NCEA Level 2Exponential modelsAS 91261

Exponential models: 7 exam questions solved step by step

What this lesson covers

  • Growth or decay by a fixed percentage each period is multiplication by the same factor each period.
  • Finding how long something takes means solving for an index, which needs logarithms.
Key idea

7 questions, one model

  • 1. Percentage to factor: 1 ± r/100.
  • 2. Model: start × factorⁿ
  • 3. How long: solve for n with logs.
  • 4. Answer in context, whole periods.
Your turn

Question 1 · Achieved, in context

  • $4000 is invested at 3.8% a year,
  • compounded annually.
  • What is it worth after 5 years?
Worked example

Q1: $4000 at 3.8% for 5 years

  • Grow 3.8%: × 1.038.
  • Round at the end.
  1. grows 3.8%: × 1.0381 + 0.038
  2. A = 4000 × 1.038ⁿthe model
  3. A = 4000 × 1.038⁵n = 5
  4. = 4000 × 1.204999...
  5. = 4819.9969...
  6. $4820.00nearest cent
Your turn

Question 2 · Achieved, in context

  • An $1800 e-bike loses 12% of its
  • value each year.
  • What is it worth after 3 years?
  • Careful: not 36% off.
Worked example

Q2: $1800, losing 12% a year

  • Lose 12%: keep 88%.
  • Each loss is 12% of less.
  1. loses 12%: × 0.881 − 0.12
  2. V = 1800 × 0.88ⁿthe model
  3. V = 1800 × 0.88³n = 3
  4. = 1800 × 0.681472
  5. = 1226.6496
  6. $1226.65nearest cent
  7. 1800 × 0.64 = $1152not 3 × 12% off
Your turn

Question 3 · Merit, in context

  • $2500 is invested at 5.2% a year,
  • compounded annually. After how many
  • whole years is it first over $4000?
Worked example

Q3: when is it over $4000?

  • n is an index: use logs.
  • Check the years either side.
  1. 2500 × 1.052ⁿ = 4000the model = 4000
  2. 1.052ⁿ = 1.6÷ 2500
  3. n log 1.052 = log 1.6log, power law
  4. n = log 1.6 ÷ log 1.052÷ log 1.052
  5. = 9.2715...
  6. n = 9: 2500 × 1.052⁹ ≈ $3945
  7. n = 10: 2500 × 1.052¹⁰ ≈ $4150
  8. After 10 whole years
Your turn

Question 4 · Merit, in context

  • $6000 is invested at 6% a year,
  • compounded monthly.
  • What is it worth after 3 years?
Worked example

Q4: 6% a year, monthly

  • Rate per month, ÷ 12.
  • Periods in months, × 12.
  1. monthly rate = 6% ÷ 12 = 0.5%
  2. factor 1.005 a month1 + 0.005
  3. periods = 3 × 12 = 36months
  4. A = 6000 × 1.005³⁶the model
  5. = 6000 × 1.196680...
  6. = 7180.0831...
  7. $7180.08nearest cent
  8. 6000 × 1.06³ = 7146.10that is yearly
Your turn

Question 5 · Merit, in context

  • A medical tracer has a half-life of 6 h.
  • A patient is given 40 mg.
  • (a) How much is left after 24 hours?
  • (b) When is 1 mg left? (1 dp)
Worked example

Q5: half-life 6 hours

  • Halves every 6 h: 0.5^(t ÷ 6).
  • Isolate the power first.
  1. M = 40 × 0.5^(t ÷ 6)halves every 6 h
  2. t = 24: 40 × 0.5⁴24 ÷ 6 = 4
  3. = 40 × 0.0625 = 2.5 mg
  4. 40 × 0.5^(t ÷ 6) = 1part (b)
  5. 0.5^(t ÷ 6) = 0.025÷ 40
  6. (t ÷ 6) log 0.5 = log 0.025log, power law
  7. t ÷ 6 = log 0.025 ÷ log 0.5÷ log 0.5
  8. = 5.3219...both logs negative
  9. t = 31.93... ≈ 31.9 hours× 6
Your turn

Question 6 · Excellence, in context

  • A predator-free island had 80 of a
  • native bird in 2020 and 125 in 2025.
  • Assume the same % growth each year.
  • Find the rate, and the first year over 300.
Worked example

Q6: find the growth rate

  • Unknown factor g.
  • Five years: g⁵.
  1. P = 80 × gⁿg: the yearly factor
  2. 80 × g⁵ = 1252025 is n = 5
  3. g⁵ = 1.5625÷ 80
  4. g = 1.5625^(1/5)fifth root
  5. = 1.093362...
  6. about 9.34% a yearg − 1 = 0.0934
Worked example

Q6: the first year over 300

  • Keep g unrounded.
  • Answer with a year.
  1. g = 1.093362...
  2. 80 × gⁿ = 300
  3. gⁿ = 3.75÷ 80
  4. n = log 3.75 ÷ log glog, power law
  5. = 14.808...g unrounded
  6. n = 14 (2034): about 279
  7. n = 15 (2035): about 305
  8. First over 300 in 2035
Your turn

Question 7 · Excellence, in context

  • Aroha invests $5000 at 4% a year.
  • Tama invests $4000 at 6.5% a year.
  • Both compound annually. After how
  • many years is Tama first ahead?
Worked example

Q7: when does Tama pass Aroha?

  • n on both sides.
  • Check either side.
  1. 4000 × 1.065ⁿ > 5000 × 1.04ⁿ
  2. (1.065 ÷ 1.04)ⁿ > 1.25÷ 1.04ⁿ, ÷ 4000
  3. n log(1.065 ÷ 1.04) > log 1.25
  4. n > log 1.25 ÷ log(1.065 ÷ 1.04)
  5. = 9.3939...
  6. n = 9: A ≈ $7117, T ≈ $7050
  7. n = 10: A ≈ $7401, T ≈ $7509
  8. Tama is first ahead after 10 years
Summary

The model that solved all seven

  • Factor, model, logs for time, then whole periods.
StepHow
Grow r%multiply by 1 + r/100
Lose r%multiply by 1 − r/100
Monthlyrate ÷ 12, periods × 12
Modelstart × factorⁿ
How longisolate, log, divide
Whole yearscheck either side, round up

Is your child on track?

Take the free placement test