Teacher page: assessing Year 11 Extended Unit 6

How to use this menu

Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The timings spread assessment through the unit rather than leaving it all to the end, and no task runs longer than one lesson.

Descriptors: Year 11 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.

The menu at a glance

Task	Criteria	When	Time	Format
A1: Quick check: logarithms, laws and equations	A	After lesson 4	25 minutes	Individual, written, in class. 20 marks.
B1: Investigation: powers and their length	B	Mid-unit, after lesson 5	45 minutes	In class. Students may collect results in pairs, but each writes up their own findings.
C1: Explain it to someone who missed it	C	After lesson 8	35 minutes	Individual, written, in class or as homework. One page of words, symbols and a diagram.
A2: End-of-unit test	A	End of unit, after lesson 8	45 minutes	Individual, written, in class. 30 marks. Scientific calculator allowed throughout.
D1: Sound at the night market	D and C	End of unit, after lesson 8	45 minutes	Individual, written, in class. The final answer is a short written recommendation.
D2: The longtail boat cooperative	D and C	End of unit, after lesson 8	45 minutes	Individual, written, in class. The final answer is a short written plan for the cooperative.

D1 and D2 are alternatives. Both assess criteria D and C, in two different real-life contexts: one is set, not both.

The unit's inquiry

Key concept. Relationships: a logarithm is the exponent that links a base to a number, and a sign diagram links an expression to the set of x-values where it is positive.
Related concepts. Models, Representation, Change
Global context. Scientific and technical innovation: logarithmic scales and growth models that people use to measure sound and acidity, plan savings and set prices.
Statement of inquiry. A logarithm names the exponent hidden in a growth or a scale, and a sign diagram shows where a quadratic or rational model is positive, so each turns a question about size into an exact answer.

MYP Year 5 descriptors, paraphrased

Each task's clarification, on its student sheet below, says what these levels look like in that task.

Level	Criterion A: Knowing and understanding
1–2	Selects suitable mathematics for simple problems in familiar situations and applies it with some success.
3–4	Selects suitable mathematics for more complex problems in familiar situations and applies it successfully.
5–6	Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts.
7–8	Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts.

Level	Criterion B: Investigating patterns
1–2	With support, applies problem-solving techniques to find simple patterns and states predictions that fit them.
3–4	Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings.
5–6	Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold.
7–8	Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules.

Level	Criterion C: Communicating
1–2	Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow.
3–4	Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organized.
5–6	Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure.
7–8	Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure.

Level	Criterion D: Applying mathematics in real-life contexts
1–2	Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it.
3–4	Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense.
5–6	Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense.
7–8	Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense.

Task A1: Quick check: logarithms, laws and equations

Why set it. A short check of lessons 1 to 4 while there is still time to reteach: logarithms of powers, the three laws, and exponential equations solved by matching bases. Questions 11 and 12 are unfamiliar, so the top band is open to everyone.

When: After lesson 4. Time: 25 minutes. Format: Individual, written, in class. 20 marks.

The student sheet, with its task-specific clarification

When: After lesson 4. Time: 25 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.

Answer every question you can and show your working: a correct method earns credit even if the final answer slips. Questions 11 and 12 are meant to make you think, so try something rather than leaving them blank.

Part 1. Evaluate

Work out log₅ 625. (1 mark)
Work out log₃ (1 ÷ 9). (1 mark)
Work out log₁₀ 0.001. (1 mark)
Work out log₆ 3 + log₆ 12. (2 marks)
Work out log₂ 24 − log₂ 3. (2 marks)

Part 2. One logarithm

Write 2 log₅ x + log₅ y as a single logarithm. (2 marks)

Part 3. Equations

Solve 3^x = 81. (1 mark)
Solve 5^{2x − 1} = 125. (2 marks)
Solve 4^x = 8. (2 marks)
Solve log₂ (x + 1) = 4. (2 marks)
Solve 2^{x + 1} + 2^x = 48. (2 marks)

Part 4. Think it through

Lena says: “log (a + b) is always the same as log a + log b.” Choose a base and values for a and b that show Lena is wrong, and show why they do. (2 marks)

How your work is judged

Level	Criterion A in task A1: what it looks like
1–2	You evaluate a few simple logarithms and solve an equation where the answer is a whole number, with some working.
3–4	You evaluate logarithms of powers and use one law correctly, and solve equations by writing both sides as a power of the same base.
5–6	You use all three laws accurately, solve each equation with clear steps, and make a start on the two unfamiliar questions.
7–8	You choose the right law each time, solve every equation including the unfamiliar ones, and give a counterexample that you show really fails.

Markscheme

Task A1 question	Answer and marking
1	4 (5⁴ = 625). 1 mark.
2	−2 (3⁻² = 1 ÷ 9). 1 mark.
3	−3 (10⁻³ = 0.001). 1 mark.
4	2. log₆ 3 + log₆ 12 = log₆ 36 = 2. 1 mark for the product law, 1 for the answer.
5	3. log₂ (24 ÷ 3) = log₂ 8 = 3. 1 mark for the quotient law, 1 for the answer.
6	log₅ (x² y). 1 mark for the power law (2 log x = log x²), 1 for the product law. The common error is log₅ (2xy).
7	x = 4. 1 mark.
8	x = 2. 125 = 5³, so 2x − 1 = 3. 1 mark for matching the bases, 1 for the solution.
9	x = 3 ÷ 2. 4^x = 2^2x and 8 = 2³, so 2x = 3. 1 mark for a common base, 1 for the solution.
10	x = 15. x + 1 = 2⁴ = 16. 1 mark for the exponential form, 1 for the solution.
11	x = 4. 2^{x + 1} = 2 × 2^x, so 3 × 2^x = 48 and 2^x = 16 = 2⁴. 1 mark for writing 2^{x + 1} as 2 × 2^x and collecting, 1 for the solution. Adding the exponents to get 2^{2x + 1} = 48 is the common error.
12	Any counterexample that is worked through. For a = b = 1 in base 2: log₂ (1 + 1) = 1 but log₂ 1 + log₂ 1 = 0. 1 mark for values that fail, 1 for showing both sides. Not a = b = 2 in base 2: both sides are 2, which is a coincidence and shows nothing. The correct law is log a + log b = log (ab).
Marks to a level	Total 20. 1 to 5 marks: 1 to 2. 6 to 10: 3 to 4. 11 to 15: 5 to 6. 16 to 20, with question 11 or 12 fully correct: 7 to 8. Use the marks as a guide, then check the profile against the clarification.

Exemplar answers

Task A1 response	What it looks like
Level 3 to 4	“1: 4. 2: −2. 3: −3. 4: log₆ 3 + log₆ 12 = log₆ 15, so I cannot work it out. 7: 4. 8: 5³ = 125 so 2x − 1 = 3 and x = 2. 10: x = 15.” Single-step evaluations and equations with a matched base are right; the product law is applied by adding inside the logarithm, and Questions 9, 11 and 12 are left blank. About 9 of 20.
Level 7 to 8	“4: log₆ 3 + log₆ 12 = log₆ 36 = 2. 6: log₅ x² + log₅ y = log₅ (x²y). 9: 4^x = 2^2x, 8 = 2³, so x = 3 ÷ 2. 11: 2^{x + 1} = 2 × 2^x, so 3 × 2^x = 48, 2^x = 16 and x = 4. 12: take base 2, a = b = 1. log₂ 2 = 1, but log₂ 1 + log₂ 1 = 0 + 0 = 0, so they are not equal. The real law multiplies: log a + log b = log (ab).” Every law named and applied. 19 or 20 of 20.

Differentiation

A way in for anyone stuck

Give a reminder card with the three laws in words, with no worked example, for Questions 4 to 6.
Suggest writing each right-hand side as a power of a small base first (81, 125 and 8 as powers of 3, 5 and 2).
For Question 12, prompt with a question, never the answer: “What is the simplest base and the simplest pair of numbers you can think of?”

Stretch

Solve 4^x − 6 × 2^x + 8 = 0. (Let y = 2^x: y = 2 or y = 4, so x = 1 or x = 2.)
Is log₂ 12 + 2 the same as log₂ 16 + 3? Justify without a calculator. (No: log₂ 12 + 2 = log₂ 48 = 4 + log₂ 3, about 5.58, while log₂ 16 + 3 = 7.)

Students learning in English as an additional language

Word in task A1	Plain meaning
logarithm	the power that a base must be raised to, to give a number: log₂ 8 = 3 because 2³ = 8
base	the number that is raised to a power
single logarithm	one log, written as one expression: log₅ (x²y), not two logs added
counterexample	one example that shows a statement is not always true
match the bases	write both sides of an equation as powers of the same number

Sentence starters for task A1:

I used the law that … because …
Both sides are a power of …, so …
This shows Lena is wrong because …

Task B1: Investigation: powers and their length

Why set it. A real pattern hunt about the size of powers. The sheet never names the pattern: students organize their results, find it, describe it as a rule, test it on a base of their own and say why it works. It uses lessons 1 and 5 without repeating either.

When: Mid-unit, after lesson 5. Time: 45 minutes. Format: In class. Students may collect results in pairs, but each writes up their own findings.

The student sheet, with its task-specific clarification

In MYP Year 5, support you give (a prompt, a table outline) is allowed at 1 to 2 and 3 to 4: note it on the student's work.

When: Mid-unit, after lesson 5. Time: 45 minutes. Calculator: scientific calculator. Criterion: B, investigating patterns.

The number 2¹⁰⁰ means 100 twos multiplied together. Written out in full it is much too long for any calculator screen. You can still say how many digits it has, without ever seeing all of them. This task is about how.

Part 1. Collect results

Work out the powers of 2 from 2¹ to 2²⁰. Record each power and how many digits it has, in a table you design yourself.
Do the same for the powers of 3, from 3¹ to 3¹².

Part 2. Describe what you find

Look at how the number of digits changes as the exponent goes up, for 2 and for 3. Describe what you notice, in words, as a rule another student could use.
Use your rule to predict how many digits 2⁴⁰ and 3³⁰ have. Then check both with a calculator.

Part 3. Test your rule

Choose a base of your own from 4 to 9. Collect enough of its powers to test whether your rule, or a rule like it, works. Say what stays the same and what changes.

Part 4. Explain

How many digits does 2¹⁰⁰ have? Show how your method gives the answer.
Explain why your method works, and whether it could ever be a little out for a particular power.

How your work is judged

Level	Criterion B in task B1: what it looks like
1–2	With help, you work out some powers, count their digits and say something you notice about how the digit count changes.
3–4	With help, you complete both tables and suggest a rule for how often the digit count goes up, for 2 or for 3.
5–6	You organize your results clearly, describe a rule that fits both bases, and use it to predict and check 2⁴⁰ and 3³⁰.
7–8	You organize your results, describe a rule that fits all your findings, make correct predictions for large powers, test it on a base you chose, and say why it works and where it could slip.

Markscheme

Task B1 question	Answer and marking
The pattern (do not share it before the task)	The number of digits of bⁿ is the whole-number part of n log b, plus 1. A new digit appears about every 3.32 powers of 2 (log 2 = 0.30103) and every 2.1 powers of 3 (log 3 = 0.47712), so the digit count of 2ⁿ goes up by three or four in every ten steps. Students may state it without logarithms; that earns the top band if it predicts correctly.
1	Digits of 2¹ to 2²⁰: 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7 (2¹⁰ = 1024, 2²⁰ = 1 048 576).
2	Digits of 3¹ to 3¹²: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6 (3¹² = 531 441).
3	Any correct description of the slow growth, for example “the digit count goes up by one about every three powers of 2, and about every two powers of 3”. A rule for one base only is evidence for 3 to 4; a rule that gives a ratio such as “about 3 digits for every 10 powers of 2” is evidence for 5 to 6 once it is used to predict; 7 to 8 also needs the test on a new base and the reason in question 7.
4	2⁴⁰ = 1 099 511 627 776 has 13 digits; 3³⁰ = 205 891 132 094 649 has 15 digits. From the rule: 40 × 0.301 = 12.04, so 13 digits; 30 × 0.477 = 14.31, so 15.
5	Any base from 4 to 9 with at least five powers collected and a clear verdict. Base 5 gets a new digit about every 1.43 powers (5, 25, 125, 625, 3125, 15 625: 1, 2, 3, 3, 4, 5 digits); base 9 about every 1.05 (9, 81, 729, 6561, 59 049: 1, 2, 3, 4, 5). Powers of 4 grow twice as fast as powers of 2 (4ⁿ = 2²ⁿ). The idea that the digit count grows steadily, at a rate set by the base, holds for every base.
6	31 digits. 100 × 0.30103 = 30.1, so the whole-number part is 30 and 2¹⁰⁰ has 31 digits. A calculator shows 1.2676506 × 10³⁰, which also has 31 digits.
7	A power of ten has a one followed by that many zeros, so a number between 10³⁰ and 10³¹ has 31 digits. The exponent n times log b is exactly where the power sits between powers of ten. Expect this only at 7 to 8. A calculator value of log 2 is rounded, so for a power that lands extremely close to a power of ten a short log could give a count one out; no power of 2 or 3 in this task does.

Exemplar answers

Task B1 response	What it looks like
Level 3 to 4	“Powers of 2 have 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4 digits. Powers of 3 have 1, 1, 2, 2, 3, 3. The number of digits goes up slowly, about every three powers. 2⁴⁰ will have about 13 digits.” Both tables right and a rule suggested, but it is not precise enough to be sure of 3³⁰, and no other base is tested.
Level 7 to 8	“In every ten powers of 2 the digit count goes up by three or four, so about 0.3 digits per power. For 3 it is about 0.48 per power. So the digits of bⁿ are about n times that rate, rounded down, plus 1. 40 × 0.3 = 12, so 2⁴⁰ has 13 digits, and 30 × 0.48 = 14.4, so 3³⁰ has 15. The calculator agrees. I tested base 5: its rate is log 5 = 0.699, so 20 × 0.699 = 13.98 and 5²⁰ has 13 + 1 = 14 digits (the calculator shows 9.54 × 10¹³). Rounding the rate to 0.7 would give 14.0 and a count one too many, so the rate needs enough decimal places. 2¹⁰⁰ has 30.1, so 31 digits. It works because every power of ten adds exactly one digit, and the rate is how far up the powers of ten each step moves you.”

Task B1 response

What it looks like

Level 3 to 4

“Powers of 2 have 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4 digits. Powers of 3 have 1, 1, 2, 2, 3, 3. The number of digits goes up slowly, about every three powers. 2⁴⁰ will have about 13 digits.” Both tables right and a rule suggested, but it is not precise enough to be sure of 3³⁰, and no other base is tested.

Level 7 to 8

“In every ten powers of 2 the digit count goes up by three or four, so about 0.3 digits per power. For 3 it is about 0.48 per power. So the digits of bⁿ are about n times that rate, rounded down, plus 1. 40 × 0.3 = 12, so 2⁴⁰ has 13 digits, and 30 × 0.48 = 14.4, so 3³⁰ has 15. The calculator agrees. I tested base 5: its rate is log 5 = 0.699, so 20 × 0.699 = 13.98 and 5²⁰ has 13 + 1 = 14 digits (the calculator shows 9.54 × 10¹³). Rounding the rate to 0.7 would give 14.0 and a count one too many, so the rate needs enough decimal places. 2¹⁰⁰ has 30.1, so 31 digits. It works because every power of ten adds exactly one digit, and the rate is how far up the powers of ten each step moves you.”

Differentiation

A way in for anyone stuck

Give a two-column table outline for Part 1 and let the student write the headings.
Suggest getting each power from the one before (double again) rather than starting from scratch, and counting digits in groups of three.
Prompt with a question, never the rule: “How many powers of 2 does it take for the digit count to go up by one?”

Stretch

Find the smallest n for which 2ⁿ has 100 digits. (n = 329, because 328 × 0.30103 = 98.7 and 329 × 0.30103 = 99.04, giving 99 then 100 digits.)
Which base makes the digit count rise by exactly one at every power, on average? Explain. (Base 10: each power of ten adds one digit.)

Students learning in English as an additional language

Word in task B1	Plain meaning
power	a number multiplied by itself again and again, written with a small raised number: 3⁴ = 3 × 3 × 3 × 3
exponent	the small raised number that says how many times the base is used
digit	one of the symbols 0 to 9 that make up a number
predict	say what will happen before you check
rule	a sentence that tells you what happens every time
standard form	a number written as a number from 1 to 10 times a power of ten, as a calculator shows a large power

Sentence starters for task B1:

I noticed that the number of digits …
My rule is: …
I tested my rule on base … and it worked / did not work because …

Task C1: Explain it to someone who missed it

Why set it. Criterion C on its own: notation, a sign diagram and a line of reasoning a classmate can follow, built around two real errors from lessons 3 and 7.

When: After lesson 8. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words, symbols and a diagram.

The student sheet, with its task-specific clarification

In MYP Year 5 the top bands expect correct, consistent notation and a justified argument; correctness of the maths is assessed in A.

When: After lesson 8. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.

Mara was away for lessons 3 to 8. She copied this working from a classmate, and it has mistakes in it.

The classmate’s working

(a) log₂ 8 + log₂ 4 = log₂ (8 + 4) = log₂ 12.

(b) Solve x² − 2x − 15 > 0. (x − 5)(x + 3) > 0, so x > 5 or x > −3, so x > −3.

Write a one-page explanation for Mara. She should be able to follow it on her own, without asking anyone.

Part 1. Put it right

Find the mistake in each part. For each one, say what was done, why it is wrong, and write the correct working.

Part 2. Teach the idea

Explain to Mara how to combine two logarithms of the same base that are added. Use the two logarithms in (a) and a check that she can do by evaluating each logarithm separately.
Explain how a sign diagram gives the solution of a quadratic inequality. Draw one for part (b), mark the roots, and say how to test which regions are positive.

Part 3. Check the answer

Show Mara one way to check her final answer to (b) by substituting a value, and say which value you chose and why.

How your work is judged

Level	Criterion C in task C1: what it looks like
1–2	You write down the corrected lines with some words; the page is hard to follow without you there to explain.
3–4	You show correct steps for each part with the main symbols right, and a diagram or check that supports them.
5–6	You lay the page out line by line, use notation consistently, and give a reason for each step so Mara could follow it alone.
7–8	You use correct notation throughout, link the diagram to the written steps, and justify each step and the check, so the page stands on its own as a short worked guide.

Markscheme

Task C1 question	Answer and marking
(a) the mistake	The two arguments were added; the product law multiplies them: log₂ 8 + log₂ 4 = log₂ (8 × 4) = log₂ 32 = 5. A check by evaluation: 3 + 2 = 5, while log₂ 12 is only 3.58.
(b) the mistake	A product is positive when its two factors have the same sign, not only when one is positive, so the answer cannot be a single inequality. Roots 5 and −3; the quadratic opens upward, so it is positive for x < −3 or x > 5. The check: x = 0 gives −15, which is not > 0, but x > −3 includes 0.
Part 2 and 3	Credit a sign diagram with the roots marked, a sign in each of three regions, and one tested value per region (for example x = −4 gives 9, x = 0 gives −15, x = 6 gives 9). A good page names the rule: add logs of the same base by multiplying inside; solve an inequality by finding where the expression changes sign.
What the page earns	Notation: log₂ written with its base each time, brackets kept, < and > the right way round, the solution written as two parts joined by or. Communication: one idea per line, a reason in words for each, and the diagram referred to in the text.

Exemplar answers

Task C1 response	What it looks like
Level 3 to 4	“(a) You multiply, not add: log₂ 8 + log₂ 4 = log₂ 32. (b) The answer is x < −3 or x > 5.” Both corrections are right and short, with a rough number line, but there is no reason given and no check, so Mara could not follow it alone.
Level 7 to 8	“(a) The classmate added 8 and 4. But log_a m + log_a n = log_a (mn): multiply. So log₂ 8 + log₂ 4 = log₂ 32 = 5. Check: log₂ 8 = 3 and log₂ 4 = 2, and 3 + 2 = 5. (b) The roots are 5 and −3. Sketch a sign diagram: in each region ask whether the product (x − 5)(x + 3) is positive. x = −4 gives (−9)(−1) = 9; x = 0 gives (−5)(3) = −15; x = 6 gives (1)(9) = 9. We want > 0, so x < −3 or x > 5. The classmate’s answer includes 0, and 0 gives −15, so it fails.”

Task C1 response

What it looks like

Level 3 to 4

“(a) You multiply, not add: log₂ 8 + log₂ 4 = log₂ 32. (b) The answer is x < −3 or x > 5.” Both corrections are right and short, with a rough number line, but there is no reason given and no check, so Mara could not follow it alone.

Level 7 to 8

“(a) The classmate added 8 and 4. But log_a m + log_a n = log_a (mn): multiply. So log₂ 8 + log₂ 4 = log₂ 32 = 5. Check: log₂ 8 = 3 and log₂ 4 = 2, and 3 + 2 = 5. (b) The roots are 5 and −3. Sketch a sign diagram: in each region ask whether the product (x − 5)(x + 3) is positive. x = −4 gives (−9)(−1) = 9; x = 0 gives (−5)(3) = −15; x = 6 gives (1)(9) = 9. We want > 0, so x < −3 or x > 5. The classmate’s answer includes 0, and 0 gives −15, so it fails.”

Differentiation

A way in for anyone stuck

Give a sentence frame for each correction: “The mistake is … The correct step is … I know because …”
Let the student draw the sign diagram on a number line with the two roots already marked.
Prompt with a question: “What value could you try in the original inequality to see whether the answer is believable?”

Stretch

Add a third error of your own, in a rational inequality, and explain it as clearly as the first two.
Write the explanation for a student two years younger. What did you change, and why?

Students learning in English as an additional language

Word in task C1	Plain meaning
product	the answer when numbers are multiplied
root	a value of x that makes an expression equal to zero
sign diagram	a number line that shows where an expression is positive and where it is negative
substitute	put a number in place of a letter
inequality	a statement with < or > instead of an equals sign

Sentence starters for task C1:

The mistake is … because …
A check you can do is …
Therefore the solution is …

Task A2: End-of-unit test

Why set it. A longer test of the whole unit in four sections. Questions 10 to 12 are unfamiliar, so the top band is open to everyone.

When: End of unit, after lesson 8. Time: 45 minutes. Format: Individual, written, in class. 30 marks. Scientific calculator allowed throughout.

The student sheet, with its task-specific clarification

When: End of unit, after lesson 8. Time: 45 minutes. Calculator: scientific calculator. Criterion: A, knowing and understanding.

Show your working. A correct method earns credit even if the final answer slips. Give answers to 3 significant figures unless a question says otherwise. Questions 10 to 12 are different from the ones you have practised, so try something rather than leaving them blank.

Part 1. Laws of logarithms

Work out log₃ (1 ÷ 27). (1 mark)
Write log 12 + log 5 − log 3 as a single logarithm. (2 marks)
Write log_a (x³ ÷ y) in terms of log_a x and log_a y. (2 marks)

Part 2. Exponential equations

Solve 8^{x − 1} = 4, giving x as an exact fraction. (3 marks)
Solve 3^x = 20. (2 marks)
Solve 5 × 2^x = 70. (3 marks)
Use the change of base law to evaluate log₄ 100. (2 marks)

Part 3. Inequalities

Solve x² − 5x − 14 ≤ 0. (3 marks)
Solve (x − 3) ÷ (x + 2) > 0. (3 marks)

Part 4. Scales and applications

A drink has pH 3.2 and a second drink has pH 5.9. How many times more acidic is the first? (3 marks)
Show that log₉ x = 0.5 log₃ x. (3 marks)
Town P has 5000 people and grows by 6% each year. Town Q has 8000 people and grows by 2% each year. After how many whole years does Town P first have more people than Town Q? (3 marks)

How your work is judged

Level	Criterion A in task A2: what it looks like
1–2	You evaluate a simple logarithm, use one law, and solve an equation or inequality where the working is mostly set out for you.
3–4	You use the laws on familiar forms, solve both kinds of exponential equation, and solve one inequality with a sign diagram.
5–6	You work accurately across all four parts, give exact and rounded answers as asked, and make real progress on the unfamiliar questions.
7–8	You choose methods without prompting, solve the rational inequality with its excluded value, and complete the proof and the growth question with correct justification.

Markscheme

Task A2 question	Answer and marking
1	−3. 1 mark.
2	log 20. log (12 × 5 ÷ 3) = log 20. 1 mark for the laws, 1 for the value.
3	3 log_a x − log_a y. 1 mark for the quotient law, 1 for the power law.
4	x = 5 ÷ 3. 8^{x − 1} = 2^{3x − 3} and 4 = 2², so 3x − 3 = 2. 1 mark for a common base, 1 for the equation, 1 for the answer.
5	x = 2.73. x = log 20 ÷ log 3 = 1.3010 ÷ 0.4771. 1 mark for the method, 1 for the answer.
6	x = 3.81. 2^x = 14, so x = log 14 ÷ log 2 = 1.1461 ÷ 0.3010. 1 mark for dividing by 5, 1 for taking logs, 1 for the answer.
7	3.32. log 100 ÷ log 4 = 2 ÷ 0.60206. 1 mark for the change of base, 1 for the answer.
8	−2 ≤ x ≤ 7. (x − 7)(x + 2) ≤ 0, positive outside the roots, so the solution lies between them. 1 mark for the roots, 1 for the sign pattern, 1 for the answer with the correct inequality signs.
9	x < −2 or x > 3. Critical values 3 and −2 (x = −2 is excluded because it makes the denominator zero). 1 mark for both values, 1 for the sign diagram, 1 for the answer.
10	501 times. The difference is 2.7, so the ratio is 10^2.7 = 501.2. 1 mark for the difference, 1 for the power of ten, 1 for the answer. Credit 10^2.7 left exact. The common error is 5.9 ÷ 3.2.
11	log₉ x = log₃ x ÷ log₃ 9 (change of base), and log₃ 9 = 2, so log₉ x = log₃ x ÷ 2 = 0.5 log₃ x. 1 mark for the change of base law, 1 for log₃ 9 = 2, 1 for the conclusion.
12	13 years. 5000 × 1.06ⁿ > 8000 × 1.02ⁿ, so (1.06 ÷ 1.02)ⁿ > 1.6 and n > log 1.6 ÷ log (1.06 ÷ 1.02) = 0.2041 ÷ 0.01671 = 12.2. 1 mark for the inequality, 1 for solving it with logarithms (or a careful table of values), 1 for rounding up to a whole year. Check: after 12 years 10 061 against 10 146; after 13 years 10 665 against 10 349.
Marks to a level	Total 30. 1 to 7 marks: 1 to 2. 8 to 15: 3 to 4. 16 to 23: 5 to 6. 24 to 30, with at least two of questions 10 to 12 fully correct: 7 to 8. Check the profile against the clarification before settling the level.

Exemplar answers

Task A2 response	What it looks like
Level 3 to 4	“1: −3. 2: log 20. 5: 3^x = 20, so x = 20 ÷ 3 = 6.67. 8: roots are −2 and 7, so x ≤ 7. 10: 5.9 ÷ 3.2 = 1.84.” Straightforward uses of a law and the roots are right; the logarithm is not used to solve 3^x = 20, the inequality answer is incomplete, and the scale is treated as linear. About 12 of 30.
Level 7 to 8	“4: 8^{x − 1} = 2^{3x − 3} = 2², so x = 5 ÷ 3. 5: x = log 20 ÷ log 3 = 2.73. 9: critical values −2 and 3; positive when both factors have the same sign: x < −2 or x > 3. 10: pH 2.7 lower, so 10^2.7 = 501 times. 11: log₉ x = log₃ x ÷ log₃ 9 = log₃ x ÷ 2. 12: (1.06 ÷ 1.02)ⁿ > 1.6 gives n > 12.2, so 13 years; after 13 years P has 10 665 and Q has 10 349.” 28 to 30 of 30.

Differentiation

A way in for anyone stuck

Give the three laws and the change of base law on a reminder card, with no worked examples.
For Question 4, supply the table of powers of 2 up to 2⁶ so the student only has to match the bases.
For Question 12, prompt with a question: “What is each town's population after n years, and when is the first one bigger?”

Stretch

Find, to 4 significant figures, the annual growth factor at which a population triples in 10 years. (3 to the power 0.1 = 1.116, a rate of 11.6%.)
Solve log₂ x + log₂ (x − 2) = 3. (x(x − 2) = 8, so x = 4; x = −2 is rejected.)

Students learning in English as an additional language

Word in task A2	Plain meaning
change of base	the law that gives a logarithm to any base from calculator logs: log_b N = log N ÷ log b
critical value	a value of x where an expression equals zero or is undefined, and so can change sign
acidic	having more acid; a lower pH means more acid
significant figures	the digits of a number that count, counting from the first non-zero digit

Sentence starters for task A2:

The law I used is …
The critical values are …, so the sign diagram is …
I round up to a whole year because …

Task D1: Sound at the night market

Why set it. Applying logarithmic scales, and the equal-sounds law from lesson 6, to a real decision about noise at a night market, then communicating the model, its assumptions and a recommendation.

When: End of unit, after lesson 8. Time: 45 minutes. Format: Individual, written, in class. The final answer is a short written recommendation.

The student sheet, with its task-specific clarification

When: End of unit, after lesson 8. Time: 45 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.

A night market in Chiang Mai has 40 food stalls along one street. The market committee has a rule that the sound level at any stall must stay at or below 85 decibels. The sound level in decibels (dB) is L = 10 log₁₀ (I ÷ I₀), where I is the sound intensity in watts per square metre and I₀ = 10⁻¹² is the quietest sound a person can hear.

A street musician plays through an amplifier. At the nearest stall the intensity is 2.5 × 10⁻⁵ watts per square metre. Several musicians have asked to play in the same area, and the committee wants to know how many identical amplifiers can play together before the rule is broken.

Part 1. One amplifier

Work out the sound level of one amplifier at the nearest stall, to 1 decimal place.
A different amplifier is measured at exactly 85 dB. Work out its intensity in watts per square metre, as a number times a power of ten.

Part 2. Several amplifiers

Two identical amplifiers play together at the same distance. Find the total sound level, and the increase compared with one amplifier.
Find the largest number of identical amplifiers that can play together at the nearest stall without breaking the 85 dB rule.

Part 3. Advise the committee

A stall holder says: “85 decibels is only a little louder than 75, so one extra amplifier cannot matter.” Use your results to say whether this is right, in a way a non-mathematician on the committee would follow.
Write a short recommendation to the committee. State the model you used, one assumption in it that might not hold in a real street, and what you would measure to check it.

How your work is judged

Level	Criterion D in task D1: what it looks like
1–2	You use the decibel formula to find a level with some help, and say what your answer means for the market.
3–4	You find the level and the intensity correctly, and use the formula for two amplifiers, linking the results to the 85 dB rule.
5–6	You solve the whole problem with a clear model, find the largest number of amplifiers, and say what an assumption means for the result.
7–8	You solve every part, justify the choice of model, test an assumption against the result, and reach a recommendation the data really support.

Level	Criterion C in task D1: what it looks like
1–2	You write the answers with units, and some working a reader can see.
3–4	You show the working in order, with the formula and units written out, and a sentence that states each answer.
5–6	You explain each step so a committee member could follow it, and the recommendation refers to your numbers.
7–8	You tailor the explanation to the audience, state the model and its limits precisely, and keep notation and units consistent throughout.

Markscheme

Task D1 question	Answer and marking
1	74.0 dB. L = 10 log (2.5 × 10⁻⁵ ÷ 10⁻¹²) = 10 log (2.5 × 10⁷) = 10 × 7.398 = 73.98, so 74.0 dB.
2	3.16 × 10⁻⁴ W/m². 85 = 10 log (I ÷ I₀), so I ÷ I₀ = 10^8.5 = 3.16 × 10⁸ and I = 3.16 × 10⁻⁴.
3	77.0 dB, an increase of 3.0 dB. Doubling the intensity adds 10 log 2 = 3.01 dB: 73.98 + 3.01 = 76.99, so 77.0 dB. Credit a method that doubles the intensity to 5.0 × 10⁻⁵ first.
4	12 amplifiers keep the level at or below 85 dB; the 13th breaks it. 73.98 + 10 log n ≤ 85 gives log n ≤ 1.102, so n ≤ 12.65 and n = 12. Check: 12 amplifiers give 73.98 + 10.79 = 84.77 dB; 13 give 85.12 dB. Credit 13 as the number that breaks the rule if it is clearly labelled.
5	Not right. Ten decibels is a factor of ten in intensity, so 85 dB carries about three times the intensity of 80 dB and ten times that of 75 dB. Here the rule allows 10^1.102 = 12.6 times the intensity of one amplifier, so going from one amplifier to twelve is eleven extra ones, not one. One extra amplifier adds 10 log (2 ÷ 1) = 3.0 dB going from one to two, but only 10 log (13 ÷ 12) = 0.3 dB going from twelve to thirteen. The reason to credit: the scale is logarithmic, so equal steps in dB are equal ratios, not equal differences.
6	A recommendation of a limit of 12 identical amplifiers (or “fewer than 13”) at the nearest stall, with the model L = 10 log (I ÷ I₀), intensities adding for equal sounds. Assumptions worth credit: all amplifiers are identical and the same distance away, there is no other noise, the sound does not reflect from walls and awnings, and 85 dB is measured at the nearest stall. A check: a sound level meter measurement with two and four amplifiers.

Exemplar answers

Task D1 response	What it looks like
Level 3 to 4	“The level of one amplifier is 10 log (2.5 × 10⁷) = 73.98 dB. Two amplifiers is 77 dB. The market should not have too many amplifiers because it is already near 85 dB.” The level is right and two amplifiers is right, but the intensity at 85 dB is not found, no largest number is reached, and no assumption is named.
Level 7 to 8	“One amplifier gives 74.0 dB. Intensities add, so n amplifiers add 10 log n dB: two give 77.0 and twelve give 84.8 dB, but thirteen give 85.1 dB, which breaks the rule. I recommend a limit of twelve. This assumes all amplifiers are identical and the same distance away, and ignores echoes from awnings, so I would measure the real level with four amplifiers before deciding. The stall holder is wrong: 10 dB is a factor of ten in intensity, so going from 75 to 85 needs about ten times the sound power, not one extra amplifier.”

Task D1 response

What it looks like

Level 3 to 4

“The level of one amplifier is 10 log (2.5 × 10⁷) = 73.98 dB. Two amplifiers is 77 dB. The market should not have too many amplifiers because it is already near 85 dB.” The level is right and two amplifiers is right, but the intensity at 85 dB is not found, no largest number is reached, and no assumption is named.

Level 7 to 8

“One amplifier gives 74.0 dB. Intensities add, so n amplifiers add 10 log n dB: two give 77.0 and twelve give 84.8 dB, but thirteen give 85.1 dB, which breaks the rule. I recommend a limit of twelve. This assumes all amplifiers are identical and the same distance away, and ignores echoes from awnings, so I would measure the real level with four amplifiers before deciding. The stall holder is wrong: 10 dB is a factor of ten in intensity, so going from 75 to 85 needs about ten times the sound power, not one extra amplifier.”

Differentiation

A way in for anyone stuck

Give the formula L = 10 log (I ÷ I₀) on a card with the value of I₀, and show how to type it on the calculator once.
For Question 4, let the student find the level for 2, 4 and 8 amplifiers first, then narrow down.
For Question 6, supply a frame: “I used … I assumed … To check, I would …”

Stretch

Two amplifiers are played: one at 75 dB and a quieter one at 70 dB. What is the combined level? (76.2 dB: the intensities, in units of I₀, are 3.16 × 10⁷ and 10⁷, which add to 4.16 × 10⁷.)
Doubling the distance from a source cuts its intensity to a quarter. By how much does the level fall? (10 log 4 = 6.0 dB.)

Students learning in English as an additional language

Word in task D1	Plain meaning
decibel (dB)	the unit for how loud a sound is, on a scale where 10 dB more means ten times the intensity
intensity	how much sound power arrives at a place, in watts per square metre
identical	exactly the same
assumption	something you decide to treat as true so you can make a model
recommendation	advice about what to do

Sentence starters for task D1:

The level of one amplifier is … because …
Each extra amplifier adds … so …
I recommend … because … This assumes …

Task D2: The longtail boat cooperative

Why set it. Applying exponential growth and decay, logarithms and a quadratic inequality to a savings and pricing decision, then communicating the plan and its limits.

When: End of unit, after lesson 8. Time: 45 minutes. Format: Individual, written, in class. The final answer is a short written plan for the cooperative.

The student sheet, with its task-specific clarification

When: End of unit, after lesson 8. Time: 45 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.

A cooperative of boat operators on a lake wants to replace an old longtail boat engine. They have 120 000 crowns in an account that pays 3.5% interest each year, added to the account at the end of each year. A new engine costs 200 000 crowns. The old engine is worth 90 000 crowns now, and loses 12% of its value each year.

The cooperative also sets the ticket price p crowns for a lake trip. The daily profit in crowns, from a model fitted to last year’s trips, is P = −2p² + 300p − 10 000.

Part 1. Saving for the engine

After how many whole years will the account first hold at least 200 000 crowns? Show the equation you solve.

Part 2. The old engine

After how many whole years will the old engine first be worth less than 30 000 crowns?

Part 3. Ticket price

Find the range of ticket prices for which the cooperative makes a profit.
The cooperative needs a daily profit of at least 800 crowns. Find the range of prices that achieve this, and the price that gives the largest profit.

Part 4. The plan

Write a short plan for the cooperative. Say when they can afford the engine, what price to set, and one thing about the models that might make your advice wrong.

How your work is judged

Level	Criterion D in task D2: what it looks like
1–2	You set up an equation for the savings or the engine value and solve it by trial or with help, then say what the answer means.
3–4	You solve the savings equation with logarithms and find the range of prices that make a profit, with the steps shown.
5–6	You solve all four parts with a clear method, link each result to the cooperative’s decision, and name a limit of a model.
7–8	You solve every part with justified methods, check the answers against the original conditions, and give a plan that weighs the two models and their limits.

Level	Criterion C in task D2: what it looks like
1–2	You write your answers with units, and some working a reader can see.
3–4	You write each equation and inequality clearly, in order, with a sentence stating each answer in context.
5–6	You explain each step to someone on the cooperative committee, use a sign diagram or sketch for the price range, and use consistent notation.
7–8	You choose a presentation that suits the audience, keep the mathematics precise and the language plain, and make the link between the plan and the models explicit.

Markscheme

Task D2 question	Answer and marking
1	15 years. 120 000 × 1.035ⁿ ≥ 200 000, so n ≥ log (5 ÷ 3) ÷ log 1.035 = 0.2218 ÷ 0.01494 = 14.85. Check: 14 years gives 194 243 crowns, 15 years gives 201 042 crowns. Credit trial and improvement shown to 14 and 15 years.
2	9 years. 90 000 × 0.88ⁿ < 30 000, so 0.88ⁿ < 1 ÷ 3 and n > log (1 ÷ 3) ÷ log 0.88 = 8.59. Both logarithms are negative, so the quotient is positive. Check: 8 years gives 32 367 crowns and 9 years gives 28 483 crowns.
3	50 < p < 100. −2(p − 50)(p − 100) > 0; the parabola opens downward so it is positive between the roots 50 and 100.
4	60 ≤ p ≤ 90; the largest profit is 1250 crowns at p = 75. −2p² + 300p − 10 000 ≥ 800 gives p² − 150p + 5400 ≤ 0, roots 60 and 90. The maximum is at the midpoint of the roots, p = 75: −11 250 + 22 500 − 10 000 = 1250.
5	A plan that states: from savings alone the engine is affordable after 15 years at 3.5%; but trading in the old engine now (90 000 crowns) with the 120 000 saved already reaches 210 000, while every year of waiting costs value (32 367 crowns after 8 years, 28 483 after 9), so the strongest plan buys now or soon with a trade-in; a price near 75 crowns, within 60 to 90; and a limit: a fixed interest rate, a profit model fitted to one year, no change in demand or in the cost of the new engine. Credit any specific limit; reward at 7 to 8 a plan that notices the trade-in.

Exemplar answers

Task D2 response	What it looks like
Level 3 to 4	“After 15 years the account has 201 000 crowns, found by trying 14 and 15 years. The old engine falls under 30 000 after about 9 years. Profit is positive when p is between 50 and 100.” The results are right and found by trial, but the range for 800 crowns is not found and there is no plan.
Level 7 to 8	“1.035ⁿ ≥ 5 ÷ 3, so n ≥ log (5 ÷ 3) ÷ log 1.035 = 14.85, so 15 years. 0.88ⁿ < 1 ÷ 3 gives n > 8.59, so 9 years. The profit is positive for 50 < p < 100 and at least 800 for 60 ≤ p ≤ 90, with the maximum 1250 crowns at p = 75. I recommend a price of 75 crowns. The engine is affordable in 15 years from savings alone; selling the old engine for about 28 500 crowns would shorten this. The models assume a fixed interest rate and demand that does not change, so I would revisit the plan every year.”

Task D2 response

What it looks like

Level 3 to 4

“After 15 years the account has 201 000 crowns, found by trying 14 and 15 years. The old engine falls under 30 000 after about 9 years. Profit is positive when p is between 50 and 100.” The results are right and found by trial, but the range for 800 crowns is not found and there is no plan.

Level 7 to 8

“1.035ⁿ ≥ 5 ÷ 3, so n ≥ log (5 ÷ 3) ÷ log 1.035 = 14.85, so 15 years. 0.88ⁿ < 1 ÷ 3 gives n > 8.59, so 9 years. The profit is positive for 50 < p < 100 and at least 800 for 60 ≤ p ≤ 90, with the maximum 1250 crowns at p = 75. I recommend a price of 75 crowns. The engine is affordable in 15 years from savings alone; selling the old engine for about 28 500 crowns would shorten this. The models assume a fixed interest rate and demand that does not change, so I would revisit the plan every year.”

Differentiation

A way in for anyone stuck

Give the growth and decay formulas (start value times the factor to the power n) on a card, and let the student fill in the numbers.
For Part 3, supply a blank sign diagram with the roots 50 and 100 already marked, and ask which regions give a positive profit.
For Part 4, supply a frame: “They can afford the engine after … I recommend a price of … This could be wrong because …”

Stretch

The cooperative also adds 2 000 crowns to the account at the end of every year. After how many years does the account first reach 200 000 crowns? (11 years: 10 years gives about 192 700 crowns and 11 years about 201 500 crowns.)
Find the prices for which the daily profit is at least 800 crowns using a sketch instead of a sign diagram, and explain why the answer is the same.

Students learning in English as an additional language

Word in task D2	Plain meaning
interest	extra money that a bank adds to an account each year
loses value	becomes worth less over time
profit	the money left after costs are taken away
range	all the values between a smallest and a largest
model	an equation that describes a real situation, with simplifying assumptions

Sentence starters for task D2:

The equation is … so n = …
A price of … crowns gives a profit because …
My advice could be wrong if …

Practicalities

Every task is original and written for this unit; none is copied or adapted from a published book.

The student sheets, with their task-specific clarifications, are public on the unit page so students know what each level looks like before they start. Markschemes and exemplars are only on this page.

To print a sheet, open it on the unit page, then print the page. A1 and C1 need no calculator; A2, B1, D1 and D2 use a scientific calculator.

The sheets are public, so a student may have seen A1 and A2. For an unseen A2 (same marks, same markscheme notes): Q1 log₂ (1 ÷ 32) (−5). Q2 log 18 + log 5 − log 3 (log 30). Q3 log_a (x⁴ ÷ y²) (4 log_a x − 2 log_a y). Q4 9^{x − 1} = 27 (x = 5 ÷ 2). Q5 5^x = 40 (2.29). Q6 3 × 2^x = 51 (4.09). Q7 log₅ 200 (3.29). Q8 x² − 3x − 10 ≤ 0 (−2 ≤ x ≤ 5). Q9 (x − 4) ÷ (x + 1) > 0 (x < −1 or x > 4). Q10 pH 3.8 and 5.6 (10^1.8 = 63.1 times). Q11 show log₈ x = (1 ÷ 3) log₂ x. Q12 a town of 4000 growing by 5% a year against one of 6000 growing by 1% (11 years: after 10 years 6516 against 6628, after 11 years 6841 against 6694). For A1, change the numbers the same way and rework the markscheme.

The Year 5 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.