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DSE Maths: indices and logarithms (tutor lesson plan)
Indices and logarithms are two sides of one relation. When students panic it is usually because they memorise logarithms as a new formula instead of hearing the question “what is the exponent?”. This lesson rebuilds the link from one question.
Objectives — Students can: (1) simplify products, quotients and powers with the same base and state where each law comes from; (2) translate between a^x = y and log_a y = x, and use change of base to evaluate values such as log base 2 of 8 = 3; (3) solve equations such as 3^(2x-1) = 27 (x = 2) and explain the parameters in 8% annual growth doubling in about 9 years and a half-life of 8 hours leaving 3.125% after 40 hours.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Meaning of a power | What is 2 to the power 10? | 1024 |
| Roots as powers | What is 9 to the power one half? | 3 |
| Solving a linear equation | If 2x – 1 = 3, what is x? | 2 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0–10 min | One question | Ask how many times 2 multiplies by itself to reach 8, then write it as log base 2 of 8 | Answer 3 and write log₂8 = 3 |
| 10–22 min | Laws of indices | Expand 2³×2⁴ and derive the laws instead of giving them | Derive aᵐ×aⁿ = aᵐ⁺ⁿ themselves |
| 22–38 min | Definition and translation | a^x = y and log_a y = x, translated ten times including fractions and negatives | Each time state “what is the exponent” |
| 38–48 min | Change of base | Compute log₂8 as log8 ÷ log2 = 3, then log₃81 | Both correct with working shown |
| 48–55 min | Equations and applications | Solve 3^(2x-1) = 27, then discuss growth and half-life parameters | x = 2 and 1.08⁹ ≈ 2.00 |
| 55–60 min | Exit check | Mark live and reteach the matching act | All three correct |
03 Teaching script (read it out)
Scene 1: Begin with “how many times”, not with a formula
Write one question on the board: how many times does 2 multiply by itself to reach 8? Students say 3.
Then write the same sentence in symbols: log₂8 = 3. Ask again: how many times for 1024? Since 2 to the 10 is 1024, log₂1024 = 10.
Say the key sentence three times: a logarithm asks what the exponent is. So log_a y = x and a^x = y are one sentence in two notations, not two formulas.
Add three boundaries: log_a 1 = 0 because a⁰ = 1, log_a a = 1, and a must be positive and not 1 while y must be positive. Domain questions in DSE always test these three conditions.
Then write the same sentence in symbols: log₂8 = 3. Ask again: how many times for 1024? Since 2 to the 10 is 1024, log₂1024 = 10.
Say the key sentence three times: a logarithm asks what the exponent is. So log_a y = x and a^x = y are one sentence in two notations, not two formulas.
Add three boundaries: log_a 1 = 0 because a⁰ = 1, log_a a = 1, and a must be positive and not 1 while y must be positive. Domain questions in DSE always test these three conditions.
Ask students: what is log base 5 of 25, and did you get it from “how many times” or from a formula?
Scene 2: Derive the laws by expanding
Write 2³ × 2⁴ = (2×2×2)(2×2×2×2), count the 2s and get 2⁷. Use the same move for aᵐ ÷ aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐⁿ.
Then handle three traps: negative powers a⁻ⁿ = 1/aⁿ (from a² ÷ a⁵), the zero power a⁰ = 1 (from aⁿ ÷ aⁿ), and fractional powers a^(1/n) = the nth root of a (from raising it to the power n).
Students must expand each one themselves once, because anyone who has expanded will not later write (aᵐ)ⁿ as a^(m+n). That error appears almost every year in DSE and starts with memorising without expanding.
Then handle three traps: negative powers a⁻ⁿ = 1/aⁿ (from a² ÷ a⁵), the zero power a⁰ = 1 (from aⁿ ÷ aⁿ), and fractional powers a^(1/n) = the nth root of a (from raising it to the power n).
Students must expand each one themselves once, because anyone who has expanded will not later write (aᵐ)ⁿ as a^(m+n). That error appears almost every year in DSE and starts with memorising without expanding.
Ask students: what is the difference between (2³)⁴ and 2³ × 2⁴, and why?
Scene 3: Change of base, equations, and what the parameters mean
Derive change of base from the definition: let log_a b = x, so a^x = b; take log base c of both sides to get x·log_c a = log_c b, hence x = log_c b ÷ log_c a. Check with log₂8 = log8 ÷ log2 = 3.
Solving needs one move: make both sides the same base. 3^(2x-1) = 27 = 3³, so 2x – 1 = 3 and x = 2. Only when the bases cannot be matched do you take logarithms of both sides.
In applications, teach the parameters, not the answer: at 8% a year, the multiplier after 9 years is 1.8⁹ ≈ 2.00, roughly double. With a half-life of 8 hours, 40 hours is 5 half-lives, leaving 0.5⁵ = 0.03125, which is 3.125%. The two situations use (1 + r) and (1 – r), and students must say which is growth and which is decay.
Solving needs one move: make both sides the same base. 3^(2x-1) = 27 = 3³, so 2x – 1 = 3 and x = 2. Only when the bases cannot be matched do you take logarithms of both sides.
In applications, teach the parameters, not the answer: at 8% a year, the multiplier after 9 years is 1.8⁹ ≈ 2.00, roughly double. With a half-life of 8 hours, 40 hours is 5 half-lives, leaving 0.5⁵ = 0.03125, which is 3.125%. The two situations use (1 + r) and (1 – r), and students must say which is growth and which is decay.
Ask students: why does a half-life use powers of 0.5 while inflation uses powers of 1.0x?
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Simplification | Simplify (2a³)² × a⁻⁴ ÷ a² | Powers and brackets first, then add and subtract indices with the same base, then clear negatives | 4a⁶ × a⁻⁴ ÷ a² = 4a⁰ = 4 |
| Logarithm evaluation | Find log₃81 and log₂(1/8) | Rewrite as powers of the same base: 81 = 3⁴ and 1/8 = 2⁻³ | 4 and −3 |
| Exponential equation | Solve 5^(x+1) = 125 | Make both sides base 5: 5^(x+1) = 5³, so x + 1 = 3 | x = 2 |
| Application | A city grows 8% a year. When does the population double? | Set 1.08ⁿ = 2 and take logs: n = log2 ÷ log1.08 | About 9 years, since 1.08⁹ ≈ 2.00 |
Sanity check (teach this): Check the numbers: 2 to the 10 is 1024; log base 10 of 1000 is 3; log₂8 is 3; 3^(2×2−1) = 27; 1.08 to the 9 is about 2.00; after 40 hours with a 8-hour half-life, 3.125% remains. All computed in python, so never replace a multiplier with “about double” in class. DSE marks the setup and the taking of logs, with the final digits worth only the last mark.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Simplify (3x²)³ ÷ (9x⁴) and name the law used at each step. | 27x⁶ ÷ 9x⁴ = 3x². | Coefficients and powers handled first, then indices subtracted in order |
| Solve 2^(3x-1) = 32 and check the answer. | 2^(3x-1) = 2⁵, so 3x – 1 = 5 and x = 2; substituting gives 32. | Both sides made base 2 and the answer substituted back |
| A drug has a half-life of 6 hours. What fraction remains after 24 hours? | 24 ÷ 6 = 4 half-lives, 0.5⁴ = 0.0625, so 6.25%. | Counts half-lives first instead of computing 0.5 × 24 |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Logarithms treated as a new formula | “I need a table for log₂8.” | “No table. Ask yourself: how many times does 2 multiply by itself to give 8? That number is the logarithm.” |
| Adding the powers in a power of a power | “(2³)⁴ = 2⁷.” | “The outer 4 means how many times you take the cube. Expand it once for me.” |
| Ignoring the domain | “log(−4) can be computed, negatives have logarithms too.” | “A logarithm asks how many times the base multiplies to give the number. A positive base never gives a negative, so the argument must be positive.” |
| Using the rate instead of the multiplier | “It grows 8% a year, so use powers of 0.08.” | “0.08 is the increase, not the new-to-old ratio. You need 1 + 0.08 = 1.08.” |
07 Homework (paper, section and question numbers)
- Redo the practice on this page — Questions 1 to 3: name the law in each simplification, substitute back in the equation, and show the setup line in the application. Question 3 is collected next lesson.
- Preview the next chapter: the circle — Find the circle equation chapter on the subject index, read the first objective and write down one question.
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| What is log₂8, and why? | 3, because 2³ = 8; a logarithm asks what the exponent is | Answering “use the table” means the definition needs reteaching |
| What are (a²)³ and a² × a³? | a⁶ and a⁵: one is a power of a power, the other a product of like bases | Giving the same answer for both means memorising without expanding |
| What is the first step in solving 3^(2x−1) = 27? | Write 27 as 3³ so both sides share a base and the exponents equate | Taking logs without saying why is acceptable only if explained |
09 Tutor FAQ
Should students still work by hand when calculators have a log key?
Yes. DSE marks the line where the logarithm is taken; the calculator only supplies the final value.
Is the natural logarithm needed?
At S5, teach base 10 and general base a. ln appears in only a few DSE items, and knowing it is the logarithm with base e is enough.
Must applications always use logarithms?
Trial and improvement works when the exponent is a whole number; once the question asks “after how many years” with a non-integer answer, logarithms are the method — which is the point of this chapter.
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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.
