Exponential growth and decay | Year 11

01 / LEARN

Understand the idea.

Constant factor, not constant difference

Exponential change multiplies the current amount by a fixed factor over each equal time interval. In $A (t) = A_{0} b^{t}$ , the initial value is $A_{0}$ because $b^{0} = 1$ . Equal time steps have equal ratios, rather than equal differences.

For a positive initial amount, $b > 1$ describes growth and $0 < b < 1$ describes decay. Convert a percentage to a decimal rate first: a rise uses $1 + r$ , while a fall uses $1 - r$ .

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

Graph of y equals two to the x, passing through (0,1), (1,2), (2,4) and (3,8).

Check the unit and the claim

The exponent counts how many multiplier-periods have passed. An hourly multiplier requires time in hours, while an annual multiplier requires years. Applying the percentage again means multiplying the new amount, not repeatedly changing the original amount.

Treat the rule as a model over an appropriate interval. Resources or changing conditions may prevent indefinite growth. With a positive decay factor, every finite-time output stays positive even though the long-term limit may be zero.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Build a growth rule

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

Worked solution

1
Convert the growth rate into a multiplier.
$b = 1 + 0.15 = 1.15$
The culture keeps its original amount and gains a further 15 percent each hour.
2
Write the model with its time unit.
$N (t) = 400 {(1.15)}^{t}, t in hours$
The time-zero count is 400, and each one-hour interval contributes one factor of 1.15.
3
Predict and round the count.
$N (3) = 400 {(1.15)}^{3} = 608.35 \approx 608 cells$
The model output is rounded to a whole cell for an interpreted count.

Answer

$N (t) = 400 {(1.15)}^{t}$ ; about 608 cells.

EXAMPLE 02

Read a decay factor

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

Worked solution

1
Interpret the remaining-value factor.
$0.82 = 82 %, 100 % - 82 % = 18 %$
Keeping 82 percent of the previous value means losing 18 percent of that value.
2
Count two annual multiplications.
$V (2) = 18000 {(0.82)}^{2}$
Time is measured in years, so two years gives exponent 2.
3
Evaluate to currency precision.
$V (2) = 18000 (0.6724) = 12103.20 dollars$
The second percentage loss acts on the reduced first-year value.

Answer

18% annual loss; 12,103.20 dollars.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

6% per time unit.

The factor $0.94$ keeps 94 percent of the quantity each time unit. The decrease is $1 - 0.94 = 0.06$ , or 6 percent per time unit.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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
$L (t + 1) - L (t) = 5$ ; $\frac{E (t + 1)}{E (t)} = 1.25$ .
$L$ 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 ↗

In this lesson

Understand the idea.

Constant factor, not constant difference

Check the unit and the claim

See the method in action.

Build a growth rule

Convert the growth rate into a multiplier.

Write the model with its time unit.

Predict and round the count.

Read a decay factor

Interpret the remaining-value factor.

Count two annual multiplications.

Evaluate to currency precision.

Try it yourself.

Check your understanding

Keep practising