Exponential growth and decay | Year 11

01 / LEARN

The idea, step by step

Constant factor, not constant difference

In $A (t) = A_{0} b^{t}$ , $A_{0}$ is the value at time zero and $b$ is the multiplier for one time unit. Equally spaced table values have a constant ratio, not a constant difference.

For positive initial values, $b > 1$ gives growth and $0 < b < 1$ gives decay. A percentage rate $r$ written as a decimal gives multiplier $1 + r$ for growth or $1 - r$ for decay.

A (t) = A_{0} b^{t}

Check the unit and the claim

The exponent counts multiplier-periods. If the factor is hourly, use hours. A factor of 2 doubles a quantity each period; it does not add 2.

A growth model can describe a useful interval without describing the distant future. Decay towards zero does not mean the formula reaches exactly zero after a finite number of periods.

A common trap

Using 0.15 as the multiplier for 15% growth.

Instead: The original 100% remains and 15% is added, giving 1.15.

02 / WORKED EXAMPLES

See the thinking

Example 1

Build a growth rule

A culture starts with 400 cells and grows by 15% each hour. Model its size and predict 3 hours ahead.

The multiplier is 1.15, giving $N (t) = 400 {(1.15)}^{t}$ .
$N (3) = 400 {(1.15)}^{3} = 608.35$ ; round to a whole cell.

N (t) = 400 {(1.15)}^{t}

; about 608 cells.

Example 2

Read a decay factor

For $V (t) = 18000 {(0.82)}^{t}$ in dollars and years, find the annual loss and value after 2 years.

It retains 82%, so loses 18% each year.
$V (2) = 18000 {(0.82)}^{2} = 12103.20$ .

18% annual loss; 12,103.20 dollars.

Check your understanding

What decrease is represented by $Q (t) = 70 {(0.94)}^{t}$ ?

Show answer

6% per time unit.

The quantity retains 94%, hence loses 6%.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A quantity is multiplied by 1.07 monthly. State its percentage change.
2 marks
Model 250 grams decreasing by 12% each day.
2 marks
For $N (t) = 300 {(1.2)}^{t}$ , find $N (2)$ .
2 marks
The equally spaced values are 5, 15, 45, 135. Does an exponential model fit?
2 marks
Compare $L (t) = 20 + 5 t$ and $E (t) = 20 {(1.25)}^{t}$ .
3 marks
Will $B (t) = 80 {(0.5)}^{t}$ equal zero at any finite real t? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A quantity is multiplied by 1.07 monthly. State its percentage change.
2 marks
1.07 is the original 1 plus 0.07.
7% monthly increase.
Model 250 grams decreasing by 12% each day.
2 marks
The retained factor is $1 - 0.12 = 0.88$ .
$M (t) = 250 {(0.88)}^{t}$ , t in days.
For $N (t) = 300 {(1.2)}^{t}$ , find $N (2)$ .
2 marks
$300 (1.44) = 432$ .
432
The equally spaced values are 5, 15, 45, 135. Does an exponential model fit?
2 marks
Each next value divided by the previous one gives 3.
Yes: common multiplier 3.
Compare $L (t) = 20 + 5 t$ and $E (t) = 20 {(1.25)}^{t}$ .
3 marks
$L (t + 1) - L (t) = 5$ ; $E (t + 1) / E (t) = 1.25$ .
L adds 5 per period; E adds 25% of its current value.
Will $B (t) = 80 {(0.5)}^{t}$ equal zero at any finite real t? Explain.
3 marks
${0.5}^{t} > 0$ for every finite real t, so the product is positive.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

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