Operations with numbers in scientific notation

IB Mathematics: Applications and Interpretation HL· 1.1 Number and algebra: Scientific notation· 25 min read

1. Multiplication and Division★☆☆☆☆HL only⏱ 10 min

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For multiplication and division, we use exponent rules to simplify the calculation: multiply/divide the mantissas (the $a$ terms) separately, then add/subtract the exponents of 10 respectively.

📐 Worked Example

Calculate $(3.2 \times 1 0^{4}) \times (2.5 \times 1 0^{2})$ and $(7.2 \times 1 0^{5}) \div (1.2 \times 1 0^{3})$ , leaving both answers in scientific notation.

1
For multiplication: Group mantissas and powers of 10 separately:
2
$(3.2 \times 2.5) \times (10^4 \times 10^2)$
3
Calculate the product of mantissas and add exponents:
4
$8 \times 10^{4+2} = 8 \times 10^6$
5
For division: Group terms and divide mantissas, subtract exponents:
6
$\frac{7.2}{1.2} \times 10^{5-3} = 6 \times 10^2$
7
Both results are already in correct scientific notation form.

2. Addition and Subtraction★★☆☆☆HL only⏱ 15 min

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📘 Definition

Addition/Subtraction Rule

To add or subtract values in scientific notation, first convert all terms to have the same power of 10, then add or subtract the mantissas, then adjust the final result to correct form if needed.

Example:

$3 \times 1 0^{2} + 5 \times 1 0^{1} = 30 \times 1 0^{1} + 5 \times 1 0^{1} = 3.5 \times 1 0^{2}$

📐 Worked Example

Calculate $(6.4 \times 1 0^{3}) + (2.1 \times 1 0^{2})$ , leaving the answer in scientific notation.

1
Step 1: Rewrite terms to share the same exponent. Rewrite $2.1 \times 1 0^{2}$ to match exponent 3:
2
$2.1 \times 10^2 = 0.21 \times 10^3$
3
Step 2: Add the mantissas, keep the common exponent:
4
$(6.4 + 0.21) \times 10^3 = 6.61 \times 10^3$
5
The result is already in correct form, so this is the final answer.

3. Adjusting Results to Correct Form★★☆☆☆HL only⏱ 10 min

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After performing operations, your result will often have a mantissa that is less than 1 or greater than or equal to 10, which does not meet the standard definition of scientific notation. You must adjust it to get full marks.

📐 Worked Example

Calculate $(5.0 \times 1 0^{3}) \times (4.0 \times 1 0^{2})$ , leaving the answer in correct scientific notation.

1
Step 1: Perform the initial multiplication:
2
$(5.0 \times 4.0) \times 10^{3+2} = 20 \times 10^5$
3
Step 2: The mantissa 20 is greater than 10, so adjust it to between 1 and 10:
4
$20 = 2.0 \times 10^1$
5
Step 3: Combine the exponents to get the final result:
6
$2.0 \times 10^1 \times 10^5 = 2.0 \times 10^6$

For a mantissa less than 1, reverse the process: multiply the mantissa by 10 and subtract 1 from the exponent. For example, $0.45 \times 1 0^{- 2} = 4.5 \times 1 0^{- 3}$ .

4. Real-World Applications★★★☆☆HL only⏱ 10 min

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Scientific notation is used for all types of real-world measurements of large (populations, astronomical distances) and small (atomic masses, microscopic lengths) values.

📐 Worked Example

The population of a country is $5.2 \times 1 0^{7}$ people, and average annual carbon emission per person is $9.5 \times 1 0^{3}$ kg. Calculate total national annual emission in scientific notation.

1
Total emission = population × emission per person, so multiply the two values:
2
$(5.2 \times 9.5) \times (10^7 \times 10^3) = 49.4 \times 10^{10}$
3
Adjust to correct scientific notation:
4
$49.4 \times 10^{10} = 4.94 \times 10^{11}$

Total annual emission is $4.94 \times 1 0^{11}$ kg.

5. Common Pitfalls

Wrong move:

Adding exponents when adding numbers in scientific notation without matching exponents first

Why:

This incorrectly applies multiplication rules to addition, leading to large errors in the result

Correct move:

Always convert all terms to the same power of 10 before adding or subtracting mantissas

Wrong move:

Forgetting to adjust the mantissa after an operation when it is outside $1 \leq a < 10$

Why:

Exam markers require answers in standard scientific notation, so you lose marks even if the numerical value is correct

Correct move:

Always check your final result and adjust the exponent and mantissa to meet the requirement

Wrong move:

Adjusting $35 \times 1 0^{4}$ to $3.5 \times 1 0^{3}$ instead of $3.5 \times 1 0^{5}$

Why:

Mistake in direction of exponent change: dividing the mantissa by 10 requires increasing the exponent by 1

Correct move:

If you divide the mantissa by $1 0^{k}$ , add $k$ to the exponent; if you multiply the mantissa by $1 0^{k}$ , subtract $k$ from the exponent

Wrong move:

Converting $2.1 \times 1 0^{- 3}$ to $0.021 \times 1 0^{- 2}$ when matching exponents for addition

Why:

Incorrect scaling of the mantissa when changing the exponent value

Correct move:

To increase the exponent by 1, divide the mantissa by 10, so $2.1 \times 1 0^{- 3} = 0.21 \times 1 0^{- 2}$

6. Quick Reference Cheatsheet

Operation	Rule	Example
Multiplication	$(a \times 1 0^{m}) (b \times 1 0^{n}) = ab \times 1 0^{m + n}$ , adjust form	$(2 \times 1 0^{3}) (3 \times 1 0^{2}) = 6 \times 1 0^{5}$
Division	$\frac{a \times 1 0 ^{m}}{b \times 1 0 ^{n}} = \frac{a}{b} \times 1 0^{m - n}$ , adjust form	$\frac{8 \times 1 0 ^{5}}{2 \times 1 0 ^{2}} = 4 \times 1 0^{3}$
Add/Subtract	Convert to same exponent, add/subtract mantissas, adjust form	$3 \times 1 0^{2} + 4 \times 1 0^{1} = 3.4 \times 1 0^{2}$
Adjustment	$a \geq 10$ : $a = a /10 \times 1 0^{1}$ , +1 to exponent; $a < 1$ : $a = 10 a \times 1 0^{- 1}$ , -1 to exponent	$25 \times 1 0^{3} = 2.5 \times 1 0^{4}$ ; $0.35 \times 1 0^{2} = 3.5 \times 1 0^{1}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Population calculation problem
2023 · 2
Astronomy distance calculation

Going deeper

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What's Next

Mastering operations with scientific notation is a foundational skill for all subsequent topics in IB AI HL, where you will regularly work with large and small real-world values in statistics, calculus, and mathematical modelling. A solid understanding of exponent rules here also supports work with logarithms and exponential growth/decay models, which are heavily weighted in both Paper 1 and Paper 2 assessments. Practising these operations now will build accuracy for more complex problems later in the course.