Algebra 2 · F.BF.B.5

Logarithmic scales

The quietest sound a person can hear and a sound loud enough to hurt differ in strength by a factor of about a million million. How can one scale with everyday-sized numbers hold both?

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Evaluating logarithms, common and naturalAlgebra 2

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Why this exists

Some quantities in nature range over many powers of ten. The intensity of a sound you can just hear is about 10^{-12} watts per square meter, and a sound strong enough to hurt the ear is about 1 watt per square meter, a trillion times stronger. The concentration of hydrogen ions in a liquid, which decides how acidic it is, runs from about 1 mole per liter down to about 10^{-14}. The shaking recorded from a small earthquake and from a huge one differ by a factor of millions.

Measured on an ordinary scale, numbers like these are awkward in two ways. They need scientific notation just to be written down, and on an ordinary axis everything except the very largest values is squashed into a speck next to zero. What people usually want to know about such quantities is also not “how much more” but “how many times more”. A scale that counts factors rather than amounts answers that question directly, and the tool that turns factors into counts is the common logarithm: it counts how many factors of ten a number holds.

Definition

A logarithmic scale measures a positive quantity by a constant multiple of the logarithm of how many times bigger it is than a fixed reference value. If the quantity is x, the reference value is x_0 > 0 and c is a nonzero constant, the reading on the scale is

S = c \log\left(\frac{x}{x_0}\right)

where \log is the common logarithm, base 10. The reference value fixes where the scale reads 0 (when x = x_0, S = c \log 1 = 0), and the constant c fixes the size and direction of one step.

Three scales in everyday scientific use have this form.

pH, for acidity. The quantity is the hydrogen ion concentration [\mathrm{H}^+] in moles per liter, the reference is 1 mole per liter and c = -1: \mathrm{pH} = -\log [\mathrm{H}^+].
Sound level, in decibels. The quantity is the intensity I in watts per square meter, the reference is I_0 = 10^{-12} and c = 10: \displaystyle L = 10 \log\left(\frac{I}{I_0}\right).
Earthquake magnitude. The quantity is the amplitude A of the wave on a seismograph, the reference is a standard amplitude A_0 and c = 1: \displaystyle M = \log\left(\frac{A}{A_0}\right).

In the pH formula the concentration is the number of moles per liter, so the logarithm is taken of a plain number. The minus sign makes the usual readings positive, since the concentrations met in practice are below 1 mole per liter and so have negative logarithms. The decibel reference I_0 = 10^{-12} watts per square meter is roughly the faintest sound a human ear can detect. The earthquake formula is the magnitude scale in its simplest form, which compares the size of the recorded wave with that of a standard small earthquake.

Intuition

On an ordinary ruler, equal distances stand for equal differences: from 2 to 3 is the same length as from 7 to 8. On a logarithmic scale, equal distances stand for equal factors: from 1 to 10 is the same length as from 10 to 100 and from 100 to 1000, because each is one more factor of ten.

Figure 1On a logarithmic axis equal distances are equal factors. Every gap between neighboring powers of ten has the same length, so 2 sits about 30\% of the way from 1 to 10, and 5 sits about 70\% of the way.

The positions inside each gap are logarithms too. The number 2 sits at \log 2 \approx 0.301 of the way from 1 to 10, so the small ticks for 2, 3, \dots, 9 are not evenly spaced: they crowd toward the top of each gap. Each gap repeats the same pattern, because 20 sits in the gap from 10 to 100 exactly where 2 sits in the gap from 1 to 10.

Reading a logarithmic scale is therefore counting. A pH of 3 against a pH of 6 is three steps, so three factors of ten in hydrogen ion concentration. A sound of 90 decibels against one of 60 is three steps of 10 decibels, so three factors of ten in intensity. The size of the quantity itself, which might be a millionth of a mole per liter, never has to be written out.

Results

The step rule. On the scale \displaystyle S = c \log\left(\frac{x}{x_0}\right), a change of d in the reading multiplies the quantity by 10^{d/c}:

S_2 - S_1 = d \quad\text{means}\quad \frac{x_2}{x_1} = 10^{\frac{d}{c}}

The factor depends only on the change d, not on where on the scale the change happens.

The factor rule. Read the other way, multiplying the quantity by a factor r > 0 changes the reading by c \log r:

x_2 = r\,x_1 \quad\text{means}\quad S_2 - S_1 = c \log r

For the three scales:

pH. Down 1 multiplies [\mathrm{H}^+] by 10; down d multiplies it by 10^d. A lower pH means a higher hydrogen ion concentration, which is a more acidic liquid.
Decibels. Up 10 decibels multiplies the intensity by 10; up d decibels multiplies it by 10^{d/10}.
Magnitude. Up 1 multiplies the wave amplitude by 10; up d multiplies it by 10^d.

Turning a reading back into the quantity. Solving \displaystyle S = c \log\left(\frac{x}{x_0}\right) for x gives

x = x_0 \cdot 10^{\frac{S}{c}}

so [\mathrm{H}^+] = 10^{-\mathrm{pH}}, I = 10^{-12} \cdot 10^{L/10} and A = A_0 \cdot 10^{M}.

Why it is true

Turning a reading back. From \displaystyle S = c \log\left(\frac{x}{x_0}\right), divide by c, which is allowed because c \neq 0: \displaystyle \log\left(\frac{x}{x_0}\right) = \frac{S}{c}. By the definition of a logarithm, \log y = k means 10^k = y, so \displaystyle \frac{x}{x_0} = 10^{S/c}, and multiplying by x_0 gives x = x_0 \cdot 10^{S/c}.

The step rule. Take two readings S_1 and S_2 of quantities x_1 and x_2. By the line above,

\frac{x_2}{x_1} = \frac{x_0 \cdot 10^{S_2/c}}{x_0 \cdot 10^{S_1/c}} = 10^{\frac{S_2 - S_1}{c}}

The x_0 cancels, and the powers of 10 divide by subtracting exponents (the quotient of powers law). With S_2 - S_1 = d this is 10^{d/c}, and the reference value has dropped out entirely, which is why the factor does not depend on where the change happens.

The factor rule. If x_2 = r\,x_1, the step rule’s equation reads 10^{(S_2 - S_1)/c} = r. By the definition of a logarithm again, \displaystyle \frac{S_2 - S_1}{c} = \log r, so S_2 - S_1 = c \log r.

Worked examples

Example 1

Finding a pH

Find the pH of each solution.

(a) A solution with [\mathrm{H}^+] = 10^{-11} moles per liter.

(b) A solution with [\mathrm{H}^+] = 3.0 \times 10^{-6} moles per liter, to two decimal places.

Solution

(a) \mathrm{pH} = -\log 10^{-11}. The common logarithm of a power of ten is its exponent, so \log 10^{-11} = -11 and \mathrm{pH} = 11. That is above 7, the pH of pure water at 25^\circ\mathrm{C}, so the solution is alkaline.

(b) The concentration is not a power of ten, so a calculator is needed: \log\left(3.0 \times 10^{-6}\right) \approx -5.52, and so \mathrm{pH} \approx 5.52. Before trusting it, place it: 10^{-6} < 3.0 \times 10^{-6} < 10^{-5}, so the logarithm is between -6 and -5 and the pH is between 5 and 6. It is, and the solution is acidic.

Example 2

Comparing two acidities

Solution A has pH 4.6 and solution B has pH 7.6. Find the hydrogen ion concentration of each, and say how many times more concentrated the hydrogen ions are in the more acidic one.

Solution

Turning a reading back, [\mathrm{H}^+] = 10^{-\mathrm{pH}}. For A, [\mathrm{H}^+] = 10^{-4.6} \approx 2.51 \times 10^{-5} moles per liter, and for B, [\mathrm{H}^+] = 10^{-7.6} \approx 2.51 \times 10^{-8} moles per liter.

The ratio does not need either decimal:

\frac{10^{-4.6}}{10^{-7.6}} = 10^{-4.6 - (-7.6)} = 10^{3} = 1000

by the quotient of powers law. Solution A, with the lower pH, has 1000 times the hydrogen ion concentration of B, so A is the more acidic. This is the step rule with c = -1: the reading went down by 3, so the concentration was multiplied by 10^{3}.

Figure 2Three steps down the pH scale multiply the hydrogen ion concentration by 10^3 = 1000, wherever on the scale the three steps are taken.

Example 3

Sound levels

(a) A sound has intensity 10^{-5} watts per square meter. Find its level in decibels.

(b) Find the intensity of a sound of 100 decibels.

Solution

(a) \displaystyle \frac{I}{I_0} = \frac{10^{-5}}{10^{-12}} = 10^{-5 - (-12)} = 10^{7}, so L = 10 \log 10^{7} = 10 \cdot 7 = 70 decibels.

(b) Turning the reading back, \displaystyle \log\left(\frac{I}{I_0}\right) = \frac{100}{10} = 10, so \displaystyle \frac{I}{I_0} = 10^{10} by the definition of a logarithm, and I = 10^{-12} \cdot 10^{10} = 10^{-2} watts per square meter.

(c) \displaystyle \frac{10^{-2}}{10^{-5}} = 10^{3} = 1000. The step rule gives the same with no intensities at all: the level rose by 30 decibels, and c = 10, so the factor is 10^{30/10} = 10^3. Thirty decibels is three factors of ten.

Example 4

Earthquake magnitudes

(a) Compare the wave amplitudes of earthquakes of magnitude 5.3 and 7.3.

(b) A second earthquake produces a wave 50 times the amplitude of the magnitude 5.3 earthquake’s wave. Find its magnitude, to one decimal place.

Solution

(a) Turning each reading back, A = A_0 \cdot 10^{5.3} and A = A_0 \cdot 10^{7.3}. Their ratio is

\frac{A_0 \cdot 10^{7.3}}{A_0 \cdot 10^{5.3}} = 10^{7.3 - 5.3} = 10^{2} = 100

so the magnitude 7.3 wave is 100 times the size of the magnitude 5.3 wave. Two steps on the scale, two factors of ten.

(b) This time the factor is known and the change in reading is wanted. Call the new magnitude M. By the step rule with c = 1, 10^{M - 5.3} = 50, so by the definition of a logarithm M - 5.3 = \log 50 \approx 1.699. Then M \approx 5.3 + 1.699 = 6.999, which is 7.0 to one decimal place. Multiplying the amplitude by 50 adds less than 2 to the magnitude, since 50 is less than 100 = 10^2.

Example 5

Star brightness: a scale with a different step

Astronomers measure how bright a star looks by its magnitude m, a scale chosen so that a difference of 5 magnitudes is a factor of exactly 100 in brightness, and larger magnitudes are dimmer stars. In the form of the definition, \displaystyle m = -2.5 \log\left(\frac{B}{B_0}\right) for a brightness B and a fixed reference B_0.

(a) By what factor does the brightness change for one step of magnitude?

(b) Compare a star of magnitude 1 with one of magnitude 6.

Solution

Here c = -2.5, so a rise of d in magnitude multiplies the brightness by 10^{d/(-2.5)} = 10^{-0.4d}.

(a) For d = 1 the brightness is multiplied by 10^{-0.4}, so it drops by a factor of 10^{0.4} \approx 2.512. Each magnitude step dims the star by about 2.5 times. Checking against the scale’s design: five steps give \left(10^{0.4}\right)^5 = 10^{2} = 100, by the power of a power law.

(b) The magnitudes differ by 5, so the brightnesses differ by 10^{0.4 \cdot 5} = 100. The magnitude 1 star, with the smaller number, is the brighter one, 100 times as bright.

(c) The difference is 3 - 0.5 = 2.5, so the factor is 10^{0.4 \cdot 2.5} = 10^{1} = 10. The magnitude 0.5 star is 10 times as bright.

Example 6

Reading a logarithmic axis

A logarithmic axis is drawn with one unit of length for each factor of ten, and 1 at position 0, so the number x sits at position \log x.

(a) Where do 2, 5 and 300 sit?

(b) Which number sits exactly halfway between 1 and 1000? Which sits halfway between 4 and 25?

Solution

(a) \log 2 \approx 0.301 and \log 5 \approx 0.699, both inside the first unit, since both numbers are between 1 and 10. For 300, 10^2 < 300 < 10^3, so it sits between 2 and 3 units, and the calculator gives \log 300 \approx 2.477.

(b) 1 sits at 0 and 1000 sits at 3, so the halfway point is position 1.5, where the number is 10^{1.5} \approx 31.62. It is not 500.5, the halfway point on an ordinary ruler.

For two numbers a and b in general, the halfway position is \displaystyle \frac{\log a + \log b}{2}, and the number there is

10^{\frac{\log a}{2} + \frac{\log b}{2}} = \left(10^{\log a}\right)^{\frac{1}{2}}\left(10^{\log b}\right)^{\frac{1}{2}} = \sqrt{a}\,\sqrt{b} = \sqrt{ab}

using the product of powers and power of a power laws, and 10^{\log a} = a. For 4 and 25 the halfway number is \displaystyle \sqrt{100} = 10. Check: 10 is 2.5 times 4, and 25 is 2.5 times 10, so the two equal distances are two equal factors.

Boundaries

Only positive quantities. The logarithm of zero or a negative number does not exist, so a logarithmic scale cannot measure a quantity that is zero. Total silence has no decibel level and a liquid with no free hydrogen ions would have no pH. The scale reaches ever further down as the quantity shrinks toward zero, without end.

Zero on the scale is not “none”. A reading of 0 means the quantity equals the reference value x_0, and negative readings are possible. A sound of 0 decibels has intensity 10^{-12} watts per square meter, and a sound fainter than that has a negative level.

Differences mean factors; sums mean nothing. Subtracting two readings gives the factor between the quantities, but adding two readings does not describe anything physical. Two sounds of 70 decibels together do not make 140 decibels: their intensities add, which doubles the intensity and raises the level by 10 \log 2 \approx 3 decibels. In the same way, a magnitude 8 earthquake is not “twice” a magnitude 4 one.

The simple models are simplified. These formulas are the textbook forms. Chemists refine the pH formula for strong solutions, and seismologists use several magnitude scales, including ones built on the energy an earthquake releases, which grows by a bigger factor per step than the wave amplitude. The mathematics of reading the scale, counting factors, stays the same.

The base sets the step. With the common logarithm, a reading changes by c for each factor of 10. A scale built on another base, such as the musical scale in the exercises, counts factors of that base instead.

Where this goes next

A logarithmic scale on one axis of a graph is a semi-log plot: exponential data, which climbs by equal factors, then lies along a straight line, and its slope reads off the growth factor. The laws of logarithms explain in general why adding on a logarithmic scale is multiplying in the quantity, which is the idea a slide rule is built on and the reason 10 \log 2 appeared for a doubled sound. The link with scientific notation runs both ways: the power of ten in a number’s scientific notation is its position on a logarithmic axis rounded down to a whole number, so 3.0 \times 10^{-6}, at position about -5.52, carries the power 10^{-6}.

Exercises

pH

Find the pH of a solution with [\mathrm{H}^+] = 10^{-1.5} moles per liter.
Find the pH of a solution with [\mathrm{H}^+] = 7.9 \times 10^{-4} moles per liter, to two decimal places.
A solution has pH 8.4. Find [\mathrm{H}^+], to three significant figures.
Compare the hydrogen ion concentrations of solutions at pH 3.3 and pH 6.1, to the nearest whole number.
A solution is diluted until its hydrogen ion concentration is \displaystyle \frac{1}{100} of what it was. By how much does its pH change, and in which direction?

Decibels

Find the level of a sound with intensity 10^{-9} watts per square meter.
Find the level of a sound with intensity 5 \times 10^{-4} watts per square meter, to one decimal place.
Find the intensity of a 45 decibel sound, to three significant figures.
How many times more intense is a 94 decibel sound than a 58 decibel sound, to the nearest whole number?
Find the intensity of a sound at -6 decibels, to three significant figures.

Earthquakes

An earthquake’s wave has 4000 times the standard amplitude A_0. Find its magnitude, to two decimal places.
Compare the wave amplitudes of earthquakes of magnitude 6.4 and 4.2, to the nearest whole number.
One earthquake’s wave is 20 times the amplitude of another’s. By how much do their magnitudes differ, to two decimal places?

Other scales

A star of magnitude 2 and a star of magnitude 9.5 are compared. Which looks brighter, and how many times as bright?
One star is 3 magnitudes brighter than another. How many times as bright is it, to one decimal place?
A musical note’s distance from the A at 440 hertz, counted in semitones, is \displaystyle n = 12 \log_2\left(\frac{f}{440}\right) for a note of frequency f hertz. (a) Find n for f = 110. (b) Find the frequency 3 semitones above the A, to two decimal places.

Reading a logarithmic axis

On these axes one unit of length stands for a factor of ten and 1 sits at position 0.

Find the positions of 1000, 0.01 and 500, the last to three decimal places.
Which number sits halfway between 2 and 200?
Which number sits one third of the way from 1 to 10, to three decimal places?

Reasoning

Two machines each run at 80 decibels. Running together, their intensities add. Show that the two together make about 83 decibels, not 160.
Prove that raising a sound level by d decibels multiplies its intensity by 10^{d/10}, whatever the starting level.
A friend says a magnitude 8 earthquake makes a wave twice the size of a magnitude 4 earthquake’s, since 8 is twice 4. Explain what is wrong, and give the true factor.
Explain why the pH formula carries a minus sign, and why that makes a lower pH mean a more acidic solution.
On a logarithmic axis, show that the number halfway between a and b is \displaystyle \sqrt{ab}, and explain why that halfway number is the same factor above a as b is above it.

Answers

\mathrm{pH} = 1.5. \log 10^{-1.5} = -1.5, so -\log [\mathrm{H}^+] = 1.5.
\mathrm{pH} \approx 3.10. \log\left(7.9 \times 10^{-4}\right) \approx -3.10; it lies between -4 and -3 as it should.
[\mathrm{H}^+] = 10^{-8.4} \approx 3.98 \times 10^{-9} moles per liter.
About 631 times. \displaystyle \frac{10^{-3.3}}{10^{-6.1}} = 10^{2.8} \approx 631; the pH 3.3 solution has the higher concentration.
Up by 2. The factor \displaystyle \frac{1}{100} = 10^{-2}, so by the factor rule with c = -1 the pH changes by -\log 10^{-2} = 2.
30 decibels. \displaystyle \frac{10^{-9}}{10^{-12}} = 10^{3}, and 10 \log 10^3 = 30.
About 87.0 decibels. \displaystyle \frac{5 \times 10^{-4}}{10^{-12}} = 5 \times 10^{8}, and 10 \log\left(5 \times 10^{8}\right) \approx 10 \cdot 8.699 = 86.99.
I = 10^{-12} \cdot 10^{4.5} = 10^{-7.5} \approx 3.16 \times 10^{-8} watts per square meter.
About 3981 times. The level rose by 36 decibels, so the factor is 10^{36/10} = 10^{3.6} \approx 3981.
I = 10^{-12} \cdot 10^{-0.6} = 10^{-12.6} \approx 2.51 \times 10^{-13} watts per square meter, a quarter of the reference intensity, roughly.
M = \log 4000 \approx 3.60. Check: 10^3 < 4000 < 10^4.
About 158 times. 10^{6.4 - 4.2} = 10^{2.2} \approx 158; the magnitude 6.4 wave is the larger.
\log 20 \approx 1.30. By the factor rule with c = 1.
The magnitude 2 star, 1000 times as bright. The difference is 7.5, and 10^{0.4 \cdot 7.5} = 10^{3}.
About 15.8 times. 10^{0.4 \cdot 3} = 10^{1.2} \approx 15.85.
(a) n = -24. \displaystyle \frac{110}{440} = \frac{1}{4} = 2^{-2}, so n = 12 \cdot (-2). (b) f = 440 \cdot 2^{3/12} = 440 \cdot 2^{1/4} \approx 523.25 hertz, from turning the reading back: \displaystyle \frac{f}{440} = 2^{n/12}.
3, -2 and about 2.699. \log 1000 = 3, \log 0.01 = -2, \log 500 \approx 2.699.
20. \displaystyle \sqrt{2 \cdot 200} = \sqrt{400} = 20; it is 10 times 2 and 200 is 10 times it.
10^{1/3} \approx 2.154. One third of the way is position \displaystyle \frac{1}{3}.
About 83.0 decibels. Each machine has I = 10^{-12} \cdot 10^{8} = 10^{-4}, so together I = 2 \times 10^{-4} and L = 10 \log\left(2 \times 10^{8}\right) \approx 10 \cdot 8.301 = 83.01. Doubling the intensity adds only about 3 decibels.
Turning the readings back, I_1 = I_0 \cdot 10^{L/10} and I_2 = I_0 \cdot 10^{(L + d)/10}. Dividing, \displaystyle \frac{I_2}{I_1} = 10^{\frac{L + d}{10} - \frac{L}{10}} = 10^{d/10} by the quotient of powers law, and L has canceled, so the factor is the same from any starting level.
Magnitude counts factors of ten, not amounts. The amplitudes are A_0 \cdot 10^{8} and A_0 \cdot 10^{4}, so the ratio is 10^{8 - 4} = 10^{4} = 10{,}000, not 2.
Hydrogen ion concentrations in practice are below 1 mole per liter, and the common logarithm of a number between 0 and 1 is negative; the minus sign makes the usual pH values positive. The logarithm rises as the concentration rises, so its negative falls: more hydrogen ions, which is more acidic, gives a smaller pH.
The positions are \log a and \log b, halfway is \displaystyle \frac{\log a + \log b}{2}, and \displaystyle 10^{\frac{\log a}{2} + \frac{\log b}{2}} = \sqrt{10^{\log a}}\sqrt{10^{\log b}} = \sqrt{ab}. Then \displaystyle \frac{\sqrt{ab}}{a} = \sqrt{\frac{b}{a}} and \displaystyle \frac{b}{\sqrt{ab}} = \sqrt{\frac{b}{a}}, the same factor, as equal distances on a logarithmic axis must be.