Which plot comes out straight tells you the model

the raw data

n/agradient, first pair

n/agradient, last pair

cannot tellverdict

The same four points every time. Only the axes change.

The figure needs JavaScript. It plots (1, 6), (2, 12), (4, 48) and (8, 768) three ways. Raw, the scatter is a steep curve. With ln y against x the points are exactly straight, gradient 0.693 and intercept 1.099. With ln y against ln x they curve, the gradient going 1, 2, 4. So the relationship is exponential, y = 3 × 2ₓ. Everything below stands on its own.

Why taking a log straightens one and not the other

Exponential. Suppose y = kaₓ. Take natural logs of both sides:

ln y = ln k + x ln a

That is a straight line in ln y against x, with gradient ln a and intercept ln k.

Power. Suppose y = kxⁿ. Take natural logs:

ln y = ln k + n ln x

That is a straight line in ln y against ln x, with gradient n and intercept ln k.

So there are two plots and one question: which one comes out straight? An exponential straightens against x. A power straightens against ln x. Whichever plot is straight names the family, and its gradient and intercept then hand you the parameters.

The data, done both ways

x	y	ln x	ln y
1	6	0	1.792
2	12	0.693	2.485
4	48	1.386	3.871
8	768	2.079	6.644

ln y against x. Check the gradient twice, at opposite ends:

From x = 1 to x = 2: (2.485 − 1.792)/1 = 0.693
From x = 4 to x = 8: (6.644 − 3.871)/4 = 0.693

The same, so it is straight. That settles the family.

ln y against ln x. The same check:

From ln x = 0 to 0.693: (2.485 − 1.792)/0.693 = 1
From ln x = 1.386 to 2.079: (6.644 − 3.871)/0.693 = 4

Four times as steep at the far end, so that plot is curved and the relationship is not a power.

Reading the parameters off

From the straight plot, ln y = 1.099 + 0.693x. Undo the logs:

Gradient 0.693 = ln a, so a = e^0.693 = 2
Intercept 1.099 = ln k, so k = e^1.099 = 3
So y = 3 × 2ₓ

Check it on a point you did not use: at x = 8, 3 × 256 = 768. Correct.

The gradient is not the base and the intercept is not the coefficient. They are the logs of them. A student who reports "a = 0.693" has given ln a, and a student who reports "k = 1.099" has given ln k. Both are one e away from the answer and both lose the mark. Write the two undoing lines out.

Why you would want a log scale anyway

The raw y values run from 6 to 768. On a linear axis tall enough for 768, the first three points are squashed into the bottom centimetre and you cannot see the pattern at all. Taking logs:

Compresses a wide range into something readable.
Turns constant multiplication into constant addition, which the eye can judge.
Makes "rate of growth" visible as a slope rather than as a shape.

That is the scaling half of this sub-topic, and it is useful well beyond deciding between two families.

On the GDC: two lists of logs

Put x and y in two lists, make two more lists of their logs, then run a linear regression on each candidate pairing and compare. The one with r² essentially 1 is the family.

When you may use it. Applications. A calculator is allowed in every paper, and this sub-topic cannot sensibly be done without one.

TI-Nspire CX II

ctrl doc → Add Lists & Spreadsheet. Name the columns x and y and type the four pairs in
In the formula cell of a third column, named lny, enter =ln(y). In a fourth, lnx, enter =ln(x)
menu → Statistics → Stat Calculations → Linear Regression (mx+b), with X as x and Y as lny: m = 0.693, b = 1.099, r² = 1
Run it again with X as lnx: r² = 0.92, which is not 1, so it is not a power relationship

Casio fx-CG50

MENU → Statistics, x into List 1 and y into List 2
Put the cursor on the List 3 heading and enter ln(List 2), then on List 4 enter ln(List 1)
F2 CALC → F6 SET to choose XList 1 and YList 3, then F3 REG → F1 X: a = 0.693, b = 1.099, r² = 1
Change XList to 4 and run it again: the fit is clearly worse, which is the answer

The mark people lose. Stopping at the straight line. The regression gives you the gradient and the intercept of the LOG plot, and the question asked for the model. Convert both: base e to the gradient, coefficient e to the intercept. The other one is using log base 10 for one list and natural log for the other: either base works throughout, and mixing them gives a gradient that is right for neither.

Your turn

1. For the straightened plot, the gradient is 0.693. What is the base a of the exponential?

2. The intercept is 1.099. What is k?

3. Using y = 3 × 2ₓ, what is y at x = 8?

4. A different set of data is straight when ln y is plotted against ln x, with gradient 2. What family is it, and what is the power? Type the power.

5. Why can a scatter plot of the raw data not decide between the two families here?

Question 5. Why can a scatter plot of the raw data not decide between the two families here? Because both families give exactly the same values at x = 2 and x = 4, and both look like a steep curve. Because a scatter plot cannot be used for curved data at all. Because there are only four data points. Because the y values are too large to plot.

Where the marks go

1 markChoosing the right pair of variables to plot, and saying why.

1 markThe gradient and intercept of the straight line.

1 markConverting them back into the model's parameters.

1 markStating the model, and ideally checking it on a point.

The third mark is the one this sub-topic exists to test. A perfect straight line with no conversion is half the answer.