Mathematics Advanced • Year 11 • Module 4 • Lesson 5

Introduction to Logarithmic Functions

Build fluency converting between log and exponential forms, evaluating logs, and finding domains.

Build · Skill Drill

1. Quick recall

Answer each question. 1 mark each

Q1.1 Complete the definition:

log_a x = y ⇔ ________ = ________ (with a > 0, a ≠ 1, x > 0).

Q1.2 Two special identities (for any valid base a):

log_a 1 = ____ . log_a a = ____.

Q1.3 Naming conventions:

"log x" (no base) usually means log_{____} x. "ln x" means log_{____} x.

Stuck? Revisit lesson § Concept and § Key Terms.

2. Worked example — evaluate log₂ 8 and log₃ (1/9)

Problem. Evaluate log₂ 8 and log₃ (1/9) using the definition log_a x = y ⇔ a^y = x.

Step 1 — Express the argument as a power of the base.

8 = 2 × 2 × 2 = 2³

Step 2 — Read off the exponent.

log₂ 8 = log₂ 2³ = 3

Reason: log_a aⁿ = n.

Step 3 — Repeat for log₃ (1/9). Express 1/9 as a power of 3.

1/9 = 1/3² = 3⁻²

Step 4 — Read off the exponent.

log₃ (1/9) = log₃ 3⁻² = −2

3. Faded example — solve log₅ x = 3

Fill in the missing pieces. 4 marks

Step 1 — Convert log form to exponential form.

log₅ x = 3 ⇒ ____^____ = x

Step 2 — Evaluate.

x = ________

Step 3 — Check the domain.

The argument of the log must be ____________ . My x = ____ satisfies this. ✓

Conclusion. x = ________ .

Stuck? Revisit lesson § Worked Example 2.

4. Graduated practice

Foundation — evaluate by inspection (4 questions)

Q	Compute	Answer (exact)
4.1 1	log₂ 16
4.2 1	log₁₀ 1000
4.3 1	log₃ (1/27)
4.4 1	ln (e⁵)

Standard — convert, solve, find domain (6 questions)

4.5 Solve log₄ x = 2. 2 marks

4.6 Solve log₂ (x + 1) = 4. 2 marks

4.7 Find the domain of f(x) = log₂ (x − 3). 2 marks

4.8 Find the domain of f(x) = ln(2x + 1). 2 marks

4.9 Simplify log₅ 5⁷ and e^{ln 4}. 2 marks

4.10 Rewrite 2⁵ = 32 in logarithmic form, and 10⁻² = 0.01 in logarithmic form. 2 marks

Extension — reasoning about domain and inverses (2 questions)

4.11 Explain in one or two sentences why log_a 0 and log_a (−5) are undefined for a > 0. 3 marks

4.12 Verify, with one line each, that f(x) = 2^x and g(x) = log₂ x are inverses by computing f(g(x)) and g(f(x)). 3 marks

Stuck on 4.11? Recall that a^y is positive for any real y when a > 0.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (log₂ 16) I wrote 16 = 2⁴, so the answer is 4 (not 8 or 16).

For 4.3 (log₃ 1/27) I wrote 1/27 = 3⁻³, so the answer is −3 (negative).

For 4.4 (ln e⁵) I used the identity ln e^x = x to write 5 — I did not leave "ln e⁵" in the answer.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Definition

log_a x = y ⇔ a^y = x.

Q1.2 — Two identities

log_a 1 = 0 (because a⁰ = 1). log_a a = 1 (because a¹ = a).

Q1.3 — Naming conventions

"log x" usually means log₁₀ x (common log). "ln x" means log_e x (natural log).

Q3 — Faded log₅ x = 3

Step 1: 5³ = x.
Step 2: x = 125.
Step 3: Argument must be positive. x = 125 > 0 ✓.
Conclusion: x = 125.

Q4.1–4.4 — Inspection

4.1: 16 = 2⁴ → log₂ 16 = 4. 4.2: 1000 = 10³ → 3. 4.3: 1/27 = 3⁻³ → −3. 4.4: ln(e⁵) = 5.

Q4.5 — log₄ x = 2

x = 4² = 16.

Q4.6 — log₂ (x + 1) = 4

x + 1 = 2⁴ = 16 ⇒ x = 15. Check: x + 1 = 16 > 0 ✓.

Q4.7 — Domain of log₂ (x − 3)

Require x − 3 > 0 ⇒ x > 3, i.e. (3, ∞).

Q4.8 — Domain of ln(2x + 1)

Require 2x + 1 > 0 ⇒ x > −1/2, i.e. (−1/2, ∞).

Q4.9 — Simplify

log₅ 5⁷ = 7 (identity log_a aⁿ = n). e^{ln 4} = 4 (identity a^{log_a x} = x).

Q4.10 — Convert to log form

2⁵ = 32 ↔ log₂ 32 = 5. 10⁻² = 0.01 ↔ log₁₀ 0.01 = −2.

Q4.11 — Why log_a 0 and log_a (−5) are undefined

By the definition log_a x = y means a^y = x. But a^y > 0 for every real y when a > 0 — exponentials of a positive base are always positive. So there is no real y satisfying a^y = 0 (the closest we get is a^y → 0 as y → −∞, but it is never reached) and no real y with a^y = −5 (a positive cannot equal a negative). Hence log_a 0 and log_a (−5) are undefined.

Q4.12 — Verify inverses

f(g(x)) = f(log₂ x) = 2^{log₂ x} = x (using a^{log_a x} = x). g(f(x)) = g(2^x) = log₂ 2^x = x (using log_a a^x = x). Both compositions equal x, so f and g are inverses.