Key Idea: Standard form is the IB's way of writing very large and very small numbers compactly. It usually appears as the last line of a question — give your answer in the form a × 10ᵏ — on both papers.
🔢 The structure: a × 10ᵏ
- the coefficient — one non-zero digit before the decimal point
- the exponent — a positive or negative whole number
| Ordinary form | Standard form |
|---|---|
| 8 000 | 8 × 10³ |
| 453 000 | 4.53 × 10⁵ |
| 0.006 | 6 × 10⁻³ |
| 0.000 21 | 2.1 × 10⁻⁴ |
Big number (≥ 10) → exponent positive, point moves left. e.g. 52 000 = 5.2 × 10⁴. Small number (< 1) → exponent negative, point moves right. e.g. 0.0007 = 7 × 10⁻⁴. Count place moves, not digits — and check 1 ≤ a < 10 every time.
🖐️ Calculating by hand (Paper 1)
| Operation | Rule | Example |
|---|---|---|
| Multiply | Multiply the coefficients, add the powers of ten. | (4 × 10⁵)(2 × 10³) = 8 × 10⁸ |
| Divide | Divide the coefficients, subtract the powers of ten. | (3 × 10⁴) ÷ (6 × 10⁷) = 5 × 10⁻⁴ |
| Power | Raise the coefficient, multiply the exponent. | (4 × 10²)³ = 6.4 × 10⁷ |
✏️ IB-style worked examples
IB-style question — write a small number in standard form
Write 0.000 805 in standard form.
Step by step:
Put the point just after the first non-zero digit (8).
The point moved 4 places right, so the exponent is negative.
Write them together.
0.000 805 = 8.05 × 10⁻⁴
IB-style question — compute, then express (Paper 2)
A sphere has radius 9.4 cm. Find its volume in the form a × 10ᵏ cm³, where 1 ≤ a < 10.
Step by step:
Use the volume formula and your GDC to work out the value.
The GDC does the arithmetic.
Rewrite in standard form — move the point 3 places.
V = 3.48 × 10³ cm³ (3 s.f.)
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
IB-style question — cube and re-normalise (Paper 1)
A cube has edge length 4 × 10² cm. Find its volume in the form a × 10ᵏ cm³, where 1 ≤ a < 10, without a calculator.
Step by step:
Volume of a cube = edge³. Cube the coefficient and multiply the exponent.
Work each part out.
64 is not between 1 and 10 — re-normalise: 64 = 6.4 × 10¹.
V = 6.4 × 10⁷ cm³
Important: The final answer must have exactly one non-zero digit before the decimal point (1 ≤ a < 10). After a calculation, re-normalise: 27 × 10⁶ → 2.7 × 10⁷, 0.5 × 10⁻³ → 5 × 10⁻⁴. With powers, raise the coefficient too: (4 × 10²)³ = 64 × 10⁶, not 4 × 10⁶.
Tap each card to reveal the answer.
Write 6 200 000 in standard form 6.2 × 10⁶ — the point moves 6 places left.
Write 0.0009 in standard form 9 × 10⁻⁴ — the point moves 4 places right.
(2 × 10³) × (4 × 10⁵) 8 × 10⁸ — multiply the coefficients (2 × 4 = 8) and add the powers (3 + 5 = 8).
(9 × 10⁶) ÷ (3 × 10²) 3 × 10⁴ — divide the coefficients (9 ÷ 3 = 3) and subtract the powers (6 − 2 = 4).
(5 × 10³)² 2.5 × 10⁷ — 5² = 25, exponent 3 × 2 = 6, then re-normalise 25 × 10⁶.
Your GDC shows 1.45ᴇ7 — write it in standard form 1.45 × 10⁷ — the ᴇ7 means × 10⁷.
Exam Tips
- Check the coefficient a is between 1 and 10 — every time, including after a calculation.
- Big number → positive exponent; small number → negative exponent. Count place moves, not digits.
- Multiply → add powers. Divide → subtract. Power → multiply (and raise the coefficient too).
- Re-normalise an untidy answer: 27 × 10⁶ → 2.7 × 10⁷.
- On Paper 2, the GDC's ᴇ means × 10 — rewrite it; never leave ᴇ in your final answer.