Rucete ✏ AP Precalculus In a Nutshell
7. Logarithmic Functions — Practice Questions 2
This chapter introduces core properties of logarithms, inverse/exponential relationships, graph transformations, equations/inequalities, and real-world modeling including semi-log plots and logarithmic regression.
(Multiple Choice — Click to Reveal Answer)
1. Evaluate log(0.001).
(A) −4
(B) −3
(C) −2
(D) 3
Answer
(B) — 0.001 = 10^(−3), so log(0.001) = −3.
2. If log_b(81) = 4, then b equals
(A) 3
(B) 4
(C) 9
(D) 27
Answer
(A) — b^4 = 81 ⇒ b = 3 (b > 0, b ≠ 1).
3. Use the product rule to simplify: log(20) + log(5).
(A) log(25)
(B) log(100)
(C) log(4)
(D) 2·log(25)
Answer
(B) — log(20) + log(5) = log(100).
4. Evaluate ln(e^7).
(A) 7e
(B) e^7
(C) 7
(D) 1/7
Answer
(C) — ln(e^k) = k.
5. The domain of y = log_5(x − 2) is
(A) (−∞, ∞)
(B) (0, ∞)
(C) (2, ∞)
(D) (−2, ∞)
Answer
(C) — x − 2 > 0 ⇒ x > 2.
6. Convert to exponential form: log_2(64) = 6.
(A) 2^6 = 64
(B) 6^2 = 64
(C) 64^2 = 6
(D) 2^64 = 6
Answer
(A) — log_b(c) = a ⇔ b^a = c.
7. Evaluate log_4(1/8).
(A) −3/2
(B) −2
(C) −1/2
(D) 2/3
Answer
(A) — log_4(1/8) = (log_2(1/8))/(log_2 4) = (−3)/2.
8. Use the quotient rule: log(75) − log(3) =
(A) log(25)
(B) log(72)
(C) log(78)
(D) 25
Answer
(A) — log(75/3) = log(25).
9. If f(x) = log_2 x, which point lies on its graph?
(A) (0, 1)
(B) (1, 0)
(C) (2, 0)
(D) (0, 0)
Answer
(B) — log_2(1) = 0.
10. The vertical asymptote of y = log_b x (b > 0, b ≠ 1) is
(A) y = 0
(B) x = 0
(C) x = 1
(D) y = 1
Answer
(B) — The y-axis is the vertical asymptote.
11. Evaluate log(10^−2).
(A) −2
(B) 2
(C) −1/2
(D) 1/2
Answer
(A) — log_10(10^−2) = −2.
12. Change of base: log_7(35) equals
(A) ln 35 / ln 7
(B) ln 7 / ln 35
(C) 7 ln 35
(D) ln(35/7)
Answer
(A) — log_b a = (ln a)/(ln b).
13. If g(x) = 3·log_5 x, this is a
(A) vertical stretch of log_5 x by factor 3
(B) horizontal stretch by factor 3
(C) shift up 3
(D) shift right 3
Answer
(A) — Coefficient scales vertically.
14. Inverses: the inverse of y = 2^x is
(A) y = log_2 x
(B) y = ln x
(C) y = e^x
(D) y = 10^x
Answer
(A) — Bases match.
15. Evaluate ln(1) + ln(e).
(A) 0
(B) 1
(C) e
(D) −1
Answer
(B) — ln 1 = 0 and ln e = 1.
16. Solve for x: log_3 x = 5.
(A) 15
(B) 125
(C) 243
(D) 3/5
Answer
(C) — 3^5 = 243.
17. For y = log_2(x − 6) + 1, the graph is the parent y = log_2 x
(A) right 6, up 1
(B) left 6, up 1
(C) right 1, up 6
(D) up 6, right 1
Answer
(A) — Replace x by (x−6) and add 1.
18. Evaluate log_5(25) + log_5(1/5).
(A) 0
(B) 1
(C) 2
(D) −1
Answer
(B) — 2 + (−1) = 1.
19. Which is true for y = log_b x with 0 < b < 1?
(A) Increasing, concave up
(B) Decreasing, concave down
(C) Increasing, concave down
(D) Decreasing, concave up
Answer
(B) — Base in (0,1) gives a decreasing, concave-down log.
20. Evaluate: log(2) + log(50) − log(5).
(A) log(20)
(B) log(25)
(C) log(10)
(D) log(100)
Answer
(A) — log(2·50/5)=log(20).
21. Convert to a single logarithm: (1/2)·log a.
(A) log(a^2)
(B) log(√a)
(C) (log a)^2
(D) log(a)/2
Answer
(B) — (1/2)log a = log(√a).
22. If log_2 x = 3 and log_2 y = −1, evaluate log_2(x^2 y).
(A) 5
(B) 3
(C) 2
(D) −5
Answer
(A) — 2·3 + (−1) = 5.
23. Which point is on y = ln x?
(A) (e, 1)
(B) (1, e)
(C) (0, 1)
(D) (−1, 0)
Answer
(A) — ln e = 1; domain x>0.
24. The inverse of y = log_3(x + 1) is
(A) y = 3^x + 1
(B) y = 3^x − 1
(C) y = 3^(x − 1)
(D) y = ln(x + 1)
Answer
(B) — Swap and solve: x = log_3(y + 1) ⇒ y = 3^x − 1.
25. Which best describes data suitable for y = a + b ln x?
(A) Linear with constant slope
(B) Rapid early increase that slows
(C) Exponential doubling
(D) Sinusoidal oscillation
Answer
(B) — Typical log-growth pattern.
26. Solve for x: log_2(5x − 4) − log_2(x − 1) = 3.
(A) 4/3
(B) 2
(C) 3
(D) 5
Answer
(A) — log_2((5x−4)/(x−1))=3 ⇒ (5x−4)/(x−1)=8 ⇒ 5x−4=8x−8 ⇒ 3x=4 ⇒ x=4/3 (domain x>1).
27. Solve: ln(x − 2) + ln(x − 3) = ln(12).
(A) 4
(B) 5
(C) 6
(D) 3
Answer
(C) — (x−2)(x−3)=12 ⇒ x^2−5x−6=0 ⇒ x=6 or −1; domain x>3 ⇒ x=6.
28. For f(x) = log_3(2x − 1), the x-intercept is
(A) x = 1/2
(B) x = 3/2
(C) x = 1
(D) x = 2
Answer
(C) — f(x)=0 ⇒ 2x−1=1 ⇒ x=1.
29. The graph of y = log_2(x) reflected across the x-axis has equation
(A) y = log_2(−x)
(B) y = −log_2(x)
(C) y = log_(1/2)(x)
(D) y = −log_(1/2)(x)
Answer
(B) — Reflection over x-axis multiplies outputs by −1.
30. Solve: 3^x = 5·2^x.
(A) x = ln 5 / (ln 3 − ln 2)
(B) x = ln(5/3) / ln 2
(C) x = ln 5 / ln 6
(D) x = ln 5 / ln(3/2)
Answer
(D) — (3/2)^x=5 ⇒ x=ln 5 / ln(3/2).
31. Which has range (−∞, ∞)?
(A) y = 2^x
(B) y = e^x
(C) y = log_7 x
(D) y = 10^x
Answer
(C) — Logarithms output all real numbers.
32. If y = a + b ln x models “fast-then-slower” growth with y increasing, which is true?
(A) b < 0 and y decreases
(B) b > 0 and slope ≈ b/x
(C) b = 0 and y is constant
(D) It models exponential doubling
Answer
(B) — dy/dx = b/x; for b>0 it increases and slows.
33. Solve: log_4(x) + log_4(x − 4) = 3.
(A) x = 8
(B) x = 2 + 2√17
(C) x = 2 − 2√17
(D) x = 16
Answer
(B) — x(x−4)=4^3=64 ⇒ x^2−4x−64=0 ⇒ x=2±2√17; domain x>4 ⇒ 2+2√17.
34. A semi-log plot (log y vs. linear x) of y = a·b^x is a line with slope
(A) a
(B) b
(C) log(b)
(D) ln(a)
Answer
(C) — log y = x·log b + log a ⇒ slope = log b (same base as the axis log).
35. For data that slow down over time, the most appropriate regression is
(A) Linear y = mx + c
(B) Exponential y = A·B^x
(C) Logarithmic y = A + B ln x
(D) Quadratic y = ax^2 + bx + c
Answer
(C) — Logarithmic regression captures diminishing increases.
36. Solve for x: log_5(3x − 2) = 2.
Answer
x = 9 — 3x−2 = 25 ⇒ x = 9.
37. Find the inverse of y = 4 + 2·log_3(x − 1).
Answer
y = 1 + 3^{(x−4)/2} — x−4 = 2 log_3(y−1) ⇒ (x−4)/2 = log_3(y−1) ⇒ y−1 = 3^{(x−4)/2}.
38. Write as a single log: 2·log a − log b + (1/2)·log c.
Answer
log( a^2 · √c / b ) — Power and quotient rules.
39. Domain and vertical asymptote of y = ln(7 − x).
Answer
Domain: (−∞, 7); asymptote: x = 7 — 7−x>0 ⇒ x<7.
40. Solve: 2·e^{3x} = 7.
Answer
x = (1/3)·ln(7/2) — e^{3x}=7/2 ⇒ 3x=ln(7/2).
41. Evaluate ln( e^5 / √e ).
Answer
4.5 — e^5 / e^{1/2} = e^{9/2}; ln = 9/2.
42. Solve for x: log_2(x + 4) = 1 + log_2(x).
Answer
x = 4 — log_2((x+4)/x)=1 ⇒ (x+4)/x=2 ⇒ x=4.
43. Solve inequality: ln(2x − 1) > 0.
Answer
x > 1 — ln y > 0 ⇔ y > 1; 2x−1 > 1 ⇒ x > 1 (and x>1/2 for domain).
44. Find x so that log_9(x) = 1/2.
Answer
x = 3 — 9^{1/2} = 3.
45. Linearize y = a·e^{kx} by taking natural logs; identify slope and intercept on (x, ln y).
Answer
ln y = kx + ln a — slope k, intercept ln a.
46. Solve: log_3(x − 1) = 2 − log_3(x).
Answer
x = (1 + √37)/2 — log_3(x−1) + log_3 x = 2 ⇒ x(x−1)=9 ⇒ x^2−x−9=0.
47. Given y = A + B ln t with A = 2, B = 4, compute y(e^3).
Answer
14 — ln(e^3)=3; y=2+4·3=14.
48. Solve for x > 0: log_5(x) + log_5(4) = 3.
Answer
x = 125/4 — log_5(4x)=3 ⇒ 4x=125 ⇒ x=125/4.
49. If ln y = 1.2x + 0.5, express y as a function of x.
Answer
y = e^{1.2x+0.5} — Exponentiate both sides.
50. On (x, log_10 y), the line for y = a·b^x has intercept 2 and slope 0.3. Find a and b.
Answer
a = 10^2 = 100, b = 10^{0.3} — log_10 y = 0.3x + 2 ⇒ log_10 a = 2, log_10 b = 0.3.