Mathematics Advanced • Year 11 • Module 4 • Lesson 4
Exponential Modelling and Equations
Build fluency in solving exponential equations by matching bases and by taking logarithms.
1. Quick recall
Answer each question. 1 mark each
Q1.1 Complete: if am = an (with a > 0, a ≠ 1), then m = ____ .
Q1.2 Write the standard "take a log" identity in two forms:
If eu = k, then u = ________ . If au = k, then u = (ln k) / ________ .
Q1.3 State the general continuous-rate exponential model in symbols:
y = ________ , where A is the initial value and k is the growth/decay constant.
2. Worked example — solve 3x = 27
Problem. Solve 3x = 27 for x.
Step 1 — Express 27 as a power of 3.
27 = 3 × 3 × 3 = 33
Step 2 — Rewrite with the same base.
3x = 33
Step 3 — Equate exponents.
x = 3
Reason: am = an ⇒ m = n when a > 0, a ≠ 1.
Conclusion. x = 3.
3. Faded example — solve e2x = 10
Fill in the missing pieces. 4 marks
Step 1 — Take natural logarithm of both sides.
ln(e2x) = ln(____)
Step 2 — Apply ln(eu) = u.
____ = ln 10
Step 3 — Divide both sides.
x = (ln ____) / ____
Step 4 — Compute numerically (ln 10 ≈ 2.303).
x ≈ ____________ (to 2 dp).
4. Graduated practice
Foundation — solve by matching bases (4 questions)
| Q | Equation | x = |
|---|---|---|
| 4.1 1 | 2x = 32 | |
| 4.2 1 | 5x = 125 | |
| 4.3 1 | 3x = 1/9 | |
| 4.4 1 | 4x = 1 |
Standard — match bases and solve (6 questions)
4.5 Solve 5x − 1 = 25. 2 marks
4.6 Solve 4x + 1 = 64. 2 marks
4.7 Solve 23x = 8x − 2. 2 marks
4.8 Solve ex = 7 to 2 dp. 2 marks
4.9 Solve 2x = 5 (use logs of any base; give exact and decimal). 2 marks
4.10 Solve 100 e−0.05 t = 50 for t (i.e. half-life). Give exact form and to 1 dp. 2 marks
Extension — modelling (2 questions)
4.11 A car worth $30 000 depreciates to $20 000 in 2 years using V = 30 000 ekt. Find k to 3 sf. Will the model give a positive or negative k? Why? 3 marks
4.12 A bacterial culture grows from 200 to 500 in 4 hours under P = A ekt. Find A and k, then predict P at t = 10 hours. 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 — Equate exponents
m = n.
Q1.2 — Take a log
u = ln k. u = (ln k) / ln a.
Q1.3 — General model
y = A ekt.
Q3 — Faded e2x = 10
Step 1: ln(e2x) = ln(10).
Step 2: 2x = ln 10.
Step 3: x = (ln 10) / 2.
Step 4: x ≈ 2.303 / 2 ≈ 1.15.
Q4.1–4.4 — Match bases
4.1: 2x = 25 ⇒ x = 5. 4.2: 5x = 53 ⇒ x = 3. 4.3: 3x = 3−2 ⇒ x = −2. 4.4: 4x = 40 ⇒ x = 0.
Q4.5 — 5x − 1 = 25
25 = 52, so 5x − 1 = 52, giving x − 1 = 2 and x = 3.
Q4.6 — 4x + 1 = 64
64 = 43, so x + 1 = 3 and x = 2.
Q4.7 — 23x = 8x − 2
8 = 23, so 8x − 2 = 23(x − 2) = 23x − 6. Equation becomes 23x = 23x − 6, so 3x = 3x − 6, giving 0 = −6 — a contradiction. So no solution.
Q4.8 — ex = 7
x = ln 7 ≈ 1.95.
Q4.9 — 2x = 5
x = ln 5 / ln 2 (exact). Numerically x ≈ 1.6094 / 0.6931 ≈ 2.32.
Q4.10 — 100 e−0.05 t = 50
e−0.05 t = 0.5 ⇒ −0.05 t = ln 0.5 = −ln 2 ⇒ t = ln 2 / 0.05 (exact) ≈ 0.6931 / 0.05 ≈ 13.9 years.
Q4.11 — V = 30 000 ekt; V(2) = 20 000
20 000 = 30 000 e2k ⇒ e2k = 2/3 ⇒ 2k = ln(2/3) ⇒ k = ½ ln(2/3) ≈ −0.203. The value is decreasing (depreciation), so k must be negative, which the computed value confirms.
Q4.12 — P = A ekt; P(0) = 200, P(4) = 500
A = P(0) = 200. 500 = 200 e4k ⇒ e4k = 2.5 ⇒ 4k = ln 2.5 ⇒ k ≈ 0.9163 / 4 ≈ 0.229. P(10) = 200 e2.29 ≈ 200 × 9.876 ≈ 1975 cells.