Fitting/interpreting exponential models | Year 11

01 / LEARN

The idea, step by step

Use two points to find the multiplier

For $A (t) = A b^{t}$ , dividing two positive data values at distinct times removes A. If observations are d time units apart, their ratio is $b^{d}$ , not necessarily b. Take the d-th root to obtain a one-unit multiplier.

After finding b, substitute either point to calculate A. Only an observation at t=0 directly gives A. State the unit of t alongside the formula.

b = {(\frac{y_{2}}{y_{1}})}^{1 / (t_{2} - t_{1})}

Check the fit and the interpretation

A rule through two points does not prove a process is exponential. Check further data, the consistency of time intervals and whether multiplying behaviour makes sense in context.

Predictions inside the observed time range are interpolation; predictions beyond it are extrapolation. A model can fit its original data yet become unreasonable outside the conditions under which the data were collected.

A common trap

Using the ratio of values three periods apart as the one-period multiplier.

Instead: The ratio is the cube of the multiplier when the gap is three periods.

02 / WORKED EXAMPLES

See the thinking

Example 1

Three periods between observations

A quantity is 120 at t=0 and 405 at t=3. Fit an exponential model.

$b^{3} = 405 / 120 = 3.375$ , so b=1.5.
The initial value is 120 because it was observed at t=0.

A (t) = 120 {(1.5)}^{t}

Example 2

Neither point is at time zero

A model takes values 36 at t=2 and 121.5 at t=5. Find its rule.

$b^{3} = 121.5 / 36 = 3.375$ , giving b=1.5.
$A {(1.5)}^{2} = 36$ , so A=16.
Check: $16 {(1.5)}^{5} = 121.5$ .

y = 16 {(1.5)}^{t}

Check your understanding

The values at t=0,1,2 are 10,15,22.5. Find the multiplier.

Show answer

1.5.

$15 / 10 = 22.5 / 15 = 1.5$ .

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Fit an exponential rule to y(0)=40 and y(2)=90.
2 marks
Fit a model through (1,24) and (3,54).
2 marks
Values at equally spaced times are 100,130,160. Does a pure exponential rule fit exactly?
3 marks
For $P (t) = 60 {(0.9)}^{t}$ , interpret 60 and 0.9.
2 marks
A count rises from 80 to 180 in two hours. Under an exponential model, predict the count after one hour.
3 marks
An exponential rule is fitted to observations from days 0 to 10. Explain why its day-100 prediction needs caution.
3 marks

Read the worked solutions

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

Fit an exponential rule to y(0)=40 and y(2)=90.
2 marks
$b^{2} = 90 / 40 = 2.25$ ; take the positive root.
$y = 40 {(1.5)}^{t}$
Fit a model through (1,24) and (3,54).
2 marks
$b^{2} = 54 / 24 = 2.25$ ; b=1.5; A=24/1.5=16.
$y = 16 {(1.5)}^{t}$
Values at equally spaced times are 100,130,160. Does a pure exponential rule fit exactly?
3 marks
The ratios are 1.3 and $160 / 130 \approx 1.231$ , not equal.
The equal differences instead suggest a linear model.
No.
For $P (t) = 60 {(0.9)}^{t}$ , interpret 60 and 0.9.
2 marks
$P (0) = 60$ and $P (t + 1) / P (t) = 0.9$ .
60 is the time-zero amount; 0.9 retains 90% per time unit.
A count rises from 80 to 180 in two hours. Under an exponential model, predict the count after one hour.
3 marks
The hourly multiplier is $\sqrt{180 / 80} = 1.5$ ; $80 (1.5) = 120$ .
120
An exponential rule is fitted to observations from days 0 to 10. Explain why its day-100 prediction needs caution.
3 marks
Growth conditions or limiting resources can change. A fit to ten days does not establish the process for one hundred days.
It extrapolates far beyond the observed data.

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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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