What is 6.7: Exponential growth and decay?

Repeated percentage change over whole periods uses Q_n=Q_0(1+r)^n, while continuous change at a rate proportional to the amount uses Q=Q_0e^kt.

How many marks are 6.7 questions worth?

6.7 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 6.7 appeared in?

We have not yet indexed a verified past-paper appearance for 6.7 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.7

Understand and use exponential growth and decay; use in modelling (e.g. compound interest, radioactive decay, drug concentration decay, population growth); consideration of limitations and refinements of exponential models.

Draft — not yet indexed

Exponential growth and decay

Worked answers and methods for 6.7 on Edexcel A-level Maths 9MA0.

Explanation

Repeated percentage change over whole periods uses $Q_n=Q_0(1+r)^n$ , while continuous change at a rate proportional to the amount uses $Q=Q_0e^{kt}$ .
For decay $k<0$ ; the half-life in $Q=Q_0e^{-\lambda t}$ is $t_{1/2}=\ln2/\lambda$ , independent of the starting amount.
Use two observations to determine an unknown multiplier or exponent, preserve full precision, and interpret a calculated time or amount in the units and practical setting of the model.
Unlimited exponential growth and a constant decay parameter may fail when resources, competition, dosage cycles or environmental conditions change; a carrying-capacity or piecewise model can be a refinement.

Worked example

A medicine concentration is modelled by $C=80e^{-0.23t}$ , where $C$ is in $\text{mg L}^{-1}$ and $t$ is in hours. Calculate the half-life and the concentration after $5$ hours, each to $3$ significant figures.

1.At half-life, $e^{-0.23t}=1/2$ , so $t=\ln2/0.23=3.0136\ldots$ hours.
2.After $5$ hours, $C=80e^{-0.23(5)}=25.3309\ldots\,\text{mg L}^{-1}$ .
3.The requested values are $3.01$ hours and $25.3\,\text{mg L}^{-1}$ .

Answer: Half-life $=3.01$ hours $C(5)=25.3\,\text{mg L}^{-1}$

Common mistakes

Don't use a positive exponent for a decay model, so the predicted quantity increases.
Don't assume unlimited growth or a constant decay rate when resources, dosage cycles or conditions change.

Exam tip

Interpret the initial value at $t=0$ , then link a stated model limitation to a concrete carrying-capacity or piecewise refinement.

Worked practice

Tier 1 · Easy

A savings balance of

\pounds1200

earns

3.5\%

compound interest each year. Find the balance after

4

years, to the nearest penny.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\pounds1377.03$	2
Notes
The annual multiplier is $1.035$ , so the balance is $1200(1.035)^4=1377.027\ldots$ . Rounded to the nearest penny, this is $\pounds1377.03$ .

(2 marks)

Tier 2 · Standard

An investment of

\pounds5000

grows by

4.2\%

each year. Find the smallest whole number of years after which its value exceeds

\pounds8000

, and state one limitation of this exponential model.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$12$ years Any valid contextual limitation, for example: the annual interest rate may change; charges, deposits or withdrawals are ignored; or interest may be compounded on a different schedule	4
Notes
The value after $n$ years is $5000(1.042)^n$ . Solving $5000(1.042)^n>8000$ gives $n>\ln(1.6)/\ln(1.042)=11.423\ldots$ , so the smallest whole number is $12$ . Indeed, the model gives about $\pounds7861.67$ after $11$ years and $\pounds8191.86$ after $12$ . A limitation is that it assumes the interest rate remains fixed (and ignores charges or withdrawals).

(4 marks)

Tier 3 · Hard

A colony is modelled by

P=600e^{kt}

. Its measured population at

t=4

days is

900

. Find

k

, predict when the model first reaches

2000

, and state one limitation with a suitable refinement.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$k=\dfrac{\ln(1.5)}{4}\approx0.101$ $t\approx11.9$ days For example, limited resources invalidate unlimited growth; a logistic model could include a carrying capacity.	6
Notes
From $900=600e^{4k}$ , $e^{4k}=1.5$ , so $k=\ln(1.5)/4=0.10137\ldots$ . For $P=2000$ , $e^{kt}=2000/600=10/3$ , hence $t=\ln(10/3)/k=11.877\ldots$ days. The model assumes a constant proportional growth rate and no resource limit; a logistic model with a carrying capacity would refine this.

(6 marks)

Tier 1 · Easy

A machine is worth

\pounds4500

and loses

12\%

of its value each year. Write down a model for its value

V

pounds after

n

years.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$V=4500(0.88)^n$	2
Notes
Losing $12\%$ leaves $88\%=0.88$ of the value after each year. Repeated multiplication therefore gives $V=4500(0.88)^n$ .

(2 marks)

Tier 2 · Standard

The value of equipment is modelled by

V=15000q^t

, where

t

is measured in years. After

4

years its value is

\pounds9800

. Find the annual percentage depreciation to

3

significant figures, and predict the value after

7

years to the nearest pound. Use unrounded values in your working.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Annual depreciation $=10.1\%$ Value after $7$ years $=\pounds7122$	4
Notes
From $9800=15000q^4$ , $q=(9800/15000)^{1/4}=0.899050\ldots$ . The annual percentage loss is $100(1-q)=10.0950\ldots\%$ , giving $10.1\%$ . After $7$ years, $V=15000q^7=7121.59\ldots$ , so the predicted value is $\pounds7122$ .

(4 marks)

Tier 3 · Hard

A chemical mass is modelled by

M=M_0e^{-kt}

, where

t

is in days. Measurements give

M=240

mg when

t=3

and

M=85

mg when

t=9

. Find

k

and

M_0

, then find when the model first predicts

M<30

mg. Give numerical answers to

3

significant figures and state one limitation of the model.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$k=0.173\text{ day}^{-1}$ $M_0=403$ mg $M<30$ mg when $t>15.0$ days For example, the model assumes a constant proportional decay rate under unchanged conditions	6
Notes
Dividing the observations gives $85/240=e^{-6k}=17/48$ , so $k=\ln(48/17)/6=0.172998\ldots$ . Then $M_0=240e^{3k}=403.281\ldots$ mg. The inequality $M_0e^{-kt}<30$ gives $t>\ln(M_0/30)/k=15.0200\ldots$ days. A limitation is that the proportional decay constant is assumed not to change with environmental conditions or composition.

(6 marks)

Tier 2 · Standard

A substance is modelled by

M=250e^{-kt}

, where

t

is measured in hours and

k>0

. After

5

hours,

82\%

of the initial mass remains. Find

k

and the half-life, each to

3

significant figures. Use unrounded values in your working.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$k=0.0397\,\text{h}^{-1}$ Half-life $=17.5$ hours	4
Notes
The observation gives $e^{-5k}=0.82$ , so $k=-\ln(0.82)/5=0.0396901\ldots\,\text{h}^{-1}$ . If $T$ is the half-life, then $e^{-kT}=1/2$ , so $T=\ln2/k=17.4639\ldots$ hours. Therefore $k=0.0397\,\text{h}^{-1}$ and the half-life is $17.5$ hours, each to $3$ significant figures.

(4 marks)

Tier 3 · Hard

Two populations are modelled by

A=900(1.08)^t

and

B=1500(0.96)^t

, where

t

is measured in years. Find when the models predict equal populations, giving

t

3

significant figures. Hence find the first whole number of years for which

A>B

. State one limitation of this comparison and a suitable refinement. Use unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$t=4.34$ years The first whole number of years for which $A>B$ is $5$ For example, both percentage changes are assumed constant; refine the models using updated or piecewise rates	6
Notes
Equality gives $900(1.08)^t=1500(0.96)^t$ , so $(1.08/0.96)^t=5/3$ . Hence $t=\ln(5/3)/\ln(1.125)=4.33700\ldots$ , giving $4.34$ years. At $t=4$ , $A/B=(3/5)(1.125)^4=0.961\ldots<1$ , while at $t=5$ the ratio is $1.081\ldots>1$ , so the first whole number is $5$ . The comparison assumes both percentage changes remain constant; a piecewise model using rates updated from later observations would be a suitable refinement.

(6 marks)

Tier 3 · Hard

A town population is modelled by

P=2000e^{kt}

, where

t

is in years. The measured population at

t=3

2500

. Find

k

exactly and use the model to predict

P

t=6

. The actual population at

t=6

2900

. Calculate the percentage by which the model overestimates this value, giving your answer to

3

significant figures, and assess the constant-rate assumption. Use unrounded values in your working.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$k=\dfrac{\ln(5/4)}3$ Model prediction $P(6)=3125$ Percentage overestimate $=7.76\%$ The constant proportional growth assumption is not supported: the successive three-year growth factors are $1.25$ and $1.16$	6
Notes
From $2500=2000e^{3k}$ , $e^{3k}=5/4$ and $k=\ln(5/4)/3$ . Hence $P(6)=2000(e^{3k})^2=2000(5/4)^2=3125$ . Relative to the actual value, the overestimate is $100(3125-2900)/2900=225/29=7.7586\ldots\%$ , giving $7.76\%$ . The observed three-year factors are $2500/2000=1.25$ and $2900/2500=1.16$ , so a constant proportional rate does not fit both intervals.

(6 marks)

Q10

Tier 3 · Hard

10.

Before a treatment begins, a population is modelled by

P=1200e^{0.18t}

for

0\leq t\leq4

, where

t

is in days. From

t=4

onwards, the treatment makes the population decay continuously at a rate of

12\%

of its current value per day. Construct a continuous model for

P

when

t\geq4

. Find the maximum modelled population and the time when the model first returns to

1200

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P=1200e^{0.72}e^{-0.12(t-4)}$ for $t\geq4$ Maximum population $1200e^{0.72}$ at $t=4$ The model first returns to $1200$ at $t=10$ days	6
Notes
At treatment time, $P(4)=1200e^{0.18(4)}=1200e^{0.72}$ . A continuous decay rate of $12\%$ gives the factor $e^{-0.12(t-4)}$ after treatment, producing the stated continuous model. The first branch increases and the second decreases, so the maximum occurs at $t=4$ . Setting the second model equal to $1200$ gives $0.72-0.12(t-4)=0$ , hence $t-4=6$ and $t=10$ .

(6 marks)

Verified exam appearances

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Other points in 6 Exponentials and logarithms

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