Mathematics Advanced • Year 11 • Module 4 • Lesson 5

Introduction to Logarithmic Functions

Practise HSC-style writing on log definitions, inverses, domain, and the geometry of log graphs.

Master · Past-Paper Style

1. Short-answer questions

1.1 Evaluate log₃ 81, log₂ (1/8), and ln(e⁴). Show a one-line justification for each. 3 marks Band 3

1.2 Solve log₂ (x + 1) = 4, and state the domain of f(x) = log₂ (x + 1). 3 marks Band 4

1.3 The function f(x) = log₅ (2x − 4) is given.
(a) Find the domain of f.
(b) Find the exact x-intercept of f.
(c) State the vertical asymptote. 4 marks Band 4

Stuck on 1.3(c)? The argument 2x − 4 = 0 sets the asymptote.

2. Extended response

2.1 Let f(x) = 2^x and g(x) = log₂ x.
(a) Show by direct computation that f(g(x)) = x for every x in the domain of g, and g(f(x)) = x for every real x.
(b) State the domain and range of f and the domain and range of g. Comment in one sentence on how these compare and why.
(c) On the same set of axes, sketch y = f(x), y = g(x), and the line y = x, clearly labelling: the asymptotes of f and g, the y-intercept of f, the x-intercept of g, and the geometric reflection symmetry between the two curves.
(d) Hence solve 2^x = x graphically (state how many solutions exist and a rough numerical estimate for each), explaining briefly why the algebraic methods of this lesson cannot give a closed-form answer. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — f(g(x)) = 2^{log₂ x} = x for x > 0 (cites identity a^{log_a x} = x).

• 1 mark — g(f(x)) = log₂ 2^x = x for all real x (cites identity log_a a^x = x).

Part (b) — 2 marks

• 1 mark — domain of f = ℝ, range = (0, ∞); domain of g = (0, ∞), range = ℝ.

• 1 mark — comments that the domain of one is the range of the other (and vice versa), because inverse functions swap inputs and outputs.

Part (c) — 2 marks

• 1 mark — both curves drawn with correct shapes and asymptotes labelled (y = 0 for f, x = 0 for g) and intercepts marked ((0,1) for f, (1, 0) for g).

• 1 mark — line y = x drawn, with the reflection symmetry between the two curves explicitly indicated (e.g. by labelling reflection arrow or stating it).

Part (d) — 1 mark

• 1 mark — observes that the curves y = 2^x and y = x do not intersect for any real x (since 2^x ≥ 1 + x ln 2 > x for "most" x; concrete check: 2⁰ = 1 > 0, 2¹ = 2 > 1, and exponential grows faster than linear), hence no solution. The lesson's algebraic tools require we isolate the variable on one side; here x appears both as a base of comparison and inside an exponential, so no log/exponential algebra reduces it cleanly — numerical/graphical methods are required for such "transcendental" equations.

Your response:

Stuck on (d)? Check intersections of y = 2^x with y = x at a few values (x = 0, 1, 2).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Evaluate three logarithms (3 marks)

Sample response.
log₃ 81 = 4, because 3⁴ = 81.
log₂ (1/8) = −3, because 2⁻³ = 1/8.
ln(e⁴) = 4, using the identity ln(e^x) = x.

Marking notes. 1 mark per correct value with the one-line justification. A bare numerical answer scores 0.5 per part. Common error: writes log₂ (1/8) = 3 (forgets the negative).

1.2 — Solve log₂ (x + 1) = 4 and state the domain (3 marks)

Sample response. Convert to exponential form: x + 1 = 2⁴ = 16, so x = 15. (Check: x + 1 = 16 > 0 ✓.) The domain of f requires x + 1 > 0, i.e. x > −1 (or (−1, ∞)).

Marking notes. 1 mark — converts log form to exponential. 1 mark — solves x = 15 and checks domain. 1 mark — explicitly states the domain x > −1 (or (−1, ∞)). Common error: writes "domain: x > 0" without considering the shift.

1.3 — f(x) = log₅ (2x − 4) (4 marks)

Sample response.
(a) Require 2x − 4 > 0 ⇒ x > 2, so domain is (2, ∞).
(b) x-intercept: f(x) = 0 ⇒ log₅ (2x − 4) = 0 ⇒ 2x − 4 = 5⁰ = 1 ⇒ 2x = 5 ⇒ x = 5/2. Point (5/2, 0).
(c) Vertical asymptote where argument → 0: 2x − 4 = 0 ⇒ x = 2.

Marking notes. (a) 1 mark — inequality 2x − 4 > 0 solved cleanly. (b) 1 mark for the equation setup; 1 mark for x = 5/2. (c) 1 mark for x = 2 (this is also the boundary of the domain). Common errors: stating the asymptote as x = 0 (forgetting the inside-transformation), or claiming x = 4 (sets argument to 4 by mistake).

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a). For x > 0:

f(g(x)) = 2^{log₂ x} = x [identity: a raised to the log_a of x recovers x]. [1 mark.]

For every real x:

g(f(x)) = log₂ 2^x = x [identity: log of a power of the same base extracts the exponent]. [1 mark.]

Part (b). Domain of f = ℝ; range of f = (0, ∞). Domain of g = (0, ∞); range of g = ℝ. [1 mark.] The domain of one is the range of the other (and vice versa) because inverse functions swap inputs and outputs — every (x, y) on f corresponds to (y, x) on g. [1 mark — clear inverse-relationship comment.]

Part (c). Sketch: y = 2^x increasing through (0, 1) with horizontal asymptote y = 0 (dashed); y = log₂ x increasing through (1, 0) with vertical asymptote x = 0 (dashed); line y = x at 45°. Both curves shown as reflections of each other in the line y = x — e.g. mark the symmetric points (0, 1) ↔ (1, 0) and (1, 2) ↔ (2, 1). [1 mark — both curves correct with asymptotes and intercepts labelled. 1 mark — line y = x drawn and reflection indicated.]

Part (d). Solve 2^x = x graphically by looking at where the curve y = 2^x meets the line y = x. From the sketch: at x = 0 we have 2⁰ = 1 > 0; at x = 1, 2¹ = 2 > 1; at x = −1, 2⁻¹ = 0.5 > −1; at x = −2, 2⁻² = 0.25 > −2. The exponential 2^x is always strictly above the line y = x — so there is no real solution. The methods of this lesson cannot isolate x because x appears once inside an exponential (2^x) and once outside (= x); no rearrangement using logs/exponentials produces a single expression "x = ...". Equations of this transcendental type require numerical methods (or special functions like the Lambert W function, beyond syllabus). [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Computes f(g(x)) but cites no identity; gives partial domain/range; sketch missing y = x. ≈ 2-3 marks.

Band 4: Completes (a) with identities. (b) lists correct domains/ranges but no inverse-relationship comment. (c) sketches both curves but no y = x line. ≈ 4-5 marks.

Band 5: Full (a)(b)(c) with reflection clearly indicated. (d) checks intersections at one or two values, identifies "no intersection" but no comment on why algebra fails. ≈ 5-6 marks.

Band 6: All parts complete. (d) explicitly tests several x-values, concludes no solution, and explains why log/exp algebra cannot solve transcendental equations of this form. 7/7.