IB DP Mathematics: Analysis and Approaches – Exponents, logarithms, the binomial theorem and simple proof Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the exponents, logarithms, binomial theorem and simple proof unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 1.5, 1.6, 1.7 and 1.9. Every question is common content for SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology and Paper 2 requires it, at both SL and HL. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give answers exactly or to 3 significant figures unless a question says otherwise.

Learn the methods first in the study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free)

(a) Simplify (2a³b⁻²)⁻² ÷ (4a⁻¹b), giving your answer with positive exponents. [3] (b) Evaluate 125^(2/3). [1] (c) Evaluate (16/81)^(−3/4). [2]

2. (calculator allowed)

(a) Use the formula nCr = n!/(r!(n − r)!) to show that 8C3 = 56. [2] (b) Using a table of values on your GDC, find the two values of r for which 12Cr = 495. [2]

3. (calculator allowed)

(a) Solve 10^x = 350. [2] (b) Solve 4e^(0.5t) = 30. [3]

4. (calculator-free) Let log_a 2 = p and log_a 3 = q, where a > 1. Write each of the following in terms of p and q.

(a) log_a 18 [2] (b) log_a (8/9) [2] (c) log_a √(12a) [2]

5. (calculator-free)

(a) Show that log₄ 8 = 3/2. [2] (b) Hence, or otherwise, solve log₂ x + log₄ x = 6. [3]

6. (calculator-free)

(a) Solve 3^(2x+1) = 27^(x−1). [3] (b) Solve 9^x − 10(3^x) + 9 = 0. [4]

7. (calculator allowed) Solve 2^(x+3) = 5^(x−1). Give your answer in the form x = ln p / ln q, where p, q ∈ ℚ, and also to 3 significant figures. [4]

8. (calculator-free)

(a) Show that 3/4 − 2/3 = 1/12. [1] (b) Show that (n + 1)/(n + 2) − n/(n + 1) ≡ 1/((n + 1)(n + 2)) for n ∈ ℤ⁺. [3] (c) Hence find the exact value of 100/101 − 99/100. [1]

9. (calculator-free)

(a) Show that (x + 2)³ − (x − 2)³ ≡ 12x² + 16. [3] (b) Hence explain why the equation (x + 2)³ = (x − 2)³ has no real solutions. [2]

10. (calculator-free)

(a) Expand (2 − x)⁵ fully, simplifying each term. [3] (b) Hence find the coefficient of x³ in the expansion of (1 + 3x)(2 − x)⁵. [3]

11. (calculator-free) Consider the expansion of (x² + k/x)⁶, where k > 0.

(a) Show that the general term can be written as 6Cr k^r x^(12 − 3r). [2] (b) The constant term is 240. Find the value of k. [4] (c) Find the coefficient of x⁶. [2]

12. (calculator allowed) The loudness L of a sound, in decibels, is given by L = 10 log(I/I₀), where I is the intensity in W m⁻² and I₀ = 10⁻¹² W m⁻².

(a) Find L when I = 3.2 × 10⁻⁵ W m⁻². [2] (b) The intensity of a sound is multiplied by k. Show that its loudness increases by 10 log k decibels. [3] (c) Find the intensity of a sound with loudness 90 decibels. [2] (d) Hence, or otherwise, find how many times more intense a 90 decibel sound is than a 60 decibel sound. [1]

Answers

1. (a) (2a³b⁻²)⁻² = 2⁻²a⁻⁶b⁴ = b⁴/(4a⁶) [1]. Dividing by 4a⁻¹b: b⁴/(4a⁶) × a/(4b) [1] = b³/(16a⁵) [1] (b) (∛125)² = 5² = 25 [1] (c) (16/81)^(−3/4) = (81/16)^(3/4) [1] = (3/2)³ = 27/8 [1] Examiner insight: In (a) the −2 must act on the 2 as well as on the letters; an answer with 4 in place of 1/4 loses the accuracy mark even if the letters are right.

2. (a) 8C3 = 8!/(3! 5!) [1] = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56 [1] (b) The table shows 12C4 = 495 [1] and 12C8 = 495, so r = 4 or r = 8 [1] Examiner insight: This is a “show that”, so writing 8C3 = 56 from a calculator earns nothing; the factorial substitution must be seen.

3. (a) x = log 350 [1] = 2.54 [1] (b) e^(0.5t) = 7.5 [1], so 0.5t = ln 7.5 [1] and t = 2 ln 7.5 = 4.03 [1] Examiner insight: Dividing by 4 must come before taking ln; ln(4e^(0.5t)) is not 4 × 0.5t, and that error loses both remaining marks.

4. (a) log_a 18 = log_a 2 + log_a 9 = log_a 2 + 2 log_a 3 [1] = p + 2q [1] (b) log_a (8/9) = 3 log_a 2 − 2 log_a 3 [1] = 3p − 2q [1] (c) log_a √(12a) = ½(log_a 4 + log_a 3 + log_a a) [1] = ½(2p + q + 1) = p + q/2 + 1/2 [1] Examiner insight: In (c) log_a a = 1 must appear; leaving it as log_a a means the answer is not fully in terms of p and q.

5. (a) By change of base, log₄ 8 = (log₂ 8)/(log₂ 4) [1] = 3/2 [1] (b) log₄ x = (log₂ x)/2 [1], so (3/2) log₂ x = 6 and log₂ x = 4 [1]. So x = 16 [1] Examiner insight: A “hence, or otherwise” allows any valid method, but the change of base must be shown; an answer of 16 found by trial with no working gets at most the final mark.

6. (a) 27^(x−1) = 3^(3x−3) [1], so 2x + 1 = 3x − 3 [1] and x = 4 [1] (b) Let y = 3^x, so y² − 10y + 9 = 0 [1]. Then (y − 1)(y − 9) = 0, so y = 1 or y = 9 [1]. 3^x = 1 gives x = 0 [1]; 3^x = 9 gives x = 2 [1] Examiner insight: Stopping at y = 1 or 9 loses both final accuracy marks; the question asks for x, so convert each value back.

7. Take ln of both sides: (x + 3) ln 2 = (x − 1) ln 5 [1]. So x(ln 5 − ln 2) = 3 ln 2 + ln 5 [1] and x = ln 40 / ln 2.5, so x = ln 40 / ln (5/2) [1] = 4.03 [1] Examiner insight: The exact form earns its own mark, so combine the logs with the laws; a 3 s.f. value alone loses it.

8. (a) LHS = 9/12 − 8/12 = 1/12 = RHS [1] (b) LHS = ((n + 1)² − n(n + 2)) / ((n + 1)(n + 2)) [1] = (n² + 2n + 1 − n² − 2n) / ((n + 1)(n + 2)) [1] = 1/((n + 1)(n + 2)) = RHS [1] (c) Put n = 99: 1/(100 × 101) = 1/10 100 [1] Examiner insight: A proof that manipulates both sides until they meet, or that ends at “1 = 1”, is not an LHS to RHS proof and can lose the final mark.

9. (a) (x + 2)³ = x³ + 6x² + 12x + 8 [1] and (x − 2)³ = x³ − 6x² + 12x − 8 [1]. Subtracting: LHS = 12x² + 16 = RHS [1] (b) The equation is (x + 2)³ − (x − 2)³ = 0, that is, 12x² + 16 = 0. Since x² ≥ 0, 12x² + 16 ≥ 16 [1]. So the left-hand side is never zero and there are no real solutions [1] Examiner insight: In (b) you must link back to the identity from (a) and give the inequality; “it has no solutions” with no reason scores zero.

10. (a) Coefficients from row 5: 1, 5, 10, 10, 5, 1 with a = 2, b = −x [1]; 2⁵ + 5(2⁴)(−x) + 10(2³)(−x)² + 10(2²)(−x)³ + 5(2)(−x)⁴ + (−x)⁵ [1] = 32 − 80x + 80x² − 40x³ + 10x⁴ − x⁵ [1] (b) The x³ terms come from 1 × (−40x³) [1] and 3x × 80x² = 240x³ [1]. The coefficient is −40 + 240 = 200 [1] Examiner insight: In (b) both products are needed; using only the x³ term from (a) gives −40 and earns just the first method mark.

11. (a) General term = 6Cr (x²)^(6−r) (k/x)^r [1] = 6Cr k^r x^(12 − 2r) x^(−r) = 6Cr k^r x^(12 − 3r) [1] (b) Constant term: 12 − 3r = 0, so r = 4 [1]. The term is 6C4 k⁴ = 15k⁴ = 240 [1], so k⁴ = 16 [1]. Since k > 0, k = 2 [1] (c) 12 − 3r = 6 gives r = 2 [1]. The coefficient is 6C2 × 2² = 15 × 4 = 60 [1] Examiner insight: k = ±2 loses the final mark in (b) because the question states k > 0; use the given condition explicitly.

12. (a) L = 10 log(3.2 × 10⁻⁵ / 10⁻¹²) = 10 log(3.2 × 10⁷) [1] = 75.1 dB [1] (b) The new loudness is 10 log(kI/I₀) [1] = 10(log k + log(I/I₀)) by the product law [1] = 10 log k + L, so the loudness increases by 10 log k [1] (c) 90 = 10 log(I/10⁻¹²), so log(I/10⁻¹²) = 9 [1]. Then I = 10⁹ × 10⁻¹² = 10⁻³ W m⁻² [1] (d) From (b), 10 log k = 90 − 60 = 30, so log k = 3 and k = 1000 [1] Examiner insight: In (b) the product law must be named or clearly used; writing the final line without the step 10(log k + log(I/I₀)) does not earn the reasoning marks.

Where marks are usually lost

• Applying a power to the letters but not the number, as in question 1(a).
• Writing a calculator value in a “show that” such as question 2(a), where the formula must be seen.
• Taking ln before isolating the exponential term, as in question 3(b).
• Leaving log_a a unsimplified, or splitting a log of a sum.
• Giving only a decimal when an exact form is requested, as in question 7.
• Solving a hidden quadratic for y and not converting back to x.
• Working on both sides of an identity instead of going from LHS to RHS.
• Losing the negative sign of b in (2 − x)⁵, so the terms do not alternate.
• Ignoring a stated condition such as k > 0 or n ∈ ℤ⁺.
• Rounding to fewer than 3 significant figures, or rounding early in a multi-step calculation.

Next steps

• Recap the key facts in the revision notes.
• Re-learn any weak method in the study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Practise more equations mixing exponentials and logarithms in the AA functions practice.
• Plan your final weeks with the AA exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).