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Week 15 · Quiz

Week 15 — Quiz (auto-graded) · Exponential & Logarithmic Equations & Applications

College Algebra · MATH 120 Fall 2026 · Prof. Calloway Fictional sample

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objective tested: Objective 8 — solving exponential and logarithmic equations; growth, decay, and compound-interest applications.
Points: 10 (1 each) · Assignment group: Quizzes (15% of grade) · Due: end of Module 15.

This is the human-readable quiz with its vetted answer key and feedback. The import-ready Classic QTI is in F-quiz-week-15-qti.xml. AI is not permitted on quizzes (course AI policy). Every numeric answer below is pre-computed and independently re-verified (Python w15_verify.py, PASS — 45 checks clean).


Blueprint

# Type Concept Objective
1 Multiple choice Solve 2ˣ = 16 (same-base method) 8
2 Multiple choice Solve 3^(x+1) = 27 (same-base method) 8
3 Multiple choice Solve 5ˣ = 125 (same-base method) 8
4 Multiple choice Solve 2ˣ = 10 (log of both sides; exact or decimal) 8
5 Multiple choice Solve log₂(x) = 5 (convert to exponential) 8
6 Multiple choice Solve log(x) = 2 (convert to exponential, base 10) 8
7 Multiple choice Solve ln(x) = 0 8
8 Multiple choice Solve log₂(x − 1) = 3 (with domain check) 8
9 Multiple choice Solve log(x) + log(x − 3) = 1 (extraneous solution) 8
10 Multiple choice Doubling time at 6% continuous compounding 8

No trick questions; distractors target the Week 15 misconceptions named in the lecture outline (treating exponents as multipliers; wrong log-of-both-sides step; accepting an extraneous log solution; application setup).


Questions, key, and feedback

Q1 (MC). Solve: 2ˣ = 16
- A. x = 8
- B. x = 4
- C. x = 14
- D. x = 32
Feedback: 16 = 2⁴, so 2ˣ = 2⁴ → x = 4 (equate exponents, same-base method). (A = 2·8, confusing base × exponent; C = 16 − 2; D = 2 × 16.)

Q2 (MC). Solve: 3^(x+1) = 27
- A. x = 1
- B. x = 2
- C. x = 8
- D. x = 3
Feedback: 27 = 3³, so 3^(x+1) = 3³ → x + 1 = 3 → x = 2. (A = solving 3·(x+1) = 27 incorrectly; C = treating exponent additively; D = confusing x with the exponent 3.)

Q3 (MC). Solve: 5ˣ = 125
- A. x = 25
- B. x = 120
- C. x = 3
- D. x = 625
Feedback: 125 = 5³, so 5ˣ = 5³ → x = 3. (A = 5²; B = 125 − 5; D = 5⁴.)

Q4 (MC). Solve: 2ˣ = 10 (give the exact expression or the best decimal approximation)
- A. x = 5
- B. x = 20
- C. x = log₂(10) ≈ 3.32
- D. x = log(10) = 1
Feedback: Take ln of both sides: x·ln 2 = ln 10 → x = ln(10)/ln(2) = log₂(10) ≈ 3.322. (A = 10/2; B = 10·2; D = log₁₀(10) = 1, ignoring the base-2 structure.)

Q5 (MC). Solve: log₂(x) = 5
- A. x = 10
- B. x = 32
- C. x = 5/2
- D. x = 25
Feedback: Convert: 2⁵ = x → x = 32. (A = base 10 confusion; C = 5÷2; D = 5².)

Q6 (MC). Solve: log x = 2 (log = base 10)
- A. x = 20
- B. x = 2/10
- C. x = 100
- D. x = 10
Feedback: Convert: 10² = x → x = 100. (A = 10·2; B = 2÷10; D = 10¹, confusing 1 for the given 2.)

Q7 (MC). Solve: ln x = 0
- A. x = 1
- B. x = 0
- C. x = e
- D. x = −1
Feedback: ln x = 0 means e⁰ = x → x = 1. (B = confusing ln(0); C = ln(e) = 1, not 0; D = undefined, ln of a negative.)

Q8 (MC). Solve: log₂(x − 1) = 3
- A. x = 7
- B. x = 9
- C. x = 4
- D. x = 2
Feedback: Convert: x − 1 = 2³ = 8 → x = 9. Domain check: x − 1 = 8 > 0. ✓ (A = 2³ − 1 = 7, but then 7+1 = 8 ≠ right setup; C = 2² + 1; D = 2¹ + 1.)

Q9 (MC). Solve: log x + log(x − 3) = 1 — select the valid solution only.
- A. x = −2 only
- B. x = 5 only
- C. Both x = 5 and x = −2
- D. x = 8
Feedback: Condense: log[x(x−3)] = 1 → x(x−3) = 10 → x² − 3x − 10 = 0 → (x−5)(x+2) = 0. x = 5: log(5) and log(2) both defined ✓. x = −2: log(−2) undefined — extraneous. Answer: x = 5 only. (A = accepting the extraneous root; C = forgetting the domain check; D = arithmetic error.)

Q10 (MC). Money is invested at 6% interest compounded continuously. About how many years does it take to double?
- A. About 6 years
- B. About 8 years
- C. About 11.6 years
- D. About 17 years
Feedback: A = Pe^(rt); doubling → 2 = e^(0.06t) → t = ln(2)/0.06 ≈ 11.55 years ≈ 11.6 yr. (A = 1/rate; B = 72/9 (wrong rate); D = ln(2)/0.04.)


Answer key (quick reference)

Q Answer
1 B (x = 4)
2 B (x = 2)
3 C (x = 3)
4 C (log₂(10) ≈ 3.32)
5 B (x = 32)
6 C (x = 100)
7 A (x = 1)
8 B (x = 9)
9 B (x = 5 only)
10 C (≈ 11.6 years)

Quality gate (self-checked, computer-verified): each single-answer item has exactly one correct option. Arithmetic pre-computed and independently re-verified (w15_verify.py, PASS — 45 checks): Q1 2⁴=16; Q2 3³=27, x=2; Q3 5³=125; Q4 ln10/ln2≈3.322; Q5 2⁵=32; Q6 10²=100; Q7 e⁰=1; Q8 2³=8, x=9; Q9 x=5 valid (domain check), x=−2 extraneous; Q10 ln2/0.06≈11.55. All checks PASS. QTI parse confirmation: F-quiz-week-15-qti.xml generated and confirmed "parses OK, 10 items."


Item-bank entries (for variants + the final)

All ten items are tagged course=MATH120 · week=15 · objective=8 · topic=exp-log-equations-applications and deposited in Item Bank: Week 15 — Exponential & Logarithmic Equations & Applications. The final (Week 16) draws Objective 8 items from this bank. (Tags: q1 same-base, q2 same-base-linear-exp, q3 same-base, q4 log-both-sides, q5 log-to-exp, q6 log-to-exp-base10, q7 ln-to-exp, q8 log-to-exp-domain, q9 extraneous-log, q10 doubling-time.)

Canvas placement block

canvas_object   = Quizzes::Quiz
title           = "Week 15 Quiz — Exponential & Logarithmic Equations & Applications"
assignment_group = "Quizzes"
points_possible = 10
grading_type    = points
due_offset_days = 6        # 6 days after module start (Sun Dec 13)
published       = true
shuffle_answers = true
provenance      = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"
This is the human-readable quiz with its vetted answer key and rationale. The import-ready Classic-QTI version (F-quiz-week-15-qti.xml) ships inside the course's .imscc package — it lands in the Canvas gradebook on import.

~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com