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1r BTL CCSS · Real numbers Notes 5 Oct 2026

12. Uses of logarithms

Where logarithms show up outside the classroom: the Richter scale, decibels and stock market charts, each with an interactive chart.

Why logarithms

Some quantities range from the tiny to the enormous: the amplitude of an earthquake, the intensity of a sound or the price of a share over decades. Written out as they are, the numbers have too many zeros to compare at a glance.

The common idea

Instead of the value, we give its logarithm. That way, multiplying the value by $10$ means adding $1$ on the scale:

$$\log(10 \cdot x) \overset{(1)}{=} \log 10 + \log x = 1 + \log x$$

This is property (1) from section 10: products become sums.

In class we looked at three examples: the Richter scale, sound intensity in decibels and stock market charts.

1. The Richter scale

In 1935 the seismologist Charles Richter proposed measuring earthquakes by the maximum amplitude $A$ of the waves traced by the seismograph, compared with a very small reference amplitude $A_0$:

Magnitude

$$M = \log \frac{A}{A_0} \overset{(2)}{=} \log A - \log A_0$$

Each extra point of magnitude means an amplitude $10$ times larger. The energy released grows even faster: it is multiplied by $10^{1.5} \approx 31.6$ for each point, so $2$ more points mean $1{,}000$ times more energy.

Example: Turkey (2023) and Lorca (2011)

The Turkey–Syria earthquake had magnitude $7.8$ and the Lorca one, $5.1$. Subtracting the two magnitudes:

$$7.8 - 5.1 = \log \frac{A_T}{A_0} - \log \frac{A_L}{A_0} \overset{(2)}{=} \log \frac{A_T}{A_L} \;\Longrightarrow\; \frac{A_T}{A_L} = 10^{2.7} \approx 501$$

The amplitude was about $500$ times larger. And the energy, $10^{1.5 \cdot 2.7} = 10^{4.05} \approx 11{,}200$ times more.

Interactive: compare earthquakes

Seismologists now use the moment magnitude ($M_w$), which is also logarithmic and calibrated to give values similar to Richter's. The magnitudes in the chart are these; the media still call it the "Richter scale".

2. Sound intensity: decibels

The human ear can hear sounds from an intensity of $10^{-12}\ \text{W/m}^2$ up to sounds a trillion times more intense. And it doesn't perceive them proportionally: a sound ten times more intense doesn't seem ten times louder. That is why sound level is measured in decibels (dB), with a logarithm:

Sound intensity level

$$\beta = 10 \cdot \log \frac{I}{I_0} \qquad I_0 = 10^{-12}\ \text{W/m}^2 \;\text{(threshold of hearing)}$$

Adding $10$ dB means multiplying the intensity by $10$. Adding $20$ dB, multiplying it by $100$.

Example: a conversation

A normal conversation is about $60$ dB. What is its intensity?

$$60 = 10 \log \frac{I}{10^{-12}} \;\Longrightarrow\; \log \frac{I}{10^{-12}} = 6 \;\Longrightarrow\; \frac{I}{10^{-12}} = 10^{6} \;\Longrightarrow\; I = 10^{-6}\ \text{W/m}^2$$

Interactive: from decibels to intensity

The reference sounds are approximate. From about $85$ dB, many hours of exposure can damage your hearing.

Example: two identical sources

If a second motorbike just like the first one ($\beta$ dB) joins it, the intensity doubles. In decibels:

$$10 \log \frac{2I}{I_0} \overset{(1)}{=} 10 \log 2 + 10 \log \frac{I}{I_0} \approx 10 \cdot 0.301 + \beta \approx \beta + 3$$

Two motorbikes don't make twice the decibels: they only add $3$. Here is the $\log 2 \approx 0.301$ from the homework again.

Interactive: how many identical sources?

The curve is the graph of a logarithm, $\beta_n = \beta + 10 \log n$: it rises quickly at first and then more and more slowly. To gain $10$ dB you need $10$ sources; to gain $20$, $100$.

3. Stock market charts: the log scale

When a stock chart covers many decades, the vertical axis is often logarithmic: the marks don't go up in steps of $100$, but $100$, $1{,}000$, $10{,}000$… Each step multiplies by $10$.

What the log scale changes

On a log axis, the same vertical distance means the same percentage. Going from $10$ to $20$ takes up the same space as going from $100$ to $200$, because both are a $100\,\%$ rise:

$$\log 20 - \log 10 \overset{(2)}{=} \log \frac{20}{10} = \log 2 = \log \frac{200}{100} = \log 200 - \log 100$$

And an investment that grows by the same percentage $r$ every year draws a straight line:

$$P(t) = P_0 (1 + r)^{t} \;\Longrightarrow\; \log P(t) \overset{(1),(3)}{=} \log P_0 + t \cdot \log(1 + r)$$

which is a linear function of $t$, with slope $\log(1 + r)$.

Interactive: linear or log scale

Two fictitious investments of $100$ € over $40$ years. A grows by the same percentage every year; B grows by the same amount every year, chosen so that both end up equal.

Investment A (fixed percentage)Investment B (fixed amount)

On the linear scale, A seems to do nothing for years and then takes off at the end. The log scale shows what is really going on: it always grows at the same rate (a straight line). B, whose percentage growth keeps shrinking, flattens out.

Example: when does an investment double?

If an investment grows by $7\,\%$ a year, how many years does it take to double? It's an exponential equation, like the one in section 10:

$$1.07^{t} = 2 \;\Longrightarrow\; t \cdot \log 1.07 = \log 2 \;\Longrightarrow\; t = \frac{\log 2}{\log 1.07} \approx \frac{0.301}{0.0294} \approx 10.2 \text{ years}$$

Try it

1Richter

One earthquake has magnitude $6$ and another, magnitude $4$. How many times larger is the amplitude of the first? And the energy?

  See the solution

The difference in magnitude is $2$:

$$\frac{A_1}{A_2} = 10^{6 - 4} = 10^{2} = 100 \qquad \frac{E_1}{E_2} = 10^{1.5 \cdot 2} = 10^{3} = 1{,}000$$
2Decibels

A motorbike makes $90$ dB. How many decibels do $10$ identical motorbikes make?

  See the solution
$$10 \log \frac{10\,I}{I_0} \overset{(1)}{=} 10 \log 10 + 10 \log \frac{I}{I_0} = 10 + 90 = 100 \text{ dB}$$
3Decibels

How many times more intense is an $85$ dB sound than a $65$ dB one?

  See the solution
$$85 - 65 = 10 \log \frac{I_1}{I_0} - 10 \log \frac{I_2}{I_0} \overset{(2)}{=} 10 \log \frac{I_1}{I_2} \;\Longrightarrow\; \log \frac{I_1}{I_2} = 2 \;\Longrightarrow\; \frac{I_1}{I_2} = 100$$
4Stock market

A share goes from $20$ € to $80$ € in $6$ years, rising by the same percentage every year. What is that annual percentage?

  See the solution
$$20(1 + r)^{6} = 80 \;\Longrightarrow\; (1 + r)^{6} = 4 \;\Longrightarrow\; 6 \log(1 + r) = \log 4$$
$$\log(1 + r) = \frac{\log 4}{6} = \frac{2 \log 2}{6} \approx \frac{0.602}{6} \approx 0.1003 \;\Longrightarrow\; 1 + r = 10^{0.1003} \approx 1.26$$

It rises by roughly $26\,\%$ a year. On a log scale, this share would draw a straight line.