Logarithmic Scales

Some quantities in nature cover an enormous range. The loudest sound you can stand is about a trillion times as intense as the quietest one you can hear. Instead of writing numbers like $0.000\,000\,000\,001$, scientists use logarithmic scales, which turn each multiplication by $10$ into a step of $1$. The pH scale, earthquake magnitudes, and decibels all work this way. Knowing the math helps you avoid the classic mistake of thinking a magnitude $6$ earthquake is “twice” a magnitude $3$.

Key ideas

What a logarithmic scale does

On a logarithmic scale, the reading is (a multiple of) the common log of the quantity. Since $\log 10 = 1$, $\log 100 = 2$, $\log 1000 = 3$, and so on:

• each step of $1$ on the scale means the quantity is multiplied by $10$
• a difference of $n$ on the scale means a ratio of $10^n$

The pH scale

The acidity of a solution depends on its concentration $C$ of hydrogen ions, in moles per litre (mol/L):

$$\text{pH} = -\log C \qquad\Longleftrightarrow\qquad C = 10^{-\text{pH}}$$

The negative sign makes pH values positive for typical solutions. Pure water has pH $7$. A lower pH means a higher concentration, so a more acidic solution. Each drop of $1$ in pH means the concentration is $10$ times as great.

Each step to the right on the pH scale divides the concentration by $10$. (Values for household items are approximate.)

Earthquake magnitude

A simple model for the magnitude $M$ of an earthquake is

$$M = \log \frac{I}{I_0}$$

where $I$ is the intensity of the earthquake and $I_0$ is the intensity of a tiny reference earthquake. (This is a simplified school version of the Richter scale, which is based on how much a seismograph needle moves.) A magnitude $6$ earthquake is $10$ times as intense as a magnitude $5$, and $100$ times as intense as a magnitude $4$.

Decibels

The sound level $L$, in decibels (dB), of a sound with intensity $I$ (in watts per square metre) is

$$L = 10\log \frac{I}{I_0}, \qquad I_0 = 10^{-12} \ \text{W/m}^2$$

$I_0$ is roughly the quietest sound a person can hear, and it has level $0$ dB. Because of the factor of $10$ in front, it’s each $10$ dB that multiplies the intensity by $10$.

Comparing two readings

Use the quotient law. For two earthquakes:

$$M_2 - M_1 = \log \frac{I_2}{I_0} - \log \frac{I_1}{I_0} = \log \frac{I_2}{I_1} \quad\Rightarrow\quad \frac{I_2}{I_1} = 10^{M_2 - M_1}$$

The same idea gives the other scales:

Scale	Ratio of the quantities
Earthquake magnitude	$\dfrac{I_2}{I_1} = 10^{M_2 - M_1}$
Decibels	$\dfrac{I_2}{I_1} = 10^{\frac{L_2 - L_1}{10}}$
pH (more acidic over less acidic)	$\dfrac{C_1}{C_2} = 10^{\text{pH}_2 - \text{pH}_1}$

Only the difference of the readings matters, not the readings themselves.

Worked examples

Example 1: pH and concentration

• (a) A sample of rainwater has $C = 3.2 \times 10^{-4}$ mol/L. Find its pH, to two decimal places.
• (b) A sample of seawater has pH $8.2$. Find its hydrogen-ion concentration.

Solution.

(a)

$$\text{pH} = -\log\left(3.2 \times 10^{-4}\right) \approx -(-3.49) = 3.49$$

That’s between $3$ and $4$, as expected, since $C$ is between $10^{-4}$ and $10^{-3}$.

(b) Solve for $C$:

$$C = 10^{-8.2} \approx 6.3 \times 10^{-9} \ \text{mol/L}$$

Example 2: Comparing earthquakes

• (a) How many times as intense is a magnitude $7.1$ earthquake as a magnitude $5.4$ earthquake?
• (b) An earthquake is $200$ times as intense as a magnitude $4.6$ earthquake. Find its magnitude.

Solution.

(a) The difference in magnitude is $7.1 - 5.4 = 1.7$:

$$\frac{I_2}{I_1} = 10^{1.7} \approx 50.1$$

It’s about $50$ times as intense. (Not ”$1.3$ times”, which is what you’d get by dividing $7.1$ by $5.4$.)

(b) The ratio is $200$, so the difference in magnitude is $\log 200$:

$$M = 4.6 + \log 200 \approx 4.6 + 2.30 = 6.9$$

Check: $10^{6.9 - 4.6} = 10^{2.3} \approx 200$. ✓

Example 3: Decibels

• (a) A busy street has a sound intensity of $3 \times 10^{-5}$ W/m². Find its sound level.
• (b) A rock concert is $95$ dB and a normal conversation is $60$ dB. How many times as intense is the concert?

Solution.

(a)

$$L = 10\log \frac{3 \times 10^{-5}}{10^{-12}} = 10\log\left(3 \times 10^{7}\right) \approx 10(7.477) \approx 74.8 \ \text{dB}$$

(b) The difference is $95 - 60 = 35$ dB, so

$$\frac{I_2}{I_1} = 10^{\frac{35}{10}} = 10^{3.5} \approx 3162$$

The concert is about $3200$ times as intense as the conversation. (That’s why hearing protection at concerts is a good idea.)

Example 4: Diluting an acid

A solution has pH $2.3$. By what factor must you dilute it (reduce its concentration) to raise its pH by $1.5$? Would the answer be different if it started at pH $3.0$?

Solution. Let $C_1$ and $C_2$ be the concentrations before and after. Then $\text{pH}_2 - \text{pH}_1 = 1.5$, so

$$\frac{C_1}{C_2} = 10^{1.5} \approx 31.6$$

You must dilute the solution by a factor of about $31.6$: the new concentration is about $\tfrac{1}{31.6}$ of the original. For example, $10$ mL of the solution would be topped up to about $316$ mL.

The answer does not depend on the starting pH, because only the change in pH appears in the calculation. Starting at pH $3.0$ (going to $4.5$), you’d still dilute by a factor of $10^{1.5} \approx 31.6$.

Check with the formula directly: $\dfrac{10^{-2.3}}{10^{-3.8}} = 10^{1.5}$. ✓

Common mistakes

Thinking the scale is linear. A magnitude $6$ earthquake is not twice as intense as a magnitude $3$. The difference is $3$, so it’s $10^3 = 1000$ times as intense.

Dividing the readings instead of subtracting them. Ratios of intensities come from the difference of the readings: $10^{M_2 - M_1}$. Never compute $\dfrac{M_2}{M_1}$ or $\dfrac{L_2}{L_1}$.

Forgetting the factor of $10$ in decibels. A $20$ dB increase is $10^{\frac{20}{10}} = 100$ times the intensity, not $10^{20}$ times and not $20$ times.

Getting the direction of pH backwards. Lower pH means more acidic, a higher concentration. And don’t drop the negative sign in $\text{pH} = -\log C$: without it, you’d get a negative pH for ordinary solutions.

Entering powers of ten incorrectly. For $C = 10^{-8.2}$, make sure the negative is part of the exponent. Estimate first: pH $8.2$ should give a concentration between $10^{-9}$ and $10^{-8}$.

Practice

1. (Warm-up) Use $\text{pH} = -\log C$.

• (a) Find the pH of a solution with $C = 1 \times 10^{-5}$ mol/L.
• (b) Find the concentration of a solution with pH $3$.

Solution

(a) $\text{pH} = -\log\left(10^{-5}\right) = 5$

(b) $C = 10^{-3} = 0.001$ mol/L

2. (Warm-up) How many times as intense is a magnitude $6$ earthquake as a magnitude $4$ earthquake?

Solution

$10^{6 - 4} = 10^2 = 100$ times as intense.

3. (Warm-up) How many times as intense is an $80$ dB sound as a $50$ dB sound?

Solution

$10^{\frac{80 - 50}{10}} = 10^3 = 1000$ times as intense.

4. (Core) A sample of tomato juice has pH $4.2$ and a sample of milk has pH $6.5$. How many times as great is the hydrogen-ion concentration of the tomato juice?

Solution

$$\frac{C_{\text{tomato}}}{C_{\text{milk}}} = 10^{6.5 - 4.2} = 10^{2.3} \approx 200$$

The tomato juice has about $200$ times the concentration.

5. (Core) Find the pH of a solution with $C = 4.5 \times 10^{-3}$ mol/L, to two decimal places.

Solution

$\text{pH} = -\log\left(4.5 \times 10^{-3}\right) \approx 2.35$. Check: $C$ is between $10^{-3}$ and $10^{-2}$, so the pH should be between $2$ and $3$. ✓

6. (Core) A leaf blower produces a sound intensity of $2.5 \times 10^{-2}$ W/m² at the operator’s ear.

• (a) Find the sound level in decibels.
• (b) Earmuffs reduce the sound level by $25$ dB. By what factor do they reduce the intensity?

Solution

(a)

$$L = 10\log \frac{2.5 \times 10^{-2}}{10^{-12}} = 10\log\left(2.5 \times 10^{10}\right) \approx 10(10.398) \approx 104 \ \text{dB}$$

(b) The ratio is $10^{\frac{25}{10}} = 10^{2.5} \approx 316$. The earmuffs cut the intensity to about $\tfrac{1}{316}$ of its original value (about $0.32\%$).

7. (Core) An earthquake is $15$ times as intense as a magnitude $5.8$ earthquake. What is its magnitude, to one decimal place?

Solution

$$M = 5.8 + \log 15 \approx 5.8 + 1.176 \approx 7.0$$

8. (Challenge) You have $50$ mL of a solution with pH $3.1$ and want to raise the pH to $4.0$ by adding water. About how much water should you add?

Solution

The pH must go up by $0.9$, so the concentration must be divided by $10^{0.9} \approx 7.94$. The amount of acid stays the same, so the volume must be multiplied by $7.94$:

$$50 \times 10^{0.9} \approx 397 \ \text{mL}$$

You’d add about $397 - 50 = 347$ mL of water.

9. (Challenge) A machine in a workshop produces a sound level of $70$ dB.

• (a) Show that two identical machines running together (twice the intensity) produce about $73$ dB, not $140$ dB.
• (b) How many identical machines would it take to reach $80$ dB?

Solution

(a) Doubling the intensity adds $10\log 2$ to the level, by the product law:

$$10\log \frac{2I}{I_0} = 10\log 2 + 10\log \frac{I}{I_0} \approx 3.0 + 70 = 73 \ \text{dB}$$

(b) $80$ dB is $10$ dB more than $70$ dB, which means $10^{\frac{10}{10}} = 10$ times the intensity. It would take $10$ machines.