NUMERICAL CONCEPTS
PART 2: LOGARITHMS
Full Teaching Notes • Worked Examples • Practice Exercises
About this lesson
This lesson develops logarithms as the inverse operation of exponentiation. It is written for Mathshub’s Uganda A-Level Principal Mathematics series, using the National Curriculum Development Centre (NCDC) 2025 Principal Mathematics syllabus as the main alignment reference. In that syllabus, Numerical Concepts (including indices, logarithms and surds) is scheduled in Senior Five, Term One. The material can also support Senior Six revision and extension. Read each explanation, follow the worked examples, and attempt the exercises before checking the answers.
Learning outcomes
Syllabus alignment: The core of this chapter addresses logarithmic notation, laws of logarithms, solving index and logarithmic equations, and applications involving exponential growth and decay. Change of base is included as a useful calculator technique/enrichment skill.
Explain the meaning of a logarithm and convert between logarithmic and exponential forms.
Apply the laws of logarithms accurately.
Evaluate logarithms and use change of base.
Solve logarithmic equations while checking domain restrictions.
Solve exponential equations using logarithms.
1. Meaning of logarithms
A logarithm answers the question: “To what power must a given base be raised to obtain a particular number?”
If aˣ = N, then logₐ N = x
Here a is the base, N is the argument, and x is the logarithm. For real logarithms, the conditions are a > 0, a ≠ 1, and N > 0.
Notation note: To avoid confusion caused by small subscript numbers displaying as boxes on some devices, this edition writes “log base 2 of 8” instead of using a small 2 after log. The meaning is exactly the same.
Examples:
2³ = 8 means log base 2 8 = 3.
10² = 100 means log base 10 100 = 2.
5⁰ = 1 means log base 5 1 = 0.
3⁻² = 1/9 means log base 3(1/9) = −2.
Common logarithm: log base 10 x is often written log x in many school texts. Natural logarithm: logₑ x is written ln x, where e ≈ 2.71828. Follow the notation used in your syllabus or examination.
Worked Example 1: Convert between forms
Write each statement in the other form.
a) 4³ = 64
log base 4 64 = 3
b) log base 3 81 = 4
3⁴ = 81
c) log base 2(1/8) = −3
2⁻³ = 1/8
2. Evaluating logarithms
To evaluate a logarithm, express the argument as a power of the base, then use the exponent.
Worked Example 2
Evaluate log base 5 125.
Since 125 = 5³, we have:
log base 5 125 = log base 5(5³) = 3
Worked Example 3
Evaluate log base 2(1/16).
Since 1/16 = 2⁻⁴:
log base 2(1/16) = −4
Worked Example 4
Evaluate log base 9 3.
Since 9 = 3², let x = log base 9 3. Then 9ˣ = 3, so (3²)ˣ = 3¹. Hence 2x = 1 and x = 1/2.
log base 9 3 = 1/2
3. The laws of logarithms
For positive M and N, and a valid base a, the following laws apply.
| Law | Rule | Example |
|---|---|---|
| Product | logₐ(MN) = logₐ M + logₐ N | log base 2(8 × 4) = 3 + 2 = 5 |
| Quotient | logₐ(M/N) = logₐ M − logₐ N | log base 3(81/3) = 4 − 1 = 3 |
| Power | logₐ(Mʳ) = r logₐ M | log base 10(100³) = 3 × 2 = 6 |
| Logarithm of 1 | logₐ 1 = 0 | log base 7 1 = 0 |
| Logarithm of base | logₐ a = 1 | log base 5 5 = 1 |
| Inverse relationship | logₐ(aˣ) = x | log base 2(2⁷) = 7 |
Important: logarithms do not split over addition or subtraction. In general, logₐ(M + N) is not equal to logₐ M + logₐ N.
Worked Example 5: Product and quotient laws
Expand log base 3(9x/√y), where x > 0 and y > 0.
log base 3(9x/√y) = log base 3 9 + log base 3 x − log base 3(√y)
Since 9 = 3² and √y = y¹/²:
= 2 + log base 3 x − ½ log base 3 y
Worked Example 6: Condense into one logarithm
Write 2 log base 5 x + log base 5 y − log base 5 z as a single logarithm, where x, y, z > 0.
Use the power law first, then the product and quotient laws.
log base 5(x²) + log base 5 y − log base 5 z = log base 5(x²y/z)
Worked Example 7: Solve a logarithmic equation
Solve log base 2 x + log base 2(x − 2) = 3.
Domain: x > 0 and x − 2 > 0, so x > 2.
log base 2[x(x − 2)] = 3
x(x − 2) = 2³ = 8
Therefore x² − 2x − 8 = 0, so (x − 4)(x + 2) = 0. The candidates are x = 4 or x = −2. Only x = 4 satisfies x > 2.
Answer: x = 4
4. Change of base
The change-of-base formula allows a logarithm to be evaluated using another base, often base 10 or base e.
logₐ N = log_b N / log_b a
For example, log base 2 7 = log 7 / log 2 ≈ 2.807 (to 3 decimal places). Use calculator notation carefully and round only at the end.
Worked Example 8: Change of base
Evaluate log base 4 10 to 3 decimal places.
log base 4 10 = log 10 / log 4 ≈ 1.661
5. Solving logarithmic equations
Combine logarithms using the laws where appropriate, convert to exponential form, solve the resulting equation, and check every candidate in the original equation. The argument of each real logarithm must be positive.
Worked Example 9
Solve log base 3(x + 1) = 2.
The domain requires x + 1 > 0. Convert to exponential form:
x + 1 = 3² = 9, so x = 8
Since x = 8 satisfies the domain, it is valid.
Worked Example 10
Solve log base 10 x + log base 10(x − 9) = 1.
Domain: x > 9. Combine the logarithms:
log base 10[x(x − 9)] = 1
x(x − 9) = 10
Thus x² − 9x − 10 = 0, so (x − 10)(x + 1) = 0. The candidates are x = 10 and x = −1. Only x = 10 satisfies x > 9.
Answer: x = 10
6. Solving exponential equations using logarithms
When the unknown appears in the exponent and both sides cannot conveniently be expressed using the same base, take logarithms of both sides.
Worked Example 11
Solve 5ˣ = 17.
x log 5 = log 17
x = log 17 / log 5 ≈ 1.760
Worked Example 12
Solve 2^(3x − 1) = 7.
(3x − 1) log 2 = log 7
3x − 1 = log 7 / log 2
x = [1 + (log 7 / log 2)] / 3 ≈ 1.269
6A. Applications to exponential growth and decay
Logarithms are useful when a quantity changes exponentially and we need to find the time or number of periods. A common model is N = N₀aᵗ, where N₀ is the initial amount, a is the growth or decay factor per period, and t is the number of periods.
Worked Example 13: Exponential growth
A quantity grows according to N = 200(1.5)ᵗ. Find t when N = 675.
Substitute N = 675: 675 = 200(1.5)ᵗ, so (1.5)ᵗ = 675/200 = 3.375.
Take logarithms of both sides:
t log 1.5 = log 3.375, hence t = log 3.375 / log 1.5 = 3.
The quantity reaches 675 after 3 periods.
Worked Example 14: Exponential decay
A substance has amount A = 800(0.8)ᵗ after t hours. Find t when A = 409.6.
409.6 = 800(0.8)ᵗ, so (0.8)ᵗ = 0.512. Taking logarithms gives t log 0.8 = log 0.512.
Therefore t = log 0.512 / log 0.8 = 3.
The amount reaches 409.6 after 3 hours.
Extension: simultaneous equations involving logarithms
Some problems combine logarithmic expressions with algebraic equations. Use the logarithm laws carefully, state the domain restrictions, solve the resulting equation, and check all candidates in the original equations.
Worked Example 15
Solve log base 2 x + log base 2 y = 3 and x − y = 2, where x > 0 and y > 0.
The first equation gives log base 2(xy) = 3, so xy = 8. From x − y = 2, x = y + 2.
Substitute: y(y + 2) = 8, so y² + 2y − 8 = 0 and (y + 4)(y − 2) = 0.
Since y > 0, y = 2, and then x = 4.
Answer: (x, y) = (4, 2).
7. Common errors and examination tips
Check that every logarithm has a positive argument.
The base must be positive and must not equal 1.
Use addition for the logarithm of a product and subtraction for the logarithm of a quotient.
The power rule brings the exponent to the front: logₐ(Mʳ) = r logₐ M.
Do not split logₐ(M + N) into separate logarithms.
When solving equations, reject any candidate that violates the original logarithm domain.
Keep extra calculator digits during working and round only at the end.
8. Practice exercises
Attempt all questions and show clear working. Give decimal answers to 3 decimal places where requested.
A. Core skills
1. Write 7³ = 343 in logarithmic form.
2. Write log base 4 64 = 3 in exponential form.
3. Evaluate log base 2 32.
4. Evaluate log base 3(1/27).
5. Evaluate log base 7 1.
6. Evaluate log base 2 8 + log base 2 4.
7. Expand log base 5(25x), where x > 0.
8. Expand log base 3(a²/b), where a > 0 and b > 0.
9. Write 3 log base 2 x − log base 2 y as a single logarithm, where x, y > 0.
B. Equations
10. Solve log base 5 x = 3.
11. Solve log base 2(x − 1) = 4.
12. Solve log base 3 x + log base 3(x − 2) = 1.
13. Solve log base 10 x + log base 10(x − 9) = 1.
14. Solve 4ˣ = 11, giving x to 3 decimal places.
15. Solve 3^(2x + 1) = 20, giving x to 3 decimal places.
C. Challenge questions
16. Solve log base 2 x + log base 2(x − 2) = 3.
17. Solve log base 3(x + 2) − log base 3 x = 1.
18. Given log base 10 2 = 0.3010 and log base 10 3 = 0.4771, find log base 10 12.
19. Evaluate log base 9 27.
20. Solve 2ˣ = 13, giving x to 3 decimal places.
21. A population follows P = 120(1.25)ᵗ. Find t when P = 375.
22. A machine’s value follows V = 5000(0.9)ᵗ. Find t when V = 3645.
9. Answers to practice exercises
1. log base 7 343 = 3.
2. 4³ = 64.
3. 5.
4. −3.
5. 0.
6. 3 + 2 = 5.
7. 2 + log base 5 x.
8. 2 log base 3 a − log base 3 b.
9. log base 2(x³/y).
10. x = 125.
11. x − 1 = 16, so x = 17.
12. Domain x > 2. x(x − 2) = 3, so x² − 2x − 3 = 0; x = 3 or −1. Valid answer: x = 3.
13. x = 10 (domain x > 9).
14. x = log 11 / log 4 ≈ 1.730.
15. (2x + 1) log 3 = log 20, so x = ½(log 20/log 3 − 1) ≈ 0.864.
16. x = 4 (domain x > 2).
17. Domain x > 0. (x + 2)/x = 3, so x = 1.
18. log base 10 12 = log base 10(3 × 4) = log base 10 3 + 2 log base 10 2 = 0.4771 + 0.6020 = 1.0791.
19. Since 9 = 3² and 27 = 3³, log base 9 27 = 3/2.
20. x = log 13 / log 2 ≈ 3.700.
21. 120(1.25)ᵗ = 375 gives (1.25)ᵗ = 3.125, so t = log 3.125 / log 1.25 = 5.
22. 5000(0.9)ᵗ = 3645 gives (0.9)ᵗ = 0.729, so t = log 0.729 / log 0.9 = 3.
10. Revision checklist
☐ I can convert between exponential and logarithmic forms.
☐ I can evaluate logarithms using powers.
☐ I can apply product, quotient and power laws.
☐ I can expand and condense logarithmic expressions.
☐ I can use the change-of-base formula.
☐ I check domain restrictions when solving logarithmic equations.
☐ I can solve exponential equations using logarithms.
Next part
Numerical Concepts — Part 3 will be planned after checking the intended A-Level syllabus and the scope of Parts 1 and 2. Possible content includes surds, standard form, and numerical approximation, subject to syllabus requirements.
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