Solutions — Graphs of Logarithmic Functions

Q1 — Domain, range, and vertical asymptote

(a) y = log₂(x): Domain: x > 0, Range: all real numbers, Vertical asymptote: x = 0

(b) y = log₃(x − 4): Domain: x > 4, Range: all real numbers, Vertical asymptote: x = 4

(c) y = log(x + 2): Domain: x > −2, Range: all real numbers, Vertical asymptote: x = −2

(d) y = ln(x − 5): Domain: x > 5, Range: all real numbers, Vertical asymptote: x = 5

Q2 — Finding the x-intercept

(a) y = log₂(x) − 3 = 0 → log₂(x) = 3 → x = 2³ = 8. x-intercept: (8, 0)

(b) y = log(x + 5) = 0 → x + 5 = 10⁰ = 1 → x = −4. x-intercept: (−4, 0)

(c) y = 2log₃(x − 2) + 4 = 0 → log₃(x − 2) = −2 → x − 2 = 3⁻² = 1/9 → x = 2 + 1/9 = 19/9. x-intercept: (19/9, 0)

(d) y = log₅(x) + 1 = 0 → log₅(x) = −1 → x = 5⁻¹ = 1/5. x-intercept: (1/5, 0)

Q3 — Describing transformations from y = log₂(x)

(a) y = log₂(x) + 5: Vertical shift up 5 units. The graph moves up by 5; the asymptote remains at x = 0.

(b) y = log₂(x − 3): Horizontal shift right 3 units. The vertical asymptote moves from x = 0 to x = 3; domain becomes x > 3.

(c) y = 3log₂(x): Vertical stretch by a factor of 3. Each y-value is multiplied by 3; the x-intercept (1, 0) is unchanged.

(d) y = −log₂(x): Reflection in the x-axis. The graph is turned upside down; it is now decreasing. The x-intercept (1, 0) is unchanged.

Q4 — Finding the y-intercept

(a) y = log₃(x + 3): substitute x = 0: log₃(3) = 1. y-intercept: (0, 1)

(b) y = log₂(x − 1) + 2: domain is x > 1, so x = 0 is not in the domain. No y-intercept.

(c) y = log(2x): domain is x > 0, so x = 0 is not in the domain (log(0) is undefined). No y-intercept.

(d) y = ln(x + e): substitute x = 0: ln(0 + e) = ln(e) = 1. y-intercept: (0, 1)

Q5 — Sketching key features

(a) y = log₂(x + 4)

Vertical asymptote: x = −4 | Domain: x > −4

x-intercept: log₂(x + 4) = 0 → x + 4 = 2⁰ = 1 → x = −3. x-intercept: (−3, 0)

y-intercept: x = 0: log₂(4) = 2. y-intercept: (0, 2)

Extra point: x = −2: log₂(2) = 1 → (−2, 1). Increasing curve approaching the asymptote x = −4 from the right.

(b) y = log₃(x) − 1

Vertical asymptote: x = 0 | Domain: x > 0

x-intercept: log₃(x) = 1 → x = 3. x-intercept: (3, 0)

No y-intercept (x > 0 required).

Extra point: x = 1: log₃(1) − 1 = 0 − 1 = −1 → (1, −1). Increasing curve approaching the asymptote x = 0 from the right.

Q6 — Increasing or decreasing, domain, x-intercept

(a) y = log_0.5(x): Base 0 < a < 1 → decreasing. Domain: x > 0. x-intercept: log_0.5(x) = 0 → x = 0.5⁰ = 1. x-intercept: (1, 0)

(b) y = −2log₃(x + 1): A = −2 < 0 → decreasing (reflection in x-axis). Domain: x + 1 > 0 → x > −1. x-intercept: −2log₃(x + 1) = 0 → log₃(x + 1) = 0 → x + 1 = 1 → x = 0. x-intercept: (0, 0)

(c) y = log₄(x − 3) + 2: A = 1 > 0 and base 4 > 1 → increasing. Domain: x − 3 > 0 → x > 3. x-intercept: log₄(x − 3) = −2 → x − 3 = 4⁻² = 1/16 → x = 3 + 1/16 = 49/16. x-intercept: (49/16, 0)

Q7 — Writing equations of transformed log₂ functions

(a) Shift right 5, up 3: replace x with (x − 5) and add 3. y = log₂(x − 5) + 3

(b) Vertical stretch ×4 and reflect in x-axis: multiply by −4. y = −4log₂(x)

(c) Shift left 1 (replace x with x + 1), compress by 1/2 (multiply by 1/2), shift down 2 (subtract 2). y = ½log₂(x + 1) − 2

Q8 — Finding equations from given information

(a) Form: y = log_b(x − 1). We need to check what points are consistent. The x-intercept occurs when log_b(x − 1) = 0, i.e. x − 1 = 1, i.e. x = 2. However the question specifies x-intercept (4, 0), which means y = 0 at x = 4: log_b(4 − 1) = log_b(3) = 0 — this is impossible for any base b (since log_b(3) = 0 would require 3 = 1). Re-examining: the form y = A log_b(x − 1) gives x-intercept where A log_b(x−1)=0 → x−1=1 → x=2. The stated x-intercept of (4,0) requires the form y = log_b(x − 1) − log_b(3) = log_b((x−1)/3). Then at x=4: log_b(1)=0 ✓. At x=10: log_b(9/3)=log_b(3)=1 → b=3. y = log₃((x − 1)/3) or equivalently y = log₃(x − 1) − 1

(b) Form: y = A log(x) + k. At (1, −3): A log(1) + k = A(0) + k = k = −3. At (10, 2): A log(10) + k = A(1) + (−3) = A − 3 = 2 → A = 5. y = 5log(x) − 3

Q9 — Comparing log graphs

(a) Expand g(x) using the log product law:

g(x) = log₂(4x) − 1 = log₂(4) + log₂(x) − 1 = 2 + log₂(x) − 1 = log₂(x) + 1

Wait — this gives g(x) = log₂(x) + 1 ≠ f(x). However if g(x) = log₂(4x) − 2:

g(x) = log₂(4) + log₂(x) − 2 = 2 + log₂(x) − 2 = log₂(x) = f(x) ✓

So f(x) = g(x) because log₂(4x) − 2 = log₂(4) + log₂(x) − 2 = 2 + log₂(x) − 2 = log₂(x).

(b) Since f(x) = g(x) algebraically for all x in the domain (x > 0), the two functions produce identical output for every input. Their graphs are therefore exactly the same curve — they cannot be distinguished.

Q10 — Decibel model

(a) L = 10log(I / 10⁻¹²) = 10[log(I) − log(10⁻¹²)] = 10[log(I) + 12] = 10log(I) + 120

(b) The argument of the logarithm must be positive and intensity is a physical quantity that is always positive. Domain: I > 0

(c) Set L = 0:
10log(I) + 120 = 0 → log(I) = −12 → I = 10⁻¹² W/m²
This is the threshold of human hearing — the minimum intensity that the average human ear can detect. A sound at this intensity is perceived as having 0 dB loudness.

(d) Key points on the graph of L vs I:
• At I = 10⁻¹²: L = 0 → point (10⁻¹², 0) — threshold of hearing
• At I = 1: L = 10log(1) + 120 = 120 → point (1, 120) — threshold of pain
The graph is an increasing logarithmic curve starting from (10⁻¹², 0), passing through (1, 120), with domain I > 0.