Week 14 — Quiz (auto-graded) · Logarithmic Functions

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objective tested: Objective 8 — logarithms as inverses; evaluating logs; common and natural logs; domain; properties.
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Module 14.

This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in F-quiz-week-14-qti.xml. AI is not permitted on quizzes (course AI policy). Every numeric answer below is pre-computed and independently re-verified (Python w14_verify.py, PASS — 42 checks).

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Blueprint

#	Type	Concept	Objective
1	Multiple choice	Evaluate log₂(8)	8
2	Multiple choice	Evaluate log₁₀(1000)	8
3	Multiple choice	Evaluate log₅(1)	8
4	Multiple choice	Evaluate ln(e)	8
5	Multiple choice	Rewrite 2³=8 in log form	8
6	Multiple choice	Rewrite log₃(9)=2 in exponential form	8
7	Multiple choice	Domain of f(x)=log(x−2)	8
8	Matching	Log property ↔ its form	8
9	Multiple choice	Expand log_b(xy)	8
10	Multiple choice	Evaluate log₂(1/4)	8

No trick questions; distractors target the Week 14 misconceptions (log(M+N) ≠ log M + log N; non-positive argument; confusing base and argument; wrong sign on negative-exponent logs).

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Questions, key, and feedback

Q1 (MC). Evaluate: log₂(8)
- A. 3 ✅
- B. 4
- C. 2
- D. 8

Feedback: log₂(8) asks "2 to what power gives 8?" Since 2³ = 8, the answer is 3. (B = wrong; 2⁴=16 ≠ 8. C = wrong; 2² = 4. D = the argument, not the log value.)

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Q2 (MC). Evaluate: log₁₀(1000)
- A. 3 ✅
- B. 4
- C. 2
- D. 10

Feedback: 10³ = 1000, so log₁₀(1000) = 3. (B = 10⁴=10,000 ≠ 1000. C = 10²=100 ≠ 1000. D = the base, not the log value.)

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Q3 (MC). Evaluate: log₅(1)
- A. 0 ✅
- B. 1
- C. 5
- D. Undefined

Feedback: Any base raised to the 0 power equals 1 (b⁰ = 1), so log_b(1) = 0 for any base b. (B = log_b(b), not log_b(1). D = 1 is positive so the log is defined.)

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Q4 (MC). Evaluate: ln(e)
- A. 1 ✅
- B. e
- C. 0
- D. 2.718

Feedback: ln(e) = logₑ(e). Since e¹ = e, ln(e) = 1. (B = the argument, not the log. C = ln(1)=0, not ln(e). D = e ≈ 2.718, not the log value.)

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Q5 (MC). Rewrite 2³ = 8 in logarithmic form.
- A. log₂(8) = 3 ✅
- B. log₈(2) = 3
- C. log₃(8) = 2
- D. log₂(3) = 8

Feedback: The base (2) stays the base of the log, the exponent (3) becomes the log value, and the result (8) becomes the argument: log₂(8) = 3. (B = base and argument switched. C = exponent confused with base. D = roles all scrambled.)

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Q6 (MC). Rewrite log₃(9) = 2 in exponential form.
- A. 3² = 9 ✅
- B. 9² = 3
- C. 2³ = 9
- D. 3⁹ = 2

Feedback: log_b(x) = y ↔ bʸ = x. Here base = 3, log value = 2, argument = 9: 3² = 9. (B = base and argument swapped. C = log value and base swapped. D = all roles mixed up.)

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Q7 (MC). What is the domain of f(x) = log(x − 2)?
- A. x > 2 ✅
- B. x < 2
- C. x > 0
- D. All real numbers

Feedback: The argument (x − 2) must be strictly > 0: x − 2 > 0 → x > 2. The vertical asymptote is at x = 2. (B = argument would be negative; no real log. C = that's the domain of log(x), not log(x−2). D = logs require positive arguments.)

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Q8 (Matching). Match each logarithm property to its rule.

Property	Rule
Product rule	log_b(MN) = log_b M + log_b N
Quotient rule	log_b(M/N) = log_b M − log_b N
Power rule	log_b(Mᵖ) = p · log_b M
Log of 1	log_b(1) = 0

Feedback: Product = multiplication inside → sum outside. Quotient = division inside → difference outside. Power = exponent inside → multiplier outside. Log of 1 = 0 because b⁰ = 1 for any base.

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Q9 (MC). Expand log_b(xy) using logarithm properties.
- A. log_b(x) + log_b(y) ✅
- B. log_b(x) · log_b(y)
- C. log_b(x) − log_b(y)
- D. log_b(x + y)

Feedback: Product rule: log_b(MN) = log_b M + log_b N. Multiplication inside the log becomes addition of separate logs. (B = logs are never multiplied by the product rule. C = that's the quotient rule. D = log(x+y) cannot be simplified; there's no log property for addition inside.)

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Q10 (MC). Evaluate: log₂(1/4)
- A. −2 ✅
- B. 2
- C. −4
- D. 1/4

Feedback: 1/4 = 2⁻², so log₂(1/4) = −2. (When the argument is a reciprocal of a power of the base, the log is negative.) (B = forgetting the negative sign; 2² = 4, not 1/4. C = confusing the argument's denominator with the log value. D = the argument, not the log.)

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Answer key (quick reference)

Q	Answer
1	A (3)
2	A (3)
3	A (0)
4	A (1)
5	A (log₂(8)=3)
6	A (3²=9)
7	A (x>2)
8	Product→log_b(MN)=log_b M+log_b N / Quotient→log_b(M/N)=log_b M−log_b N / Power→log_b(Mᵖ)=p·log_b M / Log of 1→log_b(1)=0
9	A (log_b(x)+log_b(y))
10	A (−2)

Quality gate (self-checked, computer-verified): each single-answer item has exactly one correct option; the matching item pairs all four properties 1:1. Arithmetic pre-computed and independently re-verified (w14_verify.py, PASS — 42 checks): Q1 2³=8→3; Q2 10³=1000→3; Q3 b⁰=1→0; Q4 e¹=e→1; Q5 2³=8 check; Q6 3²=9 check; Q7 x−2>0→x>2; Q9 product rule verified numerically; Q10 2⁻²=1/4→−2. All checks PASS. QTI parse confirmation: F-quiz-week-14-qti.xml parses as imsqti_xmlv1p2 with 10 items.

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Item-bank entries (for variants + the final)

All ten items are tagged course=MATH120 · week=14 · objective=8 · topic=logarithmic-functions and deposited in Item Bank: Week 14 — Logarithmic Functions. The final (Week 16) and per-term variant updates draw fresh items from this bank. (Tags: q1 evaluate-log, q2 evaluate-log-10, q3 log-of-1, q4 natural-log, q5 exp-to-log, q6 log-to-exp, q7 domain, q8 properties-matching, q9 expand-product, q10 negative-log.)

Canvas placement block

canvas_object   = Quizzes::Quiz
title           = "Week 14 Quiz — Logarithmic Functions"
assignment_group = "Quizzes"
points_possible = 10
grading_type    = points
due_offset_days = 6        # 6 days after module start (Sun Dec 6)
published       = true
shuffle_answers = true
provenance      = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"