Practice Maths

Topic Review — Exponential Functions — Solutions

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★ U2T1 — Exponential Functions Review

Index laws • Graphs of exponential functions • Solving exponential equations • Growth and decay applications

  1. Q1 — Index laws

    Simplify: (a) x3 × x5   (b) (2a3)4   (c) 6x4 ÷ (2x7)   (d) (x2y3)0   (e) 4−3/2

    (a) x8   (b) 16a12   (c) 3x−3 = 3/x3   (d) 1   (e) (22)−3/2=2−3=1/8

  2. Q2 — Scientific notation

    (a) Write 0.000047 in scientific notation.   (b) Write 3.2 × 105 as an ordinary number.   (c) Calculate (4×103) × (2×104).

    (a) 4.7 × 10−5   (b) 320 000   (c) 8×107 = 80 000 000

  3. Q3 — Graph features

    For y = 3 × 2x: (a) state the y-intercept and horizontal asymptote   (b) is it increasing or decreasing?   (c) what is y when x = 3? When x = −2?

    (a) y-int: 3×1=3. Asymptote: y=0

    (b) Base 2>1 and coefficient positive → increasing

    (c) y(3)=3×8=24. y(−2)=3×(1/4)=3/4

  4. Q4 — Solving by matching bases

    Solve: (a) 2x = 128   (b) 9x+1 = 27   (c) 52x−1 = 1/25   (d) (1/4)x = 32

    (a) 27=128 → x=7

    (b) 32(x+1)=33 → 2x+2=3 → x=1/2

    (c) 52x−1=5−2 → 2x−1=−2 → x=−1/2

    (d) 2−2x=25 → −2x=5 → x=−5/2

  5. Q5 — Growth model

    A town's population grows by 3% annually from a base of 25 000. (a) Write a model P(t). (b) Find the population after 8 years. (c) When does it exceed 40 000 (use trial and improvement)?

    (a) P(t) = 25000(1.03)t

    (b) P(8) = 25000(1.03)8 ≈ 25000×1.2668 ≈ 31 670

    (c) Trial: t=16: P≈40089. t=15: P≈38911. Exceeds 40 000 after 16 years.

  6. Q6 — Decay and half-life

    A substance has half-life 10 years, initial mass 160 g. (a) Write a model. (b) Find mass after 30 years. (c) When does mass fall below 5 g?

    (a) A = 160(0.5)t/10

    (b) A(30) = 160(0.5)3 = 160/8 = 20 g

    (c) (0.5)t/10=5/160=1/32=(0.5)5 → t/10=5 → t=50 years

  7. Q7 — Compound interest

    $4 000 invested at 5% p.a. compounded quarterly. (a) Write the model. (b) Find the balance after 6 years. (c) How many years to reach $6 000?

    (a) A = 4000(1+0.05/4)4t = 4000(1.0125)4t

    (b) A(6) = 4000(1.0125)24 ≈ 4000×1.3474 ≈ $5 390

    (c) Trial: t=9: A≈$6248. t=8: A≈$5939. Exceeds $6000 in year 9.

  8. Q8 — Comparing growth/decay

    Two populations: A(t) = 1000(1.05)t and B(t) = 2000(0.97)t. (a) Which is growing, which decaying? (b) Find when A = B (to nearest year using trial and improvement). (c) After 20 years, which is larger?

    (a) A is growing (base 1.05 >1). B is decaying (base 0.97 <1).

    (b) Trial: t=15: A≈2079, B≈1304. t=10: A≈1629, B≈1484. t=11: A≈1710, B≈1439. Approximately t≈10 to 11 years (they cross between years 10 and 11).

    (c) A(20)≈2653. B(20)=2000(0.97)20≈2000×0.5438≈1088. A is larger after 20 years.

  9. Q9 — Depreciation

    A car bought for $35 000 depreciates at 18% per year. (a) Write a model V(t). (b) Find the value after 4 years. (c) After how many years is it worth less than $10 000?

    (a) V(t) = 35000(0.82)t

    (b) V(4) = 35000(0.82)4 ≈ 35000×0.4521 ≈ $15 823

    (c) Trial: t=6: V≈$10485. t=7: V≈$8598. Falls below $10 000 after 7 years.

  10. Q10 — Fitting and interpreting models

    Data: (0, 50), (2, 200). (a) Find a and b in A = a × bt. (b) Predict A at t = 5. (c) What does the rate of change of A at t = 0 suggest about early growth?

    (a) a=50. 200=50b2 → b2=4 → b=2. Model: A = 50 × 2t

    (b) A(5) = 50×32 = 1600

    (c) The value doubles every unit of time. Early growth is rapid — exponential growth quickly accelerates (increasing rate of increase). This is characteristic of unconstrained growth.