Log & Exponential Applications

9709 Paper 3 · §3.2 Logarithmic & Exponential Functions · Lesson 5 · Exponential Growth & Decay Modelling

Quick Recall — Adaptive Quiz

Five questions from Lessons 1–4, one at a time. The quiz adapts to you: answer correctly and the next question is harder and worth more points; miss one and it gets easier. Collect your total score at the end!

The Wall

A colony of bacteria grows according to the model $P = 100e^{0.05t}$, where $t$ is time in hours. Reading the model is easy: $P_0 = 100$ cells at $t = 0$, hourly growth rate $k = 0.05$, and substituting $t = 50$ gives $P \approx 1218$. But stop for a moment: where did this model come from?

A biologist never receives a finished model — she observes facts: "the culture starts with 100 cells" and "the count doubles about every 13.9 hours." Turning those facts into a formula $P = 100e^{kt}$ means choosing $k$ so that the doubling time $\dfrac{\ln 2}{k}$ equals 13.9 hours, i.e. $k = \dfrac{\ln 2}{13.9} \approx 0.05$. Building a model from observations is this lesson's focus. And once the model exists, even "when does the population reach 1 million?" is just one more step: set $P = 10^6$ and invert with $\ln$ to get $t \approx 184$ hours.

The Need

We need three modelling skills: (1) read a model $P = P_0e^{kt}$ — spot the starting value and say whether the quantity grows or decays; (2) build and use a model — turn observed facts (a doubling time or half-life) into the constant $k$, then invert with natural logs to find when a target value is reached; (3) read bounded models like $h = a - bc^t$ — quantities that level off towards a limit. Today's lesson builds exactly this toolkit.

📖 Key Concepts
Key skill 1 — Reading a model: a quantity that grows or decays at a rate proportional to its current size is modelled by $$P = P_0\, e^{kt}$$ where $P_0$ is the initial amount (set $t = 0$ to check: the model gives $P = P_0$), $k$ is the rate constant ($k \gt 0$ means growth, $k \lt 0$ means decay — the larger $|k|$, the faster the change) and $t$ is the time. Predicting forward is pure substitution: put in the given $t$ and evaluate. The colony from The Wall, $P = 100e^{0.05t}$, reads as: start at 100, grow at rate 0.05 per hour.

Interactive: drag $P_0$ to move the starting point, drag $k$ to change how steeply the curve grows (or decays), and drag $t$ to slide along the curve. The axes stay fixed, so you can see each parameter's effect.

50
0.020
30

Current value: P=50e0.020×30≈91.11P = 50 e^{0.020 \times 30} \approx 91.11

Key skill 2 — Solving backwards: "How long until…?" questions run the model in reverse — the target $P$ is given and $t$ is unknown. Take natural logs of both sides of $P = P_0e^{kt}$ and rearrange: $$\ln P = \ln P_0 + kt \quad\Longrightarrow\quad t = \frac{\ln P - \ln P_0}{k} = \frac{\ln(P/P_0)}{k}.$$ One logarithm move replaces hours of trial-and-improvement with an exact answer — the move The Wall's "when do we reach 1 million?" was waiting for.

Interactive: the colony grows as $P = 100e^{0.05t}$. Drag the target population and watch the logarithm work backwards to find the time $t$.

1000

Key skill 3 — Models with a limit: trees do not grow forever, and hot coffee cools down to the temperature of the room. A bounded model contains a term that dies away: $$h = a - b\,c^{\,t} \ (0 \lt c \lt 1) \ \ \text{rises to the limit } a \qquad\qquad T = a + b\,e^{-kt} \ (k \gt 0) \ \ \text{falls to the limit } a$$ At $t = 0$, the quantity is $a - b$ (or $a + b$): the starting value. As $t \to \infty$, the dying term $\to 0$, so the quantity $\to a$: the curve levels off towards the limit $a$ but never quite reaches it. To find the long-run value, set the dying term to 0 and read off the constant term.

Interactive: a cup of coffee starts at $80^\circ\text{C}$ and cools in a room kept at $20^\circ\text{C}$. Its temperature is modelled by $T = 20 + 60\,e^{-0.1t}$ °C, where $t$ is the time in minutes. Drag $t$ and watch $T$ fall towards the room temperature.

10

Activity 1 · Group Growth & Decay Simulator

You are a team of modellers. The simulator below is your model $P = P_0e^{kt}$, and each task is a real-world question about a quantity that grows or decays: predict a population, work backwards to a time, then design a growth rate. Assign roles: one student adjusts parameters, one predicts the outcome, one records observations, one presents findings. Work through three phases: Free Play, Guided Inquiry, and Challenge.

100
0.10
10

Formula: P=P0×ekt=100×e0.10tP = P_0 \times e^{kt} = 100 \times e^{0.10t}

Current value: P(10)=100×e1.00≈271.8P(10) = 100 \times e^{1.00} \approx 271.8

Phase 1 Free Play (2 min)

Adjust P0P_0, kk, and tt freely. Observe how each parameter changes the curve. Try: What happens when kk changes from positive to negative? What happens when P0P_0 doubles?

Phase 2 Guided Inquiry Quest (3 min)

Solve 3 real-world modelling cases with the simulator: a bacteria culture, a drug in the bloodstream, a radioactive sample. For each round, read the model, apply the setup, predict the answer, then check it on the simulator. Score 3 points for a correct first guess, 1 point if you need a second try!

Score: 0 Round 1 / 3
Phase 3 Challenge (1 min)

A fish farm starts with 200 fish. The owner wants the stock to double in exactly 10 years, so the fish population is modelled by $P = 200e^{kt}$. Find the value of $k$ that the model needs.

Debrief

The doubling time formula is tdouble=ln⁡2kt_{\text{double}} = \dfrac{\ln 2}{k}. For decay, the half-life is thalf=ln⁡2∣k∣t_{\text{half}} = \dfrac{\ln 2}{|k|}. Notice that doubling time and half-life depend only on kk, not on P0P_0.

Activity 2 · Individual Sketch & Verify: Model Builder

Three real-world models, three different shapes: growth without a limit, decay towards a limit, and growth towards a limit. Read each scenario, then sketch what you think the graph looks like by drawing on the canvas with your mouse. Click "Verify" to see the correct curve - and its long-run limit, when it has one - overlaid on your sketch. Compare and reflect on any differences!

Scenario 1 of 3
Scenario 1 Town Population

The population of a town is modelled by $P = 10000 \times 10^{0.1t}$, where $t$ is the time in years. Sketch $P$ against $t$ for $t = 0$ to 10 years.

Debrief

Exponential growth curves start gently then rise steeply. Decay curves drop quickly then level off toward zero (or toward an asymptote like room temperature). The key shape signature: a constant ratio over equal time intervals.

✏️ Worked Examples
Worked Example 1 · Group Population Growth — P=100e0.05tP = 100e^{0.05t}
Ex 2D Q6 Medium

A colony of humans settle on a previously uninhabited planet. After tt years, their population, PP, is given by P=100e0.05tP = 100e^{0.05t}.

(i) Sketch the graph of PP against tt.
(ii) How many settlers land on the planet initially?
(iii) What is the population after 50 years?
(iv) How long does it take the population to reach 1 million?

Explorer: drag tt to read the population P=100e0.05tP = 100e^{0.05t}. Try to find the tt that makes P=1,000,000P = 1{,}000{,}000.

50

At tt: P=100 e0.05×50≈1218.2P = 100\,e^{0.05 \times 50} \approx 1218.2

Hint

For part (iv), set P=1,000,000P = 1{,}000{,}000 and divide both sides by 100 first. Then take ln⁡\ln of both sides to bring tt down from the exponent. Remember ln⁡(104)=4ln⁡10\ln(10^4) = 4\ln 10.

Part (i): Sketch

P=100e0.05tP = 100e^{0.05t} is an exponential growth curve. It passes through (0,100)(0, 100), rises slowly at first and then steeply, with no upper bound. As t→−∞t \to -\infty, P→0P \to 0 (horizontal asymptote P=0P = 0 for negative tt).

Part (ii): Initial population

At t=0t = 0: P=100e0=100×1=100P = 100e^{0} = 100 \times 1 = 100.

100 settlers\boxed{100 \text{ settlers}}

Part (iii): Population after 50 years

At t=50t = 50: P=100e0.05×50=100e2.5P = 100e^{0.05 \times 50} = 100e^{2.5}.

Using e2.5≈12.1825e^{2.5} \approx 12.1825:

P≈100×12.1825≈1218(to the nearest integer)P \approx 100 \times 12.1825 \approx 1218 \quad (\text{to the nearest integer})

Part (iv): Time to reach 1 million

Set P=1,000,000P = 1{,}000{,}000:

100e0.05t=1,000,000100e^{0.05t} = 1{,}000{,}000

e0.05t=10,000e^{0.05t} = 10{,}000

Take ln⁡\ln of both sides:

0.05t=ln⁡(10,000)=4ln⁡10≈9.21030.05t = \ln(10{,}000) = 4\ln 10 \approx 9.2103

t=9.21030.05≈184.2t = \dfrac{9.2103}{0.05} \approx 184.2

t≈184 years (3 s.f.)\boxed{t \approx 184 \text{ years (3 s.f.)}}

Worked Example 2 · Individual Working with a Bounded Model — Pine Tree Growth
Ex 2A Q4 Standard

The height $h$ m of a species of pine tree, $t$ years after planting, is modelled by the equation $h = 20 - 19 \times 0.9^t$. (i) What is the height of the trees when they are planted? (ii) Calculate the height of the trees after 2 years. (iii) Use trial and improvement to find the number of years for the height to reach 10 metres. (iv) What height does the model predict that the trees will eventually reach?

Trial and improvement: drag $t$ and watch $h(t)$ climb towards the 10 m target line. Start with $t = 6$ and $t = 7$ to bracket the answer, then zoom in.

6.00

At this $t$:

Hint

For (iii), try $t = 6$ then $t = 7$: $h(6) \approx 9.90 < 10$ and $h(7) \approx 10.91 > 10$, so the answer lies between 6 and 7 years. Keep narrowing the interval. To check exactly, solve $0.9^t = \dfrac{10}{19}$ by taking logs of both sides.

Step 1: (i) Height when planted ($t = 0$)

$$h(0) = 20 - 19 \times 0.9^0 = 20 - 19 = 1$$ The trees are 1 m tall when planted.

Step 2: (ii) Height after 2 years

$$h(2) = 20 - 19 \times 0.9^2 = 20 - 19 \times 0.81 = 20 - 15.39 = 4.61$$ After 2 years the trees are 4.61 m tall.

Step 3: (iii) Trial and improvement — when is $h = 10$?

Try $t = 6$: $h(6) = 20 - 19 \times 0.9^6 \approx 9.90$ m — just below 10 m.
Try $t = 7$: $h(7) = 20 - 19 \times 0.9^7 \approx 10.91$ m — just above 10 m.
So the answer lies between 6 and 7 years. Try $t = 6.1$: $h(6.1) \approx 10.00$ m. The tree reaches 10 m after about 6.1 years.

Logarithm shortcut: solve $20 - 19 \times 0.9^t = 10$ exactly: $$0.9^t = \frac{10}{19} \;\Longrightarrow\; t = \frac{\ln(10/19)}{\ln 0.9} \approx 6.09 \text{ years}$$

Step 4: (iv) Long-term height (the limit)

As $t \to \infty$, $0.9^t \to 0$, so $h \to 20 - 0 = 20$. The model predicts the trees will eventually approach 20 m but never exceed it.

Check

At $t = 6.09$: $h(6.09) = 20 - 19 \times 0.9^{6.09} \approx 20 - 19 \times 0.5263 \approx 10.00$ m ✓. The trial-and-improvement and logarithm routes agree.

🎯 Concept Quiz — Quick Check

Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!

Score: 0 / 4
Q1 (Objective 1) — Reading a model

A population is modelled by $P = 500e^{-0.12t}$. What does the model predict at $t = 0$, and how does the population change over time?

At $t = 0$: $P(0) = 500e^{0} = 500$. The rate $-0.12$ is negative, so the quantity decays: $P$ falls towards 0 as $t$ grows.

Q2 (Objective 2) — Designing a model

A population of 200 doubles in exactly 10 years, modelled by $P = 200e^{kt}$. Find the value of $k$.

After 10 years $P = 400$: $200e^{10k} = 400$, so $e^{10k} = 2$ and $k = \dfrac{\ln 2}{10} \approx 0.0693$.

Q3 (Objective 3) — Inverting the model

Under the model $P = 200e^{0.04t}$, how long does it take for $P$ to reach 800?

Set $200e^{0.04t} = 800$, so $e^{0.04t} = 4$. Taking logs: $0.04t = \ln 4 \approx 1.386$, giving $t \approx 34.7$ time units.

Q4 (Objective 4) — Limits of a model

Birds on an island: $n = 200 + 320 \times 2^{-t}$. What does the model predict for the long-term population?

As $t \to \infty$, $2^{-t} \to 0$, so $n \to 200$: the population settles at the food-supply limit of 200 birds.

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You scored 0 / 4
✏️ More Practice

Work from standard to challenging: read a growth model, handle a model with a limit, linearise a power model, then fit and extrapolate an exponential-decay model from data.

Exercise 2A Q3 Standard

The growth in population $P$ of a certain town after time $t$ years can be modelled by the equation $P = 10000 \times 10^{0.1t}$. (i) State the initial population of the town and calculate the population after 5 years. (ii) Sketch the graph of $P$ against $t$.

Step 1: (i) Initial population

At $t = 0$: $$P(0) = 10000 \times 10^0 = 10000$$ The initial population is 10 000.

Step 2: (i) Population after 5 years

$$P(5) = 10000 \times 10^{0.1 \times 5} = 10000 \times 10^{0.5} = 10000 \times 3.162 \approx 31623$$ After 5 years the population is about 31 600 (3 s.f.).

Step 3: (ii) Sketch

Exponential growth through $(0, 10000)$, rising ever more steeply, with no limit: the curve hugs the axes for small $t$, then climbs sharply.

Exercise 2A Q5 Standard

A flock of birds was caught in a hurricane and blown far out to sea. They landed on a small and very remote island and settled there. There was only a limited sustainable supply of suitable food for them on the island. Their population size, $n$ birds, at a time $t$ years after they arrived, can be modelled by the equation $n = 200 + 320 \times 2^{-t}$. (i) How many birds arrived on the island? (ii) How many birds were there after 5 years? (iii) Sketch a graph of $n$ against $t$. (iv) What is the long-term size of the population?

Step 1: (i) Birds that arrived

At $t = 0$: $$n(0) = 200 + 320 \times 2^0 = 200 + 320 = 520$$ 520 birds arrived.

Step 2: (ii) After 5 years

$$n(5) = 200 + 320 \times 2^{-5} = 200 + 320 \times \frac{1}{32} = 200 + 10 = 210$$ After 5 years there were 210 birds.

Step 3: (iii) Sketch

The curve starts at $(0, 520)$ and falls towards its limit: it drops steeply at first, then levels off towards 200.

Step 4: (iv) Long-term population

As $t \to \infty$, $2^{-t} \to 0$, so $n \to 200$. The population settles at 200 birds — the limit set by the food supply.

Exercise 2C Q6 Medium

An experimenter takes observations of a quantity $y$ for various values of a variable $x$. He wishes to test whether these observations conform to a formula $y = A \times x^B$ and, if so, to find the values of the constants $A$ and $B$. (i) Take logarithms of both sides of the formula and explain what he should plot, and how to find $A$ and $B$ from the graph. (ii) The observations are: $x$: 4, 7, 10, 13, 20   |   $y$: 3.00, 3.97, 4.74, 5.41, 6.71. Plot $\log y$ against $\log x$ and use the (near-)straight line to estimate $B$ and $A$, and so give the formula.

Step 1: (i) Linearise the model

Take logs of $y = Ax^B$: $$\log y = \log A + B\log x$$ Plot $\log y$ against $\log x$: the graph is a straight line with gradient $B$ and $\log y$-intercept $\log A$.

Step 2: (ii) Read the gradient $B$

Using the ends of the data: $$B = \frac{\log 6.71 - \log 3.00}{\log 20 - \log 4} \approx \frac{0.8267 - 0.4771}{1.3010 - 0.6021} \approx 0.50$$

Step 3: (ii) Read the intercept and give the formula

The intercept is $\log A \approx 0.176$, so $A = 10^{0.176} \approx 1.5$. Hence $$y = 1.5\,x^{0.5} = 1.5\sqrt{x}$$ Check: $x = 10 \Rightarrow y = 1.5\sqrt{10} \approx 4.74$ ✓, matching the data.

Exercise 2C Q7 Challenging

After the introduction of a vaccine, the numbers of new cases of a virus infection in successive weeks are: week $t$: 1, 2, 3, 4, 5   |   new cases $y$: 240, 150, 95, 58, 38. (i) Plot the points $(t, y)$ on graph paper and join them with a smooth curve. (ii) Decide on a suitable model to relate $t$ and $y$, and plot an appropriate graph to find this model. (iii) Find the value of $y$ when $t = 15$, and explain what you find.

Step 1: (i) Sketch the curve

The points fall on a smooth, steadily-decreasing curve — each week roughly 60–65% of the previous week.

Step 2: (ii) Choose and fit the model

The roughly constant ratio suggests exponential decay $y = a\,b^t$. Plotting $\ln y$ against $t$ gives a straight line; fitting it gives $\ln y = -0.464t + 5.94$, so $$y \approx 380\,e^{-0.464t} \qquad \left(\text{or } y \approx 240 \times 0.63^{\,t-1}\right)$$ The weekly factor is $e^{-0.464} \approx 0.63$.

Step 3: (iii) Extrapolate to $t = 15$

$$y(15) \approx 380\,e^{-0.464 \times 15} \approx 0.36$$ The model predicts about 0.4 new cases per week — effectively zero, so the vaccine is bringing the infection under control. (Extrapolating far beyond the data is risky: real outbreaks may level off or be affected by other factors.)

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