Fitting/interpreting exponential models | Year 11

01 / LEARN

Understand the idea.

Use two points to find the multiplier

For a positive exponential model $y = A b^{t}$ , divide the later value by the earlier value. This removes the unknown initial factor and gives $b^{t_{2} - t_{1}}$ . The observed ratio is a one-unit multiplier only when the time gap is one unit.

Find the positive root corresponding to the time gap, then substitute either observation to recover $A$ . An observation at a nonzero time is not the initial value. Verify the finished rule against both data points.

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

Check the fit and the interpretation

Two points determine a pure exponential rule but do not establish that the real process remains exponential. Compare further observations at equal time intervals and check whether a repeated percentage change makes sense in the setting.

An estimate between observed times is interpolation; an estimate beyond them is extrapolation. The latter needs particular care because physical conditions or the multiplier may change outside the measured interval.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Three periods between observations

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

Worked solution

1
Use the time-zero observation.
$A (0) = A_{0} b^{0} = A_{0} = 120$
At time zero the exponential factor is 1.
2
Use the second observation to find the multiplier.
$120 b^{3} = 405 ⟹ b^{3} = \frac{405}{120} = 3.375$
The three-unit gap contains three equal multiplications.
3
Take the positive cube root.
$b = \sqrt[3]{3.375} = 1.5$
A real-time exponential model uses a positive base.
4
Write and verify the rule.
$A (t) = 120 {(1.5)}^{t}, A (3) = 120 {(1.5)}^{3} = 405$
The rule reproduces both supplied values.

Answer

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

EXAMPLE 02

Neither point is at time zero

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

Worked solution

1
Divide the observations to remove the initial factor.
$\frac{A b^{5}}{A b^{2}} = \frac{121.5}{36} ⟹ b^{3} = 3.375$
The observation times differ by three units, not one.
2
Recover the one-unit multiplier.
$b = \sqrt[3]{3.375} = 1.5$
The cube root converts the three-unit ratio to its per-unit factor.
3
Use the earlier observation to find the initial value.
$A {(1.5)}^{2} = 36 ⟹ A = \frac{36}{2.25} = 16$
The value 36 was measured at time 2, so two multiplications must be undone.
4
Write the rule and check the later value.
$y = 16 {(1.5)}^{t}, 16 {(1.5)}^{5} = 121.5$
The calculated parameters reproduce the second observation as well.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

1.5.

The times are equally spaced by one unit, and $\frac{15}{10} = \frac{22.5}{15} = 1.5$ . Therefore the one-unit multiplier is 1.5.

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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
$b^{2} = \frac{90}{40} = 2.25$ ; take the positive root.
$y = 40 {(1.5)}^{t}$
Fit a model through $(1, 24)$ and $(3, 54)$ .
2 marks
$b^{2} = \frac{54}{24} = 2.25$ ; $b = 1.5$ ; $A = \frac{24}{1.5} = 16$ .
$y = 16 {(1.5)}^{t}$
Values at equally spaced times are 100,130,160. Does a pure exponential rule fit exactly?
3 marks
The ratios are 1.3 and $\frac{160}{130} \approx 1.231$ , 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 $\frac{P (t + 1)}{P (t)} = 0.9$ .
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 $\sqrt{\frac{180}{80}} = 1.5$ ; $80 (1.5) = 120$ .
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 ↗

In this lesson

Understand the idea.

Use two points to find the multiplier

Check the fit and the interpretation

See the method in action.

Three periods between observations

Use the time-zero observation.

Use the second observation to find the multiplier.

Take the positive cube root.

Write and verify the rule.

Neither point is at time zero

Divide the observations to remove the initial factor.

Recover the one-unit multiplier.

Use the earlier observation to find the initial value.

Write the rule and check the later value.

Try it yourself.

Check your understanding

Keep practising