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)