Powers and Roots · Study Notes · Edexcel IGCSE 4MA1 Higher

Index laws power algebra, surds, standard form, and exponential growth. Master the rules and you unlock half the Number topic.

At a glance

Product: $x^{a} \times x^{b} = x^{a+b}$ — same base, ADD indices.
Quotient: $x^{a} / x^{b} = x^{a-b}$ — same base, SUBTRACT.
Power of power: $(x^{a})^{b} = x^{ab}$ — MULTIPLY.
Zero: $x^{0} = 1$ for $x \ne 0$ .
Negative: $x^{-n} = \dfrac{1}{x^{n}}$ — reciprocal.
Fractional: $x^{m/n} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$ — root first.
Same-base only for the addition/subtraction rules.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.6 — Use index laws (product, quotient, power, zero, negative, fractional).
1.6 — Evaluate fractional and negative indices.
1.6 — Use square roots, cube roots and other roots.
1.6 — Estimate roots of non-perfect powers.

The seven index laws

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Memorise these seven. Every algebra and surd question on Edexcel uses at least one.

The index laws apply to ANY base $x$ (with $x \ne 0$ where division is involved):

Law	Rule
Product	$x^{a} \times x^{b} = x^{a+b}$
Quotient	$\dfrac{x^{a}}{x^{b}} = x^{a-b}$
Power of a power	$(x^{a})^{b} = x^{ab}$
Power of a product	$(xy)^{n} = x^{n} y^{n}$
Power of a quotient	$\left(\dfrac{x}{y}\right)^{n} = \dfrac{x^{n}}{y^{n}}$
Zero index	$x^{0} = 1$ ( $x \ne 0$ )
Negative index	$x^{-n} = \dfrac{1}{x^{n}}$

These rules ONLY apply to the SAME base. $x^{3} \times y^{4}$ does NOT simplify; $x^{3} \times x^{4} = x^{7}$ does.

Examples.

Expression	Simplified
$2^{5} \times 2^{3}$	$2^{5+3} = 2^{8} = 256$
$\dfrac{x^{7}}{x^{3}}$	$x^{7-3} = x^{4}$
$(x^{4})^{3}$	$x^{4 \times 3} = x^{12}$
$(2x)^{3}$	$2^{3} x^{3} = 8x^{3}$
$5^{0}$	$1$
$3^{-2}$	$\dfrac{1}{3^{2}} = \dfrac{1}{9}$

Worked qualitative. Why does $x^{a} \times x^{b} = x^{a+b}$ ?

$x^{a} = x \cdot x \cdot \ldots \cdot x$ ( $a$ times).
$x^{b} = x \cdot x \cdot \ldots \cdot x$ ( $b$ times).
Multiplying: $x \cdot x \cdot \ldots \cdot x$ ( $a + b$ times) $= x^{a+b}$ .
Counting matters more than memorising — if you know what $x^{n}$ means, the product rule is automatic.

Multiplying powers of the same base just stacks the factors — so the indices add.

Edexcel tip. Show every step in algebra simplification questions. Mark schemes have separate marks for each application of an index law.

Same base only.
Product → ADD indices.
Quotient → SUBTRACT.
Power of power → MULTIPLY.
Zero → 1.
Negative → reciprocal.

Fractional and negative indices

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Fractional = root + power. Negative = reciprocal. Always tackle the trickier bit first.

Fractional indices combine roots with powers.

$x^{m/n} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$

For non-calculator papers, ALWAYS take the root FIRST (smaller numbers, less arithmetic error).

Denominator gives the root, numerator gives the power — take the root first.

Examples:

Expression	Step	Result
$8^{1/3}$	cube root	$2$
$16^{1/2}$	square root	$4$
$8^{2/3}$	cube root, then square	$(2)^{2} = 4$
$27^{2/3}$	cube root, then square	$(3)^{2} = 9$
$64^{5/6}$	sixth root, then fifth power	$(2)^{5} = 32$
$32^{4/5}$	fifth root, then fourth power	$(2)^{4} = 16$

Negative indices mean RECIPROCAL.

$x^{-n} = \dfrac{1}{x^{n}}$

For fractions: flip the fraction, then apply the positive power.

$\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}$

Examples:

Expression	Step	Result
$2^{-3}$	reciprocal	$\dfrac{1}{2^{3}} = \dfrac{1}{8}$
$5^{-2}$	reciprocal	$\dfrac{1}{25}$
$\left(\dfrac{2}{3}\right)^{-2}$	flip, then square	$\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$
$\left(\dfrac{1}{4}\right)^{-3}$	flip, then cube	$4^{3} = 64$

Combining negative AND fractional.

$8^{-2/3}$ :

Negative → reciprocal: $\dfrac{1}{8^{2/3}}$ .
Fractional → root first: $8^{1/3} = 2$ , then $2^{2} = 4$ .
Final: $\dfrac{1}{4}$ .

Worked qualitative. Why does $x^{1/2}$ mean $\sqrt{x}$ ?

We want $x^{1/2}$ to follow the index laws.
$x^{1/2} \times x^{1/2} = x^{1/2 + 1/2} = x^{1}$ .
So $x^{1/2}$ is something that, multiplied by itself, gives $x$ .
That's the definition of $\sqrt{x}$ .

Edexcel tip. Memorise common fractional powers. $8^{1/3} = 2$ , $27^{1/3} = 3$ , $64^{1/3} = 4$ , $125^{1/3} = 5$ . $16^{1/4} = 2$ , $81^{1/4} = 3$ . $32^{1/5} = 2$ , $243^{1/5} = 3$ .

Fractional: root first, then power.
$x^{1/n} = \sqrt[n]{x}$ .
Negative: reciprocal.
$x^{-n} = 1/x^{n}$ .
Combined: handle one at a time.

Square, cube and other roots

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Inverse of powering. Used for surds, Pythagoras, areas, volumes.

Definition. $\sqrt[n]{x}$ is the value $y$ with $y^{n} = x$ .

Square root ( $n = 2$ ): $\sqrt{x}$ .

Edexcel uses POSITIVE root only when written as $\sqrt{}$ .
$\sqrt{16} = 4$ (not $\pm 4$ ).
$\sqrt{x}$ requires $x \ge 0$ (no real square root of negative).

Cube root ( $n = 3$ ): $\sqrt[3]{x}$ .

Works for ALL real $x$ (positive, negative, zero).
$\sqrt[3]{27} = 3$ , $\sqrt[3]{-8} = -2$ .

Higher roots. Same idea. $\sqrt[n]{x^{n}} = x$ for $x \ge 0$ .

Estimating square roots (non-calculator paper).

For $\sqrt{50}$ :

$7^{2} = 49 < 50 < 64 = 8^{2}$ .
$\sqrt{50}$ between $7$ and $8$ , much closer to $7$ .
$7.1^{2} = 50.41$ → $\sqrt{50} \approx 7.07$ (≈ $7.1$ to 1 d.p.).

Trap the number between neighbouring perfect squares to estimate its root.

Memorise common roots.

Square roots	Cube roots
$\sqrt{4} = 2$	$\sqrt[3]{8} = 2$
$\sqrt{9} = 3$	$\sqrt[3]{27} = 3$
$\sqrt{16} = 4$	$\sqrt[3]{64} = 4$
$\sqrt{25} = 5$	$\sqrt[3]{125} = 5$
$\sqrt{36} = 6$	$\sqrt[3]{216} = 6$
$\sqrt{49} = 7$	$\sqrt[3]{343} = 7$
$\sqrt{64} = 8$	$\sqrt[3]{512} = 8$
$\sqrt{81} = 9$	$\sqrt[3]{729} = 9$
$\sqrt{100} = 10$	$\sqrt[3]{1000} = 10$
$\sqrt{121} = 11$
$\sqrt{144} = 12$
$\sqrt{169} = 13$
$\sqrt{196} = 14$
$\sqrt{225} = 15$

Worked qualitative. Why is $\sqrt[3]{-8} = -2$ but $\sqrt{-8}$ undefined?

For odd roots, $(-2)^{3} = -8$ . So $\sqrt[3]{-8} = -2$ is fine.
For even roots, $y^{2} \ge 0$ always (squares of reals are non-negative). So no real $y$ has $y^{2} = -8$ .
Even roots of negatives don't exist in the real numbers (only in the complex numbers, beyond IGCSE).

Edexcel tip. Square roots in answers stay as surds when not perfect (covered in surds topic). Don't approximate $\sqrt{17}$ as $4.12$ unless asked — keep it as $\sqrt{17}$ .

$\sqrt{x} \ge 0$ in 4MA1.
$\sqrt[3]{}$ works for negatives.
Even roots: $x \ge 0$ required.
Memorise to $15^{2}$ and $10^{3}$ .

Quick recap

Index laws: product, quotient, power, zero, negative, fractional.
Same-base only for product/quotient.
Zero index: $1$ .
Negative: reciprocal.
Fractional: root first, then power.
Square roots: positive only.
Cube roots: positive AND negative work.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Product: $x^{a} \times x^{b} = x^{a+b}$ .
Quotient: $x^{a} / x^{b} = x^{a-b}$ .
Power of power: $(x^{a})^{b} = x^{ab}$ .
Zero: $x^{0} = 1$ .
Negative: $x^{-n} = 1/x^{n}$ .
Fractional: $x^{m/n} = \sqrt[n]{x^{m}}$ — root first.

How it’s examined

Index laws appear on EVERY 4MA1 paper. Standalone questions (3-5 marks) on simplifying expressions; embedded use throughout algebra (Topic 2), surds (later in Topic 1), and standard form. Examiner reports flag (1) confusing $-$ with reciprocal for negative indices, (2) computing $x^{0} = 0$ , and (3) computing fractional powers using power-first instead of root-first on Paper 1H.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2023, 4MA1/1H Jan 2024. Last reviewed 2026-05-07.

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

1Apply index laws to simplify
Foundation• index laws
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Question
Simplify $\dfrac{x^{8} \times x^{3}}{x^{5}}$ .
Step-by-step solution
1. Step 1
  $x^{8} \times x^{3} = x^{8+3} = x^{11}$ (multiplication: ADD indices).
2. Step 2
  $\dfrac{x^{11}}{x^{5}} = x^{11-5} = x^{6}$ (division: SUBTRACT indices).
Answer
$x^{6}$
Examiner tip
Edexcel mark schemes are strict: ADD for multiplication, SUBTRACT for division. Mistakes here cascade through algebraic questions.
2Evaluate a negative index
Higher• Adapted from 4MA1/1H May/Jun 2023 Q9• negative index
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Question
Evaluate $\left(\dfrac{2}{3}\right)^{-2}$ .
Step-by-step solution
1. Step 1
  Negative index = reciprocal: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}$ .
2. Step 2
  $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$ .
Answer
$\dfrac{9}{4}$ or $2\dfrac{1}{4}$
Examiner tip
$x^{-n} = \dfrac{1}{x^{n}}$ . For fractions: flip first, then power. Mark scheme awards M1 for the reciprocal step, A1 for the answer.
3Evaluate a fractional index
Higher• fractional index, roots
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Question
Evaluate $27^{2/3}$ .
Step-by-step solution
1. Step 1
  $x^{m/n} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$ .
2. Step 2
  Take the cube root first (smaller numbers): $27^{1/3} = 3$ .
3. Step 3
  Then square: $3^{2} = 9$ .
Answer
$9$
Examiner tip
ALWAYS take the root first, then power. Smaller numbers → less arithmetic error. $27^{2/3} = (27^{1/3})^{2} = 3^{2} = 9$ .
4Anything to the power of zero
Foundation• zero index
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Question
Evaluate $5^{0}$ and $(-7)^{0}$ .
Step-by-step solution
1. Step 1
  $x^{0} = 1$ for any $x \ne 0$ .
2. Step 2
  Both equal $1$ — including the negative example.
Answer
$5^{0} = 1$ , $(-7)^{0} = 1$
Examiner tip
$x^{0} = 1$ is a definition that comes from the rule $x^{a-a} = \dfrac{x^{a}}{x^{a}} = 1$ . Don't write $x^{0} = x$ — that's the most common error.
5Combine multiple index laws
Higher• Adapted from 4MA1/1H Jan 2024 Q11• mixed indices
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Question
Simplify $\dfrac{(2x^{3})^{2} \times x^{-1}}{x^{4}}$ .
Step-by-step solution
1. Step 1
  Expand the bracket: $(2x^{3})^{2} = 2^{2} \times x^{6} = 4x^{6}$ .
2. Step 2
  Substitute: $\dfrac{4x^{6} \times x^{-1}}{x^{4}}$ .
3. Step 3
  Numerator: $x^{6} \times x^{-1} = x^{5}$ . So $\dfrac{4x^{5}}{x^{4}}$ .
4. Step 4
  Divide: $\dfrac{4x^{5}}{x^{4}} = 4x^{5-4} = 4x$ .
Answer
$4x$
Examiner tip
Multi-step index questions: handle ONE rule at a time. Take care of the bracket first ( $(ab)^{n} = a^{n}b^{n}$ ), then combine in the numerator, then divide.
6Estimate a square root without a calculator
Foundation• square root, estimation
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Question
Estimate $\sqrt{50}$ to 1 decimal place WITHOUT using a calculator. Justify your estimate.
Step-by-step solution
1. Step 1
  Find the nearest perfect squares: $7^{2} = 49$ and $8^{2} = 64$ .
2. Step 2
  $\sqrt{50}$ is between $7$ and $8$ , closer to $7$ .
3. Step 3
  Estimate: $\sqrt{50} \approx 7.1$ (since $7.1^{2} = 50.41$ , very close).
Answer
$\sqrt{50} \approx 7.1$
Examiner tip
Paper 1H is non-calculator. For square roots of non-perfect squares, find the bracket of perfect squares first, then estimate the position.
7Numbers that are both square and cube [1 mark]
Getting started• Paper 1H/2H style — 1 mark• square numbers, cube numbers
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Question
Here is a list of numbers.

$1 \quad 12 \quad 25 \quad 27 \quad 48 \quad 64 \quad 81 \quad 100$

Write down the numbers from the list that are both a square number and a cube number. [1]
Step-by-step solution
1. Step 1
  The square numbers in the list are $1$ , $25$ , $64$ , $81$ and $100$ . The cube numbers are $1$ , $27$ and $64$ .
  $1 = 1^{2} = 1^{3}, \qquad 64 = 8^{2} = 4^{3}$
2. Step 2
  B1 The numbers in both lists are $1$ and $64$ .
Answer
$1$ and $64$
Examiner tip
Examiner tip: Both numbers are needed for the mark. A number that is both square and cube is a sixth power: $1^{6} = 1$ , $2^{6} = 64$ .

Common mistake: Leaving out $1$ because it feels too simple, or including $81$ — it is $3^{4}$ , a square, but not a cube.
8A cube root and a negative power of decimals [2 marks]
Getting started• Paper 1H/2H style — 2 marks• cube root, negative index, decimals
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Question
(a) Write down the value of $\sqrt[3]{0.064}$ [1]

(b) Work out the value of $0.2^{-2}$ [1]
Step-by-step solution
1. Step 1
  B1 (a) $0.064 = \frac{64}{1000}$ , and both $64$ and $1000$ are cubes.
  $\sqrt[3]{\dfrac{64}{1000}} = \dfrac{4}{10} = 0.4$
2. Step 2
  B1 (b) $0.2 = \frac{1}{5}$ . A negative index takes the reciprocal, then square.
  $0.2^{-2} = \left(\dfrac{1}{5}\right)^{-2} = 5^{2} = 25$
Answer
(a) $0.4$ (b) $25$
Examiner tip
Examiner tip: Change a decimal into a fraction before rooting or applying a negative power; the numbers you know ( $4^{3}$ , $10^{3}$ , $5^{2}$ ) then appear.

Common mistake: Giving $0.04$ in (a) — check by cubing: $0.04^{3} = 0.000064$ , not $0.064$ .
9Ordering negative and fractional powers [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• negative index, fractional index, ordering
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Question
Write these numbers in order of size. Start with the smallest.

$2^{-3} \qquad 3^{-2} \qquad 25^{-\frac{1}{2}} \qquad 64^{-\frac{1}{3}} \qquad 10^{-1}$

Show your working clearly. [3]
Step-by-step solution
1. Step 1
  B2 Write each number as a unit fraction (B1 for at least three correct).
  $2^{-3} = \tfrac{1}{8}, \quad 3^{-2} = \tfrac{1}{9}, \quad 25^{-\frac{1}{2}} = \tfrac{1}{5}, \quad 64^{-\frac{1}{3}} = \tfrac{1}{4}, \quad 10^{-1} = \tfrac{1}{10}$
2. Step 2
  B1 With the same numerator, the larger the denominator the smaller the number, so order the denominators $10, 9, 8, 5, 4$ .
  $10^{-1},\; 3^{-2},\; 2^{-3},\; 25^{-\frac{1}{2}},\; 64^{-\frac{1}{3}}$
Answer
$10^{-1}$ , $3^{-2}$ , $2^{-3}$ , $25^{-\frac{1}{2}}$ , $64^{-\frac{1}{3}}$
Examiner tip
Examiner tip: Give the final order in the original form, as the question lists them. The unit fractions are working, not the answer.

Common mistake: Ordering by denominator the wrong way round and putting $\frac{1}{4}$ first. With unit fractions, a bigger denominator means a smaller number.
10Powers of 5 written in terms of x [4 marks]
Stretch• Paper 1H/2H style — 4 marks• index laws, powers of a prime
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Question
$x = 5^{m}$

Write each of the following in terms of $x$

(a) $25^{m}$ [1]

(b) $5^{m+2}$ [1]

(c) $125^{m-1}$ [2]
Step-by-step solution
1. Step 1
  B1 (a) $25 = 5^{2}$ , so a power of a power multiplies the indices.
  $25^{m} = 5^{2m} = \left(5^{m}\right)^{2} = x^{2}$
2. Step 2
  B1 (b) Adding indices came from multiplying powers, so split the power.
  $5^{m+2} = 5^{m} \times 5^{2} = 25x$
3. Step 3
  M1 (c) Write $125$ as $5^{3}$ and expand the index.
  $125^{m-1} = 5^{3(m-1)} = 5^{3m-3} = \dfrac{5^{3m}}{5^{3}}$
4. Step 4
  A1 Replace $5^{3m}$ with $x^{3}$ .
  $\dfrac{\left(5^{m}\right)^{3}}{125} = \dfrac{x^{3}}{125}$
Answer
(a) $x^{2}$ (b) $25x$ (c) $\dfrac{x^{3}}{125}$
Examiner tip
Examiner tip: Read $5^{m+2}$ backwards through the multiplication law: a sum in the index means a product of powers. A difference means a quotient.

Common mistake: Writing $5^{m+2} = x^{2}$ or $x + 2$ . The $2$ in the index becomes a factor of $5^{2} = 25$ , not a power of $x$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
Write down the value of $5^{-2}$ [1]
Model answer
$5^{-2} = \dfrac{1}{5^{2}} = \dfrac{1}{25}$

Answer: $\dfrac{1}{25}$ (or $0.04$ )
Why this scores
Mark scheme:
- B1 for $\dfrac{1}{25}$ oe
Why this scores full marks: The reciprocal is taken before squaring. The common wrong answers, $-25$ and $-10$ , come from treating the minus sign as making the number negative.
Question 2
Paper 1H/2H style — 2 marks2 marks
Work out $\sqrt{196} - \sqrt[3]{-27}$ [2]
Model answer
$\sqrt{196} = 14, \qquad \sqrt[3]{-27} = -3 \quad \text{since } (-3)^{3} = -27$

$14 - (-3) = 17$

Answer: $17$
Why this scores
Mark scheme:
- M1 for $14$ and $-3$ seen
- A1 for $17$ cao
Why this scores full marks: The candidate knows a negative number has a real cube root (unlike a square root) and keeps the bracket in $14 - (-3)$ , so the subtraction becomes an addition. Scripts writing $14 - 3 = 11$ keep only the M mark.
Question 3
Paper 1H/2H style — 2 marks2 marks
Explain why $\sqrt[3]{100}$ lies between $4$ and $5$ [2]
Model answer
$4^{3} = 64 \qquad 5^{3} = 125$

$100$ is between $64$ and $125$ , so $\sqrt[3]{100}$ is between $\sqrt[3]{64} = 4$ and $\sqrt[3]{125} = 5$
Why this scores
Mark scheme:
- B1 for $64$ and $125$ (or $4^{3}$ and $5^{3}$ ) both seen
- B1 for a complete explanation: $64 < 100 < 125$ , so the cube root lies between $4$ and $5$
Why this scores full marks: The explanation states both cubes and says explicitly that $100$ lies between them. Quoting a calculator value of $4.64$ is not an explanation and scores $0$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
Find the value of $\left(2\dfrac{1}{4}\right)^{-\frac{1}{2}}$

Give your answer as a fraction. Show your working clearly. [3]
Model answer
$2\dfrac{1}{4} = \dfrac{9}{4}$

$\left(\dfrac{9}{4}\right)^{-\frac{1}{2}} = \left(\dfrac{4}{9}\right)^{\frac{1}{2}} = \dfrac{\sqrt{4}}{\sqrt{9}} = \dfrac{2}{3}$

Answer: $\dfrac{2}{3}$
Why this scores
Mark scheme:
- M1 for $\dfrac{9}{4}$
- M1 for $\dfrac{4}{9}$ or $\dfrac{3}{2}$ seen
- A1 for $\dfrac{2}{3}$ cao
Why this scores full marks: The mixed number is converted to an improper fraction first. Applying the power to the $2$ and the $\frac{1}{4}$ separately is the error that sinks weaker scripts.
Question 5
Paper 1H/2H style — 3 marks3 marks
$\sqrt{27} = 9^{k}$

Find the value of $k$ . Show your working clearly. [3]
Model answer
$\sqrt{27} = \left(3^{3}\right)^{\frac{1}{2}} = 3^{\frac{3}{2}}$

$9^{k} = \left(3^{2}\right)^{k} = 3^{2k}$

$2k = \dfrac{3}{2}$

Answer: $k = \dfrac{3}{4}$
Why this scores
Mark scheme:
- M1 for $3^{\frac{3}{2}}$ or $3^{2k}$
- M1 for $2k = \dfrac{3}{2}$ oe
- A1 for $k = \dfrac{3}{4}$ oe
Why this scores full marks: Neither $27$ nor $9$ is a power of the other, so the candidate rewrites both in base $3$ — the common prime — before equating indices.
Question 6
Paper 1H/2H style — 3 marks3 marks
$N = 2^{4} \times 3^{6} \times 11^{2}$

(a) Write $\sqrt{N}$ as a product of powers of its prime factors. [2]

(b) Work out the value of $\sqrt{N}$ [1]
Model answer
(a) $\sqrt{N} = 2^{\frac{4}{2}} \times 3^{\frac{6}{2}} \times 11^{\frac{2}{2}} = 2^{2} \times 3^{3} \times 11$

(b) $4 \times 27 \times 11 = 1188$

Answers: (a) $2^{2} \times 3^{3} \times 11$ (b) $1188$
Why this scores
Mark scheme:
- M1 for halving at least two of the indices
- A1 for $2^{2} \times 3^{3} \times 11$ oe
- B1 for $1188$ (ft their (a))
Why this scores full marks: The candidate halves each index rather than square-rooting each base. Scripts that take $\sqrt{2} \times \sqrt{3} \times \sqrt{11}$ confuse the base with the index.
Question 7
Paper 1H/2H style — 4 marks4 marks
Work out $\dfrac{3^{-2} \times 3^{5}}{3^{4}} + 4^{-\frac{1}{2}}$

Give your answer as a fraction. Show your working clearly. [4]
Model answer
$\dfrac{3^{-2} \times 3^{5}}{3^{4}} = 3^{-2 + 5 - 4} = 3^{-1} = \dfrac{1}{3}$

$4^{-\frac{1}{2}} = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}$

$\dfrac{1}{3} + \dfrac{1}{2} = \dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6}$

Answer: $\dfrac{5}{6}$
Why this scores
Mark scheme:
- M1 for $3^{-2 + 5 - 4}$ or $3^{-1}$
- A1 for $\dfrac{1}{3}$
- B1 for $4^{-\frac{1}{2}} = \dfrac{1}{2}$
- B1 for $\dfrac{5}{6}$ cao
Why this scores full marks: The index laws are applied to the base- $3$ term only; the $4^{-\frac{1}{2}}$ is evaluated separately, since different bases cannot be combined by adding indices.
Question 8
Paper 1H/2H style — 4 marks4 marks
$\sqrt[3]{p} = 4$ and $q^{-2} = \dfrac{1}{81}$ , where $q > 0$

Work out the value of $p^{\frac{2}{3}} \times q^{-\frac{1}{2}}$

Show your working clearly. [4]
Model answer
$p = 4^{3} = 64$

$q^{2} = 81, \ q > 0 \ \Rightarrow \ q = 9$

$p^{\frac{2}{3}} = \left(\sqrt[3]{64}\right)^{2} = 16, \qquad q^{-\frac{1}{2}} = \dfrac{1}{\sqrt{9}} = \dfrac{1}{3}$

$16 \times \dfrac{1}{3} = \dfrac{16}{3}$

Answer: $\dfrac{16}{3}$
Why this scores
Mark scheme:
- B1 for $p = 64$
- B1 for $q = 9$
- M1 for $16$ or $\dfrac{1}{3}$ from their $p$ or $q$
- A1 for $\dfrac{16}{3}$ oe
Why this scores full marks: The candidate uses the condition $q > 0$ to reject $-9$ , and notices that $p^{\frac{2}{3}}$ is simply $(\sqrt[3]{p})^{2} = 4^{2}$ without needing $p$ at all — a useful check on the value $64$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
$N = 2^{5} \times 3^{4} \times 5$

Show your working clearly.

(a) Find the smallest positive integer $k$ such that $kN$ is a square number. [2]

(b) Find the smallest positive integer $m$ such that $mN$ is a cube number. [2]

(c) Hence write down the value of $\sqrt[3]{mN}$ [1]
Model answer
(a) For a square number every index must be even. $3^{4}$ is already even; $2^{5}$ and $5^{1}$ each need one more factor.

$kN = 2^{6} \times 3^{4} \times 5^{2} \quad\Rightarrow\quad k = 2 \times 5 = 10$

(b) For a cube number every index must be a multiple of $3$ .

$mN = 2^{6} \times 3^{6} \times 5^{3} \quad\Rightarrow\quad m = 2 \times 3^{2} \times 5^{2} = 450$

(c) $\sqrt[3]{2^{6} \times 3^{6} \times 5^{3}} = 2^{2} \times 3^{2} \times 5 = 4 \times 9 \times 5 = 180$

Answers: (a) $k = 10$ (b) $m = 450$ (c) $180$
Why this scores
Mark scheme:
- M1 for identifying that the indices of $2$ and $5$ must be made even, e.g. $2^{6} \times 3^{4} \times 5^{2}$
- A1 for $k = 10$ cao
- M1 for $2^{6} \times 3^{6} \times 5^{3}$ or for $2 \times 3^{2} \times 5^{2}$ oe
- A1 for $m = 450$ cao
- B1 for $180$ (ft their (b) if every index is a multiple of $3$ )
Why this scores full marks: The candidate works entirely with the indices — raise each one to the next even number for a square, to the next multiple of $3$ for a cube — and writes the target number in index form before reading off $k$ and $m$ . The common wrong answer to (b) is $m = 2 \times 3 \times 5 = 30$ , from supplying one factor of each prime instead of enough to reach a multiple of $3$ . In (c), dividing every index by $3$ gives the cube root with no calculation of $mN = 5\,832\,000$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
A cube has a volume of $2^{9} \times 3^{6} \times 5^{3}$ cm³

(a) Find the length of one edge of the cube. [2]

(b) Find the total surface area of the cube. Give your answer as a product of powers of its prime factors. [3]
Model answer
(a) $\sqrt[3]{2^{9} \times 3^{6} \times 5^{3}} = 2^{3} \times 3^{2} \times 5 = 8 \times 9 \times 5 = 360$

Edge length $= 360$ cm

(b) Area of one face:

$\left(2^{3} \times 3^{2} \times 5\right)^{2} = 2^{6} \times 3^{4} \times 5^{2}$

Six faces, and $6 = 2 \times 3$ :

$2 \times 3 \times 2^{6} \times 3^{4} \times 5^{2} = 2^{7} \times 3^{5} \times 5^{2}$

Answers: (a) $360$ cm (b) $2^{7} \times 3^{5} \times 5^{2}$ cm²
Why this scores
Mark scheme:
- M1 for dividing each index by $3$
- A1 for $360$ or $2^{3} \times 3^{2} \times 5$
- M1 for squaring their edge in index form, e.g. $2^{6} \times 3^{4} \times 5^{2}$
- M1 for multiplying by $6$ written as $2 \times 3$
- A1 for $2^{7} \times 3^{5} \times 5^{2}$ cao
Why this scores full marks: Staying in index form throughout makes the cube root and the square immediate. Writing the $6$ as $2 \times 3$ is the step most candidates miss — they leave $6 \times 2^{6} \times 3^{4} \times 5^{2}$ , which is not a product of powers of primes and loses the final A mark.

4MA1 Powers and Roots

At a glance

What you’ll learn

The seven index laws

Fractional and negative indices

Square, cube and other roots

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Apply index laws to simplify

2Evaluate a negative index

3Evaluate a fractional index

4Anything to the power of zero

5Combine multiple index laws

6Estimate a square root without a calculator

7Numbers that are both square and cube [1 mark]

8A cube root and a negative power of decimals [2 marks]

9Ordering negative and fractional powers [3 marks]

10Powers of 5 written in terms of x [4 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10