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NotesMath AI HLTopic 1.5
Unit 1 · Number & Algebra · Topic 1.5

IB Math AI HL — Exponents and logarithms

IB Mathematics AI SL topic covering core concepts and exam-style applications.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Exam technique guidePractice questions

Key concepts in Exponents and logarithms

Key Idea: Exponents raise a number to a power. Logarithms undo that — they tell you what the power was. The two are inverses of each other.

Key skills for this topic

Switch log ↔ power

  • aˣ = b ↔ x = logₐ b
  • log₃ 81 = 4 because 3⁴ = 81
  • log = base 10, ln = base e

Common mistakes

  • Mixing log and ln in one calc
  • Forgetting log laws need same base
  • Treating log(A + B) as log A + log B

Solve aˣ = b

  • Use: x = log b ÷ log a
  • E.g. 3ˣ = 50
  • x = log 50 ÷ log 3 ≈ 3.56
ax=b  ⇔  x=log⁡aba^x = b \;\Leftrightarrow\; x = \log_a bax=b⇔x=loga​b
Log definition — given on formula sheet

Index lawRuleExample
Multiply (same base)aˣ × aʸ = aˣ⁺ʸ2³ × 2⁴ = 2⁷ = 128
Divide (same base)aˣ ÷ aʸ = aˣ⁻ʸ3⁵ ÷ 3² = 3³ = 27
Power of a power(aˣ)ʸ = aˣʸ(2³)² = 2⁶ = 64
Zero exponenta⁰ = 17⁰ = 1
Negative exponenta⁻ⁿ = 1/aⁿ2⁻³ = 1/8
Fractional exponenta¹/ⁿ = ⁿ√a8¹/³ = ∛8 = 2

🔄 Log ↔ exponential

Key Idea: log₂ 8 = 3 because 2³ = 8. log (no base) = base 10. ln = base e ≈ 2.718.
Log law (same base)RuleExample
Productlog(A × B) = log A + log Blog 6 + log 5 = log 30
Quotientlog(A ÷ B) = log A − log Blog 30 − log 3 = log 10 = 1
Powerlog(Aⁿ) = n log Alog 5³ = 3 log 5

🔑 Solving aˣ = b

Example: solve 3ˣ = 50 1. Take log of both sides: log(3ˣ) = log 50 2. Power rule drops x to the front: x · log 3 = log 50 3. Divide both sides by log 3: x = log 50 ÷ log 3 4. On the GDC: x ≈ 3.56 (3 s.f.)

✏️ Worked examples

Simplify with index laws

Simplify: (2x³)² ÷ x

Step by step:

  1. Power of a product: (2x³)² = 4x⁶

  2. Divide: 4x⁶ ÷ x = 4x⁶⁻¹

  3. Answer: 4x⁵

Final answer:

4x⁵

Solve an exponential equation

Solve: 5ˣ = 80

Step by step:

  1. Take log of both sides: log(5ˣ) = log 80

  2. Power rule: x log 5 = log 80

  3. Divide: x = log 80 ÷ log 5

  4. Calculate: x = 1.903 ÷ 0.699 = 2.72 (3 s.f.)

Final answer:

x ≈ 2.72

Use log laws to simplify

Write log 6 + log 5 − log 3 as a single value.

Step by step:

  1. Product rule: log 6 + log 5 = log(6 × 5) = log 30

  2. Quotient rule: log 30 − log 3 = log(30 ÷ 3) = log 10

  3. log 10 = 1

Final answer:

1


💡 Test yourself — tap to reveal

Evaluate log₂ 32 Ask: 2 to what power = 32? 2⁵ = 32 Answer: 5

Solve 2ˣ = 1000 x = log 1000 ÷ log 2 = 3 ÷ 0.301 Answer: x ≈ 9.97 (3 s.f.)

Simplify log 200 − log 2 Quotient rule: log(200 ÷ 2) = log 100 Answer: 2

What is log₅ 1? Any non-zero base to the power 0 = 1. Answer: 0

Rewrite 4³ = 64 in log form Base 4, result 64, power 3. Answer: log₄ 64 = 3


🎯 IB-style practice — logarithms in context

Key Idea:
  • LOG → evaluates log₁₀ (use when finding the dB / pH value).
  • 2nd + LOG → 10ˣ (use to undo a log and find the original value).

Part (i) — find loudness from intensity

The loudness of a sound (in dB) is given by <strong>L = 10 log₁₀(I / I₀)</strong>, where I is the intensity in W m⁻² and I₀ = 10⁻¹² W m⁻² is the reference intensity.<br><br>A quiet library reading room has sound of intensity I = 5 × 10⁻⁹ W m⁻². Find its loudness.

Step by step:

  1. Plug in I = 5 × 10⁻⁹ and I₀ = 10⁻¹². Drop the values into the model:

    L=10log⁡10 ⁣(5×10−910−12)L = 10 \log_{10}\!\left(\frac{5 \times 10^{-9}}{10^{-12}}\right)L=10log10​(10−125×10−9​)
  2. Divide powers of 10 by subtracting exponents: −9 − (−12) = 3:

    L=10log⁡10(5×103)L = 10 \log_{10}(5 \times 10^{3})L=10log10​(5×103)
  3. Type into the GDC with LOG, then round to 3 s.f.:

    L=36.98...≈37.0 dBL = 36.98... \approx \mathbf{37.0 \text{ dB}}L=36.98...≈37.0 dB
Final answer:

L ≈ 37.0 dB

🔒 GDC walkthrough

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Part (ii) — find intensity from loudness

Using the same model <strong>L = 10 log₁₀(I / I₀)</strong> with I₀ = 10⁻¹² W m⁻²:<br><br>A motorcycle engine produces sound of loudness L = 88 dB. Find its intensity I. Give your answer in the form a × 10ᵏ where 1 ≤ a < 10 and k is an integer.

Step by step:

  1. Set L = 88. Put the given loudness into the model — now I is the unknown:

    88=10log⁡10 ⁣(I10−12)88 = 10 \log_{10}\!\left(\frac{I}{10^{-12}}\right)88=10log10​(10−12I​)
  2. Get rid of the 10 in front of the log. Divide both sides by 10:

    8.8=log⁡10 ⁣(I10−12)8.8 = \log_{10}\!\left(\frac{I}{10^{-12}}\right)8.8=log10​(10−12I​)
  3. Undo the log with 10ˣ. Because log₁₀ and 10ˣ are inverses, the log peels away:

    108.8=I10−1210^{8.8} = \frac{I}{10^{-12}}108.8=10−12I​
  4. Isolate I by multiplying both sides by 10⁻¹², then combine powers (10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ):

    I=108.8×10−12=10−3.2I = 10^{8.8} \times 10^{-12} = 10^{-3.2}I=108.8×10−12=10−3.2
  5. Write in standard form and round to 3 s.f.:

    I=10−3.2≈6.31×10−4 W m−2I = 10^{-3.2} \approx \mathbf{6.31 \times 10^{-4}} \text{ W m}^{-2}I=10−3.2≈6.31×10−4 W m−2
Final answer:

I ≈ 6.31 × 10⁻⁴ W m⁻²

🔒 GDC walkthrough

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Same base only. 2³ × 3⁴ can't be combined — different bases. Same rule for log laws. Don't mix log and ln in one calculation. Memorise: log 1 = 0, log 10 = 1, ln 1 = 0, ln e = 1. Paper 2 check: after solving aˣ = b, plug your answer back in to verify (e.g. 5².⁷² ≈ 80 ✓).

IB-style question — solving an exponential equation [5 marks]

An investment of $2000 grows by 6% each year, so after t years its value is V = 2000 × 1.06ᵗ dollars. (a) Find the value of the investment after 10 years. (b) Find the number of complete years it takes for the investment to first exceed $5000.

Step by step:

  1. (a) Substitute t = 10 into the model.

    V=2000×1.0610=2000×1.7908=3581.7V = 2000 \times 1.06^{10} = 2000 \times 1.7908 = 3581.7V=2000×1.0610=2000×1.7908=3581.7
  2. (b) Set V > 5000 and divide both sides by 2000 to isolate the power.

    2000×1.06t>5000  ⇒  1.06t>2.52000 \times 1.06^{t} > 5000 \;\Rightarrow\; 1.06^{t} > 2.52000×1.06t>5000⇒1.06t>2.5
  3. The unknown is in the exponent, so take logs of both sides; the power law brings t down.

    tlog⁡1.06>log⁡2.5  ⇒  t>log⁡2.5log⁡1.06t \log 1.06 > \log 2.5 \;\Rightarrow\; t > \frac{\log 2.5}{\log 1.06}tlog1.06>log2.5⇒t>log1.06log2.5​
  4. Evaluate the quotient on the GDC.

    t>15.725t > 15.725t>15.725
  5. t must be a whole number of years and the total must exceed 5000, so round UP to the next year.

    t=16t = 16t=16
Final answer:

(a) $3581.70 (to the nearest cent), about $3580. (b) 16 years.

🔒 GDC walkthrough

Step through the exact calculator keystrokes, screen by screen, in study mode.

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What you'll learn in Topic 1.5

  • 1.5.1 Laws of Exponents
  • 1.5.2 Introduction to Logarithms
  • 1.5.3 Laws of Logarithms
  • 1.5.4 Solving Exponential and Logarithmic Equations
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 1.5 Exponents and logarithms

1.5.1

Laws of Exponents

Notes
1.5.2

Introduction to Logarithms

Notes
1.5.3

Laws of Logarithms

Notes
1.5.4

Solving Exponential and Logarithmic Equations

Notes

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Topic 1.5 Exponents and logarithms forms a core part of Unit 1: Number & Algebra in IB Math AI HL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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