What is a logarithm and how is it related to exponents in Cambridge IGCSE Mathematics?

A logarithm is the inverse operation of exponentiation. In simple terms, if you have an equation like b^x = y, the logarithm helps you find x, which is the power to which the base b must be raised to produce the number y. In Cambridge IGCSE Mathematics, understanding this relationship is crucial for solving problems involving exponential growth and decay.

How do you solve logarithmic equations in the Cambridge IGCSE curriculum?

To solve logarithmic equations in the Cambridge IGCSE curriculum, you often need to use properties of logarithms such as the product, quotient, and power rules. For example, the equation log_b(x) = y can be rewritten in exponential form as b^y = x. This conversion is key to solving logarithmic equations, allowing you to find the unknown variable.

What are the key properties of logarithms that I need to know for IGCSE exams?

For the Cambridge IGCSE exams, you should be familiar with several key properties of logarithms: the product rule (log_b(mn) = log_b(m) + log_b(n)), the quotient rule (log_b(m/n) = log_b(m) - log_b(n)), and the power rule (log_b(m^n) = n*log_b(m)). These properties are essential for simplifying and solving logarithmic expressions and equations.

How can I convert between logarithmic and exponential forms in Mathematics?

Converting between logarithmic and exponential forms is a fundamental skill in Mathematics. If you have a logarithmic equation like log_b(x) = y, you can convert it to its exponential form as b^y = x. Conversely, if you start with an exponential equation b^y = x, you can express it in logarithmic form as log_b(x) = y. This conversion is particularly useful in solving equations and understanding the relationship between the two forms.

Why are logarithms important in real-world applications, as taught in Cambridge IGCSE Mathematics?

Logarithms are important in real-world applications because they simplify complex calculations involving exponential growth or decay, such as in finance for calculating compound interest, in science for measuring the pH level in chemistry, and in technology for understanding algorithms in computer science. The Cambridge IGCSE Mathematics curriculum introduces these concepts to help students appreciate the practical utility of logarithms in various fields.

Why is "Treating $\log(x + y)$ as $\log x + \log y$" a common mistake in Logarithms?

Why it happens: Confusing the product law with sums. How to avoid it: Product law: (xy) = x + y. Sum law for logs of sums DOES NOT EXIST. Source: 0580/42 — recurring

Why is "Bringing the exponent down without the parentheses" a common mistake in Logarithms?

Why it happens: (x^2) becomes x^2 — ambiguous. How to avoid it: Power law gives 2 x. Always.

Why is "Computing $\log$ of a negative number" a common mistake in Logarithms?

Why it happens: Students apply log to negatives without thinking. How to avoid it: Logarithms are only defined for positive arguments. If (-3) shows up, you've made an algebra error earlier.

Why is "Forgetting the base is non-$10$" a common mistake in Logarithms?

Why it happens: Calculator's button is base 10. For \log_4 they need change-of-base. How to avoid it: If the base isn't 10 or e, use \log_b a = a/ b.

Cambridge IGCSE Mathematics (0580)

Logarithms

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Short Notes - Logarithms

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Detailed Study Notes

Detailed notes on Functions for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Logarithms — Cambridge IGCSE 0580 Maths Extended (2026)

The inverse of an exponential. $\log_{a}(b) = c$ means $a^{c} = b$ . Three log laws unlock every Paper 4 manipulation question, from solving exponentials to combining log expressions.

At a glance

$\log_{a}(b) = c \iff a^{c} = b$ .
$\log_{10}$ on a calculator is the LOG button.
Natural log ( $\ln$ ) uses base $e$ ; less common at IGCSE.
Three log laws: product, quotient, power.
$\log_{a}(xy) = \log_{a}(x) + \log_{a}(y)$ .
$\log_{a}\left(\dfrac{x}{y}\right) = \log_{a}(x) - \log_{a}(y)$ .
$\log_{a}(x^{n}) = n \log_{a}(x)$ .
$\log_{a}(1) = 0$ and $\log_{a}(a) = 1$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.17 — Use logarithms to solve simple exponential equations.
Apply the laws of logarithms to manipulate expressions.

What a logarithm is

Same statement, two notations. $\log_{a}(b) = c$ is just another way of writing $a^{c} = b$ .

Definition. $\log_{a}(b) = c$ means $a^{c} = b$ .

In words: "the log base $a$ of $b$ is the power you raise $a$ to in order to get $b$ ."

Worked translations.

$\log_{10}(100) = 2$ because $10^{2} = 100$ .
$\log_{2}(8) = 3$ because $2^{3} = 8$ .
$\log_{5}(1) = 0$ because $5^{0} = 1$ .
$\log_{3}(81) = 4$ because $3^{4} = 81$ .

Two special values.

$\log_{a}(1) = 0$ — any base raised to $0$ is $1$ .
$\log_{a}(a) = 1$ — any base raised to $1$ is itself.

A logarithm just asks for an exponent: the base stays the base, and the log answer is the power.

Calculator. The LOG button is base $10$ . Some calculators also have $\ln$ (natural log, base $e$ ). For other bases, use the change-of-base formula: $\log_{a}(b) = \dfrac{\log_{10}(b)}{\log_{10}(a)}.$

$\log_{a}(b) = c \iff a^{c} = b$ .
$\log_{a}(1) = 0$ , $\log_{a}(a) = 1$ .
Calculator LOG button is base $10$ .
Change of base: $\log_{a}(b) = \dfrac{\log b}{\log a}$ .

The three log laws

Product, quotient, power. Mirror the index laws: $\times$ becomes $+$ , $\div$ becomes $-$ , power comes down.

Product law. $\log_{a}(xy) = \log_{a}(x) + \log_{a}(y)$ .

Quotient law. $\log_{a}\left(\dfrac{x}{y}\right) = \log_{a}(x) - \log_{a}(y)$ .

Power law. $\log_{a}(x^{n}) = n \log_{a}(x)$ .

These mirror the index laws: $a^{m} \cdot a^{n} = a^{m + n}$ corresponds to $\log(\text{product}) = \text{sum of logs}$ .

Multiplication becomes addition, division becomes subtraction, and an exponent comes down as a multiplier.

Worked. Simplify $\log(8) + \log(5)$ .

Product law: $\log(8 \times 5) = \log(40)$ .

Worked. Simplify $\log(60) - \log(3)$ .

Quotient law: $\log\left(\dfrac{60}{3}\right) = \log(20)$ .

Worked. Simplify $3 \log(2)$ .

Power law in reverse: $\log(2^{3}) = \log(8)$ .

Worked (combine all three). Express $\log_{a}(x) + 2\log_{a}(y) - \log_{a}(z)$ as a single log.

Power: $\log_{a}(x) + \log_{a}(y^{2}) - \log_{a}(z)$ .
Combine: $\log_{a}\left(\dfrac{xy^{2}}{z}\right)$ .

Product → SUM. Quotient → DIFFERENCE. Power → COEFFICIENT.
All laws use the SAME base.
Reverse direction: combine sums/differences into one log.
Use to simplify expressions or to solve equations.

Solving exponential equations with logs

Take the log of both sides; use the power law to bring the exponent down.

Method. When the unknown sits IN the exponent, take logs.

Worked. Solve $5^{x} = 50$ .

Take log of both sides: $\log(5^{x}) = \log(50)$ .
Power law: $x \log(5) = \log(50)$ .
Divide: $x = \dfrac{\log(50)}{\log(5)} \approx 2.43$ .

Worked. Solve $2^{x + 1} = 7$ .

$\log(2^{x + 1}) = \log(7)$ .
$(x + 1)\log(2) = \log(7)$ .
$x + 1 = \dfrac{\log(7)}{\log(2)} \approx 2.807$ .
$x \approx 1.807$ .

Equating bases (when possible). If both sides can be written as powers of the same base, just equate exponents:

$2^{x} = 16 \Rightarrow 2^{x} = 2^{4} \Rightarrow x = 4$ .

This is faster than logs when the answer is a clean integer.

Unknown in exponent → take logs.
Power law brings the exponent down to a coefficient.
Divide to isolate.
If both sides reduce to the same base, equate exponents directly.

Quick recap

$\log_{a}(b) = c \iff a^{c} = b$ .
Three log laws: product (sum), quotient (difference), power (coefficient).
$\log_{a}(1) = 0$ , $\log_{a}(a) = 1$ .
Solve exponential equations by taking logs of both sides.
Same base on both sides → equate exponents directly.
$\log(x + y) \ne \log x + \log y$ — logs only distribute over multiplication.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Logarithm — $\log_{a}(b)$ is the power that $a$ must be raised to in order to give $b$ .
Product law — $\log_{a}(xy) = \log_{a} x + \log_{a} y$ .
Quotient law — $\log_{a}(x/y) = \log_{a} x - \log_{a} y$ .
Power law — $\log_{a}(x^{n}) = n \log_{a} x$ .
Change of base — $\log_{a}(b) = \dfrac{\log b}{\log a}$ .

How it’s examined

Logs appear most years on Paper 4 as a 4-5 mark question — usually combining the three laws to simplify, or solving an exponential equation. Paper 2 has simpler 2-3 mark items. Examiner reports flag the $\log(x + y) = \log x + \log y$ misconception every series.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.17); 0580/42 Oct/Nov 2024 — Q12 (log laws + exponential solve); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Logarithms

Step-by-step solutions to past-paper-style questions on logarithms, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Logarithms

Almost every logarithms exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — evaluate, write as a single logarithm, solve — on one log expression or one exponential equation.

How to approach it

Use the definition $\log_b x=y\iff b^y=x$ , the log laws to combine or split terms, or change-of-base $\log_b a=\frac{\log a}{\log b}$ for an awkward base. To solve $b^x=c$ , take logs of both sides.

Common trap

Treating $\log(x+y)$ as $\log x+\log y$ — no sum law exists. Examiner reports also flag forgetting that bare $\log$ means base $10$ , and computing $\log$ of a negative argument.

Multi-step problem

Recognise it by

A quadratic in $\log x$ , or an equation mixing logs of different bases — several techniques chained together.

How to approach it

Substitute $u=\log x$ to turn a log-quadratic into an ordinary quadratic, or apply change-of-base so every term shares one base; solve, then convert back to find $x$ .

Common trap

Expanding $(\log x)^2$ as $\log x^2$ , or combining different-base logs directly. Examiner reports note the substitution / change-of-base step is where the method mark is earned.

1Evaluate a logarithm
ExtendedDirect calculation• evaluation
Question
Evaluate $\log_{2} 32$ .
Step-by-step solution
1. Step 1
  Express $32$ as a power of $2$ .
  $32 = 2^{5}$
2. Step 2
  $\log_{2}(2^{5}) = 5$ .
Answer
$\log_{2} 32 = 5$
2Combine using log laws
ExtendedDirect calculation• log laws
Question
Write as a single logarithm: $\log 12 - \log 3 + 2 \log 5$ .
Step-by-step solution
1. Step 1
  Apply power law on the last term.
  $\log 12 - \log 3 + \log 25$
2. Step 2
  Apply quotient and product laws.
  $\log\left(\dfrac{12}{3} \times 25\right) = \log 100$
Answer
$\log 100\ (= 2)$
3Solve an exponential equation
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q13• exponential
Question
Solve $5^{x} = 200$ . Give your answer to $3$ significant figures.
Step-by-step solution
1. Step 1
  Take logs of both sides.
  $x \log 5 = \log 200$
2. Step 2
  Divide.
  $x = \dfrac{\log 200}{\log 5} \approx 3.29$
Answer
$x \approx 3.29$ (3 s.f.)
4Use change of base
ExtendedDirect calculation• change of base
Question
Evaluate $\log_{4} 50$ to $3$ s.f.
Step-by-step solution
1. Step 1
  Apply change-of-base.
  $\log_{4} 50 = \dfrac{\log 50}{\log 4} \approx 2.82$
Answer
$2.82$ (3 s.f.)
5Evaluate $\log_{2} 16$
CoreDirect calculation• evaluation
Question
Evaluate $\log_{2} 16$ without a calculator.
Step-by-step solution
1. Step 1
  Write $16$ as a power of $2$ .
  $16 = 2^{4}$
2. Step 2
  Apply $\log_{b}(b^{n}) = n$ .
  $\log_{2}(2^{4}) = 4$
Answer
$\log_{2} 16 = 4$
6Evaluate $\log_{10} 1000$
CoreDirect calculation• evaluation, base 10
Question
Evaluate $\log 1000$ .
Step-by-step solution
1. Step 1
  When no base is shown, base $10$ is assumed.
  $1000 = 10^{3}$
2. Step 2
  Take the log.
  $\log_{10}(10^{3}) = 3$
Answer
$\log 1000 = 3$
Examiner tip
The examiner report flags candidates often forget that " $\log$ " without a stated base means base $10$ . Always check the convention in the syllabus.
7Apply $\log a + \log b = \log(ab)$
ExtendedDirect calculation• log laws, product
Question
Write $\log 8 + \log 125$ as a single logarithm and evaluate.
Step-by-step solution
1. Step 1
  Apply the product law.
  $\log 8 + \log 125 = \log(8 \times 125) = \log 1000$
2. Step 2
  Evaluate (base $10$ ).
  $\log_{10} 1000 = 3$
Answer
$3$
8Solve $\log_{3} x = 4$
ExtendedDirect calculation• Adapted from 0580/42 Feb/Mar 2023 Q12• solve, log equation
Question
Solve $\log_{3} x = 4$ .
Step-by-step solution
1. Step 1
  Convert from log to exponential form.
  $x = 3^{4}$
2. Step 2
  Evaluate.
  $x = 81$
Answer
$x = 81$
Examiner tip
The mark scheme awards a method mark for explicitly writing the exponential form $x = 3^{4}$ . The examiner report flags candidates who try to manipulate $\log_{3}$ algebraically without the conversion.
9Solve a quadratic in $\log x$
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q17• log laws, quadratic
Question
Solve $(\log x)^{2} - 5\log x + 6 = 0$ (base $10$ ).
Step-by-step solution
1. Step 1
  Let $u = \log x$ .
  $u^{2} - 5u + 6 = 0$
2. Step 2
  Factor.
  $(u - 2)(u - 3) = 0 \implies u = 2 \text{ or } u = 3$
3. Step 3
  Convert back.
  $\log x = 2 \implies x = 100;\ \log x = 3 \implies x = 1000$
Answer
$x = 100$ or $x = 1000$
Examiner tip
The mark scheme awards a method mark for the substitution $u = \log x$ . The examiner report flags candidates often try to expand $(\log x)^{2}$ as $\log x^{2}$ , which is wrong.
10Use change of base to solve
ChallengeMulti-step problem• change of base, solve
Question
Solve $\log_{2} x + \log_{4} x = 6$ . Give the exact value.
Step-by-step solution
1. Step 1
  Use change of base on $\log_{4} x$ to base $2$ .
  $\log_{4} x = \dfrac{\log_{2} x}{\log_{2} 4} = \dfrac{\log_{2} x}{2}$
2. Step 2
  Substitute.
  $\log_{2} x + \dfrac{\log_{2} x}{2} = 6 \implies \dfrac{3}{2} \log_{2} x = 6$
3. Step 3
  Solve.
  $\log_{2} x = 4 \implies x = 2^{4} = 16$
Answer
$x = 16$
Examiner tip
The examiner report flags candidates often try to combine logs with different bases directly. Change-of-base is the recommended first step.

Key Formulae — Logarithms

The formulae you need to memorise for logarithms on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Definition of logarithm
$y = \log_{b} x \iff b^{y} = x$
$b$
base, $b > 0$ and $b \neq 1$
$x$
argument, $x > 0$
When to use
Converting between logarithmic and exponential form.
Log laws
$\log_{b}(xy) = \log_{b} x + \log_{b} y,\ \ \log_{b}\left(\tfrac{x}{y}\right) = \log_{b} x - \log_{b} y,\ \ \log_{b}(x^{n}) = n\log_{b} x$
When to use
Combining or splitting logarithms.
Change-of-base formula
$\log_{b} a = \dfrac{\log a}{\log b}$
When to use
Evaluating $\log_{b} a$ when your calculator only has $\log$ (base 10) or $\ln$ (base $e$ ).

Key Definitions and Keywords — Logarithms

Definitions to memorise and the exact keywords mark schemes credit for logarithms answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Logarithm
Examiner keyword
$\log_{b} x$ is the exponent you must raise $b$ to in order to get $x$ .
Base of a logarithm
Examiner keyword
The number $b$ in $\log_{b}$ . Common bases are $10$ (often written $\log$ ) and $e$ (written $\ln$ ).
Argument
The value $x$ in $\log_{b}(x)$ — must be positive.
Natural log ( $\ln$ )
Logarithm with base $e \approx 2.718$ . $\ln x = \log_{e} x$ .

Common Mistakes and Misconceptions — Logarithms

The traps other students keep falling into on logarithms questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Treating $\log(x + y)$ as $\log x + \log y$
0580/42 — recurring
Why it happens
Confusing the product law with sums.
How to avoid it
Product law: $\log(xy) = \log x + \log y$ . Sum law for logs of sums DOES NOT EXIST.
✕Bringing the exponent down without the parentheses
Why it happens
$\log(x^{2})$ becomes $\log x^{2}$ — ambiguous.
How to avoid it
Power law gives $2 \log x$ . Always.
✕Computing $\log$ of a negative number
Why it happens
Students apply log to negatives without thinking.
How to avoid it
Logarithms are only defined for positive arguments. If $\log(-3)$ shows up, you've made an algebra error earlier.
✕Forgetting the base is non- $10$
Why it happens
Calculator's $\log$ button is base $10$ . For $\log_{4}$ they need change-of-base.
How to avoid it
If the base isn't $10$ or $e$ , use $\log_{b} a = \dfrac{\log a}{\log b}$ .

Logarithms — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Logarithms

Short Notes - Logarithms

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Logarithms

1Evaluate a logarithm

2Combine using log laws

3Solve an exponential equation

4Use change of base

5Evaluate log⁡216\log_{2} 16log2​16

6Evaluate log⁡101000\log_{10} 1000log10​1000

7Apply log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log(ab)loga+logb=log(ab)

8Solve log⁡3x=4\log_{3} x = 4log3​x=4

9Solve a quadratic in log⁡x\log xlogx

10Use change of base to solve

Definition of logarithm

Log laws

Change-of-base formula

Logarithm

Base of a logarithm

Argument

Natural log (ln⁡\lnln)

✕Treating log⁡(x+y)\log(x + y)log(x+y) as log⁡x+log⁡y\log x + \log ylogx+logy

✕Bringing the exponent down without the parentheses

✕Computing log⁡\loglog of a negative number

✕Forgetting the base is non-101010

Logarithms — frequently asked questions

5Evaluate $\log_{2} 16$

6Evaluate $\log_{10} 1000$

7Apply $\log a + \log b = \log(ab)$

8Solve $\log_{3} x = 4$

9Solve a quadratic in $\log x$

Natural log ( $\ln$ )

✕Treating $\log(x + y)$ as $\log x + \log y$

✕Computing $\log$ of a negative number

✕Forgetting the base is non- $10$