Powers, Roots, Indices, Standard Form & Surds · Notes

1.Lesson overview

Syllabus focus

Cambridge IGCSE syllabus reference

1.3 Powers and roots
1.7 Indices I
1.8 Standard form
1.18 Surds

Edexcel IGCSE syllabus reference

1.4 Powers and roots
1.9 Standard form

AQA IGCSE syllabus reference

1.1 Structure and calculation

Powers compress repeated multiplication, while roots reverse it; index laws only apply under their stated base and domain conditions. Standard form and surds then preserve very large, very small or irrational values in exact, manageable forms.

By the end of this lesson

Calculate with the following:
• squares
• square roots
• cubes
• cube roots
• other powers and roots of numbers.
1 Understand and use indices (positive, zero, negative, and fractional).
2 Understand and use the rules of indices.
1 Use the standard form $A \times 1 0^{n}$ where $n$ is a positive or negative integer and 1 $⩽$ $A < 10$ .
2 Convert numbers into and out of standard form.
3 Calculate with values in standard form.
1 Understand and use surds, including simplifying expressions.
2 Rationalise the denominator.
3 Rationalise a denominator of the form $a + b$ or $a - b$ using the conjugate.

2.Learning map

Begin with inverse relationships between powers and roots, build the index laws from repeated multiplication, and only then apply those laws to standard form and exact surd manipulation.

Indices and surds progression

Powers and inverse operations

Secure multiplication, division and fraction skills are essential before applying index laws. Big idea. Index laws apply to factors with the same base; they do not distribute across addition.

Squares Table

A square number is the result of multiplying a number by itself: $n^{2} = n \times n$ . You must memorise the squares from $1^{2}$ to $1 5^{2}$ .

Cubes Table

A cube number is formed by multiplying the same number three times: $n^{3} = n \times n \times n$ . Know the cubes of 1, 2, 3, 4, 5 and 10.

Square Roots

Square Root: The square root of a number $n$ is the value that, when multiplied by itself, gives $n$ . $n = x means$ $x^{2} = n$

3.Core idea

Secure multiplication, division and fraction skills are essential before applying index laws.

Big idea. Index laws apply to factors with the same base; they do not distribute across addition.

Key relationship

a^{m} \times a^{n} = a^{m + n}, a^{- n} = \frac{1}{a ^{n}}, a^{\frac{m}{n}} = n a^{m}

Worked example

Example. Simplify $\frac{3 20 - 45}{5}$ .

$320 - 45 = 65 - 35 = 35$ , so the expression is 3. Simplify each surd before rationalising or cancelling.

4.Squares Table

A square number is the result of multiplying a number by itself:

n^{2} = n \times n

. You must memorise the squares from

1^{2}

1 5^{2}

Square Numbers — Must Know!

n	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$n^{2}$	1	4	9	16	25	36	49	64	81	100	121	144	169	196	225

Exam Tip

Questions like "Write down the value of $169$ " expect instant recall. The answer is 13 because $1 3^{2} = 169$ .

5.Cubes Table

A cube number is formed by multiplying the same number three times:

n^{3} = n \times n \times n

. Know the cubes of 1, 2, 3, 4, 5 and 10.

Cube Numbers — Must Know!

n	1	2	3	4	5	10
$n^{3}$	1	8	27	64	125	1000
Calculation	$1 \times 1 \times 1$	$2 \times 2 \times 2$	$3 \times 3 \times 3$	$4 \times 4 \times 4$	$5 \times 5 \times 5$	$10 \times 10 \times 10$

Why "Cube"?

Just as a square number represents the area of a square, a cube number represents the volume of a cube. A cube with side length 3 has volume $3^{3} = 27$ cubic units.

Watch Out

$4^{3} = 64$ and $8^{2} = 64$ — some numbers are both perfect squares and perfect cubes! The number 64 is one of them.

6.Square Roots

Definition

Square Root: The square root of a number $n$ is the value that, when multiplied by itself, gives $n$ .
$n = x means$ $x^{2} = n$

Key Relationship

n^{2} = n and (n)^{2} = n

Squaring and taking a square root are inverse operations — they undo each other.

Example 1

1
Write down the value of $169$
2
Solution: Since $1 3^{2} = 169$ , we know $169 = 13$

Example 2

Find $2.25$ .

1
Solution: $2.25 = \frac{225}{100} = \frac{1 5 ^{2}}{1 0 ^{2}}$
2
So $2.25 = \frac{15}{10} = 1.5$

Example 3

Find $400$ .

1
Solution: $400 = 4 \times 100 = 2^{2} \times 1 0^{2} = 2 0^{2}$ .
2
So $400 = 20$

Negative Numbers

You cannot find the square root of a negative number (in real numbers). For example, $- 9$ has no real solution, because no real number squared gives a negative result.

7.Cube Roots

Definition

Cube Root: The cube root of a number $n$ is the value that, when cubed, gives $n$ .
$3 n = x means$ $x^{3} = n$

Key Relationship

3 n^{3} = n and (3 n)^{3} = n

Example 1

Find $3125$ .

1
Solution: Since $5^{3} = 125$ , we get $3125 = 5$

Example 2

1
Work out $5^{2} \times 38$
2
Solution: $5^{2} = 25$ and $38 = 2$ (since $2^{3} = 8$ )
3
$25 \times 2 = 50$ .

Example 3

Find $3 - 27$ .

1
Solution: $(- 3)^{3} = (- 3) \times (- 3) \times (- 3) = - 27$
2
So $3 - 27 = - 3$ .

Key Difference

Unlike square roots, cube roots can be taken of negative numbers. $3 - 8 = - 2$ because $(- 2)^{3} = - 8$ .

8.Other Powers

The exponent (power) tells you how many times to multiply the base by itself:

General Power

a^{n} = a \times a \times a \times \dots \times a (n factors)

Example 1

1
$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$

Example 2

1
$3^{4} = 3 \times 3 \times 3 \times 3 = 81$

Example 3

1
$1 0^{6} = 1000000$ (1 followed by 6 zeros)

Higher roots work the same way as square and cube roots:

nth Root

$n a = b means b^{n} = a$

For example: $481 = 3$ because $3^{4} = 81$

And: $532 = 2$ because $2^{5} = 32$

Fractional Powers (Preview)

Fractional powers are another way to write roots:

$a^{1/2} = a$
$a^{1/3} = 3 a$
$a^{1/ n} = n a$

You'll use these extensively in the Indices chapter.

Useful Powers of Small Numbers

Base	^1	^2	^3	^4	^5	^6
2	2	4	8	16	32	64
3	3	9	27	81	243	729
5	5	25	125	625	3125	—
10	10	100	1000	10 000	100 000	1 000 000

9.Power Calculator

Explore how powers build up through repeated multiplication.

10.Challenge Questions

Three stepping-stone questions that combine squares, cubes and roots — harder than the single-skill examples above.

Challenge 1 — combining three operations

Question. Work out $144 + 3^{3} - 364$ .

1
$144 = 12$ (since $1 2^{2} = 144$ ).
2
$3^{3} = 27$ .
3
$364 = 4$ (since $4^{3} = 64$ ).
4
$12 + 27 - 4 = 35$ .

Challenge 2 — square AND cube at once

Question. 64 is both a perfect square and a perfect cube. Find another number between 100 and 1000 with the same property, and show why it works.

1
A number that is both a square and a cube must be a sixth power — its index has to be a multiple of 2 (for a square) and a multiple of 3 (for a cube), and the smallest number that is both is 6.
2
Try $n = 3$ : $3^{6} = 729$ .
3
Check as a square: $2 7^{2} = 729$ . Check as a cube: $9^{3} = 729$ .
4
729 works, since $729 = 2 7^{2} = 9^{3}$ .

Challenge 3 — mixing signs across root types

Question. Evaluate $3 - 8 \times 25 - (- 3)^{2}$ .

1
$3 - 8 = - 2$ — cube roots of negative numbers are allowed, unlike square roots.
2
$25 = 5$ .
3
$(- 3)^{2} = 9$ — an even power of a negative number is positive.
4
$(- 2) \times 5 - 9 = - 10 - 9 = - 19$ .

11.Positive Indices

A positive index tells you how many times to multiply the base by itself.

Definition

a^{n} = a \times a \times a \times \dots \times a (n factors)

Here

a

is the base and

n

is the index (or exponent/power).

Examples

1
$3^{4} = 3 \times 3 \times 3 \times 3 = 81$
2
$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$
3
$1 0^{3} = 10 \times 10 \times 10 = 1000$
4
$(- 2)^{3} = (- 2) \times (- 2) \times (- 2) = - 8$

Watch Out

$(- 2)^{4} = 16$ (even power → positive)
$- 2^{4} = - (2^{4}) = - 16$ (the negative is not inside the bracket).
$a^{1} = a$ — any number to the power 1 is itself

12.Zero Index

Zero Index Rule

a^{0} = 1 (a \neq = 0)

Any non-zero number raised to the power zero equals 1. Here is why:

Why does

a^{0} = 1 ?

Look at the pattern of powers of 2 as the index drops by 1 each time: $2^{3} = 8$ , $2^{2} = 4$ , $2^{1} = 2$ . Each step divides the previous value by the base, 2. Continuing the pattern from $2^{1} = 2$ gives $2^{0} = 2 \div 2 = 1$ . The same reasoning works for any non-zero base $a$ , so $a^{0} = 1$ .

This also matches the division law $a^{n} \div a^{n} = a^{n - n} = a^{0}$ , and any (non-zero) number divided by itself is 1.

Examples

1
$5^{0} = 1$
2
$(- 7)^{0} = 1$
3
$(3 x)^{0} = 1$
4
$10 0^{0} = 1$

Important

$0^{0}$ is undefined. The zero index rule only applies when the base is not zero.

13.Negative Indices

Negative Index Rule

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

A negative index means "take the reciprocal". It flips the base to the bottom of a fraction.

Why does

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

Continue the same pattern below zero: from $2^{0} = 1$ , dividing by the base again gives $2^{- 1} = 1 \div 2 = \frac{1}{2}$ , then $2^{- 2} = \frac{1}{2} \div 2 = \frac{1}{4} = \frac{1}{2 ^{2}}$ . Each drop of 1 in the index divides by $a$ again, so in general $a^{- n} = \frac{1}{a ^{n}}$ .

This also matches the division law: $a^{0} \div a^{n} = a^{0 - n} = a^{- n}$ , and since $a^{0} = 1$ , that quotient is $\frac{1}{a ^{n}}$ — so $a^{- n} = \frac{1}{a ^{n}}$ .

Worked Examples

1
Example 1: $7^{- 2} = \frac{1}{7 ^{2}} = \frac{1}{49}$
2
Example 2: $2^{- 3} = \frac{1}{2 ^{3}} = \frac{1}{8}$
3
Example 3: $1 0^{- 1} = \frac{1}{10} = 0.1$
4
Example 4: $5^{- 3} = \frac{1}{5 ^{3}} = \frac{1}{125}$

Also Works in Reverse

$\frac{1}{a ^{n}} = a^{- n}$ — moving a power from bottom to top (or vice versa) changes the sign of the index.

$\frac{1}{3 ^{4}} = 3^{- 4}$ and $\frac{1}{x ^{- 2}} = x^{2}$

14.Fractional Indices

Fractional Index — Root

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

A fractional index with numerator 1 means "take the nth root".

A fractional index combines a root and a power

1
General rule
a^(m/n) · nth root, then power m
2
Example
64^(2/3) · cube root first
3
Root
∛ $64 = 4$ · denominator gives root
4
Power
$4^{2} = 16$ · numerator gives power

For positive IGCSE examples, root then power and power then root agree; choose the easier route.

Examples: Unit Fractional Indices

1
$6^{\frac{1}{2}} = 6$ (square root)
2
$1 6^{\frac{1}{4}} = 416 = 2$ because $2^{4} = 16$ .
3
$8 1^{\frac{1}{2}} = 81 = 9$ because $9^{2} = 81$ .
4
$2 7^{\frac{1}{3}} = 327 = 3$ because $3^{3} = 27$ .
5
$1000 0^{\frac{1}{4}} = 410000 = 10$ because $1 0^{4} = 10000$ .

General Fractional Index

a^{\frac{m}{n}} = (n a)^{m} = n a^{m}

The denominator gives the root, the numerator gives the power. You can do root first, then power (usually easier) or power first, then root.

Worked Example:

8^{\frac{2}{3}}

1
Root first: $8^{\frac{1}{3}} = 38 = 2$
2
Then power: $2^{2} = 4$
3
Answer: $8^{\frac{2}{3}} = 4$

Worked Example:

2 5^{\frac{3}{2}}

1
Root first: $2 5^{\frac{1}{2}} = 25 = 5$
2
Then power: $5^{3} = 125$
3
Answer: $2 5^{\frac{3}{2}} = 125$

Worked Example:

8^{- \frac{2}{3}}

1
Combine negative and fractional index rules:
2
Deal with the fraction: $8^{\frac{2}{3}} = (38)^{2} = 2^{2} = 4$ .
3
Apply the negative: $8^{- \frac{2}{3}} = \frac{1}{8 ^{\frac{2}{3}}} = \frac{1}{4}$
4
Answer: $8^{- \frac{2}{3}} = \frac{1}{4}$

Worked Example:

1 6^{- \frac{3}{4}}

1
$1 6^{\frac{1}{4}} = 416 = 2$
2
$2^{3} = 8$
3
$1 6^{- \frac{3}{4}} = \frac{1}{8}$

Top Tip

Always do the root first then the power — it keeps the numbers small and manageable.

15.Key Rules

There are seven key rules of indices. Make sure you know all of them — they are essential for simplifying algebraic expressions.

Rule 1 — Multiplication

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

Rule 2 — Division

a^{m} \div a^{n} = a^{m - n}, a \neq = 0

Rule 3 — Power of a Power

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

Rule 4 — Zero Index

a^{0} = 1

Rule 5 — Negative Index

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

Rule 6 — Fractional Index (Root)

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

Rule 7 — General Fractional Index

a^{\frac{m}{n}} = (n a)^{m}

Summary Table

Rule	Formula	Example
Multiplication	$a^{m} \times a^{n} = a^{m + n}$	$2^{3} \times 2^{4} = 2^{7} = 128$
Division	$a^{m} \div a^{n} = a^{m - n}$ , for $a \neq = 0$	$5^{6} \div 5^{2} = 5^{4} = 625$
Power of a Power	$(a^{m})^{n} = a^{mn}$	$(2^{3})^{2} = 2^{6} = 64$
Zero Index	$a^{0} = 1$	$9 9^{0} = 1$
Negative Index	$a^{- n} = \frac{1}{a ^{n}}$	$3^{- 2} = \frac{1}{9}$
Root Index	$a^{1/ n} = n a$	$6 4^{\frac{1}{3}}$ = 4
Fractional Index	$a^{m / n} = (n a)^{m}$	$4^{3/2} = 8$

Example:

2^{- 3} \times 2^{4}

1
Rule 1 (multiplication — add indices): $2^{- 3 + 4} = 2^{1} = 2$

Example: (

2^{3}

)^2

1
Rule 3 (power of a power — multiply indices): $2^{3 \times 2} = 2^{6} = 64$ .

Example:

2^{3} \div 2^{4}

1
Rule 2 (division — subtract indices): $2^{3 - 4} = 2^{- 1} = \frac{1}{2}$

Example:

3^{5} \times 3^{- 2} \div 3^{4}

1
Combine rules: $3^{5 + (- 2) - 4} = 3^{- 1} = \frac{1}{3}$

Common Mistake

The multiplication rule only works when the bases are the same. You cannot simplify $2^{3} \times 3^{4}$ using the index laws.

Planned interactivecore

Indices, Surds and Rationalising

Expose index-law counting and preserve exact surd form step by step.

Teach positive, zero, negative and fractional indices plus surd simplification.

Inputs & controls

base and exponents
law selector
surd expression
rationalising step

Learner actions

expand factors
combine indices
extract square factors
rationalise

Visible outputs

factor count
combined index
simplified surd
exact and decimal values
error explanation

Implementation note

The catalogue currently calls this built but no HTML exists; keep this block planned until a real asset is present.

Expand and combine the seven index lawsOpen full screen

16.Negative and Fractional Indices

Three key rules

Negative index — reciprocal:

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

Fractional index $\frac{1}{n}$ — nth root:

a^{1/ n} = n a

Fractional index $\frac{m}{n}$ — both:

a^{m / n} = (n a)^{m} = n a^{m}

Worked examples

1
$7^{- 2} = \frac{1}{7 ^{2}} = \frac{1}{49}$
2
$8 1^{1/2} = 81 = 9$
3
$8^{2/3} = (38)^{2} = 2^{2} = 4$
4
$2^{- 3} \times 2^{4} = 2^{4 - 3} = 2^{1} = 2$ .
5
$(2^{3})^{2} = 2^{6} = 64$
6
$2^{3} \div 2^{4} = 2^{3 - 4} = 2^{- 1} = \frac{1}{2}$ .

Exam tip

For $a^{m / n}$ , do the root first (smaller number), then the power. For $3 2^{2/5}$ : the fifth root of 32 is 2, then $2^{2} = 4$ .

17.Surds

A surd is an exact value involving a root that cannot be written as a rational number. Keep surd answers exact unless the question asks for a decimal approximation.

Simplifying surds

Use $ab = a \times b$ and take out the largest square factor.

For example, $72 = 36 \times 2 = 62$ .

Rationalising a denominator

Do not leave a single surd in the denominator. Multiply the numerator and denominator by that surd:

$\frac{3}{5} = \frac{3 5}{5}$ .

Rationalising a Binomial Denominator (Conjugates)

When the denominator is a sum or difference involving a surd, such as $a + b$ or $a - b$ , multiplying top and bottom by the surd itself does not clear the root. Instead, multiply by the conjugate — the same expression with the middle sign flipped.

Why the conjugate works

(a + b) (a - b) = a^{2} - b

The conjugate pair multiplies out using the difference of two squares, so the surd disappears from the denominator completely.

Example: rationalise $\frac{1}{- 1 + 3}$ .

1
The conjugate of $- 1 + 3$ is $- 1 - 3$ (same terms, opposite middle sign).
2
Multiply top and bottom by it: $\frac{1}{- 1 + 3} \times \frac{- 1 - 3}{- 1 - 3} = \frac{- 1 - 3}{( - 1 ) ^{2} - ( 3 ) ^{2}}$ .
3
Denominator: $1 - 3 = - 2$ .
4
$\frac{- 1 - 3}{- 2} = \frac{1 + 3}{2}$ .

Be Careful!

A single surd denominator like $\frac{3}{5}$ is rationalised by multiplying by that surd ( $5$ ). A binomial denominator like $5 + 2$ needs the conjugate ( $5 - 2$ ), not $5$ alone — multiplying by $5$ would leave a surd term behind.

Exact form

Use a surd when an exact answer is required. For example, the diagonal of a $1 \times 1$ square is $2$ , not a rounded decimal.

1
Right triangle
legs 1 and 1 · hypotenuse h
2
Pythagoras
$h^{2} = 1^{2} + 1^{2} = 2$
3
Exact length
$h = 2$ · no rounding
4
Decimal check
$2 \approx 1.414$ · approximation only

Keep $2$ in exact form unless the question requests a decimal approximation.

Simplify surds and rationalise a denominatorOpen full screen

18.Indices I — Challenge Questions

Three stepping-stone questions that chain index rules together, including one in reverse.

Challenge 1 — multiply, then divide, with negative indices

Question. Simplify $\frac{5 ^{3} \times 5 ^{- 1}}{5 ^{- 2}}$ , giving your answer as a single power of 5, then evaluate it.

1
Numerator first (multiplying — add indices): $5^{3 + (- 1)} = 5^{2}$ .
2
Now divide by $5^{- 2}$ (dividing — subtract indices): $5^{2 - (- 2)} = 5^{4}$ .
3
$5^{4} = 625$ .

Subtracting a negative index is the same trap as subtracting a negative number anywhere else — it becomes addition.

Challenge 2 — indices with algebra

Question. Simplify $(3 x^{2})^{3} \div (9 x^{4})$ .

1
Apply the power to everything inside the bracket: $(3 x^{2})^{3} = 3^{3} x^{2 \times 3} = 27 x^{6}$ .
2
Divide the coefficients and subtract the indices: $27 \div 9 = 3$ and $x^{6 - 4} = x^{2}$ .
3
$3 x^{2}$ .

Challenge 3 — working backwards for the index

Question. (a) Find $n$ if $2^{n} = 128$ . (b) Hence find $n$ if $2^{n} = \frac{1}{32}$ .

1
(a) Write 128 as a power of 2: $2^{7} = 128$ , so $n = 7$ .
2
(b) $32 = 2^{5}$ , and a fraction of the form $\frac{1}{2 ^{5}}$ is $2^{- 5}$ .
3
$n = - 5$ .

When the base is already given, the question is really asking "what power of 2 gives this?" — express the other side as a power of the same base first.

19.What Is Standard Form?

A number is in standard form when it is written as:

Standard Form

A \times 1 0^{n}

where:

Valid Standard Form

$3.56 \times 1 0^{5}$ Correct: ( $A = 3.56$ , which is between 1 and 10)

$7 \times 1 0^{- 3}$ Correct: ( $A = 7$ , which is between 1 and 10)

$1.002 \times 1 0^{12}$ Correct:

NOT Valid Standard Form

$35.6 \times 1 0^{4}$ ✗ ( $A = 35.6$ is not less than 10)

$0.42 \times 1 0^{- 2}$ ✗ ( $A = 0.42$ is less than 1)

The Power of 10

The index $n$ tells you how many places to move the decimal point:

Positive $n$ : the number is large — move the point to the right
Negative $n$ : the number is small (between 0 and 1) — move the point to the left
$n = 0$ : the number is just A itself (since $1 0^{0} = 1$ )

20.Converting to Standard Form

To write a number in standard form, follow these steps:

Example 1: Write 356 000 in standard form

1
Place the decimal after the 3 → $A = 3.56$
2
The point moved 5 places to the left
3
The number is large, so $n = + 5$
4
Answer: $356000 = 3.56 \times 1 0^{5}$

Example 2: Write 47 000 000 in standard form

1
$A = 4.7$
2
Decimal moved 7 places left
3
Answer: $47000000 = 4.7 \times 1 0^{7}$

Example 3: Write 803 in standard form

1
$A = 8.03$ , decimal moved 2 places
2
Answer: $803 = 8.03 \times 1 0^{2}$

Example 4: Write 0.00042 in standard form

1
First non-zero digit is 4, so $A = 4.2$
2
Decimal moved 4 places to the right
3
The number is small, so $n = - 4$
4
Answer: $0.00042 = 4.2 \times 1 0^{- 4}$

Example 5: Write 0.0000071 in standard form

1
$A = 7.1$ , decimal moved 6 places right
2
Answer: $0.0000071 = 7.1 \times 1 0^{- 6}$

Example 6: Write 0.059 in standard form

1
$A = 5.9$ , decimal moved 2 places right
2
Answer: $0.059 = 5.9 \times 1 0^{- 2}$

Memory device: LARS

Left Add, Right Subtract. When converting an ordinary number into standard form, moving the decimal point left to reach $A$ adds to (increases) the power; moving it right subtracts from (decreases) it.

Common Mistake

Forgetting that A must be at least 1 and less than 10. Writing $36 \times 1 0^{3}$ instead of $3.6 \times 1 0^{4}$ is not standard form.

Move the decimal point and track the exponentOpen full screen

21.Converting from Standard Form

To convert back to an ordinary number, move the decimal point according to the power of 10.

Move the decimal point to the right by

n

places, filling with zeros as needed.

Example 1: 4.6

\times 1 0^{4}

1
Move the point 4 places right: $4.6 \to 46000$
2
Answer: 46 000

Example 2: 3.072

\times 1 0^{6}

1
Move the point 6 places right: $3.072 \to 3072000$
2
Answer: 3 072 000

Move the decimal point to the left by |n| places, filling with zeros as needed.

Example 3: 5.1

\times 1 0^{- 3}

1
Move the point 3 places left: $5.1 \to 0.0051$
2
Answer: 0.0051

Example 4: 8.45

\times 1 0^{- 5}

1
Move the point 5 places left: $8.45 \to 0.0000845$
2
Answer: 0.0000845

Quick Check

Positive power → big number. Negative power → small number. If your answer contradicts this, recheck your direction.

22.Calculating in Standard Form

Multiply the A values, then add the powers of 10. Adjust if A leaves the range

[1, 10)

Multiplication

(A_{1} \times 1 0^{m}) \times (A_{2} \times 1 0^{n}) = (A_{1} \times A_{2}) \times 1 0^{m + n}

Example:

(3 \times 1 0^{4}) \times (2 \times 1 0^{3})

A values: $3 \times 2 = 6$

1
Powers: $1 0^{4 + 3} = 1 0^{7}$
2
Answer: $6 \times 1 0^{7}$

Example:

(4.5 \times 1 0^{6}) \times (3 \times 1 0^{5})

A values: $4.5 \times 3 = 13.5$

1
Powers: $1 0^{6 + 5} = 1 0^{11}$
2
But 13.5 is not in $[1, 10)$ , so adjust: $13.5 = 1.35 \times 1 0^{1}$
3
Answer: $1.35 \times 1 0^{12}$

Divide the A values, then subtract the powers of 10.

Division

(A_{1} \times 1 0^{m}) \div (A_{2} \times 1 0^{n}) = \frac{A _{1}}{A _{2}} \times 1 0^{m - n}

Example:

(8 \times 1 0^{9}) \div (4 \times 1 0^{3})

A values: $8 \div 4 = 2$

1
Powers: $1 0^{9 - 3} = 1 0^{6}$
2
Answer: $2 \times 1 0^{6}$

Example:

(3.6 \times 1 0^{5}) \div (9 \times 1 0^{7})

A values: $3.6 \div 9 = 0.4$

1
Powers: $1 0^{5 - 7} = 1 0^{- 2}$
2
Adjust: $0.4 = 4 \times 1 0^{- 1}$
3
Answer: $4 \times 1 0^{- 3}$

To add or subtract numbers in standard form, the powers of 10 must be the same. Adjust one number so both have the same power, then add or subtract the A values.

Example:

(3.4 \times 1 0^{5}) + (2.1 \times 1 0^{4})

1
Rewrite with the same power: $2.1 \times 1 0^{4} = 0.21 \times 1 0^{5}$
2
Add: $3.4 + 0.21 = 3.61$
3
Answer: $3.61 \times 1 0^{5}$

Example:

(5 \times 1 0^{- 3}) - (3 \times 1 0^{- 4})

1
Rewrite: $3 \times 1 0^{- 4} = 0.3 \times 1 0^{- 3}$
2
Subtract: $5 - 0.3 = 4.7$
3
Answer: $4.7 \times 1 0^{- 3}$

Remember

After any calculation, always check that your answer is in proper standard form: $1 ⩽ A < 10$ . If not, adjust A and $n$ accordingly.

Planned interactiverecommended

Standard Form Converter and Calculator

Move the decimal, track the exponent and normalise arithmetic results.

Link ordinary form, standard form and multiplication/division/addition rules.

Inputs & controls

mantissa
exponent
ordinary value
operation
second number

Learner actions

drag decimal point
calculate
normalise
check magnitude

Visible outputs

ordinary and standard form
decimal-move trace
unnormalised result
normalised answer

Implementation note

Enforce 1 <= absolute mantissa < 10 and include negative-number examples.

Multiply and divide numbers in standard formOpen full screen

23.Standard Form — Challenge Questions

Three stepping-stone questions using standard form in real contexts, including a case the examples above didn't cover.

Challenge 1 — how far does light travel in a year?

Question. Light travels at $3 \times 1 0^{8}$ metres per second. A year is about $3.156 \times 1 0^{7}$ seconds. How far does light travel in one year? Give your answer in standard form, to 3 significant figures.

1
Multiply the numbers: $3 \times 3.156 = 9.468$ .
2
Add the powers: $1 0^{8 + 7} = 1 0^{15}$ .
3
$9.47 \times 1 0^{15} m$ (3 s.f.) — this distance is called a light-year.

Challenge 2 — when the answer needs adjusting the other way

Question. A single grain of table salt has a mass of about $6 \times 1 0^{- 5} g$ . How many grains would it take to make a total mass of $1.2 \times 1 0^{2} g$ ?

1
Divide the numbers: $1.2 \div 6 = 0.2$ .
2
Subtract the powers: $1 0^{2 - (- 5)} = 1 0^{7}$ , giving $0.2 \times 1 0^{7}$ .
3
$0.2$ is less than 1, so this time adjust downward: $0.2 \times 1 0^{7} = 2 \times 1 0^{6}$ .
4
About $2 \times 1 0^{6}$ grains.

Renormalising isn't only for coefficients that come out too big — a coefficient below 1, like $0.2$ , needs the decimal point moved the opposite way, which decreases the power.

Challenge 3 — comparing and combining populations

Question. Country A has a population of $4.5 \times 1 0^{7}$ . Country B has a population of $9 \times 1 0^{6}$ . (a) How many times bigger is Country A's population than Country B's? (b) Find the combined population of both countries, in standard form.

1
(a) Divide: $4.5 \times 1 0^{7} \div 9 \times 1 0^{6} = (4.5 \div 9) \times 1 0^{7 - 6} = 0.5 \times 1 0^{1} = 5$ . Country A is 5 times bigger.
2
(b) To add, match the powers first: $9 \times 1 0^{6} = 0.9 \times 1 0^{7}$ .
3
Add the A-values: $4.5 + 0.9 = 5.4$ .
4
Combined population $= 5.4 \times 1 0^{7}$ .

24.Exam-Style Worked Examples

Applying the index laws

1
Multiplying
Add the indices: $a^{m} \times a^{n} = a^{m + n}$ .
2
Dividing
Subtract the indices: $\frac{a ^{m}}{a ^{n}} = a^{m - n}$ .
3
Power of a power
Multiply the indices — and raise any coefficient too.
4
Negative index
Take the reciprocal: $a^{- n} = \frac{1}{a ^{n}}$ .
5
Fractional index
Root first, then the power: $a^{m / n} = (n a)^{m}$ .

Worked example 1 — index laws (4 marks)

Question. Simplify (a) $\frac{x ^{7} \times x ^{3}}{x ^{4}}$ , (b) $(2 y^{3})^{4}$ , (c) $1 6^{3/4}$ .

1
(a) Multiplying adds indices: $x^{7 + 3} = x^{10}$ ; dividing subtracts: $x^{10 - 4} = x^{6}$ .
2
(b) The power applies to everything inside: $2^{4} \times y^{3 \times 4} = 16 y^{12}$ .
3
(c) Do the root first: $416 = 2$ , then $2^{3} = 8$ .
4
So the answers are $x^{6}$ , $16 y^{12}$ and $8$ .

Marking. 1 mark per point. In (b) the coefficient must be raised too — $2 y^{12}$ is a very common wrong answer.

Worked example 2 — standard form (4 marks)

Question. Given $a = 3.2 \times 1 0^{5}$ and $b = 8 \times 1 0^{- 3}$ , find (a) $ab$ and (b) $\frac{a}{b}$ , each in standard form.

1
(a) Multiply the numbers and add the powers: $3.2 \times 8 = 25.6$ and $1 0^{5 + (- 3)} = 1 0^{2}$ , giving $25.6 \times 1 0^{2}$ .
2
This is not yet standard form — adjust to $2.56 \times 1 0^{3}$ .
3
(b) Divide the numbers and subtract the powers: $\frac{3.2}{8} = 0.4$ and $1 0^{5 - (- 3)} = 1 0^{8}$ , giving $0.4 \times 1 0^{8}$ .
4
Adjust again: $4 \times 1 0^{7}$ .

Marking. 1 mark per point. A standard form answer must have the first number between $1$ and $10$ — always check and adjust at the end.

Worked example 3 — surds (4 marks)

Question. Simplify (a) $72$ , (b) $50 + 18$ , (c) rationalise $\frac{6}{3}$ .

1
(a) Look for the largest square factor: $72 = 36 \times 2$ , so $72 = 62$ .
2
(b) $50 = 52$ and $18 = 32$ ; the surd parts match, so they add: $82$ .
3
(c) Multiply top and bottom by $3$ : $\frac{6 3}{3}$ .
4
$= 23$ .

Marking. 1 mark per point. Surds only add when the part under the root is identical — $50 + 18 \neq = 68$ .

25.Exam Tips & Common Misconceptions

Exam Tips

Index laws: a^m × aⁿ = a^m+n, a^m ÷ aⁿ = a^m-n, (a^m)ⁿ = a^mn.
a⁰ = 1 (any nonzero base), a^-n = 1/aⁿ, a^1/n = ⁿ√a.
Standard form: A × 10ⁿ where $1 \leq A < 10$ — exactly one non-zero digit before the decimal point.
For very small numbers, the power is negative: 0.000 045 $4.5 \times 1 0^{- 5} .$ Count the zero displacement carefully.
When multiplying standard-form numbers, multiply the A-parts and add the powers; then renormalise so $1 \leq A < 10$ .

Common Misconceptions

" $2^{3} = 6$ " — $2^{3}$ means $2 \times 2 \times 2 = 8$ , not $2 \times 3$ .
Writing $35 \times 1 0^{4}$ as standard form — the coefficient must be less than 10, so $35 \times 1 0^{4} = 3.5 \times 1 0^{5}$ .
" $a^{- 2}$ is negative" — the answer is a positive reciprocal: $\frac{1}{a ^{2}}$ , not - $a^{2}$ .
Adding indices when multiplying different bases: $2^{3} \times 5^{2} \neq = 1 0^{5}$ .

26.Worked method — Simplify a surd by extracting the largest square factor

The decimal is only a check; $62$ is the exact simplified value.

Transfer the method

Reasoning prompt. Why should you factor a surd before attempting to add or cancel it?

1
Find the largest square factor inside the root.
2
Rewrite the surd using that square factor.
3
Combine only matching simplified surd terms.

Self-check

Index laws combine multiplication and division of factors with the same base, not terms joined by addition.

27.Index conditions and exact surd control

Rules with their conditions

a^{0} = 1

a \neq = 0

The zero-index result follows from $a^{m} / a^{m} = a^{m - m}$ , which already requires a non-zero denominator.

a^{- n} = 1/ a^{n}

a \neq = 0

A negative index creates a reciprocal; it does not make the value negative.

a^{1/2} = a

a \geq 0

for real values

At this level, an even root of a negative number is not a real number.

Worked synthesis: simplify before dividing

Simplify $\frac{3 8 - 18}{2}$ .

$8 = 22$ and $18 = 32$ .

The numerator is $62 - 32 = 32$ .

Therefore $\frac{3 2}{2} = 3$ .

Standard-form reasonableness check

In $(6 \times 1 0^{5}) (4 \times 1 0^{- 3})$ , the powers give $1 0^{2}$ and the leading numbers give 24. Renormalise: $24 \times 1 0^{2} = 2.4 \times 1 0^{3}$ . A standard-form coefficient must satisfy $1 \leq A < 10$ .

28.Summary

Key Points

Central principle. Index laws apply to factors with the same base; they do not distribute across addition
A square number is the result of multiplying a number by itself: $n^{2} = n \times n$ . You must memorise the squares $f r o m 1^{2}$ to $1 5^{2}$ .
A cube number is the result of multiplying a number by itself three times: $n^{3} = n \times n \times n$ . You must memorise cubes of 1, 2, 3, 4, 5 and 10.
Just as a square number represents the area of a square, a cube number represents the volume of a cube. A cube with side length 3 has volume $3^{3} = 27$ cubic units.
$4^{3} = 64$ and $8^{2} = 64$ — some numbers are both perfect squares and perfect cubes! The number 64 is one of them.
Key relationship: $a^{m} \times a^{n} = a^{m + n}, a^{- n} = \frac{1}{a ^{n}}, a^{\frac{m}{n}} = n a^{m}$
Key Relationship: $n^{2} = n and (n)^{2} = n$
Key Relationship: $3 n^{3} = n and (3 n)^{3} = n$
General Power: $a^{n} = a \times a \times a \times \dots \times a (n factors)$
Definition: $a^{n} = a \times a \times a \times \dots \times a (n factors)$
Zero Index Rule: $a^{0} = 1 (a \neq = 0)$