## Section10.2Logarithmic Scales

### Subsection10.2.1Making a Log Scale

Because $$\log x$$ grows very slowly as $$x$$ increases, logarithms are useful for modeling phenomena that take on a very wide range of values. For example, biologists study how metabolic functions such as heart rate are related to an animal’s weight, or mass. The table shows the mass in kilograms of several mammals.

 Animal Shrew Cat Wolf Horse Elephant Whale Mass, kg $$0.004$$ $$4$$ $$80$$ $$300$$ $$5400$$ $$70,000$$

Imagine trying to scale the $$x$$-axis to show all of these values. If we set tick marks at intervals of 10,000 kg, as shown below, we can plot the mass of the whale, and maybe the elephant, but the dots for the smaller animals will be indistinguishable.

On the other hand, we can plot the mass of the cat if we set tick marks at intervals of 1 kg, but the axis will have to be extremely long to include even the wolf. We cannot show the masses of all these animals on the same scale

To get around this problem, we can plot the log of the mass, instead of the mass itself. The table below shows the base 10 log of each animal's mass, rounded to 2 decimal places.

 Animal Shrew Cat Wolf Horse Elephant Whale Mass, kg $$0.004$$ $$4$$ $$80$$ $$300$$ $$5400$$ $$70,000$$ Log (mass) $$-2.40$$ $$0.60$$ $$1.90$$ $$2.48$$ $$3.73$$ $$4.85$$

The logs of the masses range from -2.40 to 4.85. We can easily plot these values on a single scale, as shown below.

The scale above is called a logarithmic scale, or log scale. The tick marks are labeled with powers of 10, because, as you recall, a logarithm is actually an exponent. For example, the mass of the horse is 300 kg, and

\begin{equation*} \log 300 = 2.48 ~~ \text{ so } ~~ 10^{2.48} = 300 \end{equation*}

When we plot 2.48 for the horse, we are really plotting the power of 10 that gives its mass, because $$10^{2.48} = 300$$ kg. The exponents on base 10 are evenly spaced on a log scale, so we plot $$10^{2.48}$$ about halfway between $$10^2$$ and $$10^3\text{.}$$

Plot the values on a log scale.

 $$x$$ $$0.0007$$ $$0.2$$ $$3.5$$ $$1600$$ $$72,000$$ $$4 \times 10^8$$
Solution.

We actually plot the logs of the values, so we first compute the base 10 logarithm of each number.

 $$x$$ $$0.0007$$ $$0.2$$ $$3.5$$ $$1600$$ $$72,000$$ $$4 \times 10^8$$ $$\log x$$ $$-3.15$$ $$-0.70$$ $$0.54$$ $$3.20$$ $$4.86$$ $$8.60$$

Then we plot each logarithm, estimating its position between integer exponents. For example, we plot the first value, $$-3.15\text{,}$$ closer to $$-3$$ than to $$-4\text{.}$$ The finished plot is shown below.

Complete the table by estimating the logarithm of each point plotted on the log scale above. Then use a calculator to give a decimal value for each point.

 $$\log x$$ $$x$$

$$-4$$

$$-2.5$$

$$1.5$$

$$4.25$$

$$0.0001$$

$$0.00316228$$

$$31.6228$$

$$17782.8$$

Solution.
 $$\log x$$ $$-4$$ $$-2.5$$ $$1.5$$ $$4.25$$ $$x$$ $$0.0001$$ $$0.00316$$ $$31.6$$ $$17,782.8$$

Fill in the blanks.

1. We use

• logarithms

• imaginary numbers

for graphing variables that take on a very wide range of values.

2. The tick marks on a log scale are labeled with

• variables

• powers of 10

.

3. To plot values on a log scale, we first compute the

• reciprocals

• logarithms

of the values.

4. A fraction less than 1 is plotted on a log scale as a power with a

• negative

• positive

exponent.

$$\text{logarithms}$$

$$\text{powers of 10}$$

$$\text{logarithms}$$

$$\text{negative}$$

Solution.
1. logarithms

2. powers of 10

3. logs

4. negative

### Subsection10.2.2Labeling a Log Scale

Log scales allow us to plot a wide range of values, but there is a trade-off. Equal increments on a log scale do not correspond to equal differences in value, as they do on a linear scale. You can see this more clearly if we label the tick marks with their values, as well as powers of 10. The difference between $$10^1$$ and $$10^0$$ is $$10 - 1 = 9\text{,}$$ but the difference between $$10^2$$ and $$10^1$$ is $$100 - 10 = 90\text{.}$$

As we move from left to right on this scale, we multiply the value at the previous tick mark by 10. Moving up by equal increments on a log scale does not add equal amounts to the values plotted; it multiplies the values by equal factors.

If we would like to label the log scale with integers, we get a very different looking scale, one in which the tick marks are not evenly spaced.

Plot the integer values 2 through 9 and 20 through 90 on a log scale.

Solution.

We compute the logarithm of each integer value.

 $$x$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$\log x$$ $$0.301$$ $$0.477$$ $$0.602$$ $$0.699$$ $$0.778$$ $$0.845$$ $$0.903$$ $$0.954$$
 $$x$$ $$20$$ $$30$$ $$40$$ $$50$$ $$60$$ $$70$$ $$80$$ $$90$$ $$\log x$$ $$1.301$$ $$1.477$$ $$1.602$$ $$1.699$$ $$1.778$$ $$1.845$$ $$1.903$$ $$1.954$$

We plot on a log scale, as shown below.

On the log scale in Example 10.2.4, notice how the integer values are spaced: They get closer together as they approach the next power of $$10\text{.}$$ You will often see log scales labeled not with powers of $$10\text{,}$$ but with integer values, like this:

In fact, log-log graph paper scales both axes with logarithmic scales.

The opening page of Chapter 6 shows the “mouse-to-elephant” curve, a graph of the metabolic rate of mammals as a function of their mass. (The elephant does not appear on that graph, because its mass is too big.) The figure above shows the same function, graphed on log-log paper.

Use this graph to estimate the mass and metabolic rate for the following animals, labeled on the graph.

 Animal Mouse Dog Sheep Mass (kg) Metabolic rate (kcal/day)
 Animal Cow Elephant Mass (kg) Metabolic rate (kcal/day)

$$0.02$$

$$15$$

$$50$$

$$3.5$$

$$500$$

$$1500$$

$$500$$

$$4000$$

$$6000$$

$$50000$$

Solution.
 Animal Mouse Dog Sheep Cow Elephant Mass (kg) $$0.02$$ $$15$$ $$50$$ $$500$$ $$4000$$ Metabolic rate (kcal/day) $$3.5$$ $$500$$ $$1500$$ $$6000$$ $$50,000$$

True or False.

1. Equal increments on a log scale correspond to equal differences in value.

• True

• False

2. Moving up by equal increments on a log scale multiplies the values by equal factors.

• True

• False

3. If we label a log scale with integers, the tick marks are evenly spaced.

• True

• False

4. On log-log graph paper, both axes are labeled with logarithmic scales.

• True

• False

$$\text{False}$$

$$\text{True}$$

$$\text{False}$$

$$\text{True}$$

Solution.
1. False

2. True

3. False

4. True

### Subsection10.2.3Acidity and the pH Scale

You may have already encountered log scales in some everyday applications. A simple example is the pH scale, used by chemists to measure the acidity of a substance or chemical compound. This scale is based on the concentration of hydrogen ions in the substance, denoted by $$[H^+]\text{.}$$ The pH value is defined by the formula

\begin{equation*} \text{pH}=-\log_{10}[H^+] \end{equation*}

Values for pH fall between 0 and 14, with 7 indicating a neutral solution. The lower the pH value, the more acidic the substance. Some common substances and their pH values are shown in the table.

 Substance pH $$[H^+]$$ Battery acid $$1$$ $$0.1$$ Lemon juice $$2$$ $$0.01$$ Vinegar $$3$$ $$0.001$$ Milk $$6.4$$ $$10^{-6.4}$$ Baking soda $$8.4$$ $$10^{-8.4}$$ Milk of magnesia $$10.5$$ $$10^{-10.5}$$ Lye $$13$$ $$10^{-13}$$

A decrease of 1 on the pH scale corresponds to an increase in acidity by a factor of 10. Thus, lemon juice is 10 times more acidic than vinegar, and battery acid is 100 times more acidic than vinegar.

1. Calculate the pH of a solution with a hydrogen ion concentration of $$3.98 \times 10^{-5}\text{.}$$

2. The water in a swimming pool should be maintained at a pH of 7.5. What is the hydrogen ion concentration of the water?

Solution.
1. We use a calculator to evaluate the pH formula with $$[H^+] = 3.98\times10^{-5}\text{.}$$

\begin{equation*} \text{pH} = -\log_{10}{(\alert{3.98 \times 10^{-5}})} \approx 4.4 \end{equation*}

2. We solve the equation

\begin{equation*} \alert{7.5} = -\log_{10}[H^+] \end{equation*}

for $$[H^+]\text{.}$$ First, we write

\begin{equation*} -7.5 = \log_{10}[H^+] \end{equation*}

Then we convert the equation to exponential form to get

\begin{equation*} [H^+] = 10^{-7.5}\approx 3.2 \times 10^{-8} \end{equation*}

The hydrogen ion concentration of the water is $$3.2 \times 10^{-8}\text{.}$$

The pH of the water in a tide pool is 8.3. What is the hydrogen ion concentration of the water?

$$10^{-8.3}$$

Solution.

$$5.01\times 10^{-9}$$

### Subsection10.2.4Decibels

The decibel scale, used to measure the loudness or intensity of a sound, is another example of a logarithmic scale. The perceived loudness of a sound is measured in decibels, $$D\text{,}$$ by

\begin{equation*} D=10 \log_{10}\left(\frac{I}{10^{-12}}\right) \end{equation*}

where $$I$$ is the intensity of its sound waves (in watts per square meter). The table below shows the intensity of some common sounds, measured in watts per square meter.

 Sound Intensity (watts/m$$^2$$) Decibels Whisper $$10^{-10}$$ $$20$$ Background music $$10^{-8}$$ $$40$$ Loud conversation $$10^{-6}$$ $$60$$ Heavy traffic $$10^{-4}$$ $$80$$ Jet airplane $$10^{-2}$$ $$100$$ Thunder $$10^{-1}$$ $$110$$

Consider the ratio of the intensity of thunder to that of a whisper:

\begin{equation*} \frac{\text{Intensity of thunder}}{\text{Intensity of a whisper}} = \frac{10^{-1}}{10^{-10}}= 10^9 \end{equation*}

Thunder is $$10^9\text{,}$$ or one billion times more intense than a whisper. It would be impossible to show such a wide range of values on a graph. When we use a log scale, however, there is a difference of only 90 decibels between a whisper and thunder.

1. Normal breathing generates about $$10^{-11}$$ watts per square meter at a distance of 3 feet. Find the number of decibels for a breath 3 feet away.

2. Normal conversation registers at about 40 decibels. How many times more intense than breathing is normal conversation?

Solution.
1. We evaluate the decibel formula with $$I = \alert{10^{-11}}$$ to find

\begin{align*} D \amp = 10 \log_{10}\left(\dfrac{\alert{10^{-11}}} {10^{-12}}\right) \amp\amp \blert{\text{Subtract exponents: }{-11}-(-12)=1}\\ \amp = 10 \log_{10} {10^1} = 10(1)\\ \amp = 10 \text{ decibels} \end{align*}
The sound of breathing registers at 10 decibels.

2. We let $$I_b$$ stand for the sound intensity of breathing, and $$I_c$$ stand for the intensity of normal conversation. We are looking for the ratio $$I_c/I_b\text{.}$$ From part (a), we know that

\begin{equation*} I_w = 10^{-11} \end{equation*}
and we can calculate $$I_c$$ from the decibelsformula.
\begin{align*} 40 \amp = 10 \log_{10}\left(\frac{I_c}{10^{-12}}\right) \amp\amp\blert{\text{Divide both sides by 10.}}\\ 4 \amp = \log_{10}\left(\frac{I_c}{10^{-12}}\right) \amp\amp\blert{\text{Convert to exponential form.}}\\ \dfrac{I_c}{10^{-12}} \amp = 10^4 \amp\amp \blert{\text{Multiply both sides by }10^{-12}.}\\ I_c \amp = 10^4(10^{-12}) = 10^{-8} \end{align*}
Finally, we compute the ratio $$\dfrac{I_c}{I_b}\text{:}$$
\begin{equation*} \dfrac{I_c}{I_b}= \frac{10^{-8}}{10^{-11}}= 10^3 \end{equation*}
Normal conversation is 1000 times more intense than breathing.

The noise of city traffic registers at about $$70$$ decibels.

1. What is the intensity of traffic noise, in watts per square meter?

Answer: $$I=$$ watts/m$$^2$$

2. How many times more intense is traffic noise than conversation?

$$10^{\frac{70}{10}-12}$$

$$10^{\frac{70-40}{10}}$$

Solution.
1. $$I = 10^{-5}$$ watts/m$$^2$$

2. $$\displaystyle 1000$$

#### Caution10.2.11.

Both the decibel model and the Richter scale in the next example use expressions of the form $$\log\left(\dfrac{a}{b}\right)\text{.}$$ Be careful to follow the order of operations when using these models. We must compute the quotient $$\dfrac{a}{b}$$ before taking a logarithm. In particular, it is not true that $$\log\left(\dfrac{a}{b}\right)$$ can be simplified to $$\dfrac{\log a}{\log b}\text{.}$$

### Subsection10.2.5The Richter Scale

One method for measuring the magnitude of an earthquake compares the amplitude $$A$$ of its seismographic trace with the amplitude $$A_0$$ of the smallest detectable earthquake. The log of their ratio is the Richter magnitude, $$M\text{.}$$ Thus,

\begin{equation*} M=\log_{10}\left(\frac{A}{A_0} \right) \end{equation*}
1. The Northridge earthquake of January 1994 registered 6.9 on the Richter scale. What would be the magnitude of an earthquake 100 times as powerful as the Northridge quake?

2. How many times more powerful than the Northridge quake was the San Francisco earthquake of 1989, which registered 7.1 on the Richter scale?

Solution.
1. We let $$A_N$$ represent the amplitude of the Northridge quake and $$A_H$$ represent the amplitude of a quake 100 times more powerful. From the Richter model, we have

\begin{equation*} 6.9 = \log_{10}\left(\dfrac{A_N}{A_0}\right) \end{equation*}
or, rewriting in exponential form,
\begin{equation*} \dfrac{A_N}{A_0}= 10^{6.9} \end{equation*}
Now, we want $$A_H$$ to be 100 times $$A_N\text{,}$$ so
\begin{align*} A_H \amp = 100A_N\\ \amp = 100\left(10^{6.9}A_0\right) = 10^{8.9}A_0 \end{align*}
Thus, the magnitude of the more powerful quake is
\begin{align*} \log_{10}\left(\dfrac{A_H}{A_0}\right) \amp = \log_{10} 10^{8.9}\\ \amp = 8.9 \end{align*}

2. We let $$A_S$$ stand for the amplitude of the San Francisco earthquake. We are looking for the ratio $$A_S/A_N\text{.}$$ First, we use the Richter formula to compute values for $$A_S$$ and $$A_N\text{.}$$

\begin{equation*} 6.9 = \log_{10}\left(\dfrac{A_N}{A_0}\right) ~~\text{ and }~~ 7.1 = \log_{10}\left(\dfrac{A_s}{A_0}\right) \end{equation*}
Rewriting each equation in exponential form, we have
\begin{equation*} \dfrac{A_N}{A_0}= 10^{6.9} ~~\text{ and }~~ \dfrac{A_S}{A_0}= 10^{7.1} \end{equation*}
or
\begin{equation*} A_N = 10^{6.9}A_0 ~~\text{ and }~~A_S = 10^{7.1}A_0 \end{equation*}
Now we can compute the ratio we want:
\begin{equation*} \frac{A_S}{A_N}= \frac{10^{7.1}A_0}{10^{6.9}A_0}= 10^{0.2} \end{equation*}
The San Francisco earthquake was $$10^{0.2}\text{,}$$ or approximately 1.58 times as powerful as the Northridge quake.

In October 2005, a magnitude 7.6 earthquake struck Pakistan. How much more powerful was this earthquake than the 1989 San Francisco earthquake of magnitude 7.1?

$$10^{7.6-7.1}$$

Solution.

3.16

#### Note10.2.14.

An earthquake 100, or $$10^2\text{,}$$ times as strong is only two units greater in magnitude on the Richter scale. In general, a difference of $$K$$ units on the Richter scale (or any logarithmic scale) corresponds to a factor of $$10^K$$ units in the intensity of the quake.

On a log scale, the weights of two animals differ by 1.6 units. What is the ratio of their actual weights?

Solution.

A difference of 1.6 on a log scale corresponds to a factor of $$10^{1.6}$$ in the actual weights. Thus, the heavier animal is $$10^{1.6}\text{,}$$ or 39.8 times as heavy as the lighter animal.

Two points, labeled $$A$$ and $$B\text{,}$$ differ by $$2.5$$ units on a log scale. What is the ratio of their decimal values?

$$10^{2.5}$$

Solution.

316.2

True or False.

1. An increase of 1 on the pH scale corresponds to an increase in acidity by a factor of 10.

• True

• False

2. A ratio of sound intensities of one billion corresponds to a difference of 90 decibels.

• True

• False

3. The second property of logs says that $$\log \left(\dfrac{a}{b} \right) = \dfrac{\log a}{\log b}\text{.}$$

• True

• False

4. A difference of 3 on a log scale corresponds to a ratio of $$10^3\text{,}$$ or 1000.

• True

• False

$$\text{False}$$

$$\text{True}$$

$$\text{False}$$

$$\text{True}$$

Solution.
1. False

2. True

3. False

4. True

### Exercises10.2.6Problem Set 10.2

#### Warm Up

##### Exercise Group.

For Problems 1 and 2, bound the base 10 log of the number between two integers. Do not use a calculator!

###### 1.
1. $$\displaystyle 8$$

2. $$\displaystyle 137$$

3. $$\displaystyle 0.2$$

4. $$\displaystyle 1,234,567$$

1. 0 and 1

2. 2 and 3

3. $$-1$$ and 0

4. 6 and 7

###### 2.
1. $$\displaystyle 97$$

2. $$\displaystyle 0.05$$

3. $$\displaystyle 1.83$$

4. $$\displaystyle 26,125$$

##### Exercise Group.

For Problems 3 and 4, given $$\log_{10} n\text{,}$$ find $$n\text{.}$$

###### 3.
1. $$\displaystyle 3.6$$

2. $$\displaystyle 0.7$$

3. $$\displaystyle -3.1$$

4. $$\displaystyle -0.4$$

1. 3981.1

2. 5.01

3. 0.00079

4. 0.398

###### 4.
1. $$\displaystyle 1.5$$

2. $$\displaystyle 5.2$$

3. $$\displaystyle 0.18$$

4. $$\displaystyle -2.5$$

#### Skills Practice

##### 5.
1. The log scale is labeled with powers of 10. Finish labeling the tick marks in the figure with their corresponding decimal values.

2. The log scale is labeled with integer values. Label the tick marks in the figure with the corresponding powers of 10.

##### 6.
1. The log scale is labeled with powers of 10. Finish labeling the tick marks in the figure with their corresponding decimal values.

2. The log scale is labeled with integer values. Label the tick marks in the figure with the corresponding powers of 10.

##### 7.

Plot the values on a log scale.

 $$x$$ $$0.075$$ $$1.3$$ $$4200$$ $$87,000$$ $$6.5\times 10^7$$
##### 8.

Plot the values on a log scale.

 $$x$$ $$4\times 10^{-4}$$ $$0.008$$ $$0.9$$ $$27$$ $$90$$
##### 9.

Estimate the decimal value of each point on the log scale.

$$1.58\text{,}$$ $$6.31\text{,}$$ $$15.8\text{,}$$ $$63.1$$

##### 10.

Estimate the decimal value of each point on the log scale.

##### Exercise Group.

In Problems 11–18, use the appropriate formulas for logarithmic models.

###### 11.

The hydrogen ion concentration of vinegar is about $$6.3\times 10^{-4}\text{.}$$ Calculate the pH of vinegar.

$$3.2$$

###### 12.

The hydrogen ion concentration of spinach is about $$3.2\times 10^{-6}\text{.}$$ Calculate the pH of spinach.

###### 13.

The pH of lime juice is 1.9. Calculate its hydrogen ion concentration.

$$0.0126$$

###### 14.

The pH of ammonia is 9.8. Calculate its hydrogen ion concentration.

###### 15.

A lawn mower generates a noise of intensity $$10^{-2}$$ watts per square meter. Find the decibel level of the sound of a lawn mower.

$$100$$

###### 16.

A jet airplane generates 100 watts per square meter at a distance of 100 feet. Find the decibel level for a jet airplane.

###### 17.

The loudest sound emitted by any living source is made by the blue whale. Its whistles have been measured at 188 decibels and are detectable 500 miles away. Find the intensity of the blue whale's whistle in watts per square meter.

6,309,573 watts per square meter

###### 18.

The noise of a leaf blower was measured at 110 decibels. What was the intensity of the sound waves?

#### Applications

##### 19.

The log scale shows various temperatures in Kelvins. Estimate the temperatures of the events indicated.

$$1\text{,}$$ $$80\text{,}$$ $$330\text{,}$$ $$1600\text{,}$$ $$7000\text{,}$$ $$4\times 10^7$$

##### 20.

The log scale shows the size of various objects, in meters. Estimate the sizes of the objects indicated.

##### 21.

The magnitude of a star is a measure of its brightness. It is given by the formula

\begin{equation*} m = 4.83 - 2.5 \log L \end{equation*}

where $$L$$ is the luminosity of the star, measured in solar units. Calculate the magnitude of the stars whose luminosities are given in the figure.

Proxima Centauri: $$15.5\text{;}$$ Barnard: $$13.2\text{;}$$ Sirius: $$1.4\text{;}$$ Vega: $$0.6\text{;}$$ Arcturus: $$-0.4\text{;}$$ Antares: $$-4.7\text{;}$$ Betelgeuse: $$-7.2$$

##### 22.

Estimate the wavelength, in meters, of the types of electromagnetic radiation shown in the figure.

##### 23.

Plot the values of $$[H^+]$$ in the section "Acidity and the pH Scale" on a log scale.

##### 24.

Plot the values of sound intensity in the section "Decibels" on a log scale.

##### 25.

The distances to two stars are separated by 3.4 units on a log scale. What is the ratio of their distances?

$$10^{3.4} \approx 2512$$

##### 26.

The populations of two cities are separated by 2.8 units on a log scale. What is the ratio of their populations?

##### 27.

The probability of discovering an oil field increases with its diameter, defined to be the square root of its area. Use the graph to estimate the diameter of the oil fields at the labeled points, and their probability of discovery. (Source: Deffeyes, 2001)

A: $$a\approx 45\text{,}$$ $$p \approx 7.4\%\text{;}$$ B: $$a \approx 400\text{,}$$ $$p \approx 15\%\text{;}$$ C: $$a\approx 6000\text{,}$$ $$p\approx 50\%\text{;}$$ D: $$a \approx 13000\text{,}$$ $$p \approx 45\%$$

##### 28.

The order of a stream is a measure of its size. Use the graph to estimate the drainage area, in square miles, for streams of orders 1 through 4. (Source: Leopold, Wolman, and Miller)

##### 29.

The pH of normal rain is 5.6. Some areas of Ontario have experienced acid rain with a pH of 4.5. How many times more acidic is acid rain than normal rain?

12.6

##### 30.

The pH of normal hair is about 5, the average pH of shampoo is 8, and 4 for conditioner. Compare the acidity of normal hair, shampoo, and conditioner.

##### 31.

At a concert by The Who in 1976, the sound level 50 meters from the stage registered 120 decibels. How many times more intense was this than a 90-decibel sound (the threshold of pain for the human ear)?

##### 32.

A refrigerator produces 50 decibels of noise, and a vacuum cleaner produces 85 decibels. How much more intense are the sound waves from a vacuum cleaner than those from a refrigerator?

$$3160$$

##### 33.

In 1964, an earthquake in Alaska measured 8.4 on the Richter scale. An earthquake measuring 4.0 is consideredsmall and causes little damage. How many times stronger was the Alaska quake than one measuring 4.0?

$$\approx 25,000$$