Exponential Growth and Decay Models

Modelling real-world situations using exponential functions

SPEC 6.7 • EXPONENTIALS & LOGARITHMS

Real-World Exponential Change

Many quantities in the real world grow or shrink at a rate proportional to their current size. This leads to exponential models.

Growth Examples

Population growth

Bacterial colonies

Investment with compound interest

Spread of diseases

Decay Examples

Radioactive decay

Drug elimination from body

Depreciation of assets

Cooling of objects

Key Insight

The common feature is that the rate of change is proportional to the current value.

Exponential Growth Model

Growth Model

P = P_{0} e^{k t}

where

k > 0

SymbolMeaning

P

Value at time

t

P_{0}

Initial value (when

t = 0

)

k

Growth constant (

k > 0

)

t

Time

Key Feature

The graph curves upward, getting steeper as time increases. It passes through

(0, P_{0})

.

Exponential Decay Model

Decay Model

P = P_{0} e^{- k t}

where

k > 0

Important

Note the negative sign in the exponent. This causes the quantity to decrease over time.

Alternative Form

Sometimes written as

P = P_{0} e^{k t}

where

k < 0

Key Feature

The graph curves downward, approaching but never reaching zero (asymptote at

P = 0

).

01Worked Example

The population $P$ of a colony at time $t$ hours is modelled by $P = 500 e^{0.12 t}$ .

(a) State the initial population. (b) Find the population after 10 hours. (c) Find the time when $P = 2000$ .

(a)

When

t = 0

:

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

(b)

When

t = 10

:

P = 500 e^{0.12 \times 10} = 500 e^{1.2} = 500 \times 3.32 = 1660

(c)

2000 = 500 e^{0.12 t}

→

4 = e^{0.12 t}

→

ln 4 = 0.12 t

→

t = \frac{1.386}{0.12} = 11.6

hours

02Worked Example

The mass $M$ grams of a substance remaining after $t$ years is modelled by $M = 80 e^{- 0.025 t}$ .

(a) Find the mass after 20 years. (b) Find the time for $M$ to reduce to 10 grams.

(a)

M = 80 e^{- 0.025 \times 20} = 80 e^{- 0.5} = 80 \times 0.6065 = 48.5

grams

(b)

10 = 80 e^{- 0.025 t}

→

0.125 = e^{- 0.025 t}

→

ln 0.125 = - 0.025 t

t = \frac{- 2.079}{- 0.025} = 83.2

years

Finding the Constant k

Often we need to find the growth or decay constant

k

from given data.

Method

Given:

P = P_{0} e^{k t}

and a known point

(t_{1}, P_{1})

Steps:

Substitute the known values:

P_{1} = P_{0} e^{k t_{1}}

Divide both sides by

P_{0}

:

\frac{P _{1}}{P _{0}} = e^{k t_{1}}

Take natural log:

ln (\frac{P _{1}}{P _{0}}) = k t_{1}

Solve for

k

:

k = \frac{1}{t _{1}} ln (\frac{P _{1}}{P _{0}})

Remember

If

k > 0

→ growth model. If

k < 0

→ decay model.

Worked Example03

A population grows exponentially. Initially there are 200 individuals, and after 5 years there are 350. Find the value of $k$ in the model $P = 200 e^{k t}$ .

1

Substitute:

350 = 200 e^{5 k}

2

Divide by 200:

1.75 = e^{5 k}

3

Take ln:

ln 1.75 = 5 k

4

Solve:

k = \frac{l n 1.75}{5} = \frac{0.5596}{5} = 0.112

04Worked Example

A substance decays according to $M = M_{0} e^{- 0.0866 t}$ where $t$ is in days. Calculate the half-life (time for the mass to halve).

1

At half-life:

\frac{M _{0}}{2} = M_{0} e^{- 0.0866 t}

2

Simplify:

0.5 = e^{- 0.0866 t}

3

Take ln:

ln 0.5 = - 0.0866 t

→

- 0.693 = - 0.0866 t

4

Solve:

t = \frac{0.693}{0.0866} = 8.0

days

05Worked Example

A population doubles every 15 years. The model is $P = P_{0} e^{k t}$ .

(a) Find the value of $k$ . (b) Find the time for the population to triple.

(a)

At doubling time:

2 P_{0} = P_{0} e^{15 k}

→

2 = e^{15 k}

→

ln 2 = 15 k

k = \frac{l n 2}{15} = \frac{0.693}{15} = 0.0462

(b)

For tripling:

3 P_{0} = P_{0} e^{0.0462 t}

→

3 = e^{0.0462 t}

→

ln 3 = 0.0462 t

t = \frac{l n 3}{0.0462} = \frac{1.099}{0.0462} = 23.8

years

06Worked Example

£5000 is invested with continuous compound interest at 3.5% per year, modelled by $A = P e^{r t}$ .

(a) Write the equation. (b) Find the value after 8 years. (c) How long until the investment doubles?

(a)

P = 5000

,

r = 0.035

→

A = 5000 e^{0.035 t}

(b)

A = 5000 e^{0.035 \times 8} = 5000 e^{0.28} = 5000 \times 1.323 = £6615

(c)

10000 = 5000 e^{0.035 t}

→

2 = e^{0.035 t}

→

t = \frac{l n 2}{0.035} = 19.8

years

07Worked Example

The value $V$ of a car $t$ years after purchase is modelled by $V = V_{0} e^{- k t}$ . The car cost £20000 and is worth £12000 after 3 years.

(a) Find $k$ . (b) Find the value after 10 years. (c) When will it be worth £5000?

(a)

12000 = 20000 e^{- 3 k}

→

0.6 = e^{- 3 k}

→

ln 0.6 = - 3 k

k = \frac{- 0.511}{- 3} = 0.170

(b)

V = 20000 e^{- 0.170 \times 10} = 20000 e^{- 1.7} = 20000 \times 0.183 = £3660

(c)

5000 = 20000 e^{- 0.170 t}

→

0.25 = e^{- 0.170 t}

→

t = \frac{l n 0.25}{- 0.170} = 8.15

years

Practice

Question 1: The mass of a substance is modelled by

M = 120 e^{- 0.035 t}

grams, where

t

is in hours.

(a) Find the initial mass.

(b) Find the mass after 24 hours.

(c) Find the half-life.

(a) When

t = 0

:

M = 120 e^{0} = 120

grams

(b)

M = 120 e^{- 0.035 \times 24} = 120 e^{- 0.84} = 120 \times 0.432 = 51.8

grams

(c)

60 = 120 e^{- 0.035 t}

→

0.5 = e^{- 0.035 t}

→

t = \frac{l n 0.5}{- 0.035} = 19.8

hours

Practice

Question 2: A population grows according to

P = 800 e^{k t}

. After 6 years, the population is 1200.

(a) Find the value of

k

.

(b) Find the population after 15 years.

(c) When will the population reach 5000?

(a)

1200 = 800 e^{6 k}

→

1.5 = e^{6 k}

→

k = \frac{l n 1.5}{6} = 0.0676

(b)

P = 800 e^{0.0676 \times 15} = 800 e^{1.014} = 800 \times 2.757 = 2206

(c)

5000 = 800 e^{0.0676 t}

→

t = \frac{l n 6.25}{0.0676} = 27.1

years

Practice

Question 3: The half-life of a substance is 12 hours. Initially there are 500 grams.

(a) Find the decay constant

k

.

(b) Write the equation for mass

M

at time

t

hours.

(c) How long until only 50 grams remain?

(a) At half-life:

0.5 = e^{- 12 k}

→

k = \frac{l n 0.5}{- 12} = \frac{0.693}{12} = 0.0578

(b)

M = 500 e^{- 0.0578 t}

(c)

50 = 500 e^{- 0.0578 t}

→

0.1 = e^{- 0.0578 t}

→

t = \frac{l n 0.1}{- 0.0578} = 39.8

hours

Summary

Key Formulas

ModelFormulaUse When...

Growth

P = P_{0} e^{k t}

,

k > 0

Quantity increases

Decay

P = P_{0} e^{- k t}

,

k > 0

Quantity decreases

Finding

k

k = \frac{1}{t} ln (\frac{P}{P _{0}})

Given two data points

Half-life / Doubling

t = \frac{l n 2}{k}

Time to halve or double

Exam Tips

Always identify

P_{0}

by substituting

t = 0

To find time, rearrange and take natural logs

Remember:

ln 2 \approx 0.693

Check units carefully (hours, days, years)