Mathematics Advanced • Year 11 • Module 4 • Lesson 8

Change of Base & Logarithmic Equations

Build procedural fluency in change of base, solving log equations, and the all-important domain-check at the end.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the change of base formula:

logax = logb(____) / logb(____)    for any base b > 0, b ≠ 1

Most common choices for b: ____________ (natural log) or ____________ (common log).

Q1.2 When solving a log equation, what is the last step you must always perform?

________________________________________________________________________

Q1.3 Define "extraneous solution" in one sentence:

________________________________________________________________________

Stuck? Revisit lesson § Key Terms.

2. Worked example — solve log2(x + 1) + log2(x − 1) = 3

Watch every step — especially the domain check at the end.

Problem. Solve log2(x + 1) + log2(x − 1) = 3.

Step 1 — Write the domain first.

x + 1 > 0 and x − 1 > 0  ⇒  x > 1

Reason: every argument of a log must be strictly positive.

Step 2 — Combine using the product law.

log2[(x + 1)(x − 1)] = 3

Reason: logaM + logaN = loga(MN).

Step 3 — Convert to exponential form.

(x + 1)(x − 1) = 2³ = 8

Reason: log2P = 3 ⇒ P = 2³.

Step 4 — Solve the algebra and apply the domain.

x² − 1 = 8  ⇒  x² = 9  ⇒  x = ±3

Domain requires x > 1, so reject x = −3.

Reason: x = −3 makes log2(−2) undefined — extraneous.

Conclusion. The only valid solution is x = 3.

3. Faded example — fill in the missing steps

Solve log3(x − 1) + log3(x + 1) = 1. 4 marks

Step 1 — Domain:

x − 1 > 0 and x + 1 > 0 ⇒ x > ____________

Step 2 — Product law:

log3[ ____________________ ] = 1

Step 3 — Exponential form:

____________________ = 31 = ____________

Step 4 — Solve:

x² − 1 = ____________  ⇒  x² = ____________  ⇒  x = ±____________

Step 5 — Apply the domain:

Reject x = ____________ because ____________________________.

Conclusion. x = ____________.

Stuck? Revisit lesson § Worked Example 2.

4. Graduated practice — change of base, log equations, exponential equations

Show change-of-base lines and explicit domain checks where applicable.

Foundation — direct change of base (4 questions)

Give each answer to 3 decimal places using ln on your calculator.

QExpressionDecimal value (3 d.p.)
4.1 1log37
4.2 1log512
4.3 1log415
4.4 1log750

Standard — solve log/exponential equations (6 questions)

Always state the domain restriction and reject extraneous solutions.

4.5 Solve log3(x + 2) = 2.    2 marks

4.6 Solve log2x + log2(x − 2) = 3.    2 marks

4.7 Solve 2x = 7 using logarithms, giving the answer to 3 d.p.    2 marks

4.8 Solve 5x − 1 = 3x + 1, to 3 d.p.    2 marks

4.9 Solve ln(x − 2) + ln(x + 2) = ln 5.    2 marks

4.10 Solve log2(x + 6) − log2(x − 2) = 3.    2 marks

Extension — quadratic-in-disguise and double-base (2 questions)

4.11 Solve 22x − 5(2x) + 4 = 0. (Hint: let u = 2x.)    3 marks

4.12 Solve log4x = log2(x − 1). Hint: change one log to base 2 using log4x = ½ log2x.    3 marks

Stuck on 4.12? After the change of base, multiply through by 2, then exponentiate.

5. Self-check the easy 3

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

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Change of base formula

logax = logb(x) / logb(a) — argument on top, base on bottom.   Common choices: b = e (ln) or b = 10 (log).

Q1.2 — Last step in every log equation

Check each candidate solution against the original domain (every log argument must be strictly positive). Reject any solution that violates the domain — it is extraneous.

Q1.3 — Extraneous solution

A value obtained from the algebraic manipulation that does not satisfy the original equation — typically because it makes a log argument zero or negative.

Q3 — Faded equation solution

Step 1: x > 1.   Step 2: log3[(x − 1)(x + 1)] = 1.   Step 3: (x − 1)(x + 1) = 3, so x² − 1 = 3.   Step 4: x² = 4, x = ±2.   Step 5: reject x = −2 because it makes log3(−3) undefined.   x = 2.

Q4.1 — log37

= ln 7 / ln 3 ≈ 1.946 / 1.099 ≈ 1.771.

Q4.2 — log512

= ln 12 / ln 5 ≈ 2.485 / 1.609 ≈ 1.544.

Q4.3 — log415

= ln 15 / ln 4 ≈ 2.708 / 1.386 ≈ 1.953.

Q4.4 — log750

= ln 50 / ln 7 ≈ 3.912 / 1.946 ≈ 2.011.

Q4.5 — log3(x + 2) = 2

Domain: x > −2.   x + 2 = 3² = 9 ⇒ x = 7.   7 > −2 ✓.   x = 7.

Q4.6 — log2x + log2(x − 2) = 3

Domain: x > 2.   log2[x(x − 2)] = 3 ⇒ x(x − 2) = 8 ⇒ x² − 2x − 8 = 0 ⇒ (x − 4)(x + 2) = 0 ⇒ x = 4 or −2.   Reject x = −2 (violates x > 2).   x = 4.

Q4.7 — 2x = 7

ln(2x) = ln 7 ⇒ x ln 2 = ln 7 ⇒ x = ln 7 / ln 2 ≈ 2.807.

Q4.8 — 5x − 1 = 3x + 1

Take ln: (x − 1) ln 5 = (x + 1) ln 3 ⇒ x ln 5 − ln 5 = x ln 3 + ln 3 ⇒ x(ln 5 − ln 3) = ln 3 + ln 5 ⇒ x = (ln 3 + ln 5)/(ln 5 − ln 3) = ln 15 / ln(5/3) ≈ 2.708 / 0.511 ≈ 5.301.

Q4.9 — ln(x − 2) + ln(x + 2) = ln 5

Domain: x > 2.   ln[(x − 2)(x + 2)] = ln 5 ⇒ x² − 4 = 5 ⇒ x² = 9 ⇒ x = ±3.   Reject x = −3.   x = 3.

Q4.10 — log2(x + 6) − log2(x − 2) = 3

Domain: x > 2.   log2[(x + 6)/(x − 2)] = 3 ⇒ (x + 6)/(x − 2) = 8 ⇒ x + 6 = 8x − 16 ⇒ 22 = 7x ⇒ x = 22/7.   22/7 ≈ 3.14 > 2 ✓.   x = 22/7.

Q4.11 — 22x − 5(2x) + 4 = 0

Let u = 2x (so u > 0). Equation becomes u² − 5u + 4 = 0 ⇒ (u − 1)(u − 4) = 0 ⇒ u = 1 or u = 4.   Then 2x = 1 ⇒ x = 0;   2x = 4 ⇒ x = 2.   Both u-values positive, so both x-values valid.   x = 0 or x = 2.

Q4.12 — log4x = log2(x − 1)

Domain: x > 0 and x − 1 > 0 ⇒ x > 1.   log4x = (ln x)/(ln 4) = (ln x)/(2 ln 2) = ½ log2x.
Equation: ½ log2x = log2(x − 1) ⇒ log2x = 2 log2(x − 1) = log2(x − 1)².   So x = (x − 1)² = x² − 2x + 1 ⇒ x² − 3x + 1 = 0 ⇒ x = (3 ± √5)/2.
Numerical: (3 + √5)/2 ≈ 2.618 (valid, > 1); (3 − √5)/2 ≈ 0.382 (rejected, < 1).   x = (3 + √5)/2 ≈ 2.618.

HSCScience • Mathematics Advanced • Year 11 • Module 4 • Lesson 8 • Worksheet 1 (Build)