IB DP Mathematics: Applications and Interpretation – Laws of logarithms, rational exponents, infinite series and complex numbers (HL) 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 HL number and algebra unit of IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 1.9–1.13, and every question is HL only (AHL). 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 HL sessions.

All three HL papers for this course require technology, so every question here is labelled “(calculator allowed)”. Where the guide expects a skill “by hand”, the question asks you to show the working. Give answers exactly or to 3 significant figures, and arguments in radians with −π < θ ≤ π.

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

Questions

1. (calculator allowed; show working by hand) Find the exact values of 125^(2/3) and 81^(−3/4). [2]

2. (calculator allowed; show working by hand) Simplify 4x^(3/2) × 3x^(−1/2) ÷ 6x^(1/3), where x > 0. [2]

3. (calculator allowed; show working by hand) Write 3 log 2 + log 15 − log 12 as a single logarithm, and hence find its exact value. [3]

4. (calculator allowed) Solve ln x + ln (x + 4) = ln 21. [4]

5. (calculator allowed) The pH of a solution is given by pH = −log [H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³.

(a) Find the pH of a solution with [H⁺] = 3.2 × 10⁻⁵ mol dm⁻³. [1] (b) Show that multiplying [H⁺] by 100 lowers the pH by exactly 2. [2]

6. (calculator allowed) A geometric series has first term 15 and sum to infinity 60.

(a) Find the common ratio. [2] (b) Find the fourth term. [1] (c) Explain why the series 3 − 4.5 + 6.75 − … has no sum to infinity. [2]

7. (calculator allowed; show working by hand) Let z = 5 − 2i and w = 1 + 3i.

(a) Find z − 2w. [1] (b) Find zw in the form a + bi. [2] (c) Find z/w in the form a + bi. [3]

8. (calculator allowed) Let f(x) = x² + 6x + 34.

(a) Find the discriminant of f(x) and state what it tells you about the graph of y = f(x). [2] (b) Solve f(x) = 0. [2] (c) The roots are plotted on an Argand diagram. Describe how the two points are related, and link their real part to the graph of y = f(x). [2]

9. (calculator allowed; show working by hand) Let z = −2 + 2i.

(a) Find the modulus and argument of z. [2] (b) Write z in exponential form. [1] (c) Hence find z⁴ in the form a + bi. [2]

10. (calculator allowed) Extended. Let u = 1 + i and z = 3 + i. On an Argand diagram, point A represents z and point B represents uz. O is the origin.

(a) Write u in exponential form. [2] (b) Describe fully the geometric transformation that maps A to B. [2] (c) Find uz in the form a + bi. [1] (d) Find the area of triangle OAB. [2]

11. (calculator allowed) Extended. A ball is dropped from a height of 10 m. After each bounce it rises to 60% of the height it last fell from.

(a) Find the height it reaches after the third bounce. [2] (b) Find the total vertical distance travelled according to this model. [3] (c) Use logarithms to find the first bounce after which the ball rises less than 1 cm. [3] (d) State one reason why this model is not realistic. [1]

12. (calculator allowed) Extended. Two AC voltage sources in a circuit give V₁ = 6 cos(30t) and V₂ = 4 cos(30t + π/3), where t is in seconds.

(a) Write V₁ and V₂ as complex numbers and show that their sum is 8 + 2√3 i. [2] (b) Find the modulus and argument of 8 + 2√3 i. [2] (c) Hence write the total voltage V = V₁ + V₂ in the form A cos(30t + B). [1] (d) Find the smallest positive value of t at which V is a maximum. [2]

Answers

1. 125^(2/3) = (cube root of 125)² = 5² = 25 [1]. 81^(−3/4) = 1 / (fourth root of 81)³ = 1/3³ = 1/27 [1]. [2] Examiner insight: “Exact value” means 1/27, not 0.0370; a decimal loses the accuracy mark even with correct working.

2. Coefficients 12 ÷ 6 = 2; powers 3/2 − 1/2 − 1/3 = 2/3 [1]. Answer 2x^(2/3) [1]. [2] Examiner insight: The method mark needs the index laws visibly applied to the powers; writing only the final answer risks losing both marks if the power is wrong.

3. 3 log 2 = log 8 [1]. log (8 × 15 ÷ 12) = log 10 [1]. Value 1 [1]. [3] Examiner insight: “Hence” means the single logarithm must appear before the value; a GDC value of 1 on its own does not earn the first two marks.

4. ln [x(x + 4)] = ln 21 [1]. x² + 4x − 21 = 0 [1]. (x + 7)(x − 3) = 0, so x = 3 or x = −7 [1]. ln (−7) is undefined, so reject it: x = 3 [1]. [4] Examiner insight: The final accuracy mark is lost if x = −7 is left in; you must reject it and say why.

5. (a) pH = −log (3.2 × 10⁻⁵) = 4.49 (3 s.f.) [1]. (b) New pH = −log (100c) = −(log 100 + log c) [1] = −2 − log c = (old pH) − 2 [1]. [3] Examiner insight: In a “show that”, two numerical examples are not a proof; the product law must be applied to a general concentration c.

6. (a) 15 / (1 − r) = 60 [1], so 1 − r = 0.25 and r = 0.75 [1]. (b) u₄ = 15 × 0.75³ = 6.33 (exact 405/64) [1]. (c) r = −4.5 / 3 = −1.5 [1]. |r| = 1.5 ≥ 1, so the terms do not tend to zero and there is no sum to infinity [1]. [5] Examiner insight: In (c) the reason mark needs the condition |r| < 1 stated and compared with the actual ratio; “the terms get bigger” alone is not enough.

7. (a) 5 − 2i − 2 − 6i = 3 − 8i [1]. (b) 5 + 15i − 2i − 6i² [1] = 11 + 13i [1]. (c) Multiply by (1 − 3i)/(1 − 3i) [1]. Numerator 5 − 15i − 2i + 6i² = −1 − 17i; denominator 1 + 9 = 10 [1]. Answer −0.1 − 1.7i [1]. [6] Examiner insight: The guide lists these operations “by hand”, so the conjugate step must be shown; a bare GDC answer earns only the final accuracy mark at best.

8. (a) b² − 4ac = 36 − 136 = −100 [1]. It is negative, so the graph does not meet the x-axis and there are no real roots [1]. (b) x = (−6 ± √(−100)) / 2 [1] = −3 ± 5i [1]. (c) The points (−3, 5) and (−3, −5) are reflections of each other in the real axis: the roots are a conjugate pair [1]. The real part, −3, is the x-coordinate of the vertex, on the axis of symmetry x = −3 [1]. [6] Examiner insight: Write √(−100) = 10i explicitly; answers such as −3 ± 5 with the i missing score no accuracy mark.

9. (a) |z| = √(4 + 4) = 2√2 [1]. The point is in the second quadrant, so arg z = π − π/4 = 3π/4 [1]. (b) z = 2√2 e^(i3π/4) [1]. (c) z⁴ = (2√2)⁴ e^(i3π) = 64 e^(i3π) [1]. Since e^(i3π) = e^(iπ) = −1, z⁴ = −64 [1]. [5] Examiner insight: An argument of −π/4 from arctan(2/−2) is wrong, and follow-through then applies in (b) and (c) only if your working is shown.

10. (a) |u| = √2 [1]. arg u = π/4, so u = √2 e^(iπ/4) [1]. (b) An anticlockwise rotation of π/4 about O [1], combined with a stretch (enlargement) of scale factor √2 from O [1]. (c) (3 + i)(1 + i) = 3 + 3i + i − 1 = 2 + 4i [1]. (d) OA = √10, OB = √2 × √10 = √20, and angle AOB = π/4 [1]. Area = ½ × √10 × √20 × sin(π/4) = 5 [1]. [7] Examiner insight: “Describe fully” needs the angle, the direction and the centre of rotation plus the scale factor; missing “anticlockwise” or “about O” costs a mark.

11. (a) Height after the nth bounce = 10 × 0.6ⁿ [1]. After three bounces: 10 × 0.6³ = 2.16 m [1]. (b) The rises form a geometric series with u₁ = 6, r = 0.6, |r| < 1, so S_∞ = 6 / 0.4 = 15 [1]. Total = 10 + 2 × 15 [1] = 40 m [1]. (c) 10 × 0.6ⁿ < 0.01, so 0.6ⁿ < 0.001 [1]. n log 0.6 < log 0.001, and log 0.6 < 0 reverses the inequality: n > log 0.001 / log 0.6 = 13.5… [1]. So the 14th bounce [1]. (d) The model gives infinitely many bounces in a finite distance; a real ball stops after a finite number, and the fraction of height kept will not stay exactly 60% [1]. [9] Examiner insight: In (c) n must be a whole number, so 13.5 must be rounded up to 14; giving 13.5 as the answer loses the final mark.

12. (a) V₁ ↔ 6 and V₂ ↔ 4e^(iπ/3) = 4(1/2 + (√3/2)i) = 2 + 2√3 i [1]. Sum = 6 + 2 + 2√3 i = 8 + 2√3 i [1]. (b) Modulus = √(64 + 12) = √76 = 8.72 [1]. Argument = arctan(2√3 / 8) = 0.409 [1]. (c) V = 8.72 cos(30t + 0.409) [1]. (d) V is a maximum when 30t + 0.409 = 2π (the value 0 gives negative t) [1]. t = (2π − 0.40864) / 30 = 0.196 s [1]. [7] Examiner insight: Carry the unrounded argument (0.40864…) into (d) and round only at the end; rounding intermediate values early can change the third significant figure and cost the accuracy mark.

Where marks are usually lost

• Splitting log (x + 4) into log x + log 4.
• Leaving a root that makes a logarithm undefined.
• Giving a decimal where the question asks for an exact value.
• Using S_∞ without checking |r| < 1, or without stating the check when asked to explain.
• Missing the factor 2 for rebounds in bouncing-ball totals.
• Forgetting that dividing by a negative logarithm reverses an inequality.
• Writing an argument in the wrong quadrant after using arctan(b/a) directly.
• Describing a transformation without the direction, centre or scale factor.
• Adding sinusoid amplitudes (6 + 4 = 10) instead of adding complex numbers.
• Giving a GDC answer to a “by hand” operation without the algebra.

Next steps

• Revision notes for this unit
• Study guide for this unit
• IB DP Maths AI course hub
• Printable syllabus checklist
• Exam preparation for IB Maths AI
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021. Sections AHL 1.9, 1.10, 1.11, 1.12 and 1.13.