IB Math AI HL Topic 1.5: Exponents and logarithms | Notes & Flashcards | Aimnova

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

a^x = b \;\Leftrightarrow\; x = \log_a b

Log definition — given on formula sheet

🔄 Log ↔ exponential

Key Idea: log₂ 8 = 3 because 2³ = 8. log (no base) = base 10. ln = base e ≈ 2.718.

🔑 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:

Power of a product: (2x³)² = 4x⁶
Divide: 4x⁶ ÷ x = 4x⁶⁻¹
Answer: 4x⁵

Final answer:

4x⁵

Solve an exponential equation

Solve: 5ˣ = 80

Step by step:

Take log of both sides: log(5ˣ) = log 80
Power rule: x log 5 = log 80
Divide: x = log 80 ÷ log 5
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:

Product rule: log 6 + log 5 = log(6 × 5) = log 30
Quotient rule: log 30 − log 3 = log(30 ÷ 3) = log 10
log 10 = 1

Final answer:

💡 Test yourself — tap to reveal

🎯 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 L = 10 log₁₀(I / I₀), where I is the intensity in W m⁻² and I₀ = 10⁻¹² W m⁻² is the reference intensity. A quiet library reading room has sound of intensity I = 5 × 10⁻⁹ W m⁻². Find its loudness.

Step by step:

Plug in I = 5 × 10⁻⁹ and I₀ = 10⁻¹². Drop the values into the model:
$L = 10 \log_{10}\!\left(\frac{5 \times 10^{-9}}{10^{-12}}\right)$
Divide powers of 10 by subtracting exponents: −9 − (−12) = 3:
$L = 10 \log_{10}(5 \times 10^{3})$
Type into the GDC with LOG, then round to 3 s.f.:
$L = 36.98... \approx \mathbf{37.0 \text{ dB}}$

Final answer:

L ≈ 37.0 dB

Part (ii) — find intensity from loudness

Using the same model L = 10 log₁₀(I / I₀) with I₀ = 10⁻¹² W m⁻²: 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:

Set L = 88. Put the given loudness into the model — now I is the unknown:
$88 = 10 \log_{10}\!\left(\frac{I}{10^{-12}}\right)$
Get rid of the 10 in front of the log. Divide both sides by 10:
$8.8 = \log_{10}\!\left(\frac{I}{10^{-12}}\right)$
Undo the log with 10ˣ. Because log₁₀ and 10ˣ are inverses, the log peels away:
$10^{8.8} = \frac{I}{10^{-12}}$
Isolate I by multiplying both sides by 10⁻¹², then combine powers (10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ):
$I = 10^{8.8} \times 10^{-12} = 10^{-3.2}$
Write in standard form and round to 3 s.f.:
$I = 10^{-3.2} \approx \mathbf{6.31 \times 10^{-4}} \text{ W m}^{-2}$

Final answer:

I ≈ 6.31 × 10⁻⁴ W m⁻²

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:

(a) Substitute t = 10 into the model.
$V = 2000 \times 1.06^{10} = 2000 \times 1.7908 = 3581.7$
(b) Set V > 5000 and divide both sides by 2000 to isolate the power.
$2000 \times 1.06^{t} > 5000 \;\Rightarrow\; 1.06^{t} > 2.5$
The unknown is in the exponent, so take logs of both sides; the power law brings t down.
$t \log 1.06 > \log 2.5 \;\Rightarrow\; t > \frac{\log 2.5}{\log 1.06}$
Evaluate the quotient on the GDC.
$t > 15.725$
t must be a whole number of years and the total must exceed 5000, so round UP to the next year.
$t = 16$

Final answer:

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

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

a^x = b \;\Leftrightarrow\; x = \log_a b

Log definition — given on formula sheet

🔄 Log ↔ exponential

Key Idea: log₂ 8 = 3 because 2³ = 8. log (no base) = base 10. ln = base e ≈ 2.718.

🔑 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:

Power of a product: (2x³)² = 4x⁶
Divide: 4x⁶ ÷ x = 4x⁶⁻¹
Answer: 4x⁵

Final answer:

4x⁵

Solve an exponential equation

Solve: 5ˣ = 80

Step by step:

Take log of both sides: log(5ˣ) = log 80
Power rule: x log 5 = log 80
Divide: x = log 80 ÷ log 5
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:

Product rule: log 6 + log 5 = log(6 × 5) = log 30
Quotient rule: log 30 − log 3 = log(30 ÷ 3) = log 10
log 10 = 1

Final answer:

💡 Test yourself — tap to reveal

🎯 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

Step by step:

Plug in I = 5 × 10⁻⁹ and I₀ = 10⁻¹². Drop the values into the model:
$L = 10 \log_{10}\!\left(\frac{5 \times 10^{-9}}{10^{-12}}\right)$
Divide powers of 10 by subtracting exponents: −9 − (−12) = 3:
$L = 10 \log_{10}(5 \times 10^{3})$
Type into the GDC with LOG, then round to 3 s.f.:
$L = 36.98... \approx \mathbf{37.0 \text{ dB}}$

Final answer:

L ≈ 37.0 dB

Part (ii) — find intensity from loudness

Step by step:

Set L = 88. Put the given loudness into the model — now I is the unknown:
$88 = 10 \log_{10}\!\left(\frac{I}{10^{-12}}\right)$
Get rid of the 10 in front of the log. Divide both sides by 10:
$8.8 = \log_{10}\!\left(\frac{I}{10^{-12}}\right)$
Undo the log with 10ˣ. Because log₁₀ and 10ˣ are inverses, the log peels away:
$10^{8.8} = \frac{I}{10^{-12}}$
Isolate I by multiplying both sides by 10⁻¹², then combine powers (10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ):
$I = 10^{8.8} \times 10^{-12} = 10^{-3.2}$
Write in standard form and round to 3 s.f.:
$I = 10^{-3.2} \approx \mathbf{6.31 \times 10^{-4}} \text{ W m}^{-2}$

Final answer:

I ≈ 6.31 × 10⁻⁴ W m⁻²

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]

Step by step:

(a) Substitute t = 10 into the model.
$V = 2000 \times 1.06^{10} = 2000 \times 1.7908 = 3581.7$
(b) Set V > 5000 and divide both sides by 2000 to isolate the power.
$2000 \times 1.06^{t} > 5000 \;\Rightarrow\; 1.06^{t} > 2.5$
The unknown is in the exponent, so take logs of both sides; the power law brings t down.
$t \log 1.06 > \log 2.5 \;\Rightarrow\; t > \frac{\log 2.5}{\log 1.06}$
Evaluate the quotient on the GDC.
$t > 15.725$
t must be a whole number of years and the total must exceed 5000, so round UP to the next year.
$t = 16$

Final answer:

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

Key concepts in Exponents and logarithms

Key skills for this topic

🔄 Log ↔ exponential

🔑 Solving aˣ = b

✏️ Worked examples

Simplify with index laws

Solve an exponential equation

Use log laws to simplify

💡 Test yourself — tap to reveal

🎯 IB-style practice — logarithms in context

Part (i) — find loudness from intensity

Part (ii) — find intensity from loudness

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

What you'll learn in Topic 1.5

Study resources — 1.5 Exponents and logarithms

Laws of Exponents

Introduction to Logarithms

Laws of Logarithms

Solving Exponential and Logarithmic Equations

Ready to study Exponents and logarithms?

Ready to practice?

Key concepts in Exponents and logarithms

Key skills for this topic

🔄 Log ↔ exponential

🔑 Solving aˣ = b

✏️ Worked examples

Simplify with index laws

Solve an exponential equation

Use log laws to simplify

💡 Test yourself — tap to reveal

🎯 IB-style practice — logarithms in context

Part (i) — find loudness from intensity

Part (ii) — find intensity from loudness

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

What you'll learn in Topic 1.5

Study resources — 1.5 Exponents and logarithms

Laws of Exponents

Introduction to Logarithms

Laws of Logarithms

Solving Exponential and Logarithmic Equations

Ready to study Exponents and logarithms?

Ready to practice?