Logarithms and Laws

This lesson introduces logarithms and the three key logarithm laws that you need for the Edexcel 9MA0 specification. Logarithms are the inverse of exponential functions and are essential for solving exponential equations.

What Is a Logarithm?

A logarithm answers the question: "What power must I raise the base to in order to get this number?"

The statement aˣ = b is equivalent to x = log_a(b)

Read "log_a(b)" as "log base a of b."

Examples:

2³ = 8, so log₂(8) = 3
10² = 100, so log₁₀(100) = 2
5⁰ = 1, so log₅(1) = 0

Important Special Cases

Statement	Explanation
log_a(1) = 0	Because a⁰ = 1 for any base a
log_a(a) = 1	Because a¹ = a
log_a(aⁿ) = n	Applying the definition directly
a^(log_a(x)) = x	The exponential and logarithm cancel

These identities follow directly from the definition and are used constantly.

The Three Logarithm Laws

Law 1: The Addition Law

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

The log of a product equals the sum of the logs.

Proof: Let log_a(x) = m and log_a(y) = n. Then aᵐ = x and aⁿ = y. So xy = aᵐ × aⁿ = aᵐ⁺ⁿ. Therefore log_a(xy) = m + n = log_a(x) + log_a(y).

Example: log₂(8) + log₂(4) = log₂(32) = 5

Law 2: The Subtraction Law

log_a(x) - log_a(y) = log_a(x/y)

The log of a quotient equals the difference of the logs.

Example: log₃(81) - log₃(3) = log₃(27) = 3

Law 3: The Power Law

log_a(xⁿ) = n × log_a(x)

The log of a power equals the index times the log.

Example: log₁₀(1000) = log₁₀(10³) = 3 × log₁₀(10) = 3 × 1 = 3

Exam Tip: These three laws only work when the logs have the same base. You cannot combine log₂(x) + log₃(y) using these laws.

Using the Laws to Simplify Expressions

Example 1: Write 2log₃(5) + log₃(4) as a single logarithm.

2log₃(5) = log₃(5²) = log₃(25) [Power Law]
log₃(25) + log₃(4) = log₃(25 × 4) = log₃(100) [Addition Law]

Example 2: Write log₅(75) - log₅(3) as a single logarithm.

log₅(75) - log₅(3) = log₅(75/3) = log₅(25) = 2 [Subtraction Law, then evaluate]

Example 3: Express log₂(48) - log₂(3) + log₂(4) as an integer.

log₂(48/3) + log₂(4) = log₂(16) + log₂(4)
= log₂(16 × 4) = log₂(64) = 6

Writing Expressions in Terms of log_a(p), log_a(q), etc.

A common exam question gives you values of log_a(p) and log_a(q) and asks you to find log_a of some expression.

Example: Given log₂(3) = p and log₂(5) = q, find log₂(45).

45 = 9 × 5 = 3² × 5
log₂(45) = log₂(3²) + log₂(5) = 2log₂(3) + log₂(5) = 2p + q

Solving Equations Using Logarithms

To solve aˣ = b, take logs of both sides:

x × log(a) = log(b)
x = log(b) / log(a)

Example: Solve 5ˣ = 20.

x = log(20) / log(5) = 1.301.../0.699... ≈ 1.861

You can use any base for the log — log₁₀ and ln both work.

Common Mistakes to Avoid

Mistake	Correction
log(a + b) = log(a) + log(b)	There is no rule for the log of a sum
log(a) × log(b) = log(ab)	The addition law involves addition of logs, not multiplication
log(a/b) = log(a) / log(b)	log(a/b) = log(a) - log(b) — use subtraction
log(a)ⁿ = n × log(a)	Only log(aⁿ) = n × log(a); the power must be on the argument

Summary

A logarithm is the inverse of an exponential: if aˣ = b then x = log_a(b).
The three laws: Addition (product), Subtraction (quotient), Power (bring the exponent down).
All three laws require the same base.
These laws are used to simplify, combine, and expand logarithmic expressions.
To solve aˣ = b, take logs of both sides and use the power law.

A-Level Deep Dive: Logarithms and Laws

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms covers the function $a^x$ and its graph, where $a$ is positive. Know and use the definition of $\log_a x$ as the inverse of $a^x$ , where $a$ is positive and $x \geq 0$ . Understand and use the laws of logarithms: $\log_a x + \log_a y = \log_a(xy)$ ; $\log_a x - \log_a y = \log_a(x/y)$ ; $k\log_a x = \log_a x^k$ (including, for example, $k = -1$ and $k = -\tfrac{1}{2}$ ) (refer to the official specification document for exact wording). The defining identity is $\log_a x = y \Leftrightarrow a^y = x$ . The Edexcel formula booklet lists the change-of-base formula but does not list the three core log laws — they must be memorised. Synoptic mentions: solving exponential equations using logs (used in section 6 modelling and across Paper 2 statistics for half-life and decay); differentiation of $\ln x$ in section 9 ( $\frac{d}{dx}\ln x = 1/x$ ); integration of $1/x$ in section 8 ( $\int 1/x \, dx = \ln|x| + C$ ); modelling exponential growth and decay ( $N = N_0 e^{kt}$ , where taking logs linearises the model); index laws from section 2 are the parent of log laws — every log law is an index law in disguise.

Worked example with full mark scheme

Question (8 marks): Solve $\log_2(x + 1) + \log_2(x - 1) = 3$ , giving the exact value of $x$ and justifying your answer with reference to the domain of the original equation.

Solution with mark scheme:

Step 1 — combine the two logs using the product law.

$\log_2(x + 1) + \log_2(x - 1) = \log_2\bigl((x + 1)(x - 1)\bigr) = \log_2(x^2 - 1)$

M1 — correct application of the product law $\log_a P + \log_a Q = \log_a(PQ)$ . Common error: writing $\log_2(x + 1) + \log_2(x - 1) = \log_2(2x)$ (adding inside the log instead of multiplying). That is the single most common mark-loss on this kind of question.

A1 — correct combined expression $\log_2(x^2 - 1)$ .

Step 2 — convert from log form to index form.

By definition, $\log_2(x^2 - 1) = 3 \Leftrightarrow 2^3 = x^2 - 1$ .

M1 — using the defining identity $\log_a P = N \Leftrightarrow a^N = P$ .

Step 3 — solve the resulting equation.

$x^2 - 1 = 8 \implies x^2 = 9 \implies x = \pm 3$

A1 — both algebraic roots stated.

Step 4 — apply the domain constraint.

The original equation involves $\log_2(x + 1)$ and $\log_2(x - 1)$ . For both to be defined we require $x + 1 > 0$ and $x - 1 > 0$ , i.e. $x > 1$ .

M1 — explicit identification of the domain restriction. The phrase examiners reward: "since $\log_a P$ is defined only for $P > 0$ ".

A1 — applying the constraint to reject $x = -3$ . (At $x = -3$ , $x - 1 = -4 < 0$ , so $\log_2(x - 1)$ is undefined.)

Step 5 — state the final solution with justification.

The only valid solution is $x = 3$ . Verification: $\log_2(4) + \log_2(2) = 2 + 1 = 3$ . $\checkmark$

A1 — final answer with verification.

B1 — communication mark for the explicit domain argument tied back to the original equation.

Total: 8 marks (M3 A4 B1).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A culture of bacteria has population $P$ at time $t$ hours, modelled by $P = 2000 \cdot 3^{0.4 t}$ .

(a) Find the time, to 3 significant figures, at which the population reaches 10 000. (4)

(b) Express your answer to (a) in the form $\dfrac{\ln k}{\ln 3}$ for an integer $k$ to be found, and hence give the exact value of $t$ . (2)

Mark scheme decomposition by AO:

(a)

M1 (AO3.4 — modelling) — setting up $2000 \cdot 3^{0.4 t} = 10000$ from the model, dividing to obtain $3^{0.4 t} = 5$ .
M1 (AO1.1b) — taking logs of both sides (any consistent base): $0.4 t \ln 3 = \ln 5$ , or equivalently $0.4 t = \log_3 5$ .
M1 (AO1.1b) — rearranging for $t$ : $t = \dfrac{\ln 5}{0.4 \ln 3}$ .
A1 (AO1.1b) — $t \approx 3.66$ hours (3 s.f.).

(b)

M1 (AO2.5) — recognising that $0.4 = 2/5$ and rewriting $t = \dfrac{\ln 5}{(2/5) \ln 3} = \dfrac{5 \ln 5}{2 \ln 3} = \dfrac{\ln 5^5}{\ln 3^2} = \dfrac{\ln 3125}{\ln 9}$ , identifying the exact form as $t = \dfrac{\ln k}{\ln 9}$ with $k = 3125$ .
A1 (AO2.5) — exact form $t = \dfrac{\ln 3125}{\ln 9} = \dfrac{5 \ln 5}{2 \ln 3}$ , with $k = 3125$ identified.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel uses log questions in modelling contexts to test the AO3.4 step (set up the equation from the model), with AO2 reserved for the exact-form transformation.

Synoptic links

Connects to:

Section 2 — Algebra and functions (index laws): every log law is the index-law twin. $\log_a(xy) = \log_a x + \log_a y$ is the log-form of $a^{p+q} = a^p \cdot a^q$ . Any candidate weak on indices will be weak on logs, because the proofs reduce instantly to index manipulation.
Section 6 — Exponentials (solving $a^x = b$ ): logarithms are the only tool for isolating the exponent in equations like $5 \cdot 2^x = 80$ . The structural pattern "exponential equation $\to$ take logs $\to$ linear in $x$ " appears in nearly every modelling question on 9MA0-02.
Section 9 — Differentiation: $\frac{d}{dx}\ln x = \dfrac{1}{x}$ for $x > 0$ , and $\frac{d}{dx}\log_a x = \dfrac{1}{x \ln a}$ . The log laws are essential for differentiating expressions like $\ln(x^2 \sqrt{1 + x})$ — split into $2\ln x + \tfrac{1}{2}\ln(1 + x)$ before differentiating, never after.
Section 8 — Integration: $\int \dfrac{1}{x} \, dx = \ln|x| + C$ , the exception to the power rule. Also $\int \dfrac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$ (the "reverse chain" pattern), which underlies many trickier Paper 1 integrations such as $\int \tan x \, dx = -\ln|\cos x| + C$ .
Sections 6 + Statistics — exponential modelling and linearisation: taking logs of $y = ab^x$ produces $\ln y = \ln a + x \ln b$ , a straight line in $(x, \ln y)$ . This is the technique behind log-linear plots in 9MA0-02 modelling questions and underlies log-graphs in physics, chemistry, biology and economics.

Mark-scheme literacy

Logs questions on 9MA0 split AO marks more evenly than indices because real modelling tasks require AO3.4:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying log laws correctly, switching between log and index form, using change-of-base, evaluating $\log_a a^x = x$
AO2 (reasoning / interpretation)	20–30%	Justifying domain restrictions, presenting answers in the requested exact form ( $\ln k / \ln a$ ), recognising when to take logs
AO3 (problem-solving / modelling)	15–25%	Setting up the equation from a population/decay/half-life model, interpreting the constants, criticising the model

Examiner-rewarded phrasing: "since $\log_a P$ is defined only for $P > 0$ we require ..."; "taking natural logs of both sides ..."; "by the power law of logarithms, $k \log_a x = \log_a x^k$ ". Phrases that lose marks: leaving an answer as $\log_2 5 / \log_2 3$ when the question requested a single $\ln$ form; writing $\log(x + y)$ when $\log x + \log y$ was intended; "cancelling" logs as in $\log A / \log B = A/B$ (a major and common error).

A specific Edexcel pattern to watch: if the question says "exact" or "in the form $\ln k / \ln a$ ", a decimal answer scores zero on the final A1 even when correct. Read the constraint precisely.

Grade-band model answers

3-mark question

Question: Given that $\log_3 a + 2 \log_3 b = 4$ , express $a$ in terms of $b$ .

Grade C response (~210 words):

Using the power law, $2 \log_3 b = \log_3 b^2$ . Then by the product law,

$\log_3 a + \log_3 b^2 = \log_3 (a b^2) = 4$ .

So $a b^2 = 3^4 = 81$ , hence $a = \dfrac{81}{b^2}$ .

Examiner commentary: Full marks (3/3). The candidate applies the power law, then the product law, then converts to index form correctly. The work is brief but each named law is used at the right step. The final rearrangement gives $a$ explicitly in terms of $b$ as the question demands. Many candidates lose marks here by stopping at $a b^2 = 81$ without isolating $a$ , or by applying the power law in the wrong direction (writing $\log_3 b^2 = (\log_3 b)^2$ , which is wrong). This is the textbook Grade C answer — efficient, accurate, no presentation flourishes.

Logarithms and Laws

Logarithms and Laws

What Is a Logarithm?

Important Special Cases

The Three Logarithm Laws

Law 1: The Addition Law

Law 2: The Subtraction Law

Law 3: The Power Law

Using the Laws to Simplify Expressions

Writing Expressions in Terms of log_a(p), log_a(q), etc.

Solving Equations Using Logarithms

Common Mistakes to Avoid

Summary

A-Level Deep Dive: Logarithms and Laws

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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