Hard A-Level Exponential modelling Questions

Challenging, exam-style A-Level Exponential modelling questions with worked solutions. Stretch yourself on the hardest finding rate, simultaneous, prediction, cooling problems.

finding ratesimultaneouspredictioncoolingnatural logsreduction to linear form
A-Level34 questionsStep-by-step solutions
Question 1
7 markschallenging
Light intensity in a lake is I=I0e−0.15xI=I_0e^{-0.15x}, where xx is depth in metres. Find, to 22 decimal places, the depth at which the intensity is 5%5\% of the surface value.
Show worked solution

Worked solution

  1. Write the model

    I=I0e−0.15xI = I_0e^{-0.15x}

    Light intensity falls with depth x metres; μ=0.15 per metre is the attenuation constant.

  2. Set up the 5% equation

    e−0.15x=0.05e^{-0.15x} = 0.05

    We want the depth where intensity is 5% of the surface value; the I_0 cancels.

  3. Take natural logs

    −0.15x=ln⁡0.05-0.15x = \ln 0.05

    Logs bring the power down.

  4. Evaluate ln 0.05

    −0.15x=−2.9957-0.15x = -2.9957

    ln 0.05 ≈ -2.9957 (negative, since 0.05 < 1).

  5. Solve for the depth x

    x=−2.9957−0.15=19.97x = \dfrac{-2.9957}{-0.15} = 19.97

    So the intensity is 5% at about 19.97 m depth.

  6. Now find the intensity at 10 m

    II0=e−0.15×10\dfrac{I}{I_0} = e^{-0.15\times 10}

    Use the same model to check the intensity at 10 m.

  7. Work out the power

    −0.15×10=−1.5-0.15\times 10 = -1.5

    Multiply out the exponent.

  8. Evaluate

    e−1.5=0.2231e^{-1.5} = 0.2231

    e^{-1.5} ≈ 0.2231.

  9. Convert to a percentage

    22.31%22.31\%

    So about 22.31% of surface light remains at 10 m. The required answer is the depth, ≈ 19.97 m.

  10. State the final answer clearly

    x=19.97 mx = 19.97\ \text{m}

    Here is our result written out on its own. It is worth stating it clearly, with the correct units and to the accuracy the question asked for.

  11. Interpret the answer in context

    x=19.97 mx = 19.97\ \text{m}

    Always turn the number back into words for the real situation. Ask yourself: does this value make sense for the population, temperature, mass or money in the question?

  12. Why logarithms were the key tool

    ln⁡(ex)=x,log⁡(10x)=x\ln\left(e^{x}\right)=x,\qquad \log\left(10^{x}\right)=x

    The variable started trapped inside a power. Logs work because they are the exact inverse of raising a base to a power, so they bring the unknown down where we can reach it.

  13. Link back to the log and index laws

    log⁡(pq)=log⁡p+log⁡q,log⁡(pn)=nlog⁡p\log(pq)=\log p+\log q,\quad \log\left(p^{n}\right)=n\log p

    These are the same log and index rules you met earlier in the course. Exponential modelling simply uses them in a real-world setting, so revising them always helps here.

  14. Watch out for a common error

    isolate the exponential BEFORE taking logs\text{isolate the exponential BEFORE taking logs}

    A very common slip is to take logs while a number is still multiplying the exponential term. Always get the e…e^{\dots} (or the power) on its own first, then log.

  15. Check the answer is reasonable

    x=19.97 mx = 19.97\ \text{m}

    Sanity-check the size and sign: a growth model should give increasing values and a decay model decreasing ones, and a time should come out positive.

Answer
x=19.97 mx = 19.97\ \text{m}
Question 2
7 markschallenging
A country's population is P=2.4e0.035tP=2.4e^{0.035t} million, tt years after a census. Find the population after 2525 years, to 44 decimal places (in millions).
Show worked solution

Worked solution

  1. Write the model

    P=2.4e0.035tP = 2.4e^{0.035t}

    Population in millions grows continuously at 3.5% per year, so k=0.035.

  2. Substitute t = 25

    P=2.4e0.035×25P = 2.4e^{0.035\times 25}

    We want the population after 25 years.

  3. Work out the power

    0.035×25=0.8750.035\times 25 = 0.875

    Multiply the rate by the time.

  4. Evaluate the exponential

    e0.875=2.3989e^{0.875} = 2.3989

    e^{0.875} ≈ 2.3989.

  5. Compute the population

    P=2.4×2.3989=5.7573P = 2.4\times 2.3989 = 5.7573

    So the population is about 5.7574 million after 25 years.

  6. Set up the doubling condition

    e0.035T=2e^{0.035T} = 2

    Now check the doubling time as extra context.

  7. Take natural logs

    0.035T=ln⁡20.035T = \ln 2

    Logs bring the power down.

  8. Solve the doubling time

    T=ln⁡20.035=19.80T = \dfrac{\ln 2}{0.035} = 19.80

    So the population doubles roughly every 19.80 years, consistent with our answer.

  9. State the final answer clearly

    5.7573 million5.7573\ \text{million}

    Here is our result written out on its own. It is worth stating it clearly, with the correct units and to the accuracy the question asked for.

  10. Interpret the answer in context

    5.7573 million5.7573\ \text{million}

    Always turn the number back into words for the real situation. Ask yourself: does this value make sense for the population, temperature, mass or money in the question?

  11. Why logarithms were the key tool

    ln⁡(ex)=x,log⁡(10x)=x\ln\left(e^{x}\right)=x,\qquad \log\left(10^{x}\right)=x

    The variable started trapped inside a power. Logs work because they are the exact inverse of raising a base to a power, so they bring the unknown down where we can reach it.

  12. Link back to the log and index laws

    log⁡(pq)=log⁡p+log⁡q,log⁡(pn)=nlog⁡p\log(pq)=\log p+\log q,\quad \log\left(p^{n}\right)=n\log p

    These are the same log and index rules you met earlier in the course. Exponential modelling simply uses them in a real-world setting, so revising them always helps here.

  13. Watch out for a common error

    isolate the exponential BEFORE taking logs\text{isolate the exponential BEFORE taking logs}

    A very common slip is to take logs while a number is still multiplying the exponential term. Always get the e…e^{\dots} (or the power) on its own first, then log.

  14. Check the answer is reasonable

    5.7573 million5.7573\ \text{million}

    Sanity-check the size and sign: a growth model should give increasing values and a decay model decreasing ones, and a time should come out positive.

  15. Relate the result to the graph

    y=Aekty=Ae^{kt}

    Picture where this answer sits on the growth or decay curve. Seeing the point on the curve is a quick way to confirm the value looks right.

Answer
5.7573 million5.7573\ \text{million}
Question 3
5 markschallenging
For Newton's cooling model θ=θs+(θ0−θs)e−kt\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}, describe the long-term behaviour of the temperature.
Show worked solution

Worked solution

  1. Write the cooling model

    θ=θs+(θ0−θs)e−kt\theta = \theta_s + (\theta_0-\theta_s)e^{-kt}

    Here θ_s is the surrounding (room) temperature and θ_0 is the starting temperature of the object.

  2. Check the starting value

    t=0⇒θ=θs+(θ0−θs)e0t=0\Rightarrow\theta = \theta_s + (\theta_0-\theta_s)e^{0}

    Put t=0 to confirm the model behaves sensibly at the start.

  3. Simplify at t = 0

    θ=θs+(θ0−θs)=θ0\theta = \theta_s + (\theta_0-\theta_s) = \theta_0

    Since e^{0}=1 the bracket is fully present, giving θ_0. Good — the object starts at its stated temperature.

  4. Identify the changing part

    (θ0−θs)e−kt(\theta_0-\theta_s)e^{-kt}

    Only this term changes with time; θ_s is a fixed constant added on.

  5. Sign of the constant

    θ0>θs⇒θ0−θs>0\theta_0>\theta_s\Rightarrow \theta_0-\theta_s>0

    For a cooling object the start is hotter than the room, so the bracket is positive.

  6. Behaviour of the exponential

    k>0⇒e−kt is positive and decreasingk>0\Rightarrow e^{-kt}\ \text{is positive and decreasing}

    Because the power −kt is negative and getting more negative, e^{-kt} shrinks towards 0.

  7. Consider large t

    e−kt→0 as t→∞e^{-kt}\to 0\ \text{as}\ t\to\infty

    As time goes on the exponential term is divided by a larger and larger number, so it heads to zero.

  8. Take the limit of θ

    θ→θs+0\theta\to\theta_s + 0

    As the changing term vanishes, only the constant θ_s is left.

  9. State the limiting temperature

    θ→θs\theta\to\theta_s

    So in the long run the object cools to the surrounding temperature θ_s.

  10. Explain it never crosses θ_s

    (θ0−θs)e−kt>0(\theta_0-\theta_s)e^{-kt}>0

    The extra term is always positive (never zero for finite t), so θ stays just above θ_s and never drops below it.

  11. Interpret physically

    object cannot get colder than its surroundings\text{object cannot get colder than its surroundings}

    This matches real life: a hot drink cools to room temperature but no further, which is exactly what the model predicts.

  12. Name the graph feature

    θ=θs is a horizontal asymptote\theta = \theta_s\ \text{is a horizontal asymptote}

    The curve gets ever closer to the horizontal line θ=θ_s from above without ever touching it.

  13. Rule out cooling to zero

    θ→θs≠0 (unless θs=0)\theta\to\theta_s\neq 0\ (\text{unless }\theta_s=0)

    A common mistake is to think it cools to 0°C; it actually levels off at room temperature θ_s.

  14. Describe the rate

    cooling slows as θ→θs\text{cooling slows as }\theta\to\theta_s

    The temperature falls quickly at first, then more and more slowly as it nears θ_s, giving the familiar curved shape.

  15. State the conclusion

    θ→θs (asymptote)\theta\to\theta_s\ \text{(asymptote)}

    So the long-term behaviour is that θ approaches the surrounding temperature θ_s, a horizontal asymptote it never crosses.

Answer
θ\theta approaches the surrounding temperature θs\theta_s (a horizontal asymptote).
Question 4
7 markschallenging
Data follow y=axny=ax^{n} through (1,4)(1,4) and (3,36)(3,36). Find the positive value of xx for which y=400y=400.
Show worked solution

Worked solution

  1. Assume a power law

    y=axny = ax^{n}

    Two points let us find both a and n.

  2. Use the point (1,4)

    4=a×1n=a4 = a\times 1^{n} = a

    Since 1 to any power is 1, the first point gives a directly.

  3. State a

    a=4a = 4

    So the model is y = 4x^n.

  4. Use the point (3,36)

    36=4×3n36 = 4\times 3^{n}

    Substitute the second point to find n.

  5. Divide by 4

    3n=93^{n} = 9

    Isolate the power of 3.

  6. Recognise the power

    3n=32⇒n=23^{n} = 3^{2}\Rightarrow n = 2

    Since 9 = 3², the powers match, giving n = 2.

  7. Write the model

    y=4x2y = 4x^{2}

    This is the fitted power law.

  8. Set up the target equation

    4x2=4004x^{2} = 400

    We want the x for which y = 400.

  9. Divide by 4

    x2=100x^{2} = 100

    Isolate x².

  10. Take the positive square root

    x=10x = 10

    We take the positive root because x represents a positive quantity here.

  11. State the final answer clearly

    x=10x = 10

    Here is our result written out on its own. It is worth stating it clearly, with the correct units and to the accuracy the question asked for.

  12. Interpret the answer in context

    x=10x = 10

    Always turn the number back into words for the real situation. Ask yourself: does this value make sense for the population, temperature, mass or money in the question?

  13. Why logarithms were the key tool

    ln⁡(ex)=x,log⁡(10x)=x\ln\left(e^{x}\right)=x,\qquad \log\left(10^{x}\right)=x

    The variable started trapped inside a power. Logs work because they are the exact inverse of raising a base to a power, so they bring the unknown down where we can reach it.

  14. Link back to the log and index laws

    log⁡(pq)=log⁡p+log⁡q,log⁡(pn)=nlog⁡p\log(pq)=\log p+\log q,\quad \log\left(p^{n}\right)=n\log p

    These are the same log and index rules you met earlier in the course. Exponential modelling simply uses them in a real-world setting, so revising them always helps here.

  15. Watch out for a common error

    isolate the exponential BEFORE taking logs\text{isolate the exponential BEFORE taking logs}

    A very common slip is to take logs while a number is still multiplying the exponential term. Always get the e…e^{\dots} (or the power) on its own first, then log.

Answer
x=10x = 10
Question 5
7 markschallenging
A quantity following y=Aekty=Ae^{kt} quadruples every 99 hours. Find, to 22 decimal places, the time for it to increase tenfold.
Show worked solution

Worked solution

  1. Write the model

    y=Aekty = Ae^{kt}

    We only know how the quantity multiplies over time, which is enough to find k.

  2. Use the quadrupling condition

    Ae9k=4AAe^{9k} = 4A

    Quadrupling means multiplying by 4 every 9 hours.

  3. Divide by A

    e9k=4e^{9k} = 4

    The start value cancels, as always for a fixed multiplier.

  4. Take natural logs

    9k=ln⁡49k = \ln 4

    Logs bring the power down.

  5. Solve for k

    k=ln⁡49=0.1540k = \dfrac{\ln 4}{9} = 0.1540

    So k ≈ 0.1540 per hour.

  6. Set up the ten-fold condition

    ekt=10e^{kt} = 10

    Increasing by a factor of 10 means the exponential factor equals 10.

  7. Take natural logs

    kt=ln⁡10kt = \ln 10

    Apply ln to reach t.

  8. Evaluate ln 10

    kt=2.3026kt = 2.3026

    ln 10 ≈ 2.3026.

  9. Solve for t

    t=2.30260.1540=14.95t = \dfrac{2.3026}{0.1540} = 14.95

    So the quantity increases tenfold after about 14.95 hours.

  10. State the final answer clearly

    t=14.95 hourst = 14.95\ \text{hours}

    Here is our result written out on its own. It is worth stating it clearly, with the correct units and to the accuracy the question asked for.

  11. Interpret the answer in context

    t=14.95 hourst = 14.95\ \text{hours}

    Always turn the number back into words for the real situation. Ask yourself: does this value make sense for the population, temperature, mass or money in the question?

  12. Why logarithms were the key tool

    ln⁡(ex)=x,log⁡(10x)=x\ln\left(e^{x}\right)=x,\qquad \log\left(10^{x}\right)=x

    The variable started trapped inside a power. Logs work because they are the exact inverse of raising a base to a power, so they bring the unknown down where we can reach it.

  13. Link back to the log and index laws

    log⁡(pq)=log⁡p+log⁡q,log⁡(pn)=nlog⁡p\log(pq)=\log p+\log q,\quad \log\left(p^{n}\right)=n\log p

    These are the same log and index rules you met earlier in the course. Exponential modelling simply uses them in a real-world setting, so revising them always helps here.

  14. Watch out for a common error

    isolate the exponential BEFORE taking logs\text{isolate the exponential BEFORE taking logs}

    A very common slip is to take logs while a number is still multiplying the exponential term. Always get the e…e^{\dots} (or the power) on its own first, then log.

  15. Check the answer is reasonable

    t=14.95 hourst = 14.95\ \text{hours}

    Sanity-check the size and sign: a growth model should give increasing values and a decay model decreasing ones, and a time should come out positive.

Answer
t=14.95 hourst = 14.95\ \text{hours}

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