Mathematics Advanced • Year 11 • Module 4 • Lesson 15
Module Synthesis — Exponential and Logarithmic Functions
Build mixed-skill fluency across the whole module: log laws, exponential equations, derivatives of e^x, ln x, a^x and log_a x.
1. Quick recall — the core formulas
Answer each question in the space provided. 1 mark each
Q1.1 Fill in each derivative:
d/dx ( e^x ) = ____________ d/dx ( ln x ) = ____________
d/dx ( a^x ) = ____________ d/dx ( loga x ) = ____________
Q1.2 Fill in each log law (a > 0, a ≠ 1; x, y > 0):
loga(xy) = ____________________
loga(x/y) = ____________________
loga(x^n) = ____________________
Q1.3 Write the change-of-base formula: loga x = ____________________
2. Worked example — solve e^(2x) − 5 e^x + 6 = 0
Follow each line. Every step has a reason on the right.
Problem. Solve e^(2x) − 5 e^x + 6 = 0.
Step 1 — Substitute u = e^x.
u² − 5u + 6 = 0
Reason: e^(2x) = (e^x)² = u².
Step 2 — Factorise.
(u − 2)(u − 3) = 0 ⇒ u = 2 or u = 3
Reason: standard quadratic factorisation.
Step 3 — Convert back via ln.
e^x = 2 ⇒ x = ln 2 e^x = 3 ⇒ x = ln 3
Reason: ln is the inverse of e^(·); both u-values are positive, so both x-values are valid.
Conclusion. x = ln 2 or x = ln 3.
3. Faded example — differentiate y = ln(x²/(x + 1))
Use log laws before differentiating. 4 marks
Step 1 — Split using log laws:
y = ln(x²) − ln(________) = ________ ln x − ln(________)
Step 2 — Differentiate each term:
dy/dx = ________ · (1/x) − 1/(________)
= 2/x − 1/(x + 1)
Step 3 — Combine into a single fraction:
dy/dx = [2(x + 1) − ________] / [ x(x + 1) ] = (x + ________) / [ x(x + 1) ]
Conclusion. dy/dx = (x + 2) / [ x(x + 1) ].
4. Graduated practice — mixed differentiation and solving
Mix of differentiation, log simplification and equation-solving. Show working for Standard and Extension items.
Foundation — direct application (4 questions)
| Q | Question | Answer |
|---|---|---|
| 4.1 1 | Solve ln(x + 2) = 3 for x exactly. | |
| 4.2 1 | Simplify log₂ 12 + log₂ 3 − log₂ 9. | |
| 4.3 1 | Differentiate y = e^(3x). | |
| 4.4 1 | Differentiate y = ln(5x). |
Standard — typical HSC difficulty (6 questions)
Show every line of working for full marks.
4.5 Differentiate y = e^(x²) · ln x. 3 marks
4.6 Differentiate y = e^(2x) / (x + 1). 3 marks
4.7 Find the half-life if a substance decays from 200 g to 50 g in 20 days. 2 marks
4.8 Find the maximum of f(x) = x · e^(−x/2) for x ≥ 0. 3 marks
4.9 Solve log₃(x − 1) + log₃(x + 1) = 2. 3 marks
4.10 A bacterial culture grows according to P = 500 e^(0.2 t). Find dP/dt at t = 5 and interpret the result. 3 marks
Extension — combine ideas (2 questions)
4.11 Solve simultaneously: y = ln(x) and y = 1 − x, for x > 0. (Either provide an exact solution if possible, or show by graph/sign-change that there is exactly one solution in 0 < x < 1.) 3 marks
4.12 Show that y = e^x is its own derivative, and use this to explain in one sentence why y = e^x is the unique function (up to a constant multiple) satisfying dy/dx = y. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Derivatives
d/dx(e^x) = e^x. d/dx(ln x) = 1/x. d/dx(a^x) = a^x · ln a. d/dx(loga x) = 1 / (x ln a).
Q1.2 — Log laws
loga(xy) = loga x + loga y. loga(x/y) = loga x − loga y. loga(x^n) = n loga x.
Q1.3 — Change of base
loga x = ln x / ln a (or logb x / logb a for any valid base b).
Q3 — Faded example y = ln(x²/(x+1))
Step 1: y = ln(x²) − ln(x + 1) = 2 ln x − ln(x + 1).
Step 2: dy/dx = 2 · (1/x) − 1/(x + 1) = 2/x − 1/(x + 1).
Step 3: dy/dx = [2(x + 1) − x] / [x(x + 1)] = (x + 2) / [x(x + 1)].
Q4.1 — Solve ln(x + 2) = 3
x + 2 = e³ ⇒ x = e³ − 2 (≈ 18.09).
Q4.2 — log₂ 12 + log₂ 3 − log₂ 9
= log₂ (12 · 3 / 9) = log₂ 4 = 2.
Q4.3 — y = e^(3x)
dy/dx = 3 e^(3x).
Q4.4 — y = ln(5x)
Chain: dy/dx = 5/(5x) = 1/x. (Or split: ln(5x) = ln 5 + ln x, derivative 1/x.)
Q4.5 — y = e^(x²) ln x
Product rule: dy/dx = 2x e^(x²) · ln x + e^(x²) · (1/x) = e^(x²) ( 2x ln x + 1/x ).
Q4.6 — y = e^(2x) / (x + 1)
Quotient rule with u = e^(2x), v = x + 1, u' = 2 e^(2x), v' = 1:
dy/dx = [ 2 e^(2x) (x + 1) − e^(2x) · 1 ] / (x + 1)² = e^(2x) (2x + 1) / (x + 1)².
Q4.7 — Half-life from 200 g → 50 g in 20 days
200 → 50 is two halvings (200 → 100 → 50), so half-life = 20/2 = 10 days. (Algebraic check: 50 = 200 · e^(20k) ⇒ e^(20k) = 1/4 = 2⁻² ⇒ 20k = −2 ln 2 ⇒ k = −(ln 2)/10; t₁/₂ = ln 2 / |k| = 10. ✓)
Q4.8 — Max of f(x) = x e^(−x/2), x ≥ 0
f'(x) = e^(−x/2) − (x/2) e^(−x/2) = e^(−x/2)(1 − x/2). Stationary at x = 2. f(2) = 2 · e^(−1) = 2/e ≈ 0.736. Sign of f' goes + then −, so this is a max.
Q4.9 — log₃(x − 1) + log₃(x + 1) = 2
log₃ [(x − 1)(x + 1)] = 2 ⇒ x² − 1 = 9 ⇒ x² = 10 ⇒ x = ±√10. Reject x = −√10 because x − 1 must be > 0 (so x > 1). x = √10 (≈ 3.162).
Q4.10 — dP/dt for P = 500 e^(0.2 t) at t = 5
dP/dt = 500 · 0.2 · e^(0.2 t) = 100 e^(0.2 t). At t = 5: dP/dt = 100 · e¹ ≈ 100 · 2.718 ≈ 272 bacteria/hour. The population is growing at approximately 272 bacteria per hour at t = 5.
Q4.11 — Solving ln x = 1 − x for x > 0
No elementary closed-form solution. Define f(x) = ln x − (1 − x) = ln x + x − 1. f(0.5) = −0.693 + 0.5 − 1 = −1.193 < 0. f(1) = 0 + 1 − 1 = 0 ✓. So x = 1 is the (unique) solution; f is strictly increasing on x > 0 (since f'(x) = 1/x + 1 > 0), so the sign change at x = 1 is the only crossing.
Q4.12 — y = e^x is self-derivative
d/dx (e^x) = e^x = y. ✓ Why unique up to constant multiple: if dy/dx = y, then d/dx (y · e^(−x)) = y' e^(−x) − y e^(−x) = (y − y) e^(−x) = 0, so y · e^(−x) is a constant C, hence y = C e^x. The constant C is fixed by any initial condition.