Exponential regression | Year 12

01 / LEARN

Understand the idea.

Recognise multiplicative change

An exponential model $\hat{y} = A b^{x}$ has $A > 0$ and $b > 0$ . Each one-unit increase in $x$ multiplies the prediction by $b$ . Growth has $b > 1$ ; decay has $0 < b < 1$ . Unlike a line, it has a constant multiplier rather than a constant additive change.

For positive responses, fit a straight line to $ln (y)$ against $x$ . If $ln (\hat{y}) = α + β x$ , exponentiating gives $\hat{y} = e^{α} {(e^{β})}^{x}$ . Thus the original-scale initial value is $e^{α}$ and the multiplier is $e^{β}$ , not $α$ and $β$ .

\begin{array}{c} ln (\hat{y}) = α + β x \\ ⟹ \hat{y} = e^{α} e^{β x} \end{array}

Check predictions on the original scale

Check the fitted curve against the original observations and assess errors in response units. Least squares on logged responses minimises squared log errors; it generally does not produce the fit that minimises squared errors on the original response scale.

Logging requires $y > 0$ . A reported $r^{2}$ from the log regression concerns logged response variation, not automatically original response variation. State the observed predictor range and keep long-term growth or decay forecasts conditional on the model continuing to apply.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An exact growth model

A culture has counts 80,100,125 at x=0,1,2 hours. Find an exponential model.

Paired measurements
Time (h)	Count
0	80
1	100
2	125

Worked solution

1
Check consecutive multipliers.
$\frac{100}{80} = 1.25, \frac{125}{100} = 1.25$
The equal one-hour intervals have the same response ratio, supporting an exact exponential pattern for these observations.
2
Identify the initial value and multiplier.
$A = 80, b = 1.25$
The count at zero hours is the initial scale; each hour multiplies the modelled count by 1.25.
3
Write and check the fitted model.
$\begin{aligned} \hat{y} & = 80 {(1.25)}^{x} \\ \hat{y} (2) & = 80 {(1.25)}^{2} = 125 \end{aligned}$
The model reproduces the final recorded count as well as the two earlier ones.
4
Calculate the next-hour prediction included in the answer.
$\hat{y} (3) = 80 {(1.25)}^{3} = 156.25$
This is a modelled count before rounding to a suitable whole number; it extrapolates one hour beyond the observed range.

Answer

$\hat{y} = 80 {(1.25)}^{x}$ ; predicted count 156.25 before sensible rounding.

EXAMPLE 02

Undo a log fit

Technology gives ln(ŷ)=2.3-0.08x. Write the exponential model and interpret its multiplier.

Worked solution

1
Exponentiate both sides of the log equation.
$\begin{aligned} \hat{y} & = e^{2.3 - 0.08 x} \\ = e^{2.3} {(e^{- 0.08})}^{x} \end{aligned}$
Exponentiation undoes the natural logarithm, while the exponent sum separates the initial scale and multiplier.
2
Evaluate the two coefficients.
$A = e^{2.3} \approx 9.9742, b = e^{- 0.08} \approx 0.9231$
Keep the unrounded exponential values when making further predictions.
3
Write the model and interpret the multiplier.
$\begin{aligned} \hat{y} & \approx 9.9742 {(0.9231)}^{x} \\ 100 (1 - e^{- 0.08}) % & \approx 7.69 % \end{aligned}$
Each input unit retains about 92.31% of the previous predicted amount, a decrease of about 7.69%.

Answer

$\hat{y} \approx 9.9742 {(0.9231)}^{x}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A fitted model is $\hat{y} = 200 {(0.9)}^{x}$ . What percentage change occurs each x-unit?

Show answer & reasoning

A 10% decrease.

The multiplier $0.9$ retains $90 %$ of the previous predicted amount each input unit. The decrease is $100 (1 - 0.9) % = 10 %$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find an exact exponential model for $y = 50, 100, 200$ at $x = 0, 1, 2$ .
3 marks
For $\hat{y} = 120 {(1.1)}^{x}$ , predict the response at $x = 3$ .
3 marks
A log fit is $ln (\hat{y}) = 1.5 + 0.2 x$ . Find $A$ and $b$ in $\hat{y} = A b^{x}$ .
3 marks
Why cannot a zero or negative response be entered directly into a real natural-log regression?
3 marks
A model multiplies by 0.96 each week. Find its modelled change over three weeks.
3 marks
A log-scale regression has $r^{2} = 0.94$ . State a careful interpretation.
3 marks

Read the worked solutions

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

Find an exact exponential model for $y = 50, 100, 200$ at $x = 0, 1, 2$ .
3 marks
Successive values have ratio 2.
At $x = 0$ the response is 50.
$y = 50 \times 2^{x}$ .
For $\hat{y} = 120 {(1.1)}^{x}$ , predict the response at $x = 3$ .
3 marks
Calculate $120 \times {1.1}^{3}$ .
159.72.
A log fit is $ln (\hat{y}) = 1.5 + 0.2 x$ . Find $A$ and $b$ in $\hat{y} = A b^{x}$ .
3 marks
Exponentiate the intercept and slope separately.
$A = e^{1.5}$ and $b = e^{0.2}$ .
$A \approx 4.4817, b \approx 1.2214$ .
Why cannot a zero or negative response be entered directly into a real natural-log regression?
3 marks
A different justified model or transformation is needed.
Do not silently discard values.
The real logarithm $ln (y)$ is defined only for $y > 0$ .
A model multiplies by 0.96 each week. Find its modelled change over three weeks.
3 marks
Calculate ${0.96}^{3}$ .
$Percentage decrease = 100 (1 - 0.884736)$ .
Multiplier 0.884736, so a decrease of about 11.53%.
A log-scale regression has $r^{2} = 0.94$ . State a careful interpretation.
3 marks
The statistic belongs to the transformed response.
It is not automatically 94% of original $y$ -variation or evidence of causation.
94% of the observed variation in $ln (y)$ is accounted for by its fitted linear relationship with $x$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: statistical models. Original lessons and assessment practice; not an official SACE assessment.

Official curriculum source ↗

In this lesson

Understand the idea.

Recognise multiplicative change

Check predictions on the original scale

See the method in action.

An exact growth model

Check consecutive multipliers.

Identify the initial value and multiplier.

Write and check the fitted model.

Calculate the next-hour prediction included in the answer.

Undo a log fit

Exponentiate both sides of the log equation.

Evaluate the two coefficients.

Write the model and interpret the multiplier.

Try it yourself.

Check your understanding

Keep practising