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DSE Maths: Indices and Logarithms — Tutor Lesson Plan
Logarithms fail when students treat them as a formula to memorise. One sentence fixes it: a logarithm asks how many powers.
Objectives — the student can apply the three index laws, convert between a^x = y and log_a y = x, and solve exponential equations with logs.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Indices | Evaluate 2⁵ and 3⁻² | 32 and 1/9 |
| Surds | Simplify √18 | 3√2 |
| Equations | Solve 2x = 8 | x = 4 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | What a log asks | log₂8 asks 2 to what power | Answer 3 |
| 8-20 min | Index laws | Multiplying, dividing, powers of powers | Six simplifications |
| 20-34 min | Log laws | log ab = log a + log b | Four simplifications |
| 34-48 min | Exponential equations | 2^x = 8, 3^(2x) = 27 | Two alone |
| 48-56 min | Change of base | log₂5 = log 5 / log 2 | One calculation |
| 56-60 min | Exit check | Three questions marked | Finish alone |
03 Teaching script (read it out)
Scene 1: A logarithm asks for the power
log₂ 8 = 3 because 2³ = 8.
So a^x = y and log_a y = x are two ways of writing one sentence.
For that reason the argument must be positive.
So a^x = y and log_a y = x are two ways of writing one sentence.
For that reason the argument must be positive.
Student explains log₃ 81 = 4.
Scene 2: Three index laws, one habit
aᵐ × aⁿ = a^(m+n); aᵐ ÷ aⁿ = a^(m−n); (aᵐ)ⁿ = a^(mn).
Negative index means reciprocal, fractional index means root.
Make the bases the same first: 4 = 2², 8 = 2³.
Negative index means reciprocal, fractional index means root.
Make the bases the same first: 4 = 2², 8 = 2³.
Student unifies bases before solving 4^x = 8.
Scene 3: Log laws and change of base
log(ab) = log a + log b; log(a/b) = log a − log b; log aⁿ = n log a.
For different bases use log_a b = log b / log a.
So 2^x = 5 gives x = log 5 / log 2, about 2.32.
For different bases use log_a b = log b / log a.
So 2^x = 5 gives x = log 5 / log 2, about 2.32.
Warn: log 5 / log 2 is not log(5/2).
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Definition | log₃ 81 | 3 to what power is 81 | 4 |
| Index laws | 2^(x+1) × 2^(x−1) | Add the indices | 4^x |
| Log laws | log 8 + log 5 | log(8 × 5) | log 40 |
| Exponential equation | 3^(2x) = 27 | 2x = 3 | x = 1.5 |
Sanity check (teach this): A logarithm needs a positive argument: log(−3) is undefined, so a negative result means the working is wrong.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| log₂ 32 | 5 | Power of 2 |
| Simplify log 8 + log 5 | log 40 | Adding logs multiplies |
| Solve 3^(2x) = 27 | x = 1.5 | Same base |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Adding logs as numbers | log 8 + log 5 = log 13 | It is log 40. |
| Change of base reversed | log₂ 5 = log 2 / log 5 | Numerator is the argument. |
| Negative index treated as negative number | 2⁻³ = −8 | It is 1/8. |
| Argument ignored | log(−2) = 0.3 | The argument must be positive. |
07 Homework (paper, section and question numbers)
- Secondary maths chapter index — Use the three questions on this page
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| log₂ 32 | 5 | All right: move to trigonometry |
| log 8 + log 5 | log 40 | One wrong: re-teach the log laws |
| Solution of 3^(2x) = 27 | 1.5 | Two wrong: repeat scene 1 |
09 Tutor FAQ
Memorise change of base?
Yes, but stress which number is the numerator.
Which calculator key?
log or ln, as long as the base is consistent.
When are logs needed?
When the two sides cannot share a base, such as 2^x = 5.
A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
