 2

WE 2

Expand using the laws

Given that log₃(x) = p and log₃(y) = q, express log₃(81x²y) in terms of p and q.

Step 1: Split using the quotient law = log₃(81x²) − log₃(y) Step 2: Split the first term using the product law = log₃(81) + log₃(x²) − log₃(y) Step 3: Apply the power law and simplify = log₃(3⁴) + 2log₃(x) − log₃(y) = 4 + 2p − q = 4 + 2p − q log₃(81) gives a number, not a variable — always evaluate “obvious” logs

WE 3

Combine using the power law

Write 3 log(2) + log(5) − log(4) as a single logarithm, and hence find its exact value. (All logs are base 10.)

Step 1: Bring the coefficient inside using the power law 3 log(2) = log(2³) = log(8) Step 2: Combine using product and quotient laws = log(8) + log(5) − log(4) = log(8 × 54) = log(10) Step 3: Recognise log₁₀(10) = 1 = 1

WE 4

Apply the useful results

Find the exact value of: (i) log₂(2⁹) (ii) 6^log₆(11) (iii) ln(e⁻⁴).

(i) log_a(a^k) = k log₂(2⁹) = 9 (ii) a^log_a(x) = x 6^log₆(11) = 11 (iii) ln(e^k) = k ln(e⁻⁴) = −4 these come up constantly — recognise them on sight

WE 5

Change of base — non-calculator

Without using a calculator, find the exact value of log₄(8).

Step 1: Both 4 and 8 are powers of 2 — change to base 2 log₄(8) = log₂(8)log₂(4) Step 2: Evaluate both — by inspection log₂(8) = 3 (since 2³ = 8) log₂(4) = 2 (since 2² = 4) log₄(8) = 32 choosing the right base — the common prime — is the whole game

WE 6

Solve a log equation — and reject the invalid solution

Solve log₂(x) + log₂(x − 2) = 3.

Step 1: Combine the LHS using the product law log₂(x(x − 2)) = 3 Step 2: Flip into exponential form x(x − 2) = 2³ = 8 Step 3: Expand and solve the quadratic x² − 2x − 8 = 0 (x − 4)(x + 2) = 0 x = 4 or x = −2 Step 4: Check — log₂(−2) is undefined, so reject x = −2 x = 4 always check both roots against the original equation — the laws can introduce extraneous solutions

💡 Top tips

Three core laws — in the booklet. Useful results — not. Memorise the corollaries cold.
Coefficient out front? Use the power law to push it inside as an exponent. 3 log(2) = log(8).
Multiple terms with the same base? Combine them — products become +, quotients become −, then read the result.
Awkward base? Reach for change of base. Choose a base where both the argument and the original base evaluate cleanly (usually a common prime).
Always check answers when solving log equations. If any final value makes a log argument zero or negative, reject it.
The same laws apply to ln, just with base e.
If you’re combining logs of different bases, you can’t — convert them all to the same base first using change of base.

⚠ Common mistakes

Splitting a sum inside a log. log(x + y) is not log(x) + log(y). The laws are for products and quotients.
Combining logs of different bases. log₂(8) + log₃(9) cannot be merged directly — change one base first.
Forgetting log(1) = 0. When a numerator simplifies to 1 (e.g., the whole quotient is 1), the log of it is 0, not 1.
Misapplying the power law on a coefficient outside. 2 log(x) = log(x²), not log(2x).
Keeping invalid solutions after solving an equation. Always plug each root back into the original equation.
Inverting change of base. The original base goes on the bottom: log_a(x) = log_b(x) ÷ log_b(a), not the other way round.
Confusing log(x^m) with (log x)^m. Only the first one obeys the power law. (log x)² is just log x squared — no simplification.

If the flip identity is the engine, the log laws are the gears. Practise expanding and combining until you can see at a glance that 3 log(2) + log(5) − log(4) = 1. The next note (Solving Exponential Equations) leans on every rule above — the more fluent you are here, the easier that one becomes.

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