Practice Maths

Equations and Applications of Exponential Functions

Key Terms

To solve an exponential equation, write both sides as a power of the same base, then equate the exponents.
If the bases cannot be matched, logarithms are needed (covered in the next topic).
Growth model
A = A0 × at where a > 1. Value increases over time.
Decay model
A = A0 × at where 0 < a < 1. Value decreases over time. Equivalent to A = A0(1 − r)t where r is the decay rate.
Doubling time
: the time for a quantity to double. Set A = 2A0 and solve.
Half-life
: the time for a radioactive substance to halve. Set A = A0/2 and solve.
Key Formulas
Growth modelA = A0 × at   (a > 1)
Decay modelA = A0 × at   (0 < a < 1)   or   A = A0(1 − r)t
Percentage growthA = A0(1 + r)t   (r is growth rate as decimal)
Compound interestA = P(1 + r/n)nt   (n = compoundings per year)
Half-lifeA = A0 × (1/2)t/h   where h = half-life period

Exponential growth A = A0 × 2t (blue) vs decay A = A0 × (1/2)t (orange)

t A 1 2 3 A₀ 2A₀ Growth Decay y=0
Hot Tip When setting up a model, identify: (1) A0 = the initial value (at t = 0), (2) the base a (the multiplier each time period), and (3) t = the time variable. Check: if the quantity doubles each period, a = 2. If it halves each period, a = 1/2. If it increases by 5% each period, a = 1.05.

Worked Example 1 — Solving exponential equations by matching bases

Question: Solve (a) 4x = 32   (b) 9x = 27x−1

(a) Write both sides as powers of 2: (22)x = 25 → 22x = 25 → 2x = 5 → x = 5/2

(b) Write both sides as powers of 3: (32)x = (33)x−1 → 32x = 33x−3 → 2x = 3x − 3 → x = 3

Worked Example 2 — Applying an exponential growth model

Question: A population of 2000 rabbits triples every 3 years. (a) Write a model for the population P after t years. (b) Find P after 9 years. (c) When does the population reach 54 000?

(a) The population triples every 3 years, so P = 2000 × 3t/3.

(b) At t = 9: P = 2000 × 33 = 2000 × 27 = 54 000.

(c) Set P = 54 000: 54 000 = 2000 × 3t/3 → 3t/3 = 27 = 33 → t/3 = 3 → t = 9 years.

Solving Exponential Equations by Matching Bases

An exponential equation has the unknown in the exponent, for example 2x = 32. The key technique is to rewrite both sides as a power of the same base. Once the bases match, the exponents must be equal.

Example: Solve 2x = 32.

Rewrite 32 = 25. So 2x = 25, which means x = 5.

Example with a fraction base: Solve (1/3)x = 27.

Rewrite: (3−1)x = 33, so 3−x = 33, giving −x = 3, so x = −3.

Common bases to recognise: Powers of 2: 4, 8, 16, 32, 64, 128, 256. Powers of 3: 9, 27, 81, 243. Powers of 5: 25, 125, 625. If both sides cannot be written with the same base, you need logarithms.

Growth and Decay Models — The Formula

Exponential functions model situations where a quantity increases or decreases by a constant percentage each time period. The general form is:

A = A0 × at

where A0 is the initial value, a is the growth/decay factor, and t is time.

  • Growth: a > 1. If growing at rate r per period, then a = 1 + r.
  • Decay: 0 < a < 1. If decaying at rate r per period, then a = 1 − r.

Example: A population of 5 000 bacteria grows at 20% per hour. After 3 hours: A = 5000 × (1.2)3 = 5000 × 1.728 = 8 640.

Half-Life and Doubling Time

Doubling time is the time for a growing quantity to double. Set A = 2A0 and solve for t: 2A0 = A0 × at, so 2 = at. Solve this by matching bases or using logarithms.

Half-life is the time for a radioactive or decaying substance to halve. Set A = A0/2 and solve: 1/2 = at. A convenient formula is A = A0 × (1/2)t/h where h is the half-life period.

Example: Carbon-14 has a half-life of 5 730 years. After 11 460 years (= 2 half-lives), the fraction remaining = (1/2)2 = 1/4 of the original amount.

Compound Interest as an Exponential Model

Compound interest is an exponential growth model where a = (1 + r/n) and the exponent counts the total number of compoundings:

A = P(1 + r/n)nt

where P = principal, r = annual interest rate (as a decimal), n = compoundings per year, t = years.

Example: $2 000 invested at 6% p.a. compounded monthly for 5 years: A = 2000 × (1 + 0.06/12)60 = 2000 × (1.005)60 ≈ $2 697.70.

Exam tip: In application problems, always identify A0 (initial value), the factor a, and what t represents before writing your equation. Check whether the factor is written as (1 + r) for growth or (1 − r) for decay — a common sign error is using the wrong one.

Mastery Practice

See Answers ➔
  1. Fluency

    Solve each exponential equation by matching bases.

    1. (a) 2x = 64
    2. (b) 3x = 1/9
    3. (c) 5x+1 = 125
    4. (d) 4x = 1/64
    5. (e) 82x = 32
  2. Fluency

    Solve each equation by writing both sides as a power of a common base.

    1. (a) 4x = 8
    2. (b) 9x = 27
    3. (c) 25x = 125
    4. (d) 42x−1 = 8x+1
  3. Fluency

    Use the given model to find the value of A at the given time.

    1. (a) A = 1000 × 2t; find A when t = 5.
    2. (b) A = 500 × (0.5)t; find A when t = 4.
    3. (c) A = 3000 × (1.1)t; find A when t = 3 (to the nearest whole number).
    4. (d) A = 800 × (1/2)t/6; find A when t = 12.
  4. Understanding

    Population growth.

    Context: A city has a population of 120 000. The population grows at 4% per year.
    1. (a) Write an exponential model P = P0(1 + r)t for the population after t years.
    2. (b) What will the population be after 10 years? Round to the nearest 1000.
    3. (c) In which year will the population first exceed 200 000? (Use trial and improvement or technology.)
    4. (d) Explain why this model may become unrealistic over very long periods of time.
  5. Understanding

    Radioactive decay and half-life.

    Context: A radioactive substance has a half-life of 5 years. The initial mass is 240 g. The model is A = 240 × (1/2)t/5.
    1. (a) Find the mass remaining after 5 years, 10 years, and 15 years.
    2. (b) Write the mass remaining as a fraction of the original after 25 years.
    3. (c) After how many years will only 15 g remain? (Use matching bases or trial and improvement.)
  6. Understanding

    Compound interest.

    Formula: A = P(1 + r/n)nt, where P = principal, r = annual interest rate (as decimal), n = compoundings per year, t = time in years.
    1. (a) $5000 invested at 6% p.a. compounded annually for 4 years. Find A.
    2. (b) $10 000 invested at 8% p.a. compounded quarterly for 5 years. Find A to the nearest cent.
    3. (c) How long (in full years) does it take for $2000 to double at 10% p.a. compounded annually? (Use trial and improvement.)
  7. Understanding

    Fitting exponential models to data.

    1. (a) The following data was collected: t = 0, A = 100; t = 1, A = 300; t = 2, A = 900. Determine the values of A0 and a in the model A = A0 × at.
    2. (b) An exponential decay model A = A0 × at has A = 80 at t = 0 and A = 20 at t = 2. Find A0 and a, and find A when t = 4.
    3. (c) Verify your model for part (b) by checking that A halves for every increase in t by one half-life.
  8. Problem Solving

    Doubling time investigation.

    Challenge. A colony of bacteria doubles every 20 minutes. The colony starts with 50 bacteria.
    1. (a) Write a model for the number of bacteria N after t minutes.
    2. (b) How many bacteria are there after 2 hours (120 minutes)?
    3. (c) Set up and solve an equation (using matching bases) to find when the colony first reaches 51 200 bacteria.
    4. (d) A second colony starts with 200 bacteria and also doubles every 20 minutes. After how many minutes does the second colony first exceed the first colony? Explain using the models.
  9. Problem Solving

    Drug concentration decay.

    Challenge. After a patient takes a 200 mg dose of medication, the concentration in the blood decays by 25% each hour.
    1. (a) Write an exponential model C = C0(1 − r)t for the concentration C mg after t hours.
    2. (b) Find C after 4 hours. Give your answer to 1 decimal place.
    3. (c) How many full hours does it take for the concentration to fall below 50 mg?
    4. (d) A second dose of 200 mg is given after 4 hours (so C from part (b) is added to 200 mg). Write the new model for the total concentration and find it after a further 2 hours.
  10. Problem Solving

    Depreciation and salvage value.

    Challenge. A machine depreciates in value by 15% per year. It was purchased for $80 000.
    1. (a) Write an exponential model for the value V after t years.
    2. (b) Find V after 5 years, to the nearest dollar.
    3. (c) The machine is scrapped when its value falls below $20 000. After how many full years does this happen? (Use trial and improvement.)
    4. (d) Compare this to a straight-line (linear) depreciation model where the machine loses $8 000 per year. Which model reaches $20 000 first?