Exponential Modelling – A Level Maths Year 1

1 Exponential Growth Models

A quantity grows exponentially when its rate of increase is proportional to its current size. Populations, money earning compound interest and the early spread of an infection all behave this way. DfE content area F asks you to use exponential functions to model growth and decay in context, and to consider the limitations and refinements of those models.

The standard model

$N = Ae^{kt}$ $A$ = initial value (when $t = 0$), $k > 0$ growth, $k < 0$ decay
$\dfrac{\mathrm{d}N}{\mathrm{d}t} = kAe^{kt} = kN$

Two equivalent forms. $N = Ab^t$ and $N = Ae^{kt}$ describe the same models, with $b = e^k$ (so $k = \ln b$). A $3\%$ annual increase is $b = 1.03$, or $k = \ln 1.03 \approx 0.0296$.

Worked Example 1 — A population model

The population of a town $t$ years after 2020 is modelled by $P = 2000e^{0.03t}$.

(a) Find the predicted population in 2030. (b) Find how long it takes the population to double. (c) Find the rate of growth in 2030.

①(a) $t = 10$: $P = 2000e^{0.3} = 2699.7\ldots \approx 2700$.

②(b) $2000e^{0.03t} = 4000 \Rightarrow e^{0.03t} = 2 \Rightarrow t = \dfrac{\ln 2}{0.03} = 23.1$ years.

③(c) $\dfrac{\mathrm{d}P}{\mathrm{d}t} = 0.03 \times 2000e^{0.03t} = 60e^{0.03t}$. At $t = 10$: $60e^{0.3} \approx 81$ people per year.

The doubling time $\dfrac{\ln 2}{k}$ does not depend on the starting value — it takes $23.1$ years to go from $2000$ to $4000$, and another $23.1$ to reach $8000$.

2 Exponential Decay Models

With $k < 0$ the quantity falls towards a limiting value. A constant added to the model changes that limit: $\theta = C + Ae^{-kt}$ tends to $C$, not $0$.

Worked Example 2 — Cooling

The temperature of a cup of coffee, $\theta$ °C, $t$ minutes after it is made is modelled by $\theta = 20 + 70e^{-0.05t}$.

(a) State the initial temperature and the room temperature. (b) Find when the coffee reaches $50$ °C. (c) Find the rate of cooling when $t = 0$.

①(a) $t = 0$: $\theta = 20 + 70 = 90$ °C. As $t \to \infty$, $e^{-0.05t} \to 0$, so $\theta \to 20$: room temperature $20$ °C.

②(b) $70e^{-0.05t} = 30 \Rightarrow e^{-0.05t} = \tfrac37 \Rightarrow -0.05t = \ln\tfrac37$

③$t = -20\ln\tfrac37 = 20\ln\tfrac73 = 16.9$ minutes.

④(c) $\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = 70 \times (-0.05)e^{-0.05t} = -3.5e^{-0.05t}$. At $t = 0$: $-3.5$, so cooling at $3.5$ °C per minute.

Worked Example 3 — Finding the constants from data

A car is worth £$18\,000$ when new and £$12\,000$ after $3$ years. Its value is modelled by $V = Ae^{-kt}$. Find $A$ and $k$, and the age at which the car is first worth less than £$5000$.

①$t = 0$: $V = A = 18\,000$.

②$t = 3$: $18\,000e^{-3k} = 12\,000 \Rightarrow e^{-3k} = \tfrac23 \Rightarrow -3k = \ln\tfrac23$

③$k = \tfrac13\ln 1.5 = 0.135$ (3 s.f.)

④$18\,000e^{-kt} < 5000 \Rightarrow e^{-kt} < \tfrac{5}{18} \Rightarrow -kt < \ln\tfrac{5}{18} \Rightarrow t > \dfrac{\ln 3.6}{k} = 9.48$

⑤The car is first worth less than £$5000$ about $9.5$ years after it is new.

Keep $k$ exact ($\frac13\ln 1.5$) or at full calculator accuracy in step ④. Using $0.135$ gives $9.49$, a small but avoidable error.

3 Reduction to Linear Form

Real data never lies exactly on a curve, and it is hard to tell by eye whether a curve is exponential. Taking logarithms turns both standard models into straight lines, which are easy to recognise and to fit.

Model	Take logs	Plot	Gradient	Intercept
$y = ab^x$	$\log y = \log a + x\log b$	$\log y$ against $x$	$\log b$	$\log a$
$y = ax^n$	$\log y = \log a + n\log x$	$\log y$ against $\log x$	$n$	$\log a$

Compare each with $Y = c + mX$. For the exponential model only $y$ is logged; for the power model both $x$ and $y$ are logged. You may use $\log_{10}$ or $\ln$, but use the same one throughout and undo it with the matching inverse ($10^{\square}$ or $e^{\square}$).

Worked Example 4 — From a straight line to $y = ab^x$

A graph of $\log_{10} y$ against $x$ is a straight line with intercept $0.7$ and gradient $0.15$. Find $a$ and $b$ in $y = ab^x$.

①$\log_{10} y = \log_{10} a + x\log_{10} b$, so $\log_{10} a = 0.7$ and $\log_{10} b = 0.15$.

②$a = 10^{0.7} = 5.01$ and $b = 10^{0.15} = 1.41$ (3 s.f.)

③Model: $y = 5.01 \times 1.41^x$.

Worked Example 5 — From a straight line to $y = ax^n$

A graph of $\log_{10} y$ against $\log_{10} x$ is a straight line through $(0.5,\, 1.2)$ and $(1.5,\, 2.1)$. Find $a$ and $n$ in $y = ax^n$.

①Gradient $n = \dfrac{2.1 - 1.2}{1.5 - 0.5} = 0.9$.

②Intercept: $1.2 = \log_{10} a + 0.9 \times 0.5 \Rightarrow \log_{10} a = 0.75$.

③$a = 10^{0.75} = 5.62$ (3 s.f.), so $y = 5.62x^{0.9}$.

The line's intercept is the value of $\log_{10} y$ when $\log_{10} x = 0$, i.e. when $x = 1$ — not when $x = 0$.

Worked Example 6 — Fitting a model to experimental data

The number of bacteria $N$ in a sample is counted every $2$ hours.

$t$ (hours)	$0$	$2$	$4$	$6$	$8$
$N$	$120$	$190$	$290$	$470$	$730$
$\ln N$	$4.79$	$5.25$	$5.67$	$6.15$	$6.59$

Show that $N = Ae^{kt}$ is a suitable model and estimate $A$ and $k$.

①If $N = Ae^{kt}$ then $\ln N = \ln A + kt$: a straight line of $\ln N$ against $t$ with gradient $k$.

②The plotted points lie very close to a straight line (below), so the model is suitable.

③Line of best fit through about $(0,\, 4.79)$ and $(8,\, 6.59)$: gradient $k = \dfrac{6.59 - 4.79}{8} = 0.225$.

④$\ln A = 4.79 \Rightarrow A = e^{4.79} \approx 120$.

⑤Model: $N = 120e^{0.225t}$.

4 Limitations and Refinements

Every model is a simplification. Exam questions regularly ask you to criticise one, and a good answer is specific to the context.

Limitation	What it means in context
Unlimited growth	$Ae^{kt} \to \infty$, but populations are limited by food, space and disease. The model is only valid for small $t$.
Extrapolation	Using the model outside the range of the data (e.g. predicting $N$ at $t = 24$ from data up to $t = 8$) is unreliable.
Continuous vs discrete	The model gives non-integer values such as $2699.7$ people; round sensibly and remember real changes may happen in steps.
Constant rate	The model assumes $k$ never changes, but interest rates, birth rates and room temperatures can.

Refinements should address a stated limitation: e.g. "add a term so the population levels off at a maximum", or "use a different value of $k$ after the first year". Year 2 develops these ideas further with differential equations.

Worked Example 7 — Testing a model against new data

Using $N = 120e^{0.225t}$ from Example 6, predict $N$ at $t = 10$. The actual count at $t = 24$ was $15\,000$; comment.

①$t = 10$: $120e^{2.25} \approx 1140$ — a short extrapolation, probably reasonable.

②$t = 24$: $120e^{5.4} \approx 26\,600$, far more than the $15\,000$ observed.

③The growth rate has slowed, perhaps as nutrients run out. The model overestimates for large $t$ and should not be used there.

5 Quick Reference

Model

$N = Ae^{kt}$; $A$ is the value at $t = 0$.

Rate

$\dfrac{\mathrm{d}N}{\mathrm{d}t} = kN$

Doubling / halving

Time $= \dfrac{\ln 2}{|k|}$

With a constant

$C + Ae^{-kt} \to C$ in the long run.

$y = ab^x$

Plot $\log y$ against $x$: gradient $\log b$.

$y = ax^n$

Plot $\log y$ against $\log x$: gradient $n$.

Intercept

$\log a$ in both cases; undo with $10^{\square}$ or $e^{\square}$.

Finding $k$

Substitute a known point and use $\ln$.

Limitations

Unlimited growth; extrapolation; constant $k$.

6 Practice Questions

Question 1

The value of a house $t$ years after purchase is modelled by $V = 180\,000e^{0.04t}$. (a) Find its value after $5$ years. (b) Find when it will first be worth £$250\,000$.

▶ Show solution

(a) $180\,000e^{0.2} \approx$ £$219\,852$.

(b) $e^{0.04t} = \dfrac{250}{180} \Rightarrow t = 25\ln\dfrac{25}{18} = 8.21$ years.

Question 2

The mass, $m$ grams, of a radioactive substance after $t$ years is $m = 50e^{-0.02t}$. Find the half-life (the time for the mass to halve).

▶ Show solution

$50e^{-0.02t} = 25 \Rightarrow e^{-0.02t} = \tfrac12 \Rightarrow -0.02t = -\ln 2$

$t = 50\ln 2 \approx 34.7$ years.

Question 3

A cup of tea cools according to $\theta = 18 + 72e^{-0.1t}$. (a) Find the initial temperature. (b) Find the temperature after $10$ minutes. (c) What happens to $\theta$ in the long term, and what does the $18$ represent?

▶ Show solution

(a) $18 + 72 = 90$ °C.

(b) $18 + 72e^{-1} \approx 44.5$ °C.

Question 4

Data fitting $y = ab^x$ gives the line $\log_{10} y = 0.5 + 0.2x$. Find $a$ and $b$.

▶ Show solution

$\log_{10} a = 0.5 \Rightarrow a = 10^{0.5} \approx 3.16$.

$\log_{10} b = 0.2 \Rightarrow b = 10^{0.2} \approx 1.58$.

Question 5

For a model $y = ax^n$, a graph of $\ln y$ against $\ln x$ has equation $\ln y = 1.2\ln x + 0.8$. Find $a$ and $n$.

▶ Show solution

Compare with $\ln y = \ln a + n\ln x$: $n = 1.2$.

$\ln a = 0.8 \Rightarrow a = e^{0.8} \approx 2.23$.

Question 6

A population is modelled by $P = Ae^{kt}$. When $t = 0$, $P = 400$ and when $t = 5$, $P = 600$. Find $A$ and $k$, and predict $P$ when $t = 12$.

▶ Show solution

$A = 400$. $400e^{5k} = 600 \Rightarrow e^{5k} = 1.5 \Rightarrow k = \tfrac15\ln 1.5 \approx 0.0811$.

$P(12) = 400e^{12k} = 400 \times 1.5^{12/5} \approx 1058$.

Question 7

Show that if $y = 4x^3$ then the graph of $\log_{10} y$ against $\log_{10} x$ is a straight line, and state its gradient and intercept.

▶ Show solution

$\log_{10} y = \log_{10}\left(4x^3\right) = \log_{10} 4 + 3\log_{10} x$.

This is of the form $Y = c + mX$ with $Y = \log_{10} y$ and $X = \log_{10} x$: a straight line.

Gradient $3$, intercept $\log_{10} 4 \approx 0.602$.

Question 8

A colony grows according to $N = 300e^{0.15t}$. Find the rate of growth at the instant when $N = 900$.

▶ Show solution

$\dfrac{\mathrm{d}N}{\mathrm{d}t} = 0.15 \times 300e^{0.15t} = 0.15N$.

When $N = 900$: rate $= 0.15 \times 900 = 135$ per unit time.

(No need to find $t$ first — that is the power of $\frac{\mathrm{d}N}{\mathrm{d}t} = kN$.)

Question 9

The table shows experimental values.

$x$	$1$	$4$	$9$	$16$
$y$	$3$	$24$	$81$	$192$

By finding $\log_{10} x$ and $\log_{10} y$, show that the data fit a model $y = ax^n$ and find $a$ and $n$.

▶ Show solution

$\log_{10} x$: $0,\; 0.602,\; 0.954,\; 1.204$

$\log_{10} y$: $0.477,\; 1.380,\; 1.908,\; 2.283$

Gradients between successive points: $\dfrac{0.903}{0.602} = 1.50$, $\dfrac{0.528}{0.352} = 1.50$, $\dfrac{0.375}{0.250} = 1.50$. Constant, so the points are collinear and $y = ax^n$ fits.

$n = 1.5$; intercept (where $\log_{10} x = 0$) is $0.477 = \log_{10} a$, so $a = 3$.

Model: $y = 3x^{1.5}$. (Check: $3 \times 16^{1.5} = 3 \times 64 = 192$ ✓)

Question 10

A small population of rabbits is introduced to an island. Its size after $t$ months is modelled by $P = 50e^{0.4t}$.

(a) State the number of rabbits introduced. (b) Find when the population first exceeds $1000$. (c) Find the rate of growth when $t = 5$. (d) Give a reason why this model will not be suitable in the long term. (e) In fact the population after $3$ months was $150$. Keeping $A = 50$, find a revised value of $k$.

▶ Show solution

(a) $t = 0$: $50$ rabbits.

(b) $50e^{0.4t} > 1000 \Rightarrow e^{0.4t} > 20 \Rightarrow t > \dfrac{\ln 20}{0.4} = 7.49$. During the $8$th month.

(c) $\dfrac{\mathrm{d}P}{\mathrm{d}t} = 20e^{0.4t}$. At $t = 5$: $20e^{2} \approx 148$ rabbits per month.

(d) The model predicts unlimited growth, but the island has limited food and space, so the population will level off (or crash).

(e) $50e^{3k} = 150 \Rightarrow e^{3k} = 3 \Rightarrow k = \tfrac13\ln 3 \approx 0.366$.

Exponential Modelling · A Level Maths Year 1 · Created with MathJax