Last reviewed: 7 November 2026
Updated November 2026. These Grade 9 indices and logarithms notes explain how to write numbers in index form, use the laws of indices, and connect powers of 10 to common logarithms. An index (power) tells you how many times the base is multiplied by itself, so 25 = 2 × 2 × 2 × 2 × 2 = 32. A common logarithm answers the reverse question: what power of 10 gives this number? For example, log 1000 = 3 because 103 = 1000.
The examples and questions below are original practice written for Kenyan junior school learners, not an official paper. For the full list of Grade 9 maths topics, start with our Grade 9 maths revision guide and the Grade 9 notes hub.
On this page
Where Indices and Logarithms Sit in Grade 9 Maths
In the Grade 9 Mathematics curriculum design published by the Kenya Institute of Curriculum Development, indices and logarithms is a sub-strand under the Numbers strand. Learners are expected to express numbers in index form, generate and apply the laws of indices, and relate powers of 10 to common logarithms. You can find the design under the KICD Grade Nine curriculum designs. These notes stay inside that scope: there is no change of base, natural logarithm or logarithmic equation work here.
Index Form: Base and Index
In 34, the number 3 is the base and 4 is the index. The value is 3 × 3 × 3 × 3 = 81. To write a number in index form, break it into prime factors and group equal factors.
- 16 = 2 × 2 × 2 × 2 = 24
- 125 = 5 × 5 × 5 = 53
- 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32
A common mistake is reading 24 as 2 × 4 = 8. The index counts repeated multiplication, not a single multiplication.
The Laws of Indices With Worked Examples
Each law can be generated by writing the powers out in full, which is exactly how the curriculum asks learners to discover them. The same rules appear in OpenStax Prealgebra’s section on multiplication properties of exponents.
| Law | Rule | Example |
|---|---|---|
| Multiplication | am × an = am+n | 23 × 24 = 27 = 128 |
| Division | am ÷ an = am−n | 56 ÷ 54 = 52 = 25 |
| Power of a power | (am)n = amn | (32)3 = 36 = 729 |
| Zero index | a0 = 1 (a ≠ 0) | 70 = 1 |
| Negative index | a−n = 1 ÷ an | 2−3 = 1/8 |
Why does the multiplication law work?
23 × 24 = (2 × 2 × 2) × (2 × 2 × 2 × 2). Count the twos: there are seven, so the answer is 27. The laws only apply when the bases are the same. You cannot combine 23 × 34 into one power.
Why is any non-zero number to the power 0 equal to 1?
Use the division law: 43 ÷ 43 = 43−3 = 40. But any non-zero number divided by itself is 1, so 40 = 1.
Worked example: simplify using the laws
Simplify (25 × 23) ÷ 26.
- Multiply first: 25 × 23 = 28.
- Divide: 28 ÷ 26 = 22.
- Evaluate: 22 = 4.
Worked example two: find x if 3x = 81. Write 81 in index form with base 3: 81 = 34. Since the bases match, x = 4.
Powers of 10 and Common Logarithms
A common logarithm uses base 10. If 10n = N, then log N = n. This is the definition given in OpenStax College Algebra’s section on logarithmic functions, which explains that a logarithm is the exponent.
| Power of 10 | Number | Common logarithm |
|---|---|---|
| 103 | 1000 | log 1000 = 3 |
| 102 | 100 | log 100 = 2 |
| 101 | 10 | log 10 = 1 |
| 100 | 1 | log 1 = 0 |
| 10−1 | 0.1 | log 0.1 = −1 |
| 10−2 | 0.01 | log 0.01 = −2 |
For numbers that are not exact powers of 10, the logarithm is a decimal. Since 50 lies between 10 and 100, log 50 lies between 1 and 2. A calculator or mathematical tables give log 50 ≈ 1.699, which means 101.699 ≈ 50. The Grade 9 design allows either calculators (IT) or tables for this step, so follow your teacher’s instruction for tests.
Worked example: estimate without a calculator
Between which two whole numbers does log 4,500 lie? Since 1,000 = 103 and 10,000 = 104, and 4,500 lies between them, log 4,500 lies between 3 and 4. A calculator gives about 3.653, which checks the estimate.
Original Practice Questions With Worked Answers
- Write 200 in index form using prime factors. 200 = 2 × 2 × 2 × 5 × 5 = 23 × 52.
- Simplify a7 × a−2. Add indices: 7 + (−2) = 5, so the answer is a5.
- Evaluate (102)3 ÷ 104. (102)3 = 106; 106 ÷ 104 = 102 = 100.
- Find log 0.001. 0.001 = 10−3, so log 0.001 = −3.
- Solve 2x+1 = 32. 32 = 25, so x + 1 = 5 and x = 4.
- A learner writes 32 × 33 = 95. What went wrong? The base must stay 3; only the indices are added. Correct answer: 35 = 243.
Cover the answers, attempt each question, then check your working line by line. For timed practice, open the Grade 9 revision papers or test yourself in the Grade 9 CBC Challenge. These notes are free; full revision access is available with the KSh 20 Day Pass or the KSh 500 Monthly Pass.
Frequently Asked Questions
What are indices in Grade 9 maths?
Indices (powers) show repeated multiplication of the same base. In 53, 5 is the base and 3 is the index, so 53 = 5 × 5 × 5 = 125.
What is a common logarithm?
A common logarithm is a logarithm with base 10. It is the power to which 10 must be raised to give the number, so log 100 = 2 because 102 = 100.
Can I add indices when the bases are different?
No. The laws of indices only combine powers with the same base. 23 × 52 stays as it is or is evaluated as 8 × 25 = 200.
Why is log 1 equal to 0?
Because 100 = 1. Any non-zero number raised to the power 0 equals 1, so the power of 10 that gives 1 is 0.