Free printable resources · Original practice
Use base-10 spacing on both axes. Plot two exact data sets and separate the exponent of a power relation from the physical angle of a line.
Use base-10 spacing on both axes. Plot two exact data sets and separate the exponent of a power relation from the physical angle of a line.
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Original blank grid. Letter or A4. Free PDF and SVG.
Original blank grid. Letter or A4. Free PDF and SVG.
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Original task with a separate reference answer. Letter or A4.
Original task with a separate reference answer. Letter or A4.
| Task | Given pairs (x,y) | Worked result |
|---|---|---|
| LL01 | (1,1); (2,4); (5,25); (10,100) | n=2; A=1; y=x^2; y(3)=9. Slope uses log(y)/log(x), not millimetres. |
| LL02 | (1,3); (4,6); (9,9); (25,15); (100,30) | n=0.5; A=3; y=3*sqrt(x); y(16)=12. |
A value v on the 1–10 axis lies at fraction log10(v) of the decade width. Thus 2 is at about 30.103%, 5 at 69.897%, and √10 ≈ 3.162 at the physical midpoint. Repeating the same mantissas in the next decade places 20, 50 and 100 using the same pattern between 10 and 100.
For LL01, compare (2,4) and (10,100): n = log10(100/4) / log10(10/2) = log10(25)/log10(5) = 2. Since y = A x² and the point (1,1) is supplied, A = 1. At x = 3 the result is 9. For LL02, y = 3√x gives exponent 1/2 and coefficient 3; at x = 16 the result is 12.
These are exact synthetic exercises. A line through a few measured observations would not establish a universal power law by itself. If the paper covers different numbers of decades horizontally and vertically, its physical line angle also changes without changing the exponent.
When plotting on these sheets, remember that minor ticks follow a logarithmic progression rather than a linear one. The position of any value v is calculated using the formula t equals the log of v minus the log of the minimum, divided by the log of the maximum minus the log of the minimum. This means the number five does not sit halfway between one and ten. You must locate ticks based on their logarithmic distance from the decade start. This spacing ensures that equal physical distances on the paper represent equal multiplicative factors, which is essential for correctly interpreting the data points you plot on the provided grid.
The two included worksheets provide synthetic exact data rather than empirical fitted laws. Worksheet LL01 plots x values of one, two, five, and ten against y values of one, four, twenty-five, and one hundred, representing the equation y equals x squared. Worksheet LL02 uses x values of one, four, nine, twenty-five, and one hundred against y values of three, six, nine, fifteen, and thirty, representing y equals three times the square root of x. Since these are exact synthetic points, both relationships become straight lines in log-log coordinates. Plot the stated pairs, then compare the transformed slope; the worksheets are not asking you to infer a law from noisy observations.
On log-log paper, the slope of your plotted line represents the exponent of the relationship, not the physical angle of the line on the paper. For the LL01 worksheet, the slope corresponds to an exponent of two, while for LL02, it corresponds to an exponent of zero point five. The coefficient is one for the first worksheet and three for the second. Do not measure the angle with a protractor to find the exponent. Instead, calculate the change in the log of y divided by the change in the log of x to determine the correct mathematical exponent.
| Watch for | Check instead |
|---|---|
| Assuming the number five is located at the physical midpoint between one and ten on the axis. | Locate five using the logarithmic formula, as it sits closer to ten than to one due to the non-linear spacing of minor ticks within each decade. |
| Using linear interpolation to estimate values between logarithmic ticks on the grid. | Remember that interpolation within logarithmic decades is not linear. Use the logarithmic position formula to find intermediate values accurately rather than estimating by eye. |
| Measuring the physical angle of the plotted line to determine the exponent. | Calculate the exponent by dividing the change in log y by the change in log x. The physical angle varies with axis scaling and does not equal the exponent. |
Choose the file matching the paper in your printer. Print the PDF at Actual size / 100%, turning off Fit, Shrink and browser headers. An SVG editor or browser can change its scale even though the file uses millimeter dimensions. Use the supplied calibration bars where present when a task depends on physical size; a screen preview alone does not establish printed size.
The packs contain the same individual pages in collection order. They are a convenient download, not extra different exercises. On coordinate plots, uniform scaling preserves the numerical positions relative to the axes. On image-copy tasks, keep the specified proportions and any stated source-to-target enlargement. Color is optional: labels, line positions and relative lightness also explain the references.
Positive values below 1 are valid logarithmic inputs, but they are outside these supplied ranges. The one-decade blank spans 1–10 on both axes; the two-decade blank and both activity charts span 1–100. Choose a different range rather than changing the data.
No, the worksheets contain synthetic exact data generated from known equations. They are not empirical fitted laws, so your plots should match the given coordinates exactly.
No zero or negative value can be placed on either ordinary log-log axis. The supplied blanks begin at 1. A tick labeled 1 is an original data value, not a transformed coordinate of 1; log10(1) = 0 determines its position at the decade start.
Yes. Every linked worksheet, PDF, SVG and answer is free without signup. Original Zumola resources use CC BY 4.0: credit Zumola, link the license and indicate changes. Analytics consent is not required to download or use them.
On a base-10 logarithmic axis, a decade runs from 1 to 10. The position of a value x inside that decade is log10(x), so 2 sits at about 0.301 of the decade, 3 at about 0.477, and 5 at about 0.699. Equal visual gaps would put 2, 4, 6, 8 at even intervals, which is wrong: the gap from 1 to 2 is far wider than from 8 to 9.
A practical rule is to place ticks for 1 through 9 in each decade, then repeat the pattern in the next decade. The spacing compresses as values grow, and the same pattern repeats exactly one decade later. If your ticks look evenly spread, the axis is probably linear, not logarithmic.
For a power relation y = k·x^n, taking base-10 logs gives log10(y) = log10(k) + n·log10(x). On log-log paper the plotted points fall on a straight line, and the exponent n is the slope measured in decades, not in millimetres of physical angle. A line rising one decade vertically for every one decade horizontally has n = 1; two vertical decades per horizontal decade gives n = 2.
The physical angle depends on the aspect ratio of the printed grid, so never read the exponent from a protractor. Measure vertical and horizontal distances in decade units instead, then divide.
Before committing data, test the sheet with a known pair. If x = 1 and y = 1, the point should sit at the lower-left corner of the first decade. If x = 10 and y = 100, the point should sit one decade right and two decades up. Any mismatch signals a mislabelled axis or a linear scale slipped in.
Also confirm no data value is zero or negative. Logarithms of non-positive numbers are undefined, so such points cannot be placed on a true log axis and must be excluded or handled separately.
Take x = 1, 2, 5 and 10 with y = 3, 12, 75 and 300. Logs give log10(x) = 0, 0.301, 0.699, 1 and log10(y) ≈ 0.477, 1.079, 1.875, 2.477. The vertical rise from x = 1 to x = 10 is 2.477 − 0.477 = 2 decades over 1 horizontal decade, so n = 2. Using (1, 3): log10(3) = log10(k) + 2·0, so k = 3. The fit is y = 3·x^2, matching the data exactly.
Data: x = 1, 4, 10 with y = 2, 32, 200. Fit y = k·x^n on log-log paper.
Logs: log10(x) = 0, 0.602, 1; log10(y) ≈ 0.301, 1.505, 2.301. From x = 1 to x = 10 the vertical rise is 2.301 − 0.301 = 2 decades over 1 horizontal decade, so n = 2. Using (1, 2): log10(2) = log10(k) + 0, so k = 2. The relation is y = 2·x^2, which reproduces 2, 32 and 200 exactly.
Yes. The tick positions differ, but the slope in decades is unchanged because changing base only rescales both axes equally.
Check that both axes are logarithmic and that you measured in decade units, not millimetres or degrees.
A true log axis cannot show zero. Remove that point or use a different representation; do not force it onto the grid.
