Practice Maths

Solutions — Equations and Applications of Exponential Functions

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  1. Q1 — Solving by matching bases

    (a) 2x = 64 = 26 → x = 6

    (b) 3x = 1/9 = 3−2 → x = −2

    (c) 5x+1 = 125 = 53 → x+1 = 3 → x = 2

    (d) 4x = 1/64 = 4−3 → x = −3

    (e) 82x = 32 → (23)2x = 25 → 26x = 25 → 6x = 5 → x = 5/6

  2. Q2 — Common base conversions

    (a) 4x = 8 → 22x = 23 → 2x = 3 → x = 3/2

    (b) 9x = 27 → 32x = 33 → 2x = 3 → x = 3/2

    (c) 25x = 125 → 52x = 53 → 2x = 3 → x = 3/2

    (d) 42x−1 = 8x+1 → 24x−2 = 23x+3 → 4x−2 = 3x+3 → x = 5

  3. Q3 — Evaluating models

    (a) A = 1000 × 25 = 1000 × 32 = 32 000

    (b) A = 500 × (0.5)4 = 500 × 1/16 = 31.25

    (c) A = 3000 × (1.1)3 = 3000 × 1.331 ≈ 3993

    (d) A = 800 × (1/2)12/6 = 800 × (1/2)2 = 800 × 1/4 = 200

  4. Q4 — Population growth

    (a) P = 120 000 × (1.04)t

    (b) P(10) = 120 000 × (1.04)10 ≈ 120 000 × 1.4802 ≈ 177 000 (to nearest 1000)

    (c) Trial: t=13: P ≈ 192 000; t=14: P ≈ 200 000. First exceeds 200 000 in year 14 (i.e. after 14 years).

    (d) The model assumes a constant 4% growth rate indefinitely. In reality, resources are limited (food, space, water), so growth eventually slows — a logistic model is more realistic over long periods.

  5. Q5 — Radioactive decay

    (a) t=5: A = 240 × (1/2)1 = 120 g; t=10: A = 240 × (1/2)2 = 60 g; t=15: A = 240 × (1/2)3 = 30 g

    (b) t=25: A = 240 × (1/2)5 = 240/32 = 15/2 = 7.5 g. As a fraction: 1/32 of the original.

    (c) 15 = 240 × (1/2)t/5 → (1/2)t/5 = 1/16 = (1/2)4 → t/5 = 4 → t = 20 years

  6. Q6 — Compound interest

    (a) A = 5000(1+0.06)4 = 5000 × 1.2625 ≈ $6 312.38

    (b) A = 10000(1+0.08/4)20 = 10000(1.02)20 = 10000 × 1.4859 ≈ $14 859.47

    (c) Trial: t=7: A≈$3869; t=8: A=$4000 would need (1.1)t≥2. (1.1)7≈1.949, (1.1)8≈2.144. t = 8 years (first full year where balance exceeds $4 000).

  7. Q7 — Fitting exponential models

    (a) At t=0, A=100 → A0=100. At t=1, A=300 → 100a=300 → a=3. Check t=2: 100×9=900 ✓. Model: A = 100 × 3t

    (b) At t=0: A0=80. At t=2: 80a2=20 → a2=1/4 → a=1/2. Model: A = 80 × (0.5)t. At t=4: A = 80 × (0.5)4 = 80/16 = 5.

    (c) A halves with each unit increase in t: t=0:80, t=1:40, t=2:20 ✓, t=3:10, t=4:5. Half-life is 1 unit. ✓

  8. Q8 — Bacteria doubling time

    (a) Doubles every 20 min: N = 50 × 2t/20

    (b) t=120: N = 50 × 26 = 50 × 64 = 3 200 bacteria

    (c) 51200 = 50 × 2t/20 → 2t/20 = 1024 = 210 → t/20=10 → t = 200 minutes

    (d) Colony 2: N2 = 200 × 2t/20. Colony 1: N1 = 50 × 2t/20. Since 200 > 50, N2 > N1 for ALL t ≥ 0. The second colony is always larger (it starts with 4 times as many bacteria and grows at the same rate). It always exceeds the first colony from t=0.

  9. Q9 — Drug concentration

    (a) C = 200(1 − 0.25)t = 200(0.75)t

    (b) C(4) = 200(0.75)4 = 200 × 0.3164 ≈ 63.3 mg

    (c) Trial: t=5: C≈47.5 mg (below 50). t=4: C≈63.3 (above). After 5 full hours the concentration falls below 50 mg.

    (d) At t=4: C≈63.3 mg. New dose: total = 63.3 + 200 = 263.3 mg. After further 2 hours: 263.3 × (0.75)2 = 263.3 × 0.5625 ≈ 148.1 mg.

  10. Q10 — Depreciation

    (a) V = 80 000(0.85)t

    (b) V(5) = 80 000(0.85)5 = 80 000 × 0.4437 ≈ $35 497

    (c) Trial: t=7: V≈$24 900; t=8: V≈$21 200; t=9: V≈$18 000. Falls below $20 000 after 9 full years.

    (d) Linear: V = 80 000 − 8 000t. Reaches $20 000 when 8 000t = 60 000 → t = 7.5, so after 8 years (first full year below $20 000). The linear model reaches $20 000 first (year 8 vs year 9 for exponential). This is because exponential depreciation slows as the value decreases, while linear depreciation removes the same amount each year.