IB Maths AI HL Exponentials & Logs Paper 1 & 2 log laws ~6 min read

Laws of Logarithms

The laws of logarithms turn products into sums, quotients into differences and powers into products. They let you combine several logarithms into one — the key move for simplifying log expressions and solving logarithmic equations.

📘 What you need to know

Product law: log_a(xy) = log_ax + log_ay.
Quotient law: log_a(x/y) = log_ax − log_ay.
Power law: log_a(x^m) = m log_ax.
The three laws are in the formula booklet; they hold only for a, x, y > 0.
Useful results: log_a1 = 0, log_aa = 1, log_aa^k = k, a^log_ax = x.
You cannot take the log of zero or a negative number — check solutions against this.

The laws of logarithms

Because a logarithm is an exponent, the index laws carry over directly. Each log law steps an operation down a level: multiplying becomes adding, dividing becomes subtracting, and raising to a power becomes multiplying.

The three laws lower every operation by one level. Read right to left, they instead combine separate logs into a single logarithm.

The three laws of logarithms log_axy = log_ax + log_ay log_axy = log_ax − log_ay log_ax^m = m log_ax these three are in the formula booklet — valid for a, x, y > 0

Useful results

A few special values follow straight from the laws. They are not in the formula booklet, so learn them — they short-cut a great many questions.

Results worth memorising log_a1 = 0 · log_aa = 1 · log_aa^k = k a^log_ax = x · ln e^x = x · e^{ln x} = x logs and powers of the same base are inverse operations — they undo each other

Solving equations and the domain

To solve a logarithmic equation, use the laws to write each side as a single logarithm. If you reach log(…) = log(…), the insides must be equal; if you reach log(…) = a number, rewrite it as a power.

A logarithm is only defined for a positive input: log_ax needs x > 0. So an equation like log_a(x − 5) requires x > 5.

Always check: substitute each solution back into the original equation and reject any that take the log of zero or a negative number. Also note log_a(x + y) ≠ log_ax + log_ay.

🧭 Recipe — simplifying or solving with log laws

Power law first: bring any coefficient up as a power, m log x = log x^m.
Product law: combine a sum of logs into one, log x + log y = log xy.
Quotient law: combine a difference into one, log x − log y = log(x/y).
Equate or unpack: if log(…) = log(…), set the insides equal; if log(…) = number, rewrite as a power.
Check the domain: reject any solution that takes a log of zero or a negative number.

Worked examples

WE 1

Combining into a single log

Write log 6 + log 5 − log 3 as a single logarithm.

product law on the sum log 6 + log 5 = log(6 × 5) = log 30 quotient law on the difference log 30 − log 3 = log(30 ÷ 3) = log 10 work left to right — sums combine with ×, differences with ÷.

WE 2

Using the power law

Express 3 log 2 + log 5 as a single logarithm.

power law on the coefficient 3 3 log 2 = log 2³ = log 8 product law to combine log 8 + log 5 = log(8 × 5) = log 40 deal with the coefficient first — turn it into a power before combining.

WE 3

Expanding a single log

Write log_ax³yz in terms of log_ax, log_ay and log_az.

quotient law splits the fraction = log_a(x³y) − log_az product law splits x³y = log_ax³ + log_ay − log_az power law on log_ax³ 3 log_ax + log_ay − log_az the same laws in reverse — break one log into several.

WE 4

Useful results, no calculator

Evaluate, without a calculator: (a) log₅5, (b) log₂1, (c) log₃3⁷, (d) ln e⁴.

apply the useful results directly (a) log_aa = 1 ⇒ log₅5 = 1 (b) log_a1 = 0 ⇒ log₂1 = 0 (c) log_aa^k = k ⇒ log₃3⁷ = 7 (d) ln e^x = x ⇒ ln e⁴ = 4 (a) 1 · (b) 0 · (c) 7 · (d) 4 when the base matches the number, the log just reads off the power.

WE 5

Solving a logarithmic equation

Solve log₂x + log₂(x − 2) = 3.

product law combines the left side log₂(x(x − 2)) = 3 rewrite log = 3 as a power of 2 x(x − 2) = 2³ = 8 x² − 2x − 8 = 0 ⇒ (x − 4)(x + 2) = 0 domain needs x > 2 — reject x = −2 x = 4 log₂(x − 2) is undefined for x ≤ 2, so −2 cannot be a solution.

WE 6

Full question: simplify, then solve

(a) Write 2 log 3 + log 2 in the form log k, where k ∈ ℤ. (b) Hence solve 2 log 3 + log 2 = log(2x).

(a) power law: 2 log 3 = log 3² = log 9 log 9 + log 2 = log(9 × 2) = log 18 (b) so log 18 = log(2x) both sides are a single log ⇒ 2x = 18 (a) k = 18 · (b) x = 9 once each side is one log, equate the insides — then check x = 9 keeps 2x > 0. ✓

💡 Top tips

Combining logs into one is the key move — sum → ×, difference → ÷.
Clear coefficients with the power law before using the product or quotient law.
Spot the useful results — log_aa^k = k and a^log_ax = x save real time.
If log(…) = log(…) with the same base, just equate the insides.
Always check solutions — a value that makes any log input ≤ 0 must be rejected.

⚠ Common mistakes

Thinking log_a(x + y) = log_ax + log_ay — the product law needs xy, not x + y.
Forgetting the power law — 2 log 3 is log 9, not log 6.
Keeping an invalid solution that makes a log input zero or negative.
Misreading log_aa^k — the answer is the exponent k, not a^k.
Mixing bases — the laws only combine logs that share the same base.

That completes Exponentials & Logs. The index laws, logarithms and the log laws all describe one relationship from different angles — and together they unlock exponential models, where logarithms are the tool for solving for an unknown power.

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