What is Standard Form in Mathematics and why is it used?

Standard Form, also known as Scientific Notation, is a way of expressing very large or very small numbers in a concise format. In Mathematics, particularly in the Cambridge IGCSE curriculum, it is written as a product of a number between 1 and 10 and a power of 10. This form is used to simplify calculations and make it easier to read and write numbers that are otherwise cumbersome.

How do you convert a number into Standard Form?

To convert a number into Standard Form, first identify the significant digits and place the decimal point after the first non-zero digit. Count the number of places the decimal has moved to determine the power of 10. If the original number is greater than 1, the power is positive; if it's less than 1, the power is negative. For example, 4500 becomes 4.5 x 10^3 in Standard Form.

What are the key steps to multiply numbers in Standard Form?

When multiplying numbers in Standard Form, multiply the significant figures and add the exponents of the powers of 10. For instance, if you have (2 x 10^3) and (3 x 10^4), multiply 2 and 3 to get 6, and add the exponents 3 and 4 to get 7. The result is 6 x 10^7.

How can Standard Form be useful in real-world applications?

Standard Form is particularly useful in fields like science and engineering, where it is common to work with extremely large or small numbers. For example, it is used to express the distance between stars or the size of microscopic organisms. In the Cambridge IGCSE Mathematics curriculum, understanding Standard Form helps students handle such numbers efficiently in calculations and data analysis.

What are common mistakes to avoid when working with Standard Form?

Common mistakes include misplacing the decimal point, incorrectly calculating the power of 10, and not maintaining the number between 1 and 10. It's essential to carefully count the decimal places moved and ensure the final number is correctly formatted. Practicing these steps can help avoid errors in converting and calculating with Standard Form.

Why is "Leaving the coefficient outside $[1, 10)$" a common mistake in Standard Form?

Why it happens: After multiplying, students get e.g. 14 × 10^4 and stop without normalising. How to avoid it: Always check 1 ≤ a < 10. Adjust by moving the decimal and changing the power. Source: 0580/42 — every recent series

Why is "Adding standard-form numbers without equalising the indices" a common mistake in Standard Form?

Why it happens: Students mechanically add coefficients, ignoring that the powers differ. How to avoid it: Match the powers first, then add coefficients.

Why is "Confusing the direction of the decimal shift for negative exponents" a common mistake in Standard Form?

Why it happens: Negative powers of 10 make small numbers, but students still shift left. How to avoid it: Negative exponent → small number → shift the decimal **left** (more zeros).

Why is "Misreading calculator output (e.g. $4.5\text{E}^{-3}$) as $4.5^{-3}$" a common mistake in Standard Form?

Why it happens: The E or ×10 symbol is small on calculator screens. How to avoid it: Calculator 4.5E^-3 means 4.5 × 10^-3, not 4.5 to the power -3.

NumberCambridge IGCSE Maths 0580 Extended

Standard Form

Q: Why is "Adding standard-form numbers without equalising the indices" a common mistake in Standard Form?

Why it happens: Students mechanically add coefficients, ignoring that the powers differ. How to avoid it: Match the powers first, then add coefficients.

Q: Why is "Confusing the direction of the decimal shift for negative exponents" a common mistake in Standard Form?

Why it happens: Negative powers of 10 make small numbers, but students still shift left. How to avoid it: Negative exponent → small number → shift the decimal **left** (more zeros).

Q: Why is "Misreading calculator output (e.g. $4.5\text{E}^{-3}$) as $4.5^{-3}$" a common mistake in Standard Form?

Why it happens: The E or ×10 symbol is small on calculator screens. How to avoid it: Calculator 4.5E^-3 means 4.5 × 10^-3, not 4.5 to the power -3.

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Standard Form — Cambridge IGCSE 0580 Maths Extended (2026)

Write very large and very small numbers as $a \times 10^{n}$ where $1 \le a < 10$ and $n$ is an integer. The form is non-negotiable; calculator skills and index laws do the rest.

What you’ll learn

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

E1.4 — Use the standard form $a \times 10^{n}$ where $n$ is an integer and $1 \le a < 10$ .
Convert between standard form and ordinary form.
Calculate with numbers in standard form, including addition, subtraction, multiplication, and division.

What standard form looks like

$a \times 10^{n}$ with $a$ between 1 (inclusive) and 10 (exclusive). One non-zero digit before the decimal point.

Standard form writes any positive number as $a \times 10^{n}$ where $1 \le a < 10$ and $n$ is an integer.

The constraint $1 \le a < 10$ is what makes it "standard". $20 \times 10^{4}$ is a correct numerical value but NOT standard form — convert to $2 \times 10^{5}$ .

Examples in standard form.

Ordinary	Standard form
$4{,}500$	$4.5 \times 10^{3}$
$3{,}200{,}000$	$3.2 \times 10^{6}$
$0.000\,072$	$7.2 \times 10^{-5}$
$0.6$	$6 \times 10^{-1}$

Conversion procedure.

Big number → standard form. Move the decimal point LEFT until exactly one non-zero digit is in front. Count the moves — that's $n$ , positive.
Small number → standard form. Move the decimal point RIGHT until exactly one non-zero digit is in front. Count the moves — that's $n$ , negative.

Reverse direction (standard → ordinary). Move the decimal point right by $n$ (if positive) or left by $|n|$ (if negative).

Count the decimal-point moves to get n — left moves give a positive exponent, right moves a negative one.

Form: $a \times 10^{n}$ with $1 \le a < 10$ .
Big numbers: positive $n$ .
Small numbers (less than 1): negative $n$ .
Count decimal moves to find $n$ .

Multiplying and dividing in standard form

Multiply the $a$ values, then handle the powers of 10 with index laws.

Multiplication. $(a \times 10^{m}) \times (b \times 10^{n}) = (a b) \times 10^{m+n}.$

If $a b$ is no longer between $1$ and $10$ , ADJUST: shift one decimal and bump the exponent.

Worked. $(3 \times 10^{4}) \times (2.5 \times 10^{6}) = 7.5 \times 10^{10}$ .

Worked needing adjustment. $(4 \times 10^{5}) \times (5 \times 10^{3}) = 20 \times 10^{8} = 2 \times 10^{9}$ .

Division. $\dfrac{a \times 10^{m}}{b \times 10^{n}} = \dfrac{a}{b} \times 10^{m-n}.$

Worked. $\dfrac{8 \times 10^{12}}{2 \times 10^{4}} = 4 \times 10^{8}$ .

Worked needing adjustment. $\dfrac{3 \times 10^{6}}{6 \times 10^{2}} = 0.5 \times 10^{4} = 5 \times 10^{3}$ .

Handle the a-values with ordinary arithmetic and the powers of 10 with index laws — then adjust if a leaves [1, 10).

Powers and roots in standard form.

$(a \times 10^{n})^{k} = a^{k} \times 10^{nk}$ .
$\sqrt{a \times 10^{n}}$ — adjust $n$ to even and use $\sqrt{a \times 10^{n}} = \sqrt{a} \times 10^{n/2}$ .

Worked. $\sqrt{4 \times 10^{6}} = 2 \times 10^{3}$ . Worked. $\sqrt{2.5 \times 10^{5}}$ — odd exponent, rewrite as $25 \times 10^{4}$ , then $\sqrt{25} \times 10^{2} = 5 \times 10^{2}$ .

Multiply: $(a \cdot b) \times 10^{m+n}$ .
Divide: $(a / b) \times 10^{m-n}$ .
Adjust if $a$ falls outside $[1, 10)$ .
For square roots, force the exponent to be even first.

Adding and subtracting in standard form

You can't add directly unless the powers of 10 match. Convert to a common power, then add.

Adding/subtracting numbers in different powers of $10$ requires writing them with the SAME power first.

Worked. $5.2 \times 10^{4} + 3.1 \times 10^{3}$ .

Make both have $\times 10^{4}$ :

$3.1 \times 10^{3} = 0.31 \times 10^{4}$ .
Sum: $(5.2 + 0.31) \times 10^{4} = 5.51 \times 10^{4}$ .

Worked. $4.8 \times 10^{-3} - 2 \times 10^{-4}$ .

Make both have $\times 10^{-3}$ :

$2 \times 10^{-4} = 0.2 \times 10^{-3}$ .
Difference: $(4.8 - 0.2) \times 10^{-3} = 4.6 \times 10^{-3}$ .

Tip. Choose the LARGER power of $10$ as the common one — that's the one most likely to leave the answer in standard form without further adjustment.

Same power of 10 first, then add/subtract.
Choose the larger exponent as the common one.
Adjust at the end if $a \not\in [1, 10)$ .

Standard form on a calculator

Use the dedicated key. Typing '× 10 ^ n' by hand multiplies by 10 first and then raises — wrong order.

Casio classpad / fx calculators have a dedicated key for entering $\times 10^{n}$ . It's labelled EXP, EE, $\times 10^{x}$ , or similar.

To enter $3.2 \times 10^{6}$ : type 3.2 → EXP → 6. (NOT 3.2 × 10 ^ 6 — that does the right thing for this case but goes wrong for negative exponents because of operator precedence.)

To enter $4.7 \times 10^{-5}$ : type 4.7 → EXP → ( - 5 ).

Calculator displays usually show $3.2 \times 10^{6}$ as 3.2$\boxed{06}$ or 3.2E6. Always rewrite the result back to proper standard form on your answer line — examiners deduct marks for 3.2E6 written down.

Use the EXP / $\times 10^{x}$ key — don't type by hand.
Wrap negative exponents in brackets.
Calculator's $E$ notation is NOT standard form.
Rewrite as $a \times 10^{n}$ on the answer line.

How it’s examined

Standard-form questions are guaranteed on every paper. Paper 2 typically asks to convert to/from standard form (1 mark) and to calculate $(a \times 10^{m}) \times (b \times 10^{n})$ in standard form (2 marks). Paper 4 hides standard form inside science contexts (mass of atoms, distance to stars). Examiner reports flag two recurring slips: leaving the answer in non-standard form like $20 \times 10^{6}$ , and writing the calculator's $E$ notation on the answer line.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.4); 0580/22 May/Jun 2024 — Q9 (multiplication in standard form); 0580/42 Oct/Nov 2024 — Q8 (mixed-power addition); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Standard Form

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

1Convert ordinary numbers to standard form
Core• conversion
Question
Write in standard form: (a) $4{,}530{,}000$ , (b) $0.000\,082$ , (c) $73$ .
Step-by-step solution
1. Step 1
  (a) Move the point so $4.53$ is between $1$ and $10$ : 6 places left → $4.53 \times 10^{6}$ .
2. Step 2
  (b) $8.2$ is between $1$ and $10$ : 5 places right → $8.2 \times 10^{-5}$ .
3. Step 3
  (c) $7.3$ between $1$ and $10$ : 1 place left → $7.3 \times 10^{1}$ .
Answer
(a) $4.53 \times 10^{6}$ (b) $8.2 \times 10^{-5}$ (c) $7.3 \times 10^{1}$
2Convert standard form back to ordinary
Core• conversion
Question
Write as ordinary numbers: (a) $6.04 \times 10^{4}$ , (b) $1.27 \times 10^{-3}$ .
Step-by-step solution
1. Step 1
  (a) Move $6.04$ four places right → $60{,}400$ .
2. Step 2
  (b) Move $1.27$ three places left → $0.00127$ .
Answer
(a) $60{,}400$ (b) $0.00127$
3Multiply numbers in standard form
Extended• Adapted from 0580/42 Oct/Nov 2023 Q11• multiplication
Question
Calculate $(3.5 \times 10^{6}) \times (4 \times 10^{-2})$ in standard form.
Step-by-step solution
1. Step 1
  Multiply the coefficients and add the indices.
  $(3.5 \times 4) \times 10^{6 + (-2)} = 14 \times 10^{4}$
2. Step 2
  Adjust so coefficient is between $1$ and $10$ .
  $14 \times 10^{4} = 1.4 \times 10^{5}$
Answer
$1.4 \times 10^{5}$
Examiner tip
An answer of $14 \times 10^{4}$ is not in standard form. Always check that $1 \leq a < 10$ .
4Add numbers in standard form
Extended• addition
Question
Calculate $(2.3 \times 10^{5}) + (4.1 \times 10^{4})$ in standard form.
Step-by-step solution
1. Step 1
  Equalise indices: $4.1 \times 10^{4} = 0.41 \times 10^{5}$ .
2. Step 2
  Add: $2.3 + 0.41 = 2.71$ .
3. Step 3
  Result: $2.71 \times 10^{5}$ .
Answer
$2.71 \times 10^{5}$
5Divide numbers in standard form
Extended• Adapted from 0580/22 May/Jun 2024 Q15• division
Question
Calculate $\dfrac{6.0 \times 10^{8}}{1.5 \times 10^{3}}$ , giving your answer in standard form.
Step-by-step solution
1. Step 1
  Divide the coefficients and subtract the indices.
  $\dfrac{6.0}{1.5} \times 10^{8 - 3} = 4 \times 10^{5}$
2. Step 2
  Coefficient $4$ satisfies $1 \le 4 < 10$ , so the answer is already in standard form.
Answer
$4 \times 10^{5}$
Examiner tip
The 2024 mark scheme awards a method mark for showing the subtraction of indices ( $10^{8-3}$ ). Candidates who jump straight to the final number lose the method mark even when the answer is correct.
6Subtract numbers in standard form with different indices
Extended• Adapted from 0580/42 Oct/Nov 2022 Q5• subtraction
Question
Calculate $(5.6 \times 10^{6}) - (8.4 \times 10^{5})$ , giving your answer in standard form.
Step-by-step solution
1. Step 1
  Equalise indices to $10^{6}$ : $8.4 \times 10^{5} = 0.84 \times 10^{6}$ .
2. Step 2
  Subtract: $5.6 - 0.84 = 4.76$ .
  $(5.6 - 0.84) \times 10^{6} = 4.76 \times 10^{6}$
Answer
$4.76 \times 10^{6}$
Examiner tip
The examiner report flags that candidates often subtract the coefficients without equalising the powers, producing nonsense like $-2.8 \times 10^{?}$ . Always rewrite both numbers to share the same power of $10$ first.
7Apply standard form to a population context
Extended• Adapted from 0580/42 May/Jun 2023 Q4• applied, context
Question
The population of a country is $4.8 \times 10^{7}$ . The total land area is $3.2 \times 10^{5}\,\text{km}^{2}$ . Find the population density (people per km²), giving your answer in standard form.
Step-by-step solution
1. Step 1
  Population density $= \dfrac{\text{population}}{\text{area}}$ .
  $\dfrac{4.8 \times 10^{7}}{3.2 \times 10^{5}}$
2. Step 2
  Compute coefficient ratio and index difference.
  $= \dfrac{4.8}{3.2} \times 10^{7-5} = 1.5 \times 10^{2}$
3. Step 3
  Coefficient $1.5$ is in range, so the answer is in standard form: $1.5 \times 10^{2}$ people per km².
Answer
$1.5 \times 10^{2}\,\text{people/km}^{2}$ (i.e. $150$ people/km²)
Examiner tip
Examiners reward candidates who explicitly state the formula "population density = population ÷ area" before substituting. The 2023 mark scheme awarded a method mark for the formula even when the arithmetic was wrong.
8Order three numbers given in standard form
Extended• Adapted from 0580/22 Feb/Mar 2024 Q8• ordering, comparison
Question
Place these numbers in ascending order: $A = 3.6 \times 10^{-4}$ , $B = 9.1 \times 10^{-5}$ , $C = 4.0 \times 10^{-4}$ .
Step-by-step solution
1. Step 1
  Equalise the powers. Rewrite $B = 9.1 \times 10^{-5} = 0.91 \times 10^{-4}$ .
2. Step 2
  Now compare the coefficients with the same power: $0.91, 3.6, 4.0$ .
3. Step 3
  Ascending: $0.91 < 3.6 < 4.0$ , which gives $B < A < C$ .
Answer
$B < A < C$ (i.e. $9.1 \times 10^{-5} < 3.6 \times 10^{-4} < 4.0 \times 10^{-4}$ ).
Examiner tip
The examiner report flags that candidates routinely treat a larger coefficient as a larger number — ignoring the power. Always equalise powers first; then compare coefficients.
9Square a number in standard form
Extended• powers
Question
Calculate $(2.5 \times 10^{-3})^{2}$ , giving your answer in standard form.
Step-by-step solution
1. Step 1
  Square the coefficient and double the index.
  $(2.5)^{2} \times 10^{2 \times (-3)} = 6.25 \times 10^{-6}$
2. Step 2
  Coefficient $6.25$ is in range — answer is in standard form.
Answer
$6.25 \times 10^{-6}$
Examiner tip
Examiners reward candidates who write the index law $(\times 10^{k})^{n} = \times 10^{kn}$ explicitly. Doubling the index (rather than squaring it: $10^{(-3)^{2}} = 10^{9}$ ) is a recurring error flagged in examiner reports.
10How many atoms? (Stretch context)
Challenge• Adapted from 0580/42 Oct/Nov 2023 Q15• context, stretch, applied
Question
The mass of one carbon atom is approximately $2.0 \times 10^{-23}\,\text{g}$ . A diamond has mass $0.4\,\text{g}$ . Estimate the number of carbon atoms in the diamond, giving your answer in standard form to 2 significant figures.
Step-by-step solution
1. Step 1
  Number of atoms $= \dfrac{\text{total mass}}{\text{mass per atom}}$ .
  $N = \dfrac{0.4}{2.0 \times 10^{-23}}$
2. Step 2
  Rewrite $0.4 = 4 \times 10^{-1}$ to make the division easy.
  $N = \dfrac{4 \times 10^{-1}}{2.0 \times 10^{-23}} = 2 \times 10^{-1 - (-23)} = 2 \times 10^{22}$
3. Step 3
  Answer is already in standard form to 2 s.f.
Answer
$2.0 \times 10^{22}$ atoms
Examiner tip
The 2023 mark scheme awards a stretch mark for the sign-handling on $10^{-1 - (-23)} = 10^{22}$ . Double-subtraction errors (writing $10^{-24}$ ) are flagged as the top mistake in the examiner report.
11Order-of-magnitude ratio between two physical quantities
Challenge• Adapted from 0580/42 May/Jun 2024 Q15• ratio, stretch, applied
Question
The distance from the Earth to the Sun is $1.5 \times 10^{8}\,\text{km}$ . The distance from the Earth to the Moon is $3.84 \times 10^{5}\,\text{km}$ . How many times further is the Sun than the Moon? Give your answer in standard form to 3 significant figures.
Step-by-step solution
1. Step 1
  Ratio $= \dfrac{\text{distance to Sun}}{\text{distance to Moon}}$ .
  $\dfrac{1.5 \times 10^{8}}{3.84 \times 10^{5}}$
2. Step 2
  Divide coefficients and subtract indices.
  $= \dfrac{1.5}{3.84} \times 10^{8 - 5} = 0.390625 \times 10^{3}$
3. Step 3
  Normalise: $0.390625 \times 10^{3} = 3.90625 \times 10^{2}$ .
4. Step 4
  Round to 3 s.f.: $3.91 \times 10^{2}$ .
Answer
$3.91 \times 10^{2}$ (the Sun is about $391$ times further than the Moon)
Examiner tip
The 2024 examiner report flags that candidates often skip the standard-form normalisation, leaving the answer as $0.391 \times 10^{3}$ . That form loses the final accuracy mark even though the value is correct — the coefficient must satisfy $1 \le a < 10$ .

Key Formulae — Standard Form

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

Standard form definition
$n = a \times 10^{k},\quad 1 \leq a < 10,\ k \in \mathbb{Z}$
$a$
coefficient between $1$ (inclusive) and $10$ (exclusive)
$k$
integer power of $10$
When to use
Always when expressing a number in standard form.
Multiplication rule
$(a \times 10^{m}) \times (b \times 10^{n}) = (ab) \times 10^{m+n}$
When to use
Multiplying two numbers in standard form. Adjust coefficient afterwards.
Division rule
$\dfrac{a \times 10^{m}}{b \times 10^{n}} = \dfrac{a}{b} \times 10^{m-n}$
When to use
Dividing two numbers in standard form.

Key Definitions and Keywords — Standard Form

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

Standard form (scientific notation)
Examiner keyword
A number written as $a \times 10^{k}$ where $1 \leq a < 10$ and $k$ is an integer.
Coefficient
The number $a$ in $a \times 10^{k}$ .
Index (exponent / power)
Examiner keyword
The integer $k$ that determines the size scale.

Common Mistakes and Misconceptions — Standard Form

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

✕Leaving the coefficient outside $[1, 10)$
0580/42 — every recent series
Why it happens
After multiplying, students get e.g. $14 \times 10^{4}$ and stop without normalising.
How to avoid it
Always check $1 \leq a < 10$ . Adjust by moving the decimal and changing the power.
✕Adding standard-form numbers without equalising the indices
Why it happens
Students mechanically add coefficients, ignoring that the powers differ.
How to avoid it
Match the powers first, then add coefficients.
✕Confusing the direction of the decimal shift for negative exponents
Why it happens
Negative powers of $10$ make small numbers, but students still shift left.
How to avoid it
Negative exponent → small number → shift the decimal left (more zeros).
✕Misreading calculator output (e.g. $4.5\text{E}^{-3}$ ) as $4.5^{-3}$
Why it happens
The E or ×10 symbol is small on calculator screens.
How to avoid it
Calculator $4.5\text{E}^{-3}$ means $4.5 \times 10^{-3}$ , not $4.5$ to the power $-3$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Standard Form — frequently asked questions

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

Ordinary

Standard form

4{,}500

4.5 \times 10^{3}

3{,}200{,}000

3.2 \times 10^{6}

0.000\,072

7.2 \times 10^{-5}

0.6

6 \times 10^{-1}

Standard Form

Short Study Notes in the form of Slides

Short Study Notes — Standard Form

Detailed Notes

What you’ll learn

How it’s examined

1Convert ordinary numbers to standard form

2Convert standard form back to ordinary

3Multiply numbers in standard form

4Add numbers in standard form

5Divide numbers in standard form

6Subtract numbers in standard form with different indices

7Apply standard form to a population context

8Order three numbers given in standard form

9Square a number in standard form

10How many atoms? (Stretch context)

11Order-of-magnitude ratio between two physical quantities

Standard form definition

Multiplication rule

Division rule

Standard form (scientific notation)

Coefficient

Index (exponent / power)

✕Leaving the coefficient outside [1,10)[1, 10)[1,10)

✕Adding standard-form numbers without equalising the indices

✕Confusing the direction of the decimal shift for negative exponents

✕Misreading calculator output (e.g. 4.5E−34.5\text{E}^{-3}4.5E−3) as 4.5−34.5^{-3}4.5−3

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Standard Form — frequently asked questions

Standard Form

Short Study Notes in the form of Slides

Short Study Notes — Standard Form

Detailed Notes

What you’ll learn

How it’s examined

1Convert ordinary numbers to standard form

2Convert standard form back to ordinary

3Multiply numbers in standard form

4Add numbers in standard form

5Divide numbers in standard form

6Subtract numbers in standard form with different indices

7Apply standard form to a population context

8Order three numbers given in standard form

9Square a number in standard form

10How many atoms? (Stretch context)

11Order-of-magnitude ratio between two physical quantities

Standard form definition

Multiplication rule

Division rule

Standard form (scientific notation)

Coefficient

Index (exponent / power)

✕Leaving the coefficient outside [1,10)[1, 10)[1,10)

✕Adding standard-form numbers without equalising the indices

✕Confusing the direction of the decimal shift for negative exponents

✕Misreading calculator output (e.g. 4.5E−34.5\text{E}^{-3}4.5E−3) as 4.5−34.5^{-3}4.5−3

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Standard Form — frequently asked questions

✕Leaving the coefficient outside $[1, 10)$

✕Misreading calculator output (e.g. $4.5\text{E}^{-3}$ ) as $4.5^{-3}$

✕Leaving the coefficient outside $[1, 10)$

✕Misreading calculator output (e.g. $4.5\text{E}^{-3}$ ) as $4.5^{-3}$