Mathematics · AQA 8300 §N5

Powers and Roots

AQA 8300 §N5 ⭐⭐⭐ ⏰ 45 mins AQA · Edexcel · OCR Grade 9

Apply all five index laws correctly to numerical and algebraic expressions
Evaluate expressions with zero and negative indices, including $a^0=1$ and $a^{-n}=\tfrac{1}{a^n}$
Work with fractional indices — evaluate $a^{m/n} = \bigl(\sqrt[n]{a}\bigr)^m$
Simplify algebraic index expressions involving multiple index laws in one problem
Solve equations involving indices by converting to the same base and equating indices

🔑 Core Concepts

Powers and Indices — The Basics

📚

DEFINITION — Index Notation

In $a^n$, the number $a$ is the base and $n$ is the index (also called the power or exponent). The index tells you how many times the base is multiplied by itself: $$a^n = \underbrace{a \times a \times a \times \cdots \times a}_{n \text{ times}}$$

The two most important cases are squares (index 2) and cubes (index 3):

$5^2 = 5 \times 5 = 25$ — "5 squared" $4^3 = 4 \times 4 \times 4 = 64$ — "4 cubed"

Square and Cube Numbers

$$n^2 = n \times n \qquad n^3 = n \times n \times n$$

$n^2$ = $n$ squared (a perfect square) $n^3$ = $n$ cubed (a perfect cube)

🎯

EXAM TIP

Know your square numbers up to $15^2 = 225$ and cube numbers up to $5^3 = 125$. Being fluent with these saves critical time in non-calculator papers.
Key cubes: $1^3=1,\ 2^3=8,\ 3^3=27,\ 4^3=64,\ 5^3=125$.

✗

COMMON MISTAKE

$3^2 \neq 3 \times 2 = 6$. The correct value is $3^2 = 3 \times 3 = 9$. The index says how many copies of the base are multiplied together — it does not multiply base by index.

Square Roots and Cube Roots

📚

DEFINITION — Roots

The square root of $a$, written $\sqrt{a}$, is the positive number that when squared gives $a$. The cube root of $a$, written $\sqrt[3]{a}$, is the number that when cubed gives $a$. More generally, the $n$th root of $a$ is $\sqrt[n]{a}$, satisfying $\bigl(\sqrt[n]{a}\bigr)^n = a$.

Roots as Fractional Indices

$$\sqrt{a} = a^{1/2} \qquad \sqrt[3]{a} = a^{1/3} \qquad \sqrt[n]{a} = a^{1/n}$$

$\sqrt{a}$ = square root (positive value) $\sqrt[n]{a}$ = $n$th root of $a$

✗

COMMON MISTAKE — Positive vs Negative Roots

$\sqrt{25} = 5$ only (the $\sqrt{\phantom{0}}$ symbol always gives the positive root). However, when solving $x^2 = 25$, the answer is $x = \pm 5$ because both $5^2 = 25$ and $(-5)^2 = 25$. These are different questions.

🎯

EXAM TIP — Simplifying Surds

Non-integer square roots can be simplified: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$. Always look for the largest perfect-square factor.

The Five Index Laws

These five laws govern all arithmetic with indices. They must be memorised and applied with precision — Grade 9 questions chain multiple laws in a single problem.

Law 1 — Multiplying Powers (Same Base)

Index Law 1 — Multiply

$$a^m \times a^n = a^{m+n}$$

$a$ = base (must be identical on both sides) $m, n$ = indices to be added

Why it works: $a^m \times a^n = \underbrace{(a \times \cdots \times a)}_{m} \times \underbrace{(a \times \cdots \times a)}_{n} = \underbrace{a \times \cdots \times a}_{m+n} = a^{m+n}$

Example: $x^4 \times x^3 = x^{4+3} = x^7$

✗

COMMON MISTAKE

Law 1 requires identical bases. You cannot simplify $2^3 \times 3^4$ using this law. The bases (2 and 3) are different.

Law 2 — Dividing Powers (Same Base)

Index Law 2 — Divide

$$a^m \div a^n = a^{m-n}$$

$a$ = base (must be identical) Subtract: top index minus bottom index

Example: $y^9 \div y^4 = y^{9-4} = y^5$ Also: $\dfrac{x^3}{x^7} = x^{3-7} = x^{-4} = \dfrac{1}{x^4}$

🎯

EXAM TIP

Write fractions as $a^{m-n}$ first, then decide if the result is positive or negative. If $m < n$ you'll get a negative index (which you may need to convert).

Law 3 — Power of a Power

Index Law 3 — Power of a Power

$$(a^m)^n = a^{mn}$$

$a$ = base $m, n$ = indices to be multiplied together

Why it works: $(a^m)^n = \underbrace{a^m \times a^m \times \cdots \times a^m}_{n} = a^{m+m+\cdots+m} = a^{mn}$

Example: $(x^3)^5 = x^{15}$ Careful with coefficients: $(2x^3)^4 = 2^4 \times (x^3)^4 = 16x^{12}$

✗

COMMON MISTAKE — Add vs Multiply

$(x^3)^4 = x^{12}$ (multiply), not $x^{3+4} = x^7$ (add). A power outside brackets means multiply the indices. A power next to a power on the same level means add (Law 1).

🎯

EXAM TIP

The power outside a bracket applies to every factor inside: $(3ab^2)^3 = 3^3 \cdot a^3 \cdot (b^2)^3 = 27a^3b^6$. Students frequently forget to apply the power to the numerical coefficient.

Law 4 — Zero Index

Index Law 4 — Zero Index

$$a^0 = 1 \qquad (a \neq 0)$$

$a$ = any non-zero number $0^0$ is undefined

Proof using Law 2: $a^n \div a^n = a^{n-n} = a^0$. But also $a^n \div a^n = 1$ (any non-zero value divided by itself). Therefore $a^0 = 1$.

This holds for any non-zero base: $\pi^0 = 1$, $(-7)^0 = 1$, $(2x+3)^0 = 1$.

✗

COMMON MISTAKE

$5^0 \neq 0$. The result is always $1$, never $0$. It is the index that is zero, not the answer.

Law 5 — Negative Index

Index Law 5 — Negative Index

$$a^{-n} = \frac{1}{a^n}$$

$a^{-n}$ = reciprocal of $a^n$ Negative index means "flip"

Proof using Law 2: $a^0 \div a^n = a^{0-n} = a^{-n}$. But $a^0 \div a^n = \dfrac{1}{a^n}$. Therefore $a^{-n} = \dfrac{1}{a^n}$.

Examples: $2^{-3} = \dfrac{1}{8}$ $x^{-4} = \dfrac{1}{x^4}$ $\dfrac{1}{5^{-2}} = 5^2 = 25$ $\left(\dfrac{2}{3}\right)^{-1} = \dfrac{3}{2}$

🧠

MEMORY TRICK

A negative index says "move me to the other side of the fraction line." $a^{-n}$ in the numerator becomes $a^n$ in the denominator, and vice versa. $$\frac{a^{-2}}{b^{-3}} = \frac{b^3}{a^2}$$

🎯

EXAM TIP — Negative Fractional Index

For a fraction with a negative index: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}$. Flip the fraction, then make the index positive. $$\left(\frac{4}{9}\right)^{-3/2} = \left(\frac{9}{4}\right)^{3/2}$$

Fractional Indices

📚

DEFINITION — Fractional Index

A fractional index combines a root and a power. The denominator of the fraction tells you which root to take; the numerator tells you which power to raise to. $$a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$$ Both forms give the same result; taking the root first keeps the numbers smaller.

Fractional Index — General Rule

$$a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$$

$n$ (denominator) = which root to take $m$ (numerator) = which power to raise to

Step-by-step evaluation of $a^{m/n}$:

Take the $n$th root of the base: $\sqrt[n]{a}$
Raise the result to the power $m$: $\bigl(\sqrt[n]{a}\bigr)^m$

Example: Evaluate $32^{3/5}$

$\sqrt[5]{32} = 2$ (since $2^5 = 32$), then $2^3 = 8$. So $32^{3/5} = 8$.

Combining negative and fractional: For $a^{-m/n}$, take the reciprocal first (or at any stage): $a^{-m/n} = \dfrac{1}{a^{m/n}}$

✗

COMMON MISTAKE

$a^{2/3} \neq a \times \dfrac{2}{3}$. A fractional index is not multiplication. $8^{2/3} = \bigl(\sqrt[3]{8}\bigr)^2 = 4$, not $8 \times \frac{2}{3} \approx 5.3$.

🎯

EXAM TIP — Root Before Power

Always take the root first, then raise to the power. Compare:

Root first: $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$ ✓ (small numbers)
Power first: $27^{2/3} = \sqrt[3]{27^2} = \sqrt[3]{729} = 9$ ✓ (harder arithmetic)

Simplifying Algebraic Index Expressions

When an expression contains a number coefficient and one or more variable bases, simplify in three steps: (1) simplify the number parts; (2) apply index laws to each variable base separately; (3) write the final answer with positive indices unless instructed otherwise.

✏

WORKED EXAMPLE — Multiplication

Simplify $4x^3y^2 \times 3x^{-1}y^4$:
Numbers: $4 \times 3 = 12$
$x$ terms: $x^3 \times x^{-1} = x^{3+(-1)} = x^2$
$y$ terms: $y^2 \times y^4 = y^{6}$
Answer: $12x^2y^6$

✏

WORKED EXAMPLE — Division (Fraction)

Simplify $\dfrac{6a^5b^{-2}}{2a^3b^4}$, giving your answer with positive indices: $$\frac{6a^5b^{-2}}{2a^3b^4} = \frac{6}{2} \cdot a^{5-3} \cdot b^{-2-4} = 3a^2b^{-6} = \frac{3a^2}{b^6}$$

⚠

IMPORTANT — Expressions Raised to a Power

When an entire fraction is raised to a power, apply that power to every factor in the numerator and denominator: $$\left(\frac{2x^3}{y^2}\right)^4 = \frac{2^4 \cdot (x^3)^4}{(y^2)^4} = \frac{16x^{12}}{y^8}$$

🗺 Visual Notes

Powers & Roots

Index Laws 1–3

$a^m \times a^n = a^{m+n}$ — add
$a^m \div a^n = a^{m-n}$ — subtract
$(a^m)^n = a^{mn}$ — multiply
All require the same base

Special Indices

$a^0 = 1$ (any non-zero $a$)
$a^{-n} = \dfrac{1}{a^n}$ — reciprocal
$a^{1/2} = \sqrt{a}$ — square root
$a^{1/n} = \sqrt[n]{a}$ — $n$th root

Fractional Indices

$a^{m/n} = \bigl(\sqrt[n]{a}\bigr)^m$
Denominator $\rightarrow$ root
Numerator $\rightarrow$ power
Take root first — stay small!

Algebraic Simplification

Separate numbers from variables
Apply index law to each variable
Convert negative indices at end
Power outside bracket: apply to ALL

Solving Index Equations

Express both sides as same base
Equate the exponents
Solve the resulting equation
Check: substitute answer back

All Six Index Rules — Summary Table

Rule	Formula	Numerical Example	Algebraic Example
Multiply (same base)	$a^m \times a^n = a^{m+n}$	$2^3 \times 2^5 = 2^8 = 256$	$x^4 \times x^3 = x^7$
Divide (same base)	$a^m \div a^n = a^{m-n}$	$5^7 \div 5^3 = 5^4 = 625$	$y^9 \div y^4 = y^5$
Power of a power	$(a^m)^n = a^{mn}$	$(3^2)^4 = 3^8 = 6561$	$(x^3)^5 = x^{15}$
Zero index	$a^0 = 1$	$17^0 = 1$	$(2x+1)^0 = 1$
Negative index	$a^{-n} = \dfrac{1}{a^n}$	$2^{-5} = \dfrac{1}{32}$	$x^{-3} = \dfrac{1}{x^3}$
Fractional index	$a^{m/n} = \bigl(\!\sqrt[n]{a}\bigr)^m$	$8^{2/3} = (\sqrt[3]{8})^2 = 4$	$x^{3/2} = (\sqrt{x})^3$

Decision Process — Evaluating $a^{m/n}$

Read $a^{m/n}$: identify $a$, $m$, $n$

→

Is the index negative?
If yes → take reciprocal: $\dfrac{1}{a^{m/n}}$

→

Take the $n$th root of $a$: $\sqrt[n]{a}$

→

Raise to power $m$: $\bigl(\sqrt[n]{a}\bigr)^m$

→

Write final simplified answer

Comparison Table — Base 4, Various Indices

Expression	Rewrite as	Step	Answer
$4^2$	$4 \times 4$	Multiply	$16$
$4^0$	$1$	Zero index law	$1$
$4^{-2}$	$\dfrac{1}{4^2}$	Negative index	$\dfrac{1}{16}$
$4^{1/2}$	$\sqrt{4}$	Square root	$2$
$4^{3/2}$	$(\sqrt{4})^3 = 2^3$	Root then power	$8$
$4^{-1/2}$	$\dfrac{1}{\sqrt{4}} = \dfrac{1}{2}$	Reciprocal of root	$\dfrac{1}{2}$
$4^{-3/2}$	$\dfrac{1}{(\sqrt{4})^3} = \dfrac{1}{8}$	Reciprocal of root-then-power	$\dfrac{1}{8}$

✏️ Worked Examples

Grade 4–5

Simplify: (a) $3^5 \times 3^{-2}$ (b) $\dfrac{2^8}{2^3}$ (c) $(x^4)^3$ (d) $7^0 \times 7^2$

Part (a) — Law 1: add the indices

Both bases are 3. The indices are 5 and $-2$: $$3^5 \times 3^{-2} = 3^{5+(-2)} = 3^3 = 27$$

Part (b) — Law 2: subtract the indices

Both bases are 2. Subtract the denominator index from the numerator index: $$\frac{2^8}{2^3} = 2^{8-3} = 2^5 = 32$$

Part (c) — Law 3: multiply the indices

The power outside the bracket multiplies the index inside: $$(x^4)^3 = x^{4 \times 3} = x^{12}$$

Part (d) — Zero index then Law 1

$7^0 = 1$ by the zero-index law. Then: $$7^0 \times 7^2 = 1 \times 49 = 49$$ Or equivalently, $7^{0+2} = 7^2 = 49$.

(a) $3^3 = 27$ | (b) $2^5 = 32$ | (c) $x^{12}$ | (d) $49$

Grade 6–7

(a) Evaluate $27^{2/3}$. (b) Evaluate $\left(\dfrac{4}{9}\right)^{-3/2}$. (c) Simplify $\dfrac{8x^5y^{-2}}{4x^{-1}y^3}$ with positive indices.

Part (a) — Fractional index: root first, then power

Index $\tfrac{2}{3}$: denominator 3 means cube root; numerator 2 means square. $$27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$$

Part (b) — Negative fractional index: flip the fraction first

Negative index on a fraction: flip numerator and denominator, then make positive: $$\left(\frac{4}{9}\right)^{-3/2} = \left(\frac{9}{4}\right)^{3/2} = \left(\sqrt{\frac{9}{4}}\right)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$$

Part (c) — Simplify algebraically: separate numbers, then each variable

Number part: $\dfrac{8}{4} = 2$. $x$ part: $x^{5-(-1)} = x^6$. $y$ part: $y^{-2-3} = y^{-5}$. $$\frac{8x^5y^{-2}}{4x^{-1}y^3} = 2 \cdot x^{5-(-1)} \cdot y^{-2-3} = 2x^6y^{-5} = \frac{2x^6}{y^5}$$

(a) $9$ | (b) $\dfrac{27}{8}$ | (c) $\dfrac{2x^6}{y^5}$

Grade 9

Given that $8^x \times 4^y = 2^{10}$ and $27^x \div 9^y = 3^2$, find the values of $x$ and $y$.

Express 8 and 4 as powers of 2

$8 = 2^3$ and $4 = 2^2$. Apply Law 3 to rewrite the first equation: $$8^x \times 4^y = (2^3)^x \times (2^2)^y = 2^{3x} \times 2^{2y} = 2^{3x+2y}$$

Form the first simultaneous equation by equating indices

Since the bases are both 2 and $2^{3x+2y} = 2^{10}$, the indices must be equal: $$3x + 2y = 10 \qquad \cdots (1)$$

Express 27 and 9 as powers of 3

$27 = 3^3$ and $9 = 3^2$. Apply Laws 2 and 3 to the second equation: $$27^x \div 9^y = (3^3)^x \div (3^2)^y = 3^{3x} \div 3^{2y} = 3^{3x-2y}$$

Form the second simultaneous equation

Since $3^{3x-2y} = 3^2$, equate the indices: $$3x - 2y = 2 \qquad \cdots (2)$$

Solve the simultaneous equations

Add equations (1) and (2): $6x = 12$, so $x = 2$.
Substitute $x = 2$ into (1): $3(2) + 2y = 10 \Rightarrow 6 + 2y = 10 \Rightarrow y = 2$.

Verify both equations

Check (1): $8^2 \times 4^2 = 64 \times 16 = 1024 = 2^{10}$ ✓
Check (2): $27^2 \div 9^2 = 729 \div 81 = 9 = 3^2$ ✓

$x = 2, \quad y = 2$

❓ Exam Questions

Q1 1 mark

Write down the value of $4^0 + 4^{-1}$.

Mark Scheme:
$4^0 = 1$ and $4^{-1} = \dfrac{1}{4}$ [implied by correct answer]
$1 + \dfrac{1}{4} = \dfrac{5}{4}$ [B1]
Answer: $\dfrac{5}{4}$

Q2 2 marks

Evaluate $27^{\frac{2}{3}}$.

Mark Scheme:
$\sqrt[3]{27} = 3$ [M1]
$3^2 = 9$ [A1]
Answer: $9$

Q3 3 marks

Simplify $\dfrac{6a^5b^{-2}}{2a^3b^4}$. Give your answer with positive indices only.

Mark Scheme:
Number part: $\dfrac{6}{2} = 3$ [M1]
$a$ part: $a^{5-3} = a^2$ [M1]
$b$ part: $b^{-2-4} = b^{-6}$, written as $\dfrac{1}{b^6}$ [A1]
Answer: $\dfrac{3a^2}{b^6}$

Q4 4 marks

Solve $4^{x-1} = 8^x$.

Mark Scheme:
Write as powers of 2: $4 = 2^2$, $8 = 2^3$ [M1]
$(2^2)^{x-1} = (2^3)^x \Rightarrow 2^{2(x-1)} = 2^{3x}$ [M1]
Equate indices: $2x - 2 = 3x$ [M1]
$x = -2$ [A1]
Answer: $x = -2$

Q5 4 marks

(a) Simplify $(3x^2y)^3$. [2 marks]
(b) Simplify $\dfrac{x^5 \cdot x^{-3/2}}{x^{7/2}}$. [2 marks]

Part (a):
$(3x^2y)^3 = 3^3 \cdot (x^2)^3 \cdot y^3$ [M1]
$= 27x^6y^3$ [A1]

Part (b):
Numerator: $x^5 \cdot x^{-3/2} = x^{5-3/2} = x^{7/2}$ [M1]
$\dfrac{x^{7/2}}{x^{7/2}} = x^{7/2 - 7/2} = x^0 = 1$ [A1]
Answers: (a) $27x^6y^3$ (b) $1$

Q6 6 marks

(a) Simplify $\dfrac{(2x^2)^3}{4x^3 \cdot x^{-3}}$. [3 marks]
(b) Hence solve $\dfrac{(2x^2)^3}{4x^3 \cdot x^{-3}} = 128$. [3 marks]

Part (a):
Numerator: $(2x^2)^3 = 8x^6$ [M1]
Denominator: $4x^3 \cdot x^{-3} = 4x^{3+(-3)} = 4x^0 = 4$ [M1]
$\dfrac{8x^6}{4} = 2x^6$ [A1]

Part (b):
Set $2x^6 = 128$ [M1]
$x^6 = 64$ [M1]
$x = \pm\,64^{1/6} = \pm\,2$ (since $2^6 = 64$; both roots required) [A1]
Answers: (a) $2x^6$ (b) $x = \pm 2$

⭐ Grade 9 Model Answers

Full annotated solution to Q6 — the highest-demand question. Each annotation identifies exactly which examiner mark is earned and why.

Grade 9 — Full Mark Annotated Solution

(a) Simplify $\dfrac{(2x^2)^3}{4x^3 \cdot x^{-3}}$. [3 marks] (b) Hence solve $\dfrac{(2x^2)^3}{4x^3 \cdot x^{-3}} = 128$. [3 marks]

Expand the numerator — Law 3 applied to ALL factors [M1]

The bracket $(2x^2)^3$ has two factors: the coefficient 2 and the variable $x^2$. The power 3 applies to both: $$(2x^2)^3 = 2^3 \cdot (x^2)^3 = 8 \cdot x^{2 \times 3} = 8x^6$$ M1 is awarded here: correctly expanding a bracketed expression raised to a power.

Simplify the denominator — Law 1 produces a zero index [M1]

$$4x^3 \cdot x^{-3} = 4 \cdot x^{3+(-3)} = 4 \cdot x^0 = 4 \times 1 = 4$$ Key insight: $x^3 \times x^{-3} = x^0 = 1$, so the $x$ terms cancel completely. This earns M1. A common error is writing $4x^0$ without simplifying to $4$.

Divide to obtain the simplified expression [A1]

$$\frac{8x^6}{4} = 2x^6$$ A1 awarded for the correct fully simplified answer. The expression must contain no fractions and no compound bases.

Set the simplified expression equal to 128 — "hence" means use part (a) [M1]

From part (a): the expression equals $2x^6$. Set up the equation: $$2x^6 = 128$$ "Hence" is a compulsory instruction. Starting afresh without using $2x^6$ will not earn method marks even with a correct final answer.

Isolate $x^6$ [M1]

Divide both sides by 2: $$x^6 = 64$$ M1 awarded for correctly isolating $x^6$.

Solve for $x$ — both roots required for A1 [A1]

$$x^6 = 64 = 2^6 \implies x = \pm\,2$$ Since 6 is even, both $+2$ and $-2$ satisfy $x^6 = 64$ (check: $(-2)^6 = 64$ ✓). Grade 9 candidates must recognise even-powered equations have two solutions. Writing only $x = 2$ loses the A1. Both solutions are required for full marks.

(a) $2x^6$ (b) $x = \pm 2$

🎯

EXAM TIP — Responding to "Hence"

"Hence" is an examiner instruction to use the result of the previous part. If you ignore it and redo the calculation from scratch, you will not earn the method marks even if your final answer is correct. Always reference your previous answer explicitly.

⚠

IMPORTANT — Grade 9 Differentiator: Even Power Equations

When $n$ is even, $x^n = k$ (with $k > 0$) always has two solutions: $x = +k^{1/n}$ and $x = -k^{1/n}$. This is a deliberate Grade 9 trap. The A mark for this type of question is almost always awarded for including the negative solution.

📋 Revision Sheet

Key Definitions

Term	Meaning
Index / Power / Exponent	Tells you how many times the base is multiplied by itself
Base	The number being raised to a power
Square number	The result of $n^2$ for integer $n$: 1, 4, 9, 16, 25 …
Cube number	The result of $n^3$ for integer $n$: 1, 8, 27, 64, 125 …
Square root $\sqrt{a}$	Positive number that squares to give $a$
Cube root $\sqrt[3]{a}$	Number that cubes to give $a$
Fractional index $a^{m/n}$	Root ($n$th) then power ($m$); denominator = root
Negative index $a^{-n}$	Reciprocal of $a^n$: equals $\frac{1}{a^n}$

Essential Formulae

$a^m \times a^n = a^{m+n}$

$a^m \div a^n = a^{m-n}$

$(a^m)^n = a^{mn}$

$a^0 = 1 \quad (a \neq 0)$

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

$a^{1/n} = \sqrt[n]{a}$

$a^{m/n} = \bigl(\sqrt[n]{a}\bigr)^m$

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

Memory Hooks

Multiply → ADD the indices (same base only)
Divide → SUBTRACT the indices (same base only)
Power of a power → MULTIPLY the indices
Anything to the 0 = 1 (except $0^0$)
Negative index = FLIP (take reciprocal)
Fraction index: BOTTOM = root, TOP = power
Root FIRST — keeps numbers manageable
Even power = TWO solutions ($\pm$)

Exam Tips

Always verify bases are identical before using Laws 1 or 2
Power outside brackets applies to the coefficient too: $(2x)^3 = 8x^3$
To solve index equations, rewrite all terms as powers of the same base
"Hence" means use the previous answer — don't start over
Write answers with positive indices unless the question says otherwise
For $x^n = k$ with $n$ even: always write $x = \pm k^{1/n}$
$\sqrt{a} = a^{1/2}$ links the radical and index notation
Check your answer: substitute back into the original expression

🔄 Flashcards

Click a card to flip it and reveal the answer. Work through all 15 cards until you can answer each without hesitation.

✗ Common Mistakes

✗

MISTAKE 1 — Multiplying base by index

Wrong: $3^4 = 3 \times 4 = 12$
Correct: $3^4 = 3 \times 3 \times 3 \times 3 = 81$
Why marks are lost: This shows a fundamental misunderstanding of index notation and will lose all marks on any powers question.
How to avoid: The index is a count of how many copies of the base are multiplied together, never the multiplier itself.

✗

MISTAKE 2 — Applying index laws across different bases

Wrong: $2^3 \times 3^4 = 6^7$
Correct: $2^3 \times 3^4 = 8 \times 81 = 648$ (no simplification by index laws)
Why marks are lost: Laws 1 and 2 only work when the bases are identical. $2^3 \times 3^4$ cannot be simplified because the bases (2 and 3) differ.
How to avoid: Before applying any law, confirm all bases match.

✗

MISTAKE 3 — Adding instead of multiplying in $(a^m)^n$

Wrong: $(x^3)^4 = x^{3+4} = x^7$
Correct: $(x^3)^4 = x^{3 \times 4} = x^{12}$
Why marks are lost: Law 3 requires multiplication of indices, not addition. Confusing this with Law 1 (which does add) is a very common error.
How to avoid: A power outside a bracket means multiply; powers on the same level being multiplied together means add.

✗

MISTAKE 4 — Thinking $a^0 = 0$

Wrong: $9^0 = 0$
Correct: $9^0 = 1$
Why marks are lost: One mark is dedicated to the zero-index rule on many papers. Confusing zero power with zero result is a direct mark loss.
How to avoid: Prove it to yourself: $a^n \div a^n = 1$ (anything divided by itself). By Law 2, $a^n \div a^n = a^{n-n} = a^0$. Therefore $a^0 = 1$.

✗

MISTAKE 5 — Treating fractional indices as division

Wrong: $8^{2/3} = 8^2 \div 3 = 64 \div 3 \approx 21.3$
Correct: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$
Why marks are lost: A fractional index is about roots and powers, not arithmetic division. This mistake almost always produces a non-integer answer that should be a warning sign.
How to avoid: Remember: denominator = root, numerator = power. Take the $n$th root first, then raise to the power $m$.

✗

MISTAKE 6 — Omitting the negative solution in $x^n = k$ (even $n$)

Wrong: $x^4 = 81 \Rightarrow x = 3$ only
Correct: $x^4 = 81 \Rightarrow x = \pm 3$ (since $3^4 = 81$ and $(-3)^4 = 81$)
Why marks are lost: The A mark in these questions is almost always specifically for giving both the positive and negative root.
How to avoid: Whenever the index is an even integer and the right-hand side is positive, there are always two solutions. Always write $\pm$.

✅ Final Checklist

Click each item once you are confident with it. Your progress is saved automatically.

I can state what index notation means: $a^n$ = base $a$ multiplied by itself $n$ times
I know square numbers to $15^2 = 225$ and cube numbers to $5^3 = 125$
I can apply Law 1: $a^m \times a^n = a^{m+n}$ (same base, add indices)
I can apply Law 2: $a^m \div a^n = a^{m-n}$ (same base, subtract indices)
I can apply Law 3: $(a^m)^n = a^{mn}$ (power of a power, multiply indices)
I know $a^0 = 1$ and can prove it using Law 2
I can evaluate and simplify negative indices: $a^{-n} = \frac{1}{a^n}$
I can evaluate $a^{1/n}$ as an $n$th root (e.g. $64^{1/3} = 4$)
I can evaluate $a^{m/n}$ by taking the root first, then raising to power $m$
I can evaluate negative fractional indices such as $a^{-m/n} = \frac{1}{(\sqrt[n]{a})^m}$
I can simplify algebraic expressions combining multiple index laws in one problem
I can solve equations of the form $a^{f(x)} = a^{g(x)}$ by equating indices
I can set up simultaneous equations from two index equations with different bases
I always give both $\pm$ solutions when solving $x^n = k$ with even $n$
I have completed at least three past-paper questions on this topic under timed conditions

0 / 15