Exponential modelling Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Exponential modelling questions. See exactly how to solve problems on growth model, initial value, substitution, decay model.

growth modelinitial valuesubstitutiondecay modely=ab^tsolving
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A population of insects is modelled by P=500e0.03tP=500e^{0.03t}, where tt is the time in years. State the initial population.

Worked solution

  1. Write down the model

    P=500e0.03tP = 500e^{0.03t}

    This models a population P after t years. The number in front, 500, sits outside the exponential.

  2. Substitute t = 0

    P=500e0.03×0P = 500e^{0.03 \times 0}

    The initial population means the value at the very start, when time t is zero.

  3. Simplify the power

    P=500e0=500×1P = 500e^{0} = 500 \times 1

    Anything to the power 0 equals 1, so e^{0}=1. This is a key index law from earlier work on indices.

  4. State the initial value

    P=500P = 500

    So the model starts at 500. The constant A in P=Ae^{kt} always gives the starting amount.

Answer
500500
Question 2
3 markseasy
A population of insects is modelled by P=500e0.03tP=500e^{0.03t}, where tt is the time in years. Find the population after 1010 years, giving your answer to the nearest whole number.

Worked solution

  1. Write down the model

    P=500e0.03tP = 500e^{0.03t}

    This is our exponential model. Here 500 is the starting amount and the number multiplying t in the power controls how fast it changes.

  2. Substitute the given value

    P=500e0.03×10P = 500e^{0.03 \times 10}

    We replace t with 10. Everything else in the formula stays exactly the same.

  3. Work out the power and evaluate

    e0.03×10=e0.3=1.3499e^{0.03 \times 10} = e^{0.3} = 1.3499

    Multiply the two numbers in the exponent, then use the e^x button on your calculator. Remember e is just a fixed number (about 2.718), so this is a single value.

  4. Multiply by the starting amount

    P=500×1.3499=675P = 500 \times 1.3499 = 675

    Multiplying the starting amount by the growth/decay factor gives the final value. Round sensibly for the context.

Answer
675675
Question 3
3 markseasy
A radioactive sample has mass mm grams after tt days, modelled by m=80e−0.05tm=80e^{-0.05t}. Find the mass after 2020 days, giving your answer to 22 decimal places.

Worked solution

  1. Write down the model

    m=80e−0.05tm = 80e^{-0.05t}

    This is our exponential model. Here 80 is the starting amount and the number multiplying t in the power controls how fast it changes.

  2. Substitute the given value

    m=80e−0.05×20m = 80e^{-0.05 \times 20}

    We replace t with 20. Everything else in the formula stays exactly the same.

  3. Work out the power and evaluate

    e−0.05×20=e−1=0.3679e^{-0.05 \times 20} = e^{-1} = 0.3679

    Multiply the two numbers in the exponent, then use the e^x button on your calculator. Remember e is just a fixed number (about 2.718), so this is a single value.

  4. Multiply by the starting amount

    m=80×0.3679=29.43m = 80 \times 0.3679 = 29.43

    Multiplying the starting amount by the growth/decay factor gives the final value. Round sensibly for the context.

Answer
29.43 g29.43\ \text{g}
Question 4
3 markseasy
The value £V\pounds V of a car after tt years is modelled by V=18000×0.85tV=18000\times 0.85^{t}. Find its value after 33 years.

Worked solution

  1. Write down the model

    V=18000×0.85tV = 18000\times 0.85^{t}

    The car starts at £18000 and keeps 85% of its value each year, so we multiply by 0.85 repeatedly.

  2. Substitute t = 3

    V=18000×0.853V = 18000\times 0.85^{3}

    After 3 years the factor 0.85 is applied three times, which is written as a power of 3.

  3. Work out the power

    0.853=0.6141250.85^{3} = 0.614125

    Use the power button on your calculator. This is the same index idea as in the surds and indices topic.

  4. Multiply by the starting value

    V=18000×0.614125=11054.25V = 18000\times 0.614125 = 11054.25

    Multiplying the start value by the decay factor gives the value after 3 years, in pounds.

Answer
£11054.25\pounds 11054.25
Question 5
3 markseasy
Solve 5e2x=405e^{2x}=40, giving your answer to 44 decimal places.

Worked solution

  1. Write the equation to solve

    5e2x=405e^{2x} = 40

    We want the time when the quantity reaches 40. Setting the model equal to this target gives an equation in x.

  2. Divide both sides by 5

    e2x=8e^{2x} = 8

    We isolate the exponential part first. Dividing undoes the multiplication by 5, leaving e to a power on its own.

  3. Take natural logs of both sides

    ln⁡(e2x)=ln⁡(8)\ln\left(e^{2x}\right) = \ln\left(8\right)

    The variable is stuck in the power, so we use logs to bring it down. Natural log (ln) is chosen because it is the inverse of e^x.

  4. Cancel the logs and solve for x

    2x=2.0794,x=2.07942=1.03972x = 2.0794,\quad x = \dfrac{2.0794}{2} = 1.0397

    Because ln and e cancel, the left side becomes just the power. Dividing by the number in front of x then leaves x on its own — this is the value we were looking for.

Answer
x=1.0397x = 1.0397

Unlock 65 more Exponential modelling questions

Create a free account to work through every A-Level Exponential modelling question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Exponential modelling practice

Hard QuestionsPractice Questions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles