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UNIT 3
Exponential & Logarithmic Functions
LESSON 13
Exponential & Logarithmic Functions Unit Test
432000 álgebra 2 b
20 questions
11th grade
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Unit 3 — Exponential & Logarithmic Functions
Lesson 13 — Unit Test (Algebra 2, 11th grade)
20 questions. Show all work. Calculators allowed where noted.
1. Evaluate: log_2 32
2. Solve: 3^x = 81
3. Solve: ln x = 3
4. Solve: log_5(x − 2) = 2
5. Write in logarithmic form: 2^(−3) = 1/8
6. Solve: 5^(2x − 1) = 125
7. Express log_7 50 using natural logarithms (ln).
8. For f(x) = log_2(x + 3) + 1, state the domain and range.
9. For g(x) = 2^(x − 2) + 3, state the horizontal asymptote and the range.
10. Financial: $1000 is invested at 5% annual interest, compounded annually. What is the account balance after 10 years? (Give a decimal rounded to two decimal places.)
11. Financial: $5000 is invested at 4% annual interest, compounded continuously. What is the balance after 5 years? (Give a decimal rounded to two decimal places.)
12. Half-life: A 80 g sample has half-life 3 years. How much remains after 7 years? (Give decimal rounded to two decimal places.)
13. Solve: log(x) + log(x − 3) = 1 (common log, base 10). State any extraneous roots.
14. Solve: e^(2x) = 7
15. Solve: ln(5x + 1) = 2
16. Find the inverse function: f(x) = 3^x
17. Evaluate: log(0.001) (common log)
18. Describe the domain and the vertical asymptote of h(x) = log_3(−x) + 2. Also describe the transformation from y = log_3 x.
19. Solve: 2 · log_3(x) = 4
20. Application: A bacterial culture has 500 bacteria and doubles every 6 hours. How many bacteria will there be after 48 hours?
Answer key (brief solutions)
1. log_2 32 = 5 (because 2^5 = 32)
2. 3^x = 81 → x = 4 (81 = 3^4)
3. ln x = 3 → x = e^3
4. log_5(x − 2) = 2 → x − 2 = 25 → x = 27
5. 2^(−3) = 1/8 converts to log_2(1/8) = −3
6. 5^(2x − 1) = 125 → 125 = 5^3 → 2x − 1 = 3 → 2x = 4 → x = 2
7. log_7 50 = ln 50 / ln 7
8. f(x) = log_2(x + 3) + 1
- Domain: x + 3 > 0 → x > −3
- Range: all real numbers (ℝ)
9. g(x) = 2^(x − 2) + 3
- Horizontal asymptote: y = 3
- Range: y > 3
10. A = 1000(1.05)^10 ≈ 1000 · 1.6288946 ≈ $1,628.89
11. A = 5000 e^(0.04·5) = 5000 e^0.2 ≈ 5000 · 1.2214028 ≈ $6,107.01
12. A = 80(1/2)^(7/3) ≈ 80 · 2^(−7/3) ≈ 80 · 0.19828 ≈ 15.86 g (≈ 15.9 g)
13. log(x) + log(x − 3) = 1
→ log[x(x − 3)] = 1 → x(x − 3) = 10
→ x^2 − 3x − 10 = 0 → (x − 5)(x + 2) = 0
→ x = 5 or x = −2. Reject x = −2 because arguments of logs must be positive (x and x−3). Final: x = 5.
14. e^(2x) = 7 → 2x = ln 7 → x = (ln 7)/2
15. ln(5x + 1) = 2 → 5x + 1 = e^2 → x = (e^2 − 1)/5
16. f(x) = 3^x → f^(−1)(x) = log_3 x
17. log(0.001) = log(10^(−3)) = −3
18. h(x) = log_3(−x) + 2
- Domain: −x > 0 → x < 0
- Vertical asymptote: x = 0
- Transformation from y = log_3 x: reflect across the y-axis (x → −x), then shift up 2 units.
19. 2·log_3(x) = 4 → log_3(x) = 2 → x = 3^2 = 9
20. Doubling every 6 hours. Number after 48 hours: n = 500 · 2^(48/6) = 500 · 2^8 = 500 · 256 = 128,000
If you want the test formatted for printing (with answer lines or in Spanish), tell me and I’ll prepare that version.
Lesson 13 — Unit Test (Algebra 2, 11th grade)
20 questions. Show all work. Calculators allowed where noted.
1. Evaluate: log_2 32
2. Solve: 3^x = 81
3. Solve: ln x = 3
4. Solve: log_5(x − 2) = 2
5. Write in logarithmic form: 2^(−3) = 1/8
6. Solve: 5^(2x − 1) = 125
7. Express log_7 50 using natural logarithms (ln).
8. For f(x) = log_2(x + 3) + 1, state the domain and range.
9. For g(x) = 2^(x − 2) + 3, state the horizontal asymptote and the range.
10. Financial: $1000 is invested at 5% annual interest, compounded annually. What is the account balance after 10 years? (Give a decimal rounded to two decimal places.)
11. Financial: $5000 is invested at 4% annual interest, compounded continuously. What is the balance after 5 years? (Give a decimal rounded to two decimal places.)
12. Half-life: A 80 g sample has half-life 3 years. How much remains after 7 years? (Give decimal rounded to two decimal places.)
13. Solve: log(x) + log(x − 3) = 1 (common log, base 10). State any extraneous roots.
14. Solve: e^(2x) = 7
15. Solve: ln(5x + 1) = 2
16. Find the inverse function: f(x) = 3^x
17. Evaluate: log(0.001) (common log)
18. Describe the domain and the vertical asymptote of h(x) = log_3(−x) + 2. Also describe the transformation from y = log_3 x.
19. Solve: 2 · log_3(x) = 4
20. Application: A bacterial culture has 500 bacteria and doubles every 6 hours. How many bacteria will there be after 48 hours?
Answer key (brief solutions)
1. log_2 32 = 5 (because 2^5 = 32)
2. 3^x = 81 → x = 4 (81 = 3^4)
3. ln x = 3 → x = e^3
4. log_5(x − 2) = 2 → x − 2 = 25 → x = 27
5. 2^(−3) = 1/8 converts to log_2(1/8) = −3
6. 5^(2x − 1) = 125 → 125 = 5^3 → 2x − 1 = 3 → 2x = 4 → x = 2
7. log_7 50 = ln 50 / ln 7
8. f(x) = log_2(x + 3) + 1
- Domain: x + 3 > 0 → x > −3
- Range: all real numbers (ℝ)
9. g(x) = 2^(x − 2) + 3
- Horizontal asymptote: y = 3
- Range: y > 3
10. A = 1000(1.05)^10 ≈ 1000 · 1.6288946 ≈ $1,628.89
11. A = 5000 e^(0.04·5) = 5000 e^0.2 ≈ 5000 · 1.2214028 ≈ $6,107.01
12. A = 80(1/2)^(7/3) ≈ 80 · 2^(−7/3) ≈ 80 · 0.19828 ≈ 15.86 g (≈ 15.9 g)
13. log(x) + log(x − 3) = 1
→ log[x(x − 3)] = 1 → x(x − 3) = 10
→ x^2 − 3x − 10 = 0 → (x − 5)(x + 2) = 0
→ x = 5 or x = −2. Reject x = −2 because arguments of logs must be positive (x and x−3). Final: x = 5.
14. e^(2x) = 7 → 2x = ln 7 → x = (ln 7)/2
15. ln(5x + 1) = 2 → 5x + 1 = e^2 → x = (e^2 − 1)/5
16. f(x) = 3^x → f^(−1)(x) = log_3 x
17. log(0.001) = log(10^(−3)) = −3
18. h(x) = log_3(−x) + 2
- Domain: −x > 0 → x < 0
- Vertical asymptote: x = 0
- Transformation from y = log_3 x: reflect across the y-axis (x → −x), then shift up 2 units.
19. 2·log_3(x) = 4 → log_3(x) = 2 → x = 3^2 = 9
20. Doubling every 6 hours. Number after 48 hours: n = 500 · 2^(48/6) = 500 · 2^8 = 500 · 256 = 128,000
If you want the test formatted for printing (with answer lines or in Spanish), tell me and I’ll prepare that version.
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