Logarithmic Scales
Discover logarithmic scales for Victorian Year 11 Mathematical Methods (VCAA) — scales such as the Richter scale, decibels and pH, where each step stands for multiplying by a fixed amount.
You will learn to apply base-ten logarithm laws to these scales, convert between a scale reading and the quantity behind it, and compare two readings, seeing how logarithms tame very large and very small numbers.
Theory
In Year 11 Mathematical Methods (Unit 2), a logarithmic scale uses the base-10 logarithm to compress a huge range of values into a manageable one. Because \(\log_{10}(10^{k})=k\), every ten-fold change in a quantity moves just one step along the scale. This page covers the three scales you meet — the Richter scale, sound decibels, and pH — how to compute a scale value, how to reverse it with a power of ten, and how to compare two readings.
A logarithmic scale measures a quantity by its base-10 logarithm rather than by the quantity itself. This turns multiplication into addition: because \(\log_{10}(10^{k})=k\), each time the underlying quantity is multiplied by \(10\), the scale reading increases by exactly \(1\).
The key relationships used throughout are the power law \(\log_{10}(10^{k})=k\), the product law \(\log_{10}(ab)=\log_{10}a+\log_{10}b\), and the fact that \(\log_{10}\) and \(10^{x}\) are inverses, so \(y=\log_{10}x\) rearranges to \(x=10^{y}\).
Three logarithmic scales appear in the course:
- Richter magnitude of an earthquake: \(M=\log_{10}\!\left(\dfrac{A}{A_0}\right)\), where \(A\) is the ground-motion amplitude and \(A_0\) a reference amplitude.
- Sound level in decibels: \(L=10\log_{10}\!\left(\dfrac{I}{I_0}\right)\), where \(I\) is the intensity and \(I_0=10^{-12}\ \text{W/m}^2\) the threshold of hearing.
- Acidity as pH: \(\mathrm{pH}=-\log_{10}[\mathrm{H}^{+}]\), where \([\mathrm{H}^{+}]\) is the hydrogen-ion concentration in \(\text{mol/L}\).
The base-10 logarithm undoes a power of ten:
The product law splits a coefficient off a power of ten — the workhorse for non-round numbers:
The three scales, and their rearranged (inverse) forms:
Finding a value on a logarithmic scale
- Substitute the quantity (or its ratio to the reference) into the scale formula.
- Write as a power of ten. If the quantity is a round power \(10^{k}\), use \(\log_{10}(10^{k})=k\) directly. If it is \(a\times 10^{k}\), use the product law: \(\log_{10}a+k\), and evaluate \(\log_{10}a\) on your calculator.
- Finish the arithmetic — multiply by \(10\) for decibels, or negate for pH — and round as asked.
Reversing the scale (given the reading, find the quantity)
- Isolate the logarithm (divide by \(10\) for decibels, change sign for pH).
- Raise \(10\) to each side using \(x=10^{y}\), then solve for the quantity.
Comparing two readings
- Subtract the two scale values; the reference cancels.
- Convert the difference back to a ratio with a power of ten (\(10^{d}\), or \(10^{d/10}\) for decibels).
Substitute the amplitude ratio:
| \(M\) | \(=\) | \(\log_{10}\!\left(\dfrac{A}{A_0}\right)\) |
| \(=\) | \(\log_{10}(100\,000)\) |
Write \(100\,000\) as a power of ten:
| \(100\,000\) | \(=\) | \(10^{5}\) |
Evaluate with \(\log_{10}(10^{k})=k\):
| \(M\) | \(=\) | \(\log_{10}(10^{5})\) |
| \(=\) | \(5\) |
Richter magnitude: \(M=5\).
Substitute the intensity ratio:
| \(L\) | \(=\) | \(10\log_{10}\!\left(\dfrac{I}{I_0}\right)\) |
| \(=\) | \(10\log_{10}(100)\) |
Evaluate the base-10 logarithm:
| \(\log_{10}(100)\) | \(=\) | \(\log_{10}(10^{2})\) |
| \(=\) | \(2\) |
Multiply by ten:
| \(L\) | \(=\) | \(10\times 2\) |
| \(=\) | \(20\) |
Loudness: \(L=20\) decibels.
Substitute the concentration:
| \(\mathrm{pH}\) | \(=\) | \(-\log_{10}(2.5\times10^{-4})\) |
Split with the product law \(\log_{10}(ab)=\log_{10}a+\log_{10}b\):
| \(\mathrm{pH}\) | \(=\) | \(-\big(\log_{10}(2.5)+\log_{10}(10^{-4})\big)\) |
| \(=\) | \(-\big(0.3979\ldots+(-4)\big)\) | |
| \(=\) | \(3.6020\ldots\) |
Round to two decimal places.
pH \(\approx 3.60\).
Subtract the two levels; the reference \(I_0\) cancels:
| \(L_1-L_2\) | \(=\) | \(10\log_{10}\!\left(\dfrac{I_1}{I_0}\right)-10\log_{10}\!\left(\dfrac{I_2}{I_0}\right)\) |
| \(=\) | \(10\log_{10}\!\left(\dfrac{I_1}{I_2}\right)\) |
Substitute \(L_1-L_2=80-60=20\) and solve for the ratio:
| \(20\) | \(=\) | \(10\log_{10}\!\left(\dfrac{I_1}{I_2}\right)\) |
| \(2\) | \(=\) | \(\log_{10}\!\left(\dfrac{I_1}{I_2}\right)\) |
| \(\dfrac{I_1}{I_2}\) | \(=\) | \(10^{2}\) |
| \(=\) | \(100\) |
The \(80\ \text{dB}\) sound is \(100\) times more intense.
Common pitfalls
Frequently asked questions
What is a logarithmic scale?
A scale that measures a quantity by its base-10 logarithm, so each \(\times 10\) in the quantity adds \(1\) to the reading. The Richter, decibel and pH scales all work this way.
Why does one step mean ten times as much?
Because \(\log_{10}(10^{k})=k\): raising the quantity by a factor of \(10\) raises the exponent, and hence the scale value, by exactly \(1\).
How do I find the pH from a hydrogen-ion concentration?
Substitute into \(\mathrm{pH}=-\log_{10}[\mathrm{H}^{+}]\). For \(a\times 10^{k}\), use the product law: \(-\big(\log_{10}a+k\big)\).
How do I go from a decibel level back to intensity?
Divide by \(10\), then raise \(10\) to that power: \(\dfrac{I}{I_0}=10^{L/10}\), so \(I=I_0\times 10^{L/10}\).
How many times louder is a 60 dB sound than a 40 dB sound?
Subtract the levels: \(60-40=20\ \text{dB}\), so the intensity ratio is \(10^{20/10}=10^{2}=100\) times.
Do these pages use natural logarithms?
No. Logarithmic scales in Units 1 & 2 use base 10 only, so every logarithm here is \(\log_{10}\).