IB DP Mathematics: Applications and Interpretation – Laws of logarithms, rational exponents, infinite series and complex numbers (HL) Study Guide

This study guide teaches the HL number and algebra unit of IB Diploma Programme Mathematics: Applications and Interpretation from scratch. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 1.9–1.13, and all of it 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.

When you have worked through it, condense it with the revision notes and test yourself with the practice questions. The IB DP Maths AI course hub and the printable syllabus checklist show where the unit sits in the course.

What this unit covers

Syllabus section	What you must be able to do	SL/HL
1.9	Use the three laws of logarithms, with base 10 or e	HL only
1.10	Simplify numerical and algebraic expressions with rational exponents	HL only
1.11	Find the sum of an infinite geometric sequence	HL only
1.12	Work with complex numbers in Cartesian form, draw Argand diagrams, solve quadratics with b² − 4ac < 0	HL only
1.13	Use polar and exponential forms; products, quotients and integer powers; add sinusoids; interpret operations geometrically	HL only

All three HL papers require technology, so you will always have your GDC. The guide still says some skills are done “by hand and with technology”, so you must be able to show the algebra as well as press the buttons.

1.9 Laws of logarithms

For a, x, y > 0:

• log_a xy = log_a x + log_a y
• log_a (x/y) = log_a x − log_a y
• log_a xᵐ = m log_a x

The guide says that in examinations a will be 10 or e. So you will see log x (base 10) and ln x (base e). The laws only work when x and y are positive, which matters when you solve equations.

Worked example 1. Write ln 12 + ln 5 − ln 15 as a single logarithm.

ln 12 + ln 5 − ln 15 = ln (12 × 5 ÷ 15)
                     = ln 4
                     = 2 ln 2     (since 4 = 2²)

Worked example 2. Solve log x + log (x − 15) = 2.

log [x(x − 15)] = 2
x(x − 15) = 10² = 100
x² − 15x − 100 = 0
(x − 20)(x + 5) = 0
x = 20 or x = −5

Check each root in the original equation. With x = −5, log x is undefined, so reject it. The only solution is x = 20.

Logarithms scale very large and very small numbers (the guide links this to AHL 2.10). The chemistry link is pH = −log [H⁺]. Because log (100c) = log 100 + log c = 2 + log c, multiplying the hydrogen ion concentration by 100 lowers the pH by exactly 2.

1.10 Rational exponents

A rational exponent combines a root and a power:

• a^(1/n) = ⁿ√a
• a^(m/n) = (ⁿ√a)ᵐ
• a^(−p) = 1/aᵖ

The usual index laws still hold: aᵖ × a^q = a^(p+q), aᵖ ÷ a^q = a^(p−q), (aᵖ)^q = a^(pq).

Worked example 3 (numerical).

27^(2/3)  = (cube root of 27)² = 3² = 9
16^(−3/4) = 1 / 16^(3/4) = 1 / (fourth root of 16)³ = 1 / 2³ = 1/8

Take the root first. It keeps the numbers small.

Worked example 4 (algebraic). Simplify (8x⁶)^(2/3), and x^(2/3) × x^(1/2) ÷ x^(1/6).

(8x⁶)^(2/3) = 8^(2/3) × x^(6 × 2/3) = 4x⁴

x^(2/3) × x^(1/2) ÷ x^(1/6) = x^(4/6 + 3/6 − 1/6) = x^(6/6) = x

Use a common denominator when you add fractional powers.

1.11 Infinite geometric series

A geometric sequence has first term u₁ and common ratio r. From SL you know S_n = u₁(1 − rⁿ)/(1 − r). If |r| < 1, then rⁿ → 0 as n → ∞, so the sum approaches a limit:

S_∞ = u₁ / (1 − r), valid only for |r| < 1.

If |r| ≥ 1 the terms do not shrink to zero and there is no sum to infinity. Always state the condition when you use the formula.

Worked example 5. A geometric series has u₁ = 10 and S_∞ = 40. Find r.

10 / (1 − r) = 40
1 − r = 0.25
r = 0.75      (|r| < 1, so the sum exists)

Worked example 6 (bouncing ball). A ball is dropped from 8 m. After each bounce it rises to 75% of its previous height. Find the total vertical distance it travels.

The first fall is 8 m. After that the ball goes up and then down the same distance each time. The rises form a geometric series with u₁ = 8 × 0.75 = 6 and r = 0.75.

Total = 8 + 2 × S_∞(rises)
      = 8 + 2 × 6 / (1 − 0.75)
      = 8 + 2 × 24
      = 56 m

The factor 2 counts each rise and its matching fall. Forgetting it, or doubling the first drop too, are the two usual errors.

Worked example 7. Write 0.272727… as a fraction.

0.272727… = 0.27 + 0.0027 + … with u₁ = 0.27 and r = 0.01, so S_∞ = 0.27 / 0.99 = 27/99 = 3/11.

1.12 Complex numbers in Cartesian form

The number i satisfies i² = −1. A complex number in Cartesian form is z = a + bi, where a, b are real.

• Real part: Re(z) = a. Imaginary part: Im(z) = b (a real number, not bi).
• Conjugate: z* = a − bi.
• Modulus: |z| = √(a² + b²), the distance from the origin on an Argand diagram.
• Argument: arg z = θ, the angle from the positive real axis to the point. On this page arguments are in radians with −π < θ ≤ π.

Worked example 8. Let z = 3 + 2i and w = 1 − 4i. Find z + w, z − w, zw and z/w by hand.

z + w = 4 − 2i
z − w = 2 + 6i
zw    = 3 − 12i + 2i − 8i² = 3 − 10i + 8 = 11 − 10i

z/w = (3 + 2i)(1 + 4i) / [(1 − 4i)(1 + 4i)]
    = (3 + 12i + 2i + 8i²) / (1 + 16)
    = (−5 + 14i) / 17
    = −5/17 + (14/17)i

To divide, multiply top and bottom by the conjugate of the denominator. The denominator becomes a real number.

The guide expects you to calculate powers of complex numbers in Cartesian form with technology. Put your GDC into complex (a + bi) mode and type the expression directly.

The complex plane

On an Argand diagram the horizontal axis is real and the vertical axis is imaginary. The point (a, b) represents a + bi. The conjugate z* is the reflection of z in the real axis.

Quadratics with complex roots

For ax² + bx + c = 0 with real coefficients and b² − 4ac < 0, the quadratic formula gives two complex roots. They are always a conjugate pair.

Worked example 9. Solve x² − 4x + 13 = 0.

b² − 4ac = 16 − 52 = −36
x = (4 ± √(−36)) / 2 = (4 ± 6i) / 2 = 2 ± 3i

The graph of f(x) = x² − 4x + 13 never meets the x-axis, because the discriminant is negative. Its vertex is at (2, 9), and the real part of the roots, 2, is the x-coordinate of the vertex.

1.13 Polar and exponential forms

A complex number with modulus r and argument θ can be written as

• polar form: z = r(cos θ + i sin θ) = r cis θ
• exponential (Euler) form: z = re^(iθ)

To convert from Cartesian form, find r = √(a² + b²), then find θ. Sketch the point first so you pick the right quadrant. To convert back, use a = r cos θ and b = r sin θ.

Worked example 10. Write z = −3 + 3i in polar and exponential forms.

r = √(9 + 9) = √18 = 3√2
The point (−3, 3) is in the second quadrant.
Reference angle = arctan(3/3) = π/4, so θ = π − π/4 = 3π/4
z = 3√2 cis(3π/4) = 3√2 e^(i3π/4)

Products, quotients and powers

For z₁ = r₁e^(iθ₁) and z₂ = r₂e^(iθ₂):

• z₁z₂ = r₁r₂ e^(i(θ₁ + θ₂)) — multiply moduli, add arguments
• z₁/z₂ = (r₁/r₂) e^(i(θ₁ − θ₂)) — divide moduli, subtract arguments
• zⁿ = rⁿ e^(inθ) for integer n

Worked example 11. Find (−3 + 3i)⁴.

(3√2 e^(i3π/4))⁴ = (3√2)⁴ e^(i3π) = 324 e^(i3π)
3π is the same direction as π, and e^(iπ) = −1
So (−3 + 3i)⁴ = −324

The guide states that in examinations you will not be asked to find roots of complex numbers, so this unit stops at integer powers.

Adding sinusoids with the same frequency

Two waves with the same frequency but different phase add to give one wave of that frequency. Represent each as a complex number: A cos(ωt + α) corresponds to Ae^(iα). Add the complex numbers, then read off the modulus (amplitude) and argument (phase).

Worked example 12. Two AC voltages are V₁ = 12 cos(50t) and V₂ = 5 cos(50t + π/2). Write the total as V = A cos(50t + B).

V₁ ↔ 12e^(i0)   = 12
V₂ ↔ 5e^(iπ/2)  = 5i
Sum = 12 + 5i
A = √(144 + 25) = 13
B = arctan(5/12) = 0.395 (3 s.f.)
V = 13 cos(50t + 0.395)

Your GDC can do the addition and the conversion in one step in polar mode. Write down the complex number you entered so your method is visible.

Geometric interpretation

• Adding or subtracting complex numbers is vector addition or subtraction on the Argand diagram.
• Multiplying by re^(iθ) rotates a point anticlockwise by θ about the origin and stretches its distance from the origin by the factor r.

For example, multiplying by i = e^(iπ/2) is a rotation of π/2 anticlockwise: (2 + i) × i = −1 + 2i. Multiplying by 2e^(iπ/6) = √3 + i rotates by π/6 and doubles the modulus.

Common errors

• Writing log (x + y) = log x + log y. The law is for a product, not a sum.
• Keeping a root that makes a log argument zero or negative.
• Using S_∞ when |r| ≥ 1, or not stating that |r| < 1.
• In bouncing-ball problems, counting the first drop twice or each rebound only once.
• Giving Im(3 − 5i) as −5i. The imaginary part is −5.
• Taking arctan(b/a) as the argument without checking the quadrant (a calculator gives −π/4 for −3 + 3i, which is wrong).
• Mixing degrees and radians in exponential form. e^(iθ) needs θ in radians.
• Adding amplitudes of sinusoids directly (12 + 5 = 17) instead of adding complex numbers.

Where to go next

Condense this into the revision notes, then try the practice set. For the rest of the course, see the AI syllabus guide, the AI subject guide and exam preparation for AI.

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.