Rucete ✏ AP Precalculus In a Nutshell
7. Logarithmic Functions — Practice Questions 3
This chapter covers exponential–logarithmic relationships, inverse functions, graph transformations, equations and inequalities, modeling (half-life, pH, semi-log lines), and parameter interpretation.
(Multiple Choice — Click to Reveal Answer)
1. Evaluate log_2(1/32).
(A) −6
(B) −5
(C) −4
(D) −3
Answer
(B) — 1/32 = 2^(−5) ⇒ log_2(1/32) = −5.
2. If 3^x = 81, then x equals
(A) 2
(B) 3
(C) 4
(D) 5
Answer
(C) — 81 = 3^4 ⇒ x = 4.
3. Change of base: log_3(45) equals
(A) ln 3 / ln 45
(B) ln 45 / ln 3
(C) 45 / 3
(D) 3 ln 45
Answer
(B) — log_b a = (ln a)/(ln b).
4. The domain of y = ln(7 − 2x) is
(A) (−∞, 7/2)
(B) (−∞, 3.5)
(C) (−∞, 7) ∪ (7, ∞)
(D) (0, ∞)
Answer
(B) — 7 − 2x > 0 ⇒ x < 3.5.
5. The range of y = e^x is
(A) (−∞, ∞)
(B) (0, ∞)
(C) [0, ∞)
(D) (−1, 1)
Answer
(B) — Exponential outputs are positive.
6. The vertical asymptote of y = log_4(x + 2) is
(A) x = −2
(B) x = 0
(C) y = −2
(D) y = 0
Answer
(A) — Set argument zero: x + 2 = 0 ⇒ x = −2.
7. Solve for x: log_5(x) = −2.
(A) 1/25
(B) 1/5
(C) 5
(D) 25
Answer
(A) — x = 5^(−2) = 1/25.
8. Solve for x: 2^x = 10.
(A) x = ln 10
(B) x = ln 10 / ln 2
(C) x = ln 2 / ln 10
(D) x = 10 ln 2
Answer
(B) — x = log_2(10) = ln 10 / ln 2.
9. If 4 = log_a(64) and a > 0, a ≠ 1, then a equals
(A) 2√2
(B) 4
(C) 8
(D) 16
Answer
(A) — a^4 = 64 ⇒ a = 64^(1/4) = 2^(6/4) = 2^(3/2) = 2√2.
10. Simplify: log(50) − log(2).
(A) log(25)
(B) log(100)
(C) log(48)
(D) 25
Answer
(A) — log(50/2) = log(25).
11. Evaluate ln(e^(−2) / √e).
(A) −2.5
(B) −1.5
(C) −2
(D) −0.5
Answer
(A) — e^(−2) / e^(1/2) = e^(−2.5); ln = −2.5.
12. Solve for x: log_3(9x) = 5.
(A) 9
(B) 18
(C) 27
(D) 45
Answer
(C) — 9x = 3^5 = 243 ⇒ x = 27.
13. The inverse of y = ln(x − 1) is
(A) y = e^x + 1
(B) y = e^x − 1
(C) y = 1 + e^x
(D) y = 1 − e^x
Answer
(C) — x = ln(y − 1) ⇒ y − 1 = e^x ⇒ y = 1 + e^x.
14. Reflecting y = log_10 x across the x-axis gives
(A) y = log_10(−x)
(B) y = −log_10 x
(C) y = log_(1/10) x
(D) y = −log_(1/10) x
Answer
(B) — Multiply outputs by −1.
15. If log_b(16) = 4, then b equals
(A) 2
(B) 4
(C) 8
(D) 16
Answer
(A) — b^4 = 16 ⇒ b = 2.
16. Which point lies on y = 10^x?
(A) (−1, 10)
(B) (0, 1)
(C) (1, 0)
(D) (1, 10)
Answer
(B) — 10^0 = 1.
17. Solve for x: ln x = 3.
(A) e^3
(B) 3e
(C) 3
(D) 1/3
Answer
(A) — x = e^3.
18. Simplify: log_a(a^(−3/2)).
(A) −3/2
(B) −3
(C) 3/2
(D) 1/(a^(3/2))
Answer
(A) — log_a(a^k) = k.
19. Solve the inequality: log_2(x − 3) ≥ 1.
(A) x ≥ 3
(B) x ≥ 4
(C) x ≥ 5
(D) x > 1
Answer
(C) — x − 3 ≥ 2 ⇒ x ≥ 5 (and x > 3 for domain).
20. The vertical asymptote of y = ln(2x + 1) is
(A) x = −1/2
(B) x = 0
(C) y = −1/2
(D) y = 0
Answer
(A) — 2x + 1 = 0 ⇒ x = −1/2.
21. If f(x) = log_3 x and g(x) = 3^x, then g(f(9)) equals
(A) 2
(B) 3
(C) 6
(D) 9
Answer
(D) — f(9) = 2 ⇒ g(2) = 9.
22. Evaluate log_8(2).
(A) 1/2
(B) 1/3
(C) 2/3
(D) 3/2
Answer
(B) — 8 = 2^3 ⇒ log_8 2 = 1/3.
23. Solve for x: 5^(x − 1) = 1/25.
(A) −3
(B) −2
(C) −1
(D) 0
Answer
(B) — 1/25 = 5^(−2) ⇒ x − 1 = −2 ⇒ x = −1.
24. If log_b(1/√b) = k, then k equals
(A) −1/2
(B) 1/2
(C) −1
(D) 1
Answer
(A) — log_b(b^(−1/2)) = −1/2.
25. For y = a·b^x, the line on (x, log_10 y) has y-intercept −1. Then a equals
(A) 10
(B) 1
(C) 0.1
(D) −1
Answer
(C) — log_10 y = x log_10 b + log_10 a; intercept = log_10 a = −1 ⇒ a = 10^(−1) = 0.1.
26. Solve for x: log_3(x − 1) + log_3(x + 1) = 2 (x > 1).
(A) √8
(B) √9
(C) √10
(D) √12
Answer
(C) — log_3((x − 1)(x + 1)) = 2 ⇒ x^2 − 1 = 9 ⇒ x = √10 (domain x > 1).
27. Solve for x: 2^x + 2^(x − 1) = 48.
(A) 4
(B) 5
(C) 6
(D) 7
Answer
(B) — 2^(x − 1)(1 + 2) = 3·2^(x − 1) = 48 ⇒ 2^(x − 1) = 16 ⇒ x = 5.
28. The x-intercept of y = ln(5x − 20) is
(A) 3
(B) 4
(C) 21/5
(D) 25/5
Answer
(C) — Set y = 0 ⇒ 5x − 20 = 1 ⇒ x = 21/5.
29. If log_b(27) = 3 and log_b(2) = p, then log_b(216) equals
(A) 3 + p
(B) 3 + 2p
(C) 3 + 3p
(D) 6p
Answer
(C) — 216 = 27·8 = 3^3·2^3 ⇒ log = 3 + 3p.
30. Solve for x: ln x − ln(x − 4) = ln 3 (x > 4).
(A) 4
(B) 5
(C) 6
(D) 7
Answer
(C) — ln(x/(x − 4)) = ln 3 ⇒ x/(x − 4) = 3 ⇒ x = 6.
31. On the (x, log_10 y) plane, the line has slope 0.3 and intercept 2. Find y when x = 1.
(A) 100
(B) 200
(C) 300
(D) 1000
Answer
(B) — log_10 y = 0.3·1 + 2 = 2.3 ⇒ y = 10^2.3 = 200.
32. Solve for x: 3^(2x − 1) = 27·√3.
(A) 7/4
(B) 9/4
(C) 5/2
(D) 11/4
Answer
(B) — 27·√3 = 3^3·3^(1/2) = 3^(7/2) ⇒ 2x − 1 = 7/2 ⇒ x = 9/4.
33. The function y = a + b ln x passes (1, 4) and (e, 6). Then b equals
(A) 1
(B) 2
(C) 3
(D) 4
Answer
(B) — ln 1 = 0 ⇒ a = 4; at x = e: 4 + b·1 = 6 ⇒ b = 2.
34. Solve: log_(1/2)(x) > −3 (x > 0).
(A) x < 8
(B) x > 8
(C) x = 8
(D) 0 < x < 1/8
Answer
(A) — Base < 1 ⇒ inequality reverses: x < (1/2)^(−3) = 8.
35. The inverse of y = 2 − log_5(x + 4) is
(A) y = 5^(2 − x) + 4
(B) y = 5^(x − 2) − 4
(C) y = 5^(2 − x) − 4
(D) y = 5^(x + 2) − 4
Answer
(C) — x = 2 − log_5(y + 4) ⇒ log_5(y + 4) = 2 − x ⇒ y = 5^(2 − x) − 4.
36. Solve for x: 4^x = 7.
Answer
x = ln 7 / ln 4 — Take natural logs and divide.
37. Solve for x: log_3(2x + 1) = log_3(5x − 11).
Answer
x = 4 — Equal bases ⇒ 2x + 1 = 5x − 11 ⇒ x = 4 (domain satisfied).
38. Write as a single logarithm: log a + 2 log b − log √c.
Answer
log( a · b^2 / c^{1/2} ) — Product, power, and quotient rules.
39. Find the domain and vertical asymptotes of y = ln(x^2 − 6x + 8).
Answer
Domain: (−∞, 2) ∪ (4, ∞); asymptotes: x = 2 and x = 4 — Factor to (x − 2)(x − 4) > 0.
40. A substance has half-life 12 hours: N(t) = N_0 (1/2)^{t/12}. When is 25% remaining?
Answer
t = 24 hours — 0.25 = (1/2)^2 ⇒ t/12 = 2 ⇒ t = 24.
41. Solve for x: ln(3x − 2) = 2.
Answer
x = (e^2 + 2)/3 — 3x − 2 = e^2.
42. Find the inverse of y = 3 ln(2x + 1) − 5.
Answer
y = ( e^{(x + 5)/3} − 1 ) / 2 — Swap, isolate ln, exponentiate, solve for y.
43. Solve the inequality: log_5(x − 1) ≤ 2.
Answer
x ≤ 26, with x > 1 — x − 1 ≤ 25 ⇒ x ≤ 26 and domain x > 1.
44. Find the x-intercept of y = 10^x − 7.
Answer
x = log_10(7) — Set y = 0 ⇒ 10^x = 7.
45. Simplify: ln x − (1/2) ln(x − 1) + ln 3.
Answer
ln( 3x / √(x − 1) ) — Combine using log rules (x > 1).
46. Solve for x: 5·2^x = 3^x.
Answer
x = ln(1/5) / ln(2/3) — (2/3)^x = 1/5 ⇒ x = ln(1/5)/ln(2/3).
47. Model identification: y = a·b^x passes through (0, 12) and (2, 48). Find a and b.
Answer
a = 12, b = 2 — a = y(0) = 12; 12·b^2 = 48 ⇒ b = 2.
48. Solve for x > 0, x ≠ 1: log_x(16) = 4.
Answer
x = 2 — x^4 = 16 ⇒ x = 2 (positive base ≠ 1).
49. Given p = log_2 x and q = log_2 y, compute log_2( x^3 / (4y) ).
Answer
3p − q − 2 — log_2 x^3 − log_2 4 − log_2 y = 3p − 2 − q.
50. A semi-log line is given by log_10 y = 1.2x − 0.5. Find y when x = 2.
Answer
y = 10^{1.9} ≈ 79.43 — Substitute x = 2 ⇒ log_10 y = 1.9.