Skip to main content

Logarithmic (base 10) scales for VCE General Mathematics Unit 3 Data analysis

Syllabus dot point

“Use a logarithmic (base 10) scale to display data ranging over several orders of magnitude, and interpret the display in terms of powers of ten”

VCEGeneral MathematicsUnit 3 Data analysis15 min read

Quick answer

A log⁡10\log_{10} scale gives each power of ten equal width, so data spanning several orders of magnitude can be displayed without most values crowding into one bar. Convert a value with log⁡10x\log_{10} x and convert back with 10x10^x. The interval aa to bb on the log axis holds values from 10a10^a to 10b10^b; one unit is a factor of 10 and half a unit is a factor of about 3.16.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Practice questions

What this dot point is asking

VCAA wants you to handle numerical data that is spread over several orders of magnitude: values like 18 and 48 000 in the same data set, or population densities from a few people per square kilometre to several thousand. You need to know why such data is displayed on a logarithmic (base 10) scale, how to convert a data value into its log⁡10\log_{10} value and back again, how to read a histogram (or dot plot or boxplot) drawn on a log scale, and how to interpret what you read in terms of powers of ten. The study design lists this explicitly under "Investigating data distributions", and it appears most years in Examination 1.

You do not need any algebra of logarithms (no log laws). You only need the meaning of log⁡10\log_{10} and the ability to use the log⁡\log and 10x10^x keys on your calculator.

The answer

What log⁡10\log_{10} means

The base 10 logarithm of a positive number is the power you raise 10 to in order to get that number:

log⁡10x=pmeans10p=x.\log_{10} x = p \quad\text{means}\quad 10^p = x.

So log⁡101000=3\log_{10} 1000 = 3 because 103=100010^3 = 1000, log⁡101=0\log_{10} 1 = 0 because 100=110^0 = 1, and log⁡100.01=−2\log_{10} 0.01 = -2 because 10−2=0.0110^{-2} = 0.01. Numbers between exact powers of ten have logs between whole numbers: 5050 lies between 1010 and 100100, so log⁡1050\log_{10} 50 lies between 1 and 2 (it is about 1.699).

Definition

A logarithmic (base 10) scale is an axis on which equal distances represent equal multiplications by ten (or equal ratios), rather than equal additions. Each whole unit on the scale is a power of ten: 1 is 1010, 2 is 100100, 3 is 10001000, and so on.

Two conversions are all you need:

  • Data value to log value: press log⁡\log. log⁡108925≈3.951\log_{10} 8925 \approx 3.951.
  • Log value to data value: raise 10 to it. 102.6≈398.110^{2.6} \approx 398.1.
Data value 0.01 0.1 1 10 100 1000 10 000 100 000
log⁡10\log_{10} value −2-2 −1-1 0 1 2 3 4 5

A useful shortcut for values of 1 or more: the whole-number part of the log is one less than the number of digits before the decimal point. 89258925 has four digits, so its log is 3.…3.\ldots; 105 600105\,600 has six digits, so its log is 5.…5.\ldots.

Why use a log scale?

On an ordinary (linear) scale, equal distances mean equal differences. That works well when values are of similar size, but it fails when a few values are thousands of times larger than the rest. To fit the biggest value on the axis, the intervals have to be so wide that nearly all the data piles into the first bar, and the display hides the shape of most of the distribution.

On a log scale, equal distances mean equal ratios. The step from 10 to 100 takes up the same width as the step from 1000 to 10 000, because each is a factor of ten. Small values are stretched out and large values are compressed, so all of the data becomes visible at once. A distribution that is strongly positively skewed on a linear scale often looks roughly symmetric on a log scale.

Typical contexts: incomes and house prices, populations and population densities of countries, the body masses of animals, the size of earthquakes, the brightness of stars, the number of followers on social media accounts, the concentration of a pollutant. Recent VCAA contexts include the prices of cars in a used-car yard (2023 Examination 1) and the population densities of countries (2025 Examination 1).

Reading a log-scale histogram

The histogram below shows the prices of 20 lots at an auction, from $18 to $48 000. The horizontal axis is log⁡10(price)\log_{10}(\text{price}) and each bar covers half a unit.

Histogram of auction prices on a logarithmic scaleHistogram of the prices of 20 auction lots with a log base 10 horizontal axis from 1.0 to 5.0 in intervals of width 0.5. Frequencies are 2 for 1.0 to 1.5, 3 for 1.5 to 2.0, 4 for 2.0 to 2.5, 5 for 2.5 to 3.0, 3 for 3.0 to 3.5, and 1 each for 3.5 to 4.0, 4.0 to 4.5 and 4.5 to 5.0. Below the log values the matching prices are shown: 10, 100, 1000, 10000 and 100000 dollars at log values 1, 2, 3, 4 and 5. The distribution is roughly symmetric with a slight tail to the right, and the modal interval is 2.5 to 3.0, which is about 316 to 1000 dollars.0123451.01.52.02.53.03.54.04.55.010100100010 000100 000log10(price), with the matching price in dollars underneathfrequencymodal: about 316 to 1000Each half unit is a factor of about 3.16; each whole unit is a factor of 10.

To read it, always translate the log values back into data values:

  • The interval 2.5 to 3.0 covers prices from 102.5≈31610^{2.5} \approx 316 dollars up to 103=100010^3 = 1000 dollars. It holds 5 lots, the most of any interval, so it is the modal interval.
  • The interval 2.0 to 3.0 covers $100 to $1000 and holds 4+5=94 + 5 = 9 lots, which is 45%45\% of the auction.
  • Prices above $1000 are those with a log above 3: 3+1+1+1=63 + 1 + 1 + 1 = 6 lots, or 30%30\%.
  • The single lot in 4.5 to 5.0 sold for somewhere between 104.5≈31 62310^{4.5} \approx 31\,623 dollars and $100 000.
Key fact

On a log⁡10\log_{10} scale, a whole unit is a factor of 10 and half a unit is a factor of 100.5≈3.1610^{0.5} \approx 3.16. The interval from aa to bb on the log axis contains data values from 10a10^a to 10b10^b. Half-way along a log interval is not half-way between the data values: the point 3.5 is about 3162, not 5000.

Placing a value on the scale

Exam questions often give a raw value, or information to calculate one, and ask which interval or bar it falls in. The routine is:

  1. Calculate the quantity in the original units if needed (for example, density == population ÷\div area).
  2. Take log⁡10\log_{10} of it.
  3. Find the interval on the axis that contains the log value, watching the boundary convention (intervals usually include their lower end and exclude their upper end).

For example, a city of 5 300 000 people living on 12 400 km² has a density of 5 300 000÷12 400≈427.45\,300\,000 \div 12\,400 \approx 427.4 people per km², and log⁡10427.4≈2.63\log_{10} 427.4 \approx 2.63, so it sits in the 2.5 to 3.0 interval. A mistake at step 1 (for example dividing the wrong way) or skipping step 2 (looking for 427 on an axis that only goes to 5) is exactly what the distractors are built around.

Describing a distribution on a log scale

Describe shape, centre and spread as you would for any histogram, but report them in the original units where possible.

  • Shape. The auction histogram is roughly symmetric with a slight positive tail on the log scale. On a linear scale the same prices would be strongly positively skewed. It is normal for a log transformation to "pull in" a long right tail, and that is usually the reason it was used.
  • Centre. The median of the log values is the log of the median value, because taking logs does not change the order of the data. Here the middle two prices are $340 and $420, so the median price is $380 (the median of the logs is 2.58, and 102.58≈38010^{2.58} \approx 380). You can also report the modal interval as a range of prices.
  • Spread. Describe it in orders of magnitude: "the prices range over more than three orders of magnitude, from about 101.310^{1.3} to about 104.710^{4.7} dollars", or give the range of the original values ($18 to $48 000).
  • Outliers. A value that looks extreme on a linear scale may not be an outlier on a log scale. If a question asks about outliers, work on the scale the display uses.

One caution: the mean of the log values does not back-transform to the mean of the original values. For the auction data the ordinary mean price is $3809.75, pulled up by the $48 000 lot, while 10mean of logs≈46610^{\text{mean of logs}} \approx 466. For skewed data spanning several orders of magnitude, the median is the better summary of a typical value.

Log scales on other displays

The same idea applies to any numerical display:

  • Dot plots and stem plots can use log⁡10\log_{10} values instead of the raw values.
  • Boxplots can be drawn on a log axis; read the five-number summary on the log axis and convert each value back with 10x10^x if asked.
  • Scatterplots with one variable on a log scale are the "logarithmic transformation" used to linearise data in the bivariate part of the course. That is a separate key skill (see the data transformation page), but the reading skill is identical.

Some displays label the log axis with the log values (1, 2, 3, ...), others label it with the powers of ten themselves (1010, 10210^2, 10310^3 or 10, 100, 1000). They mean the same thing; check the axis label before you start.

How exam questions ask about log scales

  • "In which interval / labelled column does the value for X lie?" Compute the value, take its log, locate the interval.
  • "How many of these values are in the modal interval?" Convert the modal interval's end points to data values (10a10^a and 10b10^b), then count the listed values between them, or take the log of each listed value.
  • "What percentage of the data is greater than / less than V?" Take log⁡10V\log_{10} V, then add the frequencies of the relevant bars and divide by the total.
  • "The number of countries with a population density between 100 and 1000 people per km² is..." 100 and 1000 are 10210^2 and 10310^3: add the bars from 2 to 3.
  • "Describe the shape of the distribution" on a log scale: describe what you see on the display (often approximately symmetric) and, if relevant, what it implies for the original data (positively skewed).
Worked examples: converting, placing, counting and interpreting

Converting between values and log values

Find log⁡10\log_{10} of 0.02, 7, 3000 and 250 000, and find the values whose logs are −1.3-1.3, 0.8 and 3.6.

Values to logs. log⁡100.02≈−1.699\log_{10} 0.02 \approx -1.699 (between −2-2 and −1-1, since 0.01<0.02<0.10.01 < 0.02 < 0.1); log⁡107≈0.845\log_{10} 7 \approx 0.845; log⁡103000≈3.477\log_{10} 3000 \approx 3.477; log⁡10250 000≈5.398\log_{10} 250\,000 \approx 5.398.

Logs to values. 10−1.3≈0.05010^{-1.3} \approx 0.050; 100.8≈6.3110^{0.8} \approx 6.31; 103.6≈398110^{3.6} \approx 3981.

Marker's note: a sanity check with powers of ten catches most keying errors. If log⁡10x=3.6\log_{10} x = 3.6, xx must be between 1000 and 10 000.

Counting values in the modal interval

A histogram of car prices on a log⁡10\log_{10} scale has intervals of width 0.5 and the modal interval is 3.5 to 4.0. Six of the prices are $2450, $3175, $4999, $8925, $10 250 and $105 600. How many of these are in the modal interval?

Convert the interval
3.5 to 4.0 is 103.5≈316210^{3.5} \approx 3162 dollars to 104=10 00010^4 = 10\,000 dollars.
Compare each price
$2450 is below 3162 (log 3.389). $3175 is just above 3162 (log 3.502). $4999 (log 3.699) and $8925 (log 3.951) are inside. $10 250 is above 10 000 (log 4.011). $105 600 is far above (log 5.024).
Answer
Three prices ($3175, $4999 and $8925) are in the modal interval.

Marker's note: $3175 and $10 250 are deliberately close to the boundaries. Taking the log of each value removes any doubt.

Placing a calculated quantity

A country has a population of 6 028 460 and an area of 720 km². Which interval of a log⁡10(population density)\log_{10}(\text{population density}) histogram with intervals of width 0.5 contains it?

Step 1, the quantity
Density =6 028 460÷720≈8372.9= 6\,028\,460 \div 720 \approx 8372.9 people per km².
Step 2, the log
log⁡108372.9≈3.923\log_{10} 8372.9 \approx 3.923.
Step 3, the interval
3.923 is in 3.5 to 4.0 (which is about 3162 to 10 000 people per km²).

Marker's note: this is the structure of 2025 VCAA Examination 1 Question 4, where the country was Singapore and the answer choice was one of four labelled columns on the histogram.

Interpreting a log-scale histogram in words

Using the auction histogram above, describe the distribution of prices.

Shape
On the log⁡10\log_{10} scale the distribution is approximately symmetric, with a slight tail towards higher prices. (On a linear price scale it would be strongly positively skewed.)
Centre
The modal interval is 2.5 to 3.0, which is about $316 to $1000. The median price is $380.
Spread
Prices range from $18 to $48 000, more than three orders of magnitude; the middle half of prices lie between about $100 and about $1900 (Q1=102.5Q_1 = 102.5, Q3=1900Q_3 = 1900), roughly 10210^{2} to 103.310^{3.3} dollars.

Marker's note: examiners reward descriptions that convert back to the original units. "The median log price is 2.58" is correct but incomplete; "the median price is about $380" is what a reader actually wants.

Common traps
Reading log values as data values
A bar from 3.5 to 4.0 does not contain prices of $3.50 to $4.00. It contains prices from about $3162 to $10 000.
Treating half a log unit as a linear midpoint
3.5 on the log scale is about 3162, not 5000. Each half unit multiplies by about 3.16.
Forgetting to calculate the quantity first
If the question gives population and area, compute the density before taking the log.
Using the wrong log
Use base 10 (log⁡\log on most calculators), not the natural log (ln⁡\ln).
Back-transforming the mean of the logs
10mean of logs10^{\text{mean of logs}} is not the ordinary mean. The median, however, does back-transform correctly.
Boundary errors
A value whose log is exactly 3.0 belongs to the interval that starts at 3.0 under the usual "includes the lower end" convention.
Exam technique

Write the conversion line every time: "3.5 to 4.0 on the log scale is 103.5≈316210^{3.5} \approx 3162 to 104=10 00010^4 = 10\,000." It takes five seconds and removes almost every error. When a multiple-choice question lists raw values near an interval boundary, take the log of each one on your CAS rather than estimating. When describing a log-scale display, give the shape as seen on the display and state centre and spread in the original units.

Note

Some data mixes tiny and huge numbers, like prices from $18 to $48 000. If you draw a normal graph big enough to fit the $48 000, all the cheap items get squashed into one bar at the start. A log scale fixes this by giving the same amount of space to each "times ten": 10 to 100 gets the same room as 1000 to 10 000. The number on the scale tells you how many zeros the value has, roughly: 2 means about 100, 3 means about 1000, 4 means about 10 000. So to read the graph, you just turn the scale number back into a real value by working out 10 to that power.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
Without a calculator where possible, find (a) log⁡101000\log_{10} 1000, (b) log⁡100.01\log_{10} 0.01, (c) log⁡1050\log_{10} 50 correct to three decimal places.
Show worked solution →

(a) 1000=1031000 = 10^3, so log⁡101000=3\log_{10} 1000 = 3. (1 mark)

(b) 0.01=1100=10−20.01 = \dfrac{1}{100} = 10^{-2}, so log⁡100.01=−2\log_{10} 0.01 = -2. (1 mark)

(c) 5050 is between 10=10110 = 10^1 and 100=102100 = 10^2, so its log is between 1 and 2. By calculator, log⁡1050≈1.699\log_{10} 50 \approx 1.699. (1 mark)

A quick check for any log: the whole-number part of log⁡10x\log_{10} x is one less than the number of digits before the decimal point for x≥1x \ge 1. 5050 has 2 digits, so its log starts with 1.

foundation2 marks
A value xx has log⁡10x=2.6\log_{10} x = 2.6. (a) Between which two consecutive powers of ten does xx lie? (b) Find xx correct to one decimal place.
Show worked solution →

(a) 2<2.6<32 < 2.6 < 3, so 102<x<10310^2 < x < 10^3, that is xx is between 100 and 1000. (1 mark)

(b) Undo the log by raising 10 to the power:

x=102.6≈398.1.x = 10^{2.6} \approx 398.1.

(1 mark)

foundation2 marks
A histogram of used-car prices has a logarithmic (base 10) scale with class intervals of width 0.5 starting at 3.0 (so 3.0 to less than 3.5, 3.5 to less than 4.0, and so on). In which class interval does a car priced at $8925 lie? What range of prices does that interval represent?
Show worked solution →

log⁡108925≈3.951\log_{10} 8925 \approx 3.951, which lies in the interval 3.5 to less than 4.0. (1 mark)

That interval represents prices from 103.5≈316210^{3.5} \approx 3162 dollars up to 104.0=10 00010^{4.0} = 10\,000 dollars. (1 mark)

core3 marks
The prices, in dollars, of 20 lots at an auction are summarised on a histogram with a log⁡10(price)\log_{10}(\text{price}) scale. The frequencies for the intervals 1.0 to 1.5, 1.5 to 2.0, 2.0 to 2.5, 2.5 to 3.0, 3.0 to 3.5, 3.5 to 4.0, 4.0 to 4.5 and 4.5 to 5.0 are 2, 3, 4, 5, 3, 1, 1, 1. (a) How many lots sold for between $100 and $1000? (b) What percentage of lots sold for more than $1000? (c) State the modal interval as a range of prices.
Show worked solution →

(a) $100 is 10210^2 and $1000 is 10310^3, so the lots between them are in the intervals 2.0 to 2.5 and 2.5 to 3.0: 4+5=94 + 5 = 9 lots. (1 mark)

(b) More than $1000 means log⁡10(price)>3\log_{10}(\text{price}) > 3: intervals from 3.0 upward hold 3+1+1+1=63 + 1 + 1 + 1 = 6 lots, and 620×100%=30%\dfrac{6}{20} \times 100\% = 30\%. (1 mark)

(c) The tallest bar (5 lots) is 2.5 to 3.0, which is 102.5≈31610^{2.5} \approx 316 dollars up to 103=100010^{3} = 1000 dollars. (1 mark)

core2 marks
A city has a population of 5 300 000 and an area of 12 400 km². A histogram shows log⁡10(population density)\log_{10}(\text{population density}) in intervals of width 0.5. Find the city's population density and the interval containing its log value.
Show worked solution →

Population density is people per square kilometre:

5 300 00012 400≈427.4 people per km2.\frac{5\,300\,000}{12\,400} \approx 427.4 \text{ people per km}^2.

(1 mark)

log⁡10427.4≈2.631\log_{10} 427.4 \approx 2.631, which is in the interval 2.5 to less than 3.0. (1 mark)

This is exactly the two-step structure of the 2025 VCAA Examination 1 question on population density: calculate the quantity first, then take its log to place it on the scale.

core2 marks
The same 20 auction prices range from $18 to $48 000. Explain why a histogram with a logarithmic (base 10) scale displays them better than a histogram with an ordinary (linear) price scale.
Show worked solution →

The prices span more than three orders of magnitude (from about 10110^1 to about 104.710^{4.7}). On a linear scale with equal-width intervals wide enough to reach $48 000, almost every price would fall in the first one or two bars and the single large price would sit far out on its own, so the shape of most of the data would be hidden and the distribution would look extremely positively skewed. (1 mark)

A log scale gives each power of ten the same width, so small and large prices are spread out evenly, the shape of the whole distribution becomes visible (here roughly symmetric), and the centre and spread can be described. (1 mark)

exam1 mark
On a histogram with a log⁡10\log_{10} scale, the class interval 3.5 to less than 4.0 contains values from: A. 3.5 to 4.0 B. 350 to 400 C. about 3162 to 10 000 D. 5000 to 10 000
Show worked solution →

The scale shows powers of ten, so the interval boundaries are 103.510^{3.5} and 104.010^{4.0}:

103.5≈3162.3,104=10 000.10^{3.5} \approx 3162.3, \qquad 10^{4} = 10\,000.

The answer is C. Option A reads the log values as the data values, option B multiplies by 100, and option D wrongly assumes 3.5 on the log scale means 5000 (the midpoint of 1000 and 10 000 on a linear scale). Half a unit on a log scale is a factor of 100.5≈3.1610^{0.5} \approx 3.16, not a halfway point.

exam3 marks
For a set of 20 values, the median of the log⁡10\log_{10} values is 2.58. (a) Estimate the median of the original values, to the nearest whole number. (b) The mean of the log⁡10\log_{10} values is 2.67. Explain why 102.6710^{2.67} is **not** the mean of the original values. (c) State one advantage of reporting the median rather than the mean for data like this.
Show worked solution →

(a) Taking logs keeps values in the same order, so the middle log value belongs to the middle original value: median ≈102.58≈380\approx 10^{2.58} \approx 380. (1 mark)

(b) The log scale compresses large values much more than small ones, so averaging logs and then back-transforming does not give the ordinary average. For positively skewed data the ordinary mean is pulled up by the large values and is much bigger than 102.67≈46810^{2.67} \approx 468. (1 mark)

(c) Data spread over several orders of magnitude is usually strongly skewed on the original scale, with a few very large values. The median is resistant to those values, so it is a better measure of a typical value. (1 mark)

Practise this

Sources & how we know this

ExamExplained