Section 4.5 Logarithmic Scales; Modeling

Before Starting this Section, Review

  1. 1 Exponential form for logarithms (Section 4.2 , page 445)

  2. 2 Properties of logarithms (Section 4.2 , page 446)

  3. 3 Rules for logarithms (Section 4.3 , page 471)

Objectives

  1. 1 Define pH.

  2. 2 Define the Richter scale for measuring earthquake intensity.

  3. 3 Define scales for measuring sound.

  4. 4 Define magnitude of star brightness.

  5. 5 Build models from data

A logarithmic scale is a scale in which logarithms are used in the measurement of quantities. Suppose we have a quantity that has a small positive range of variation (for example, from 0.00001=10−50.00001=10−5 to 0.0001=10−40.0001=10−4); then the common logarithms of such numbers would be between −5−5 and −4.−4. Similarly, if a quantity has a large positive range of variation from 10,000=10410,000=104 to 100,000=105,100,000=105, the common logarithms of these numbers would be between 4 and 5. So a logarithmic scale serves to make data more manageable by expanding small variations and compressing large ones.

In Example 2 we will see how a small difference on a logarithmic scale results in an impressive increase in the acidity of acid rain compared to an ordinary rain.

pH Scale

  1. 1 Define pH.

In chemistry, acidity of a substance is related to the presence of positively charged hydrogen ions. pH represents the effective concentration (activity) of hydrogen ions H+H+ in water. Because H+H+ concentrations are much smaller than other dissolved species in water, the activity of hydrogen ions is expressed most conveniently in logarithmic units.

The pH value is a measure of the acidity or alkalinity of a solution. The pH value for pure water at 25°C (77°F)25°C (77°F) is 7.

Because pH is defined as −log[H+],−log[H+], pH decreases as [H+][H+] increases (which will happen if acid is added to water). Acidic solutions have pH values less than 7, and alkaline (also called basic) solutions have pH values greater than 7. The smaller the number on the pH scale, the more acidic the substance is.

Example 1 Calculating pH and [H+][H+]

  1. Calculate to the nearest tenth the pH value of grapefruit juice if [H+][H+] of grapefruit juice is 6.32×10−4.6.32×10−4.

  2. Find the hydrogen concentration [H+][H+] in beer if its pH value is 4.82.

Solution

  1. pH=−log[H+]Definition of pH=−log(6.32×10−4)Given [H+]=6.32×10−4=−(log 6.32+log 10−4)Product rule=−log 6.32+4−log 10−4=−(−4)log 10=4≈−0.8007+4Use a calculator.=3.1993≈3.2Simplify.pH====≈=−log[H+]−log(6.32×10−4)−(log 6.32+log 10−4)−log 6.32+4−0.8007+43.1993≈3.2Definition of pHGiven [H+]=6.32×10−4Product rule−log 10−4=−(−4)log 10=4Use a calculator.Simplify.

  2. 4.82=−log[H+]Replace pH with 4.82.log[H+]=−4.82Rewrite equation.[H+]=10−4.82Exponential form=100.18×10−5−4.82=−5+0.18≈1.51×10−5Use a calculator.4.82log[H+][H+]====≈−log[H+]−4.8210−4.82100.18×10−51.51×10−5Replace pH with 4.82.Rewrite equation.Exponential form−4.82=−5+0.18Use a calculator.

Practice Problem 1

    1. The [H+][H+] of a solution is 2.68×10−6.2.68×10−6. Find its pH value.

    2. Find [H+][H+] of seawater if its pH is 8.47.

Example 2 Acid Rain

How much more acidic is acid rain with a pH value of 3 than an ordinary rain with a pH value of 6?

Solution

We have

3=pHacid rain=−log[H+]acid rainDefinition of pH
3=pHacid rain=−log[H+]acid rainDefinition of pH

so that

[H+]acid rain=10−3Exponential form of log [H+]=−3
[H+]acid rain=10−3Exponential form of log [H+]=−3

Similarly,

[H+]ordinary rain=10−6.
[H+]ordinary rain=10−6.

Therefore,

[H+]acid rain[H+]ordinary rain=10−310−6=10−3+6=103=1000.
[H+]acid rain[H+]ordinary rain=10−310−6=10−3+6=103=1000.

So the hydrogen ion concentration in this acid rain is 1000 times greater than that in ordinary rain. That is, this acid rain is 1000 times more acidic than the ordinary rain.

Practice Problem 2

  1. How much more acidic is an acid rain with a pH value of 2.8 than an ordinary rain with a pH value of 6.2?

Table 4.8 lists a few typical pH values.

Table 4.8

Solution pH Value
Battery Acid 1
Lemon Juice 2
Stomach Acid 2–3
Vinegar 3
Milk 6–7
Baking Soda, Seawater 8–9
Milk of Magnesia 9–10
Ammonia 10–11
Drain opener 10–12
Lye 13

Earthquake Intensity

  1. 2 Define the Richter scale for measuring earthquake intensity.

An earthquake is the vibration, sometimes violent, of Earth’s surface that follows a release of energy in Earth’s crust. Earth’s crust is composed of about 20 huge plates that float on the molten material beneath the crust. These plates slowly move over, under, and past each other. Sometimes the movement is gradual. At other times, the plates are locked together, unable to release the accumulating energy. When this energy grows strong enough, the plates break free and vibrations called “seismic waves” or earthquakes are generated.

The Richter scale was invented in the 1930s by the American scientist Dr. Charles Richter to measure the magnitude of an earthquake. It is based on the idea of comparing the intensity of an earthquake with that of a zero-level earthquake. He defined the zero-level earthquake as an earthquake whose seismographic reading measures 1 micron (1000 microns=1 millimeter,1000 microns=1 millimeter, or 1,000,000 microns=1 meter1,000,000 microns=1 meter) at a distance of 100 kilometers (about 62 miles) from the epicenter of the earthquake.

Let us denote the intensity of a zero-level earthquake by I0.I0. A zero-level earthquake is just noticeable and is the threshold level below which we would not be aware of a quake.

Source: US Geological Survey (USGS).

Example 3 Intensity of an Earthquake

The magnitude of an earthquake is 4.0 on the Richter scale. What is the intensity of this earthquake?

Solution

M=log(II0)Definition of magnitude4=log(II0)Replace M with 4.II0=104Exponential formI=104I0=10,000 I0Multiply both sides by I0.
M4II0I====log(II0)log(II0)104104I0=10,000 I0Definition of magnitudeReplace M with 4.Exponential formMultiply both sides by I0.

The intensity of this earthquake is 10,000 times the intensity of I0I0 (the zero-level earthquake). Notice that a 1 point increase on the Richter scale corresponds to a factor of 10 increase in intensity.

Practice Problem 3

  1. The magnitude of an earthquake is 6.5 on the Richter scale. What is the intensity of this earthquake?

Example 4 Comparing Two Earthquakes

Compare the intensity of the Mexico City earthquake of 1985, which registered 8.1 on the Richter scale, to that of the San Francisco area earthquake of 1989, which measured 6.9 on the Richter scale.

Solution

Let IMIM and ISIS denote the intensities of the Mexico City and the San Francisco earthquakes, respectively. We have

8.1=log(IMI0)IMI0=108.1IM=108.1 I06.9=log(ISI0)Definition of magnitudeISI0=106.9Exponential formIS=106.9I0Multiply both sides by I0.
8.1IMI0IM===log(IMI0)108.1108.1 I06.9ISI0IS===log(ISI0)106.9106.9I0Definition of magnitudeExponential formMultiply both sides by I0.

Divide IMIM by ISIS and simplify to get

IMIS=108.1106.9I0I0=108.1106.9=108.1−6.9=101.2IM=101.2 IS≈16IS
IMISIM==108.1106.9I0I0=108.1106.9=108.1−6.9=101.2101.2 IS≈16IS

This equation shows that the intensity of the Mexico City earthquake was about 16 times that of the San Francisco area earthquake.

Practice Problem 4

  1. The magnitudes of the earthquakes of Mozambique (2006) and Southern California (2005) were 7.0 and 5.2, respectively. Compare their intensities.

Energy of an Earthquake

The magnitude M of an earthquake is related to its released energy E (measured in joules) and is approximated by the equation

log E=4.4+1.5 M.
log E=4.4+1.5 M.

We can rewrite this equation in the exponential form as

E=104.4+1.5M=100.4×104×101.5M
E=104.4+1.5M=100.4×104×101.5M

or

E≈(2.5×104)×101.5M100.4≈2.5
E≈(2.5×104)×101.5M100.4≈2.5

The energy E0E0 released by a zero-level earthquake can be calculated by substituting M=0M=0 in the formula. We have

E0=2.5×104 joules.
E0=2.5×104 joules.

Table 4.9 Major Earthquakes (M≥7.5)(M≥7.5) of the 20th & 21st Century

Date Location Magnitude
April 1906 San Francisco 7.8
December 1908 Messina, Italy 7.5
December 1920 Gansu, China 8.6
September 1923 Sagami Bay, Japan 8.3
February 1931 New Zealand 7.9
August 1950 Assam, India 8.7
July 1976 Tangshan, China 8.0
September 1985 Mexico City 8.1
June 1990 Iran 7.7
May 1997 Iran 7.5
December 2004 Sumatra, Indonesia 9.0
March 2005 Sumatra, Indonesia 8.6
February 2010 Maule, Chile 8.8
March 2011 Tohoku, Japan 9.0
Source: National Earthquake Information Center, U.S. Geological Survey.

If two earthquakes have magnitude M1M1 and M2M2, we can compute the ratio of their corresponding energies E1E1 and E2E2 as follows:

E1E2=(2.5×104)×101.5M1(2.5×104)×101.5M2=101.5M1101.5M2=101.5M1−1.5M2=101.5(M1−M2).
E1E2=(2.5×104)×101.5M1(2.5×104)×101.5M2=101.5M1101.5M2=101.5M1−1.5M2=101.5(M1−M2).

Example 5 Comparing Two Earthquakes

Compare the estimated energies released by the San Francisco earthquake of 1906 and the Northridge earthquake of 1994 with magnitude of 6.7.

Solution

Let M1906M1906 and E1906E1906 represent the magnitude and energy released by the 1906 quake of San Francisco. Attach similar meaning to M1994M1994 and E1994.E1994.

With M1906=7.8M1906=7.8 (Table 4.9) and M1994=6.7M1994=6.7 and the equation E1E2=101.5(M1−M2),E1E2=101.5(M1−M2), we have

E1906E1994=101.5(7.8−6.7)=101.5(1.1)≈45Use a calculator.
E1906E1994=101.5(7.8−6.7)=101.5(1.1)≈45Use a calculator.

The 1906 earthquake released more than 45 times as much energy as the 1994 earthquake.

Practice Problem 5

  1. Compare the energies released by the Japan earthquake of 2011 and Iran earthquake of 1997 (see Table 4.9 ).

Loudness of Sound

  1. 3 Define scales for measuring sound.

The intensity of a sound wave is defined as the amount of power the wave transmits through a given area. The intensity of a sound is measured in watts per square meters (abbreviated W /m2W /m2). The faintest sound that the human ear can detect has an intensity of 10−12 W /m2.10−12 W /m2. This faintest sound is known as the threshold of hearing (TOH). The most intense sound that the human ear can safely detect without suffering any physical damage is more than one billion times the TOH.

Because the range of intensities that the human ear can detect is so large, a logarithmic scale called decibels (abbreviated dB) is used to measure the loudness of a sound. A decibel (as the name suggests), is one-tenth of a bel, a unit named after the telephone inventor Alexander Graham Bell.

If we substitute I=I0I=I0 in the formula for loudness, we obtain

Loudness of TOH=10 log I0I0=10 log 1=10(0)=0.
Loudness of TOH=10 log I0I0=10 log 1=10(0)=0.

So the threshold of hearing has 0 dB loudness.

Table 4.10 lists approximate loudness of some common sounds.

Table 4.10 Approximate loudness of common sounds

Source Intensity in W/m2W/m2 Loudness in dB
Threshold of hearing 10−1210−12 0
Rustling leaves 10−1110−11 10
Whisper 10−1010−10 20
Background noise in average home 10−810−8 40
Normal conversation 10−610−6 60
Busy street traffic 10−510−5 70
Vacuum cleaner 10−410−4 80
Portable audio player at maximum level 10−210−2 100
Front rows of rock concert 10−110−1 110
Threshold of pain 10 130
Jet aircraft 50 m away 102102 140

Example 6 Computing Loudness

Find the decibel level of a TV that has an intensity of 250×10−7 W/m2250×10−7 W/m2.

Solution

We are given I=250×10−7 W/m2,I=250×10−7 W/m2, and we know that I0=10−12 W/m2.I0=10−12 W/m2. We have

L=10 log(II0)Formula for loudness=10 log(250×10−710−12)I=250⋅10−7 and I0=10−12=10 log(250×105)10−710−12=10−7+12=105=10[log 250+log 105]Product rule for logarithms=10[log 250+5 log 10]Power rule for logarithms=10[log 250+5]log 10=log10 10=1≈74Use a calculator, rounding.
L======≈10 log(II0)10 log(250×10−710−12)10 log(250×105)10[log 250+log 105]10[log 250+5 log 10]10[log 250+5]74Formula for loudnessI=250⋅10−7 and I0=10−1210−710−12=10−7+12=105Product rule for logarithmsPower rule for logarithmslog 10=log10 10=1Use a calculator, rounding.

So the decibel level of this TV is approximately 74 dB.

Practice Problem 6

  1. Find the decibel level of a TV that has an intensity of 200×10−7W/m2200×10−7W/m2.

Example 7 Comparing Intensities

How much more intense is a 65-dB sound than a 42-dB sound?

Solution

I=I0×10L10Formula for intensity, exponential formI65=I0×106510I65 is the intensity of a 65-dB sound.I42=I0×104210I42 is the intensity of a 42-dB sound.I65I42=I0×106510I0×104210Divide I65 by I42.=I0×106.5I0×104.2=106.5−4.2Remove I0 and use the quotient rule for exponents.=102.3Simplify.≈199.53Use a calculator.≈200
II65I42I65I42=======≈≈I0×10L10I0×106510I0×104210I0×106510I0×104210I0×106.5I0×104.2106.5−4.2102.3199.53200Formula for intensity, exponential formI65 is the intensity of a 65-dB sound.I42 is the intensity of a 42-dB sound.Divide I65 by I42.Remove I0 and use the quotient rule for exponents.Simplify.Use a calculator.

So a sound of 65-dB is about 200 times more intense than a sound of 42-dB.

Practice Problem 7

  1. How much more intense is a 75-dB sound than a 55-dB sound?

Example 8 Computing Sound Intensity

Calculate the intensity in watts per square meter of a sound of 73 dB.

Solution

I=I0×10L10Formula for intensity, exponential formI=10−12×107310Replace I0 with 10−12 and L with 73.=10−12×107.3=10−4.7Simplify.=100.3×10−5−4.7=0.3−5≈1.995×10−5Use a calculator.≈2×10−5 W/m2
II=====≈≈I0×10L1010−12×10731010−12×107.310−4.7100.3×10−51.995×10−52×10−5 W/m2Formula for intensity, exponential formReplace I0 with 10−12 and L with 73.Simplify.−4.7=0.3−5Use a calculator.

Practice Problem 8

  1. What is the intensity of a 48-dB sound?

Musical Pitch

Pitch of a sound is determined by its frequency. For example, on a piano, the pitch of the note A above middle C may be denoted by A440. This means that A vibrates at 440 Hertz (cycles per second). In music, an interval whose higher note frequency is twice that of its lower note is an octave. Two notes that are an octave apart sound essentially “the same,” but one has a higher pitch. For this reason, notes an octave apart are given the same name. An octave jump is usually divided into 12 approximately equal semitones on a log scale. In an equal tempered scale, a semitone is further divided into 100 cents. So there are 1200 cents in an octave.

We note that the pitch difference of 1 octave means f=2f0,f=2f0, so

P(2f0)=1200 log2 (2f0f0)=1200 log22=1200 cents,
P(2f0)=1200 log2 (2f0f0)=1200 log22=1200 cents,

which agrees with our previous assertion.

Most people are not sensitive to pitch differences of less than about 2 cents.

Example 9 Equal Tempered Scale

In an “equal tempered” scale for a given reference frequency f0f0 of a note, find the frequency of the next higher-pitched semitone.

Solution

For the given reference frequency f0,f0, let f be the frequency of the next higher-pitched semitone. Because in an equal tempered scale every semitone is 100 cents, we have P(f)=100.P(f)=100.

P(f)=1200 log2 (ff0)Change in pitch formula100=1200 log2 (ff0)Replace P(f) with 100.112=log2 (ff0)Divide by 1200 and simplify.ff0=2112=12√2Exponential formf=12√2 f0Multiply by f0.
P(f)100112ff0f=====1200 log2 (ff0)1200 log2 (ff0)log2 (ff0)2112=2–√122–√12 f0Change in pitch formulaReplace P(f) with 100.Divide by 1200 and simplify.Exponential formMultiply by f0.

Practice Problem 9

  1. In an equal tempered scale, the notes are denoted as follows:

    A A# B C C# D D# E F F# G G# A
    A A# B C C# D D# E F F# G G# A

    If the frequency of A is 440 Hz, estimate the frequency of B.

The next example shows that an equal tempered scale does not always produce the best harmonics.

Example 10 Harmony in Music

A jump of one perfect fifth gives a frequency increase of 50%. An equal tempered fifth is 700 cents. Find the difference in cents (rounded to the nearest cent). Is the difference noticeable?

Solution

If the reference frequency is f0f0 and it increases by 50%, then f=f0+0.5f0=(1+0.5)f0=1.5f0.f=f0+0.5f0=(1+0.5)f0=1.5f0. So

P(f)=1200 log2(1.5f0/f0)Replace f with 1.5f0.=1200 log2(1.5)Simplify.=1200log(1.5)log 2Change to base 10.≈701.955Use a calculator.≈702
P(f)===≈≈1200 log2(1.5f0/f0)1200 log2(1.5)1200log(1.5)log 2701.955702Replace f with 1.5f0.Simplify.Change to base 10.Use a calculator.

Because a fifth on an equal tempered scale is 700 cents, the difference is a little less than 2 cents. This difference is barely noticeable to most listeners.

Practice Problem 10

  1. Repeat Example 10 if a perfect major third gives a frequency increase of 25% and an equal tempered major third is 400 cents.

Star Brightness

  1. 4 Define magnitude of star brightness.

Two thousand years ago Greek astronomers Hipparchus and Ptolemy created a system to quantify the apparent brightness (wattage received on the surface of the Earth) of stars. They classified star brightness on a scale from magnitude 1 for the brightest visible star to magnitude 6 for the dimmest visible star. Other stars were assigned magnitudes based on their brightness compared with those of magnitude 1 or 6. This system was refined considerably, leading to a much larger range of magnitude values and the following definition.

Although this definition of apparent magnitude was originally intended for comparing the brightness of stars, it can also be used for objects such as planets and their moons. The Hubble telescope can detect stars with apparent magnitude 30.

Example 11 Comparing Star Brightness

  1. Compare the brightness of a magnitude 1 star with that of a magnitude 6 star.

  2. Find the magnitude m of a star that is 650 times as bright as one of magnitude 7.25.

Solution

  1. m2−m1=2.5 log(b1b2)Apparent magnitude formula6−1=52log(b1b2)m2=6, m1=1, and 2.5=522=log(b1b2)Multiply both sides by 25 and simplify.b1b2=102Exponential formb1=102b2=100b2Multiply both sides by b2.m2−m16−12b1b2b1=====2.5 log(b1b2)52log(b1b2)log(b1b2)102102b2=100b2Apparent magnitude formulam2=6, m1=1, and 2.5=52Multiply both sides by 25 and simplify.Exponential formMultiply both sides by b2.

    So a star of magnitude 1 is 100 times brighter than a star of magnitude 6.

  2. Letting m2=m, m1=7.25m2=m, m1=7.25 and b2=650b1b2=650b1 in the formula, we have

    m−7.25=2.5 log (b1650b1)m−7.25=2.5 log(1650)=2.5 log(650)−1m−7.25=−2.5 log(650)Power rule for logsm=7.25−2.5 log(650)Add 7.25 to both sides.m≈0.2177Use a calculator.
    m−7.25m−7.25m−7.25mm====≈2.5 log (b1650b1)2.5 log(1650)=2.5 log(650)−1−2.5 log(650)7.25−2.5 log(650)0.2177Power rule for logsAdd 7.25 to both sides.Use a calculator.

Practice Problem 11

    1. Compare the brightness of a magnitude 0 star with that of a magnitude 2 star.

    2. Find the magnitude m of a star that is 50% brighter than a star of magnitude 4.6.

Modeling

  1. 5 Build models from data.

In Section 2.3, we discussed modeling the relationship between two variables (x and y) when the scatterplot of the corresponding data points appears to be linear. The linear least-squares technique (“regression line” y=ax+by=ax+b) is applied to best fit a line for such a set of points. However, if the scatterplot of the corresponding data points does not appear to be linear, then it is common practice to transform the data in such a way that the resulting plot resembles a straight line. Logarithms can be helpful in transforming the data.

We can apply logarithm transformations to either the y-variable, the x-variable, or both the x- and y-variables at the same time. If we find that one of these transformations results in a linear relationship that produces a regression line, we can deduce that the relationship between the original variables is exponential, logarithmic, or power, respectively. The transformations and resulting models are summarized in the following table:

x variable y variable Regression line Model
x y y=ax+by=ax+b Linear: y=ax+by=ax+b
x ln y ln y=ax+bln y=ax+b Exponential: y=ceax,y=ceax, with c=ebc=eb
ln x y y=a ln x +by=a ln x +b Logarithmic: y=a ln x +by=a ln x +b
ln x ln y ln y=a ln x +bln y=a ln x +b Power: y=cxay=cxa with c=ebc=eb

The graphical representations of scattered data and the corresponding models are shown in Figure 4.22.

Figure 4.22

The resulting models can be used to summarize the data, to predict unobserved values, or to understand the mechanisms that produce the observed values.

Example 12 Temperature Inside a Parked Car

The data in Table 4.11 show the temperature inside a parked car on a hot day (90° F)(90° F) with the windows closed and taken in 5-minute intervals. The initial temperature in the cabin was 75° F.75° F.

Table 4.11

Elapsed Time 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
Temperature 101 110 118 122 126 130 133 136 136 138 140 141 142 142 144 144 145 145
ln (Elapsed Time) 1.61 2.30 2.71 3.00 3.22 3.40 3.56 3.69 3.81 3.91 4.01 4.09 4.17 4.25 4.32 4.38 4.44 4.50
Source: Based on RACQ study.

Set x as the elapsed time and y as the temperature inside a car. Apply the logarithmic transformation to x. Create a scatterplot of data (y versus x) as well as a scatterplot of transformed data (y vs. ln x) and find the logarithmic model that fits these data.

Solution

The corresponding scatterplots of the data are presented in Figure 4.23.

Figure 4.23

From the scatterplot of the transformed data, we see that the logarithmic model was a good choice. Using a graphing calculator (see page 208), we find the regression line (in the variables Y=yY=y and X=ln xX=ln x) for the transformed data:

Y=74.92+15.98X.
Y=74.92+15.98X.

From that we find the coefficients of the logarithmic model that fit the original data:

y=74.92+15.98 ln x.
y=74.92+15.98 ln x.

The regression line and logarithmic model are shown on their corresponding scatterplots in Figure 4.23. This type of model is called the logarithmic growth model.

Practice Problem 12

  1. Repeat Example 12 with the data shown in Table 4.12 . Use natural log values for the elapsed time rounded to two decimal places. The data show the temperature inside a parked car in identical conditions as in Example 12 , except now the car windows are left rolled down 2 inches.

    Table 4.12

    Elapsed Time 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
    Temperature 92 98 103 108 111 114 116 118 116 118 121 121 122 122 123 123 123 125
    Source: Based on RACQ study.

In the next example, we investigate Kleiber’s Law. In 1932, Max Kleiber observed that, for the vast majority of animals, an animal’s metabolic rate q varies directly with the 3434 power of the animal’s mass M, that is, q=c⋅M3/4.q=c⋅M3/4.

Side Note

The basal metabolic rate is the amount of energy expended while at rest to support only the functioning of the vital organs.

Example 13 Body Weight Versus Metabolic Rate for Small Animals

The data in Table 4.13 show the body weight (in grams, g) and the corresponding basal metabolism (in watts, W) for 15 different species of small animals.

Table 4.13

Body Weight (g) 16 67 120 206 336 490 598 828 1120 1551 1782 2250 2750 3257 3600
Metabolism (W) 0.13 0.33 0.58 1.08 1.19 1.24 1.78 2.21 2.84 3.23 3.01 5.07 6.81 6.05 7.74
ln (Body Weight) 2.77 4.20 4.79 5.33 5.82 6.19 6.39 6.72 7.02 7.35 7.48 7.72 7.92 8.09 8.18
ln (Metabolism) −2.04−2.04 −1.11−1.11 −0.545−0.545 0.077 0.174 0.215 0.577 0.793 1.04 1.17 1.10 1.62 1.92 1.80 2.05
Source: Based on data from “Size and Power ln Mammals” by A. Hausner, J. Exp. Biol. 160, 25–54 (1991).

Set x as the body weight and y as the metabolism rate. Apply the logarithmic transformation to both x and y. Create a scatterplot of data (y vs. x) as well as a scatterplot of transformed data (ln y vs. ln x) and find the power model that fits these data.

Solution

The corresponding scatterplots of the data are shown in Figure 4.24.

Figure 4.24

From the scatterplot of the transformed data, we see that the power model is a good choice. Using a graphing calculator (see page 208), we find the regression line (in variables Y=ln yY=ln y and X=ln x)X=ln x) for the transformed data:

Y=−4.09+0.73X.
Y=−4.09+0.73X.

From that, we find the coefficients of the power model that fit the original data:

y=e−4.09x 0.73≈0.0167x0.73.
y=e−4.09x 0.73≈0.0167x0.73.

The regression line and power model are shown with their corresponding scatterplots in Figure 4.24. Note that the power exponent 0.73 we obtain is very close to 34=0.7534=0.75, the exponent in Kleiber’s Law. This type of model is called the power growth model.

Practice Problem 13

  1. Repeat Example 13 with the data shown in Table 4.14 .

    Table 4.14

    Body Weight (kg) 1.1 1.5 2.4 3.6 4.0 4.3 5.7 6.5 6.9 7.7 13.6 15.9 25 30 45.2
    Metabolism (W) 2.23 2.79 3.65 7.74 5.63 6.49 5.11 6.52 8.23 11.03 14.98 12.54 14.08 34.46 46.85
    Source: Based on data from “Size and Power ln Mammals” by Hausner, J. Exp. Biol. 160, 25–54 (1991).

Section 4.5 Exercises

Concepts and Vocabulary

  1. On the Richter scale the magnitude of an earthquake M=_.M=−−−−−−−−−−−. The energy E released by an earthquake of magnitude M is given by log E=_log E=−−−−−−−−−−−.

  2. The loudness L of sound of intensity I is given by L=_L=−−−−−−−−−−−, and I0=10−12W /m2.I0=10−12W /m2.

  3. If f0f0 is a reference frequency and f is the frequency of a note, then the change in pitch P(f)=_P(f)=−−−−−−−−−−−.

  4. If two stars have magnitudes m1m1 and m2m2 with apparent brightness b1b1 and b2,b2, respectively, then m2−m1=_m2−m1=−−−−−−−−−−−.

  5. True or False. If the pH value of a solution is bigger than 7, the solution is acidic.

  6. True or False. The acidity of a solution increases as its pH value increases.

  7. True or False. For a one-unit increase in the magnitude of an earthquake, its intensity increases tenfold.

  8. True or False. The apparent brightness of a star increases if its magnitude decreases.

Building Skills

In Exercises 9–12, the concentration [H+][H+] of a substance is given. Calculate the pH value of each substance and classify each substance as an acid or a base.

  1. [H+]=10−8[H+]=10−8

  2. [H+]=10−4[H+]=10−4

  3. [H+]=2.3×10−5[H+]=2.3×10−5

  4. [H+]=4.7×10−9[H+]=4.7×10−9

In Exercises 13–16, use the following information. Negative ions, designated by the notation [OH−][OH−], are always present in any acid or base. The concentration, [OH−],[OH−], of these ions is related to [H+][H+] by the equation [OH−]⋅[H+]=10−14[OH−]⋅[H+]=10−14 moles per liter. Find the concentrations of [OH−][OH−] and [H+][H+] in moles per liter for the substances with the following pH values.

  1. pH=6pH=6

  2. pH=8pH=8

  3. pH=9.5pH=9.5

  4. pH=3.7pH=3.7

In Exercises 17–20, the magnitude M of an earthquake is given.

  1. Find the earthquake intensity I in terms of the zero-level earthquake intensity I0.I0.

  2. Find the energy released by the earthquake.

  1. M=5M=5

  2. M=2M=2

  3. M=7.8M=7.8

  4. M=3.7M=3.7

In Exercises 21–24, the energy E released by an earthquake is given.

  1. Find the earthquake’s magnitude.

  2. Find the earthquake’s intensity.

  1. E=1013.4 joulesE=1013.4 joules

  2. E=1010.4 joulesE=1010.4 joules

  3. E=1012 joulesE=1012 joules

  4. E=109 joulesE=109 joules

In Exercises 25–28, the intensity I of a sound is given. Find the loudness L of the sound (I0=10−12 W/m2).(I0=10−12 W/m2).

  1. I=10−8 W/m2I=10−8 W/m2

  2. I=10−10 W/m2I=10−10 W/m2

  3. I=3.5×10−7 W/m2I=3.5×10−7 W/m2

  4. I=2.37×10−5 W/m2I=2.37×10−5 W/m2

In Exercises 29–32, a loudness L is given. Find the intensity of the sound.

  1. L=80 dBL=80 dB

  2. L=90 dBL=90 dB

  3. L=64.7 dBL=64.7 dB

  4. L=37.4 dBL=37.4 dB

In Exercises 33–36, use the equal tempered scale with the twelve semitones A, A#, B, C, C#, D, D#, E, F, F#, G, and G# and the frequency of A is 440 Hz. On a piano keyboard, the distance between two white keys that are side-by-side is a whole note if there is a black key between them and is a semitone if there is no black key between them. A black key (such as the one between C and D) is called a sharp. The black key between C and D is denoted by C#.

Remember that the change in frequency P(f) between any two consecutive semitones is 100 cents.

  1. Find the frequencies of A# and C.

  2. Find the frequencies of D and E.

  3. A whole tone is a frequency ratio of 10:9. That is, ff0=109.ff0=109. An equal tempered whole tone is 200 cents. Find the difference in cents between a whole tone and an equal tempered whole tone. Is the difference noticeable?

  4. Repeat Exercise 35 if a whole tone is a frequency ratio of 9 : 8.

  5. The magnitudes of two stars, A and B, are 4 and 20, respectively. Compare the brightness of these stars.

  6. Repeat Exercise 37 for two stars with magnitudes −2−2 and 5.

  7. The magnitude of star A is 2 more than that of star B. How is their corresponding brightness related?

  8. The magnitude of star A is 1 more than that of star B. How is their corresponding brightness related?

Applying the Concepts

  1. pH of human blood. The hydrogen ion concentration [H+][H+] of blood is approximately 3.98×10−83.98×10−8 moles per liter. Find the pH value of human blood to the nearest tenth. Is human blood acidic or basic?

  2. pH of milk. The hydrogen ion concentration [H+][H+] of milk is approximately 3.16×10−73.16×10−7 moles per liter. Find the pH value of milk to the nearest tenth. Is milk acidic or basic?

  3. pH of some common substances. The pH value of a sample of each substance is given. Find [H+][H+] of the substance.

    1. Wine grapes:

    		 pH=3.15pH=3.15

    1. Water:

    		 pH=7.2pH=7.2

    1. Eggs:

    		 pH=7.78pH=7.78

    1. Vinegar:

    		 pH=3pH=3

  4. pH of some common substances. The pH value of a sample of each substance is given. Find [H+][H+] of the substance.

    1. Battery acid:

    		 pH=1.0pH=1.0

    1. Baking soda:

    		 pH=8.7pH=8.7

    1. Ammonia:

    		 pH=10.6pH=10.6

    1. Stomach acid:

    		 pH=2.3pH=2.3

  5. Acid rain. In the Netherlands, the average pH value of the rainfall is 3.8. Find the average concentration of hydrogen ions [H+][H+] in the rainfall. How much more acidic is this rain than ordinary rain with a pH value of 6?

  6. pH of a solution. Suppose the pH value of a solution A is 1.0 more than the pH value of a solution B. How are the concentrations of hydrogen ions in the two solutions related?

  7. Comparing pH of solutions. Suppose the hydrogen ion concentration in a solution A is 100 times that in solution B. How are the pH values of the two solutions related?

  8. pH of a solution. Suppose the pH value of a solution is increased by 1.5. How much change does this represent in the hydrogen ion concentration of this solution? Does the increase in pH make the solution more acidic or more basic?

  9. pH of a solution. Suppose the hydrogen ion concentration of a solution is increased 50 times. How much change does this represent in the pH value of this solution? Does this increase in [H+][H+] make the solution more acidic or more basic?

  10. China earthquake. The Great China Earthquake of 1920 registered 8.6 on the Richter scale.

    1. What was the intensity of this earthquake?

    2. How many joules of energy were released?

  11. San Francisco earthquake. Repeat Exercise 50 for the Great San Francisco Earthquake of 1906, which registered 7.8 on the Richter scale.

  12. Comparing earthquakes. Table 4.9 lists two earthquakes in Japan.

    1. Compare the intensities of these earthquakes.

    2. Compare the energies released by these quakes.

  13. Comparing earthquakes. Suppose earthquake A registers 1 more point on the Richter scale than earthquake B.

    1. How are their corresponding intensities related?

    2. How are their released energies related?

  14. Comparing earthquakes. Repeat Exercise 53, if earthquake A registers 1.5 more points on the Richter scale than earthquake B.

  15. Comparing magnitudes. If one earthquake is 150 times as intense as another, what is the difference in the Richter scale readings of the two earthquakes?

  16. Comparing magnitudes. If the energy released by one earthquake is 150 times that of another, what is the difference in the Richter scale readings of the two earthquakes?

In Exercises 57–64, use Table 4.10 as necessary.

  1. dB and intensity. What is the loudness, in decibels, of a radio with intensity 5.2 ×10−5 W/m25.2 ×10−5 W/m2?

  2. dB of a jet. What is the loudness of a jet aircraft with intensity 2.5×102 W/m22.5×102 W/m2?

  3. Comparing two sounds. How much more intense is a sound of the threshold of pain (130 dB) than a conversation at 65 dB?

  4. Comparing two sounds. How much more intense is a 75 dB sound than a 62 dB sound?

  5. Comparing two sounds. Suppose a sound is 1000 times as intense as one at the threshold of pain for the human ear. Find the loudness of this sound in decibels.

  6. Comparing intensities. Show that if two sounds of intensity I1I1 and I2I2 register decibel levels of L1L1 and L2L2, respectively, then I2I1=100.1(L2−L1).I2I1=100.1(L2−L1).

  7. Comparing intensities. Suppose one sound registers 1 decibel more than another. How are the intensities of the two sounds related?

  8. Comparing sounds. The noise level of a street sound outside a new office building in downtown Chicago is measured to be approximately 70 dB. With special insulation material, the noise level inside the office building is reduced to 29 dB. Compare the intensities of sounds outside and inside the building.

  9. Violin tuning. A violinist is tuning her A-string to match concert pitch. The concert’s A is a frequency of 440 Hz. Her A-string is improperly tuned to 441 Hz. Find the difference in cents.

  10. Violin tuning. Repeat Exercise 65 if her A-string is improperly tuned to 438 Hz.

In Exercises 67–72, use the following table of approximate apparent magnitudes of some celestial objects.

Object The Sun Full Moon Venus Sirius Vega Saturn Neptune
Magnitude −27−27 −13−13 −4−4 −1−1 0.03 1.47 7.8

  1. Sun-moon How many times brighter is the sun than the full moon?

  2. Moon-Vega. How many times brighter is the full moon than Vega?

  3. Sun-Venus. How many times brighter is the sun than Venus?

  4. Venus-Neptune. How many times brighter is Venus than Neptune?

  5. Saturn-Star. What is the magnitude of a star that is 560 times brighter than Saturn?

  6. Neptune-Star. What is the magnitude of a star that is 1 billion times brighter than Neptune?

  7. Solar panels. The data in the following table show the number of annual solar PV cells installations (in megawatts [MW]) over the period 2004–2014.

    Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014
    PV Installations 58 79 105 160 298 382 852 1922 3369 4776 6201
    Source: Solar Energy Industries Association, www.seia.org

    Find the exponential model that fits the data, letting the year 2004 correspond to x=1x=1. This type of model is called the exponential growth model.

  8. CD sales. The data in this table show annual CD album sales in the United States (in millions of units) over the period, 2004–2014.

    Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014
    CD Sales 651 599 553 449 361 295 237 224 193 165 141
    Source: Nielsen Soundscan.

    Find the exponential model that fits the data, letting the year 2004 correspond to x=1x=1. This type of model is called the exponential decay model.

  9. NFL draft. The data in the following table show the estimated Career Value Rating in relation to the Draft Pick Number for the NFL averaged over the past 25 years.

    Pick Number 1 5 10 15 20 25 30 40 50 100 200
    Career Value Rating 64.8 51.4 44 38.2 36 32.1 33.1 27.7 27.4 18.2 10.5
    Source: http://www.pro-football-reference.com

    Find the logarithmic model that fits the data. This type of model is called the logarithmic decay model.

  10. Heart rate. The data in this table show the body weight (in grams g) and corresponding heart rate (in beats per minute) for nine different species of warm-blooded animals at rest.

    Weight 25 60 200 341 1100 2000 5000 90000
    Heart Rate 670 450 420 378 190 150 120 60

    Find the power model that fits the data. This type of model is called the power decay model.

Beyond the Basics

  1. Decibel levels. The decibel level of Professor Stout’s voice decreases with the distance from the professor according to the formula

    L=10 log (4×104r2),
    L=10 log (4×104r2),

    where L is the decibel level and r is the distance in feet from the professor to the listener.

    1. Find the decibel levels at distances of 10, 25, 50, and 100 feet.

    2. How far must a listener be from the professor so that the decibel level drops to 0?

    3. Express L in the form L=a+b log r,L=a+b log r, for suitable constants a and b.

    4. Express r as a function of L.

  2. Sound intensity. The intensity of sound is inversely proportional to the square of the distance. That is, I=Kr2,I=Kr2, where I is the intensity at a distance r from the source and K is a constant.

    1. Show that if L1L1 and L2L2 are decibel readings of a sound at distances r1r1 and r2r2 (in meters), respectively, then L1−L2=20 log (r2r1).L1−L2=20 log (r2r1).

    2. A military jet at 30 meters has 140 decibel level of loudness. How loud will you hear this jet sound at a distance of 100 meters? 300 meters?

    3. Express r2r1r2r1 in the exponential form.

  3. Nuclear bomb. Compare the energy released by a 1 megaton nuclear bomb (about 5×10155×1015 joules) to the energy released by a great earthquake of magnitude 8.

  4. Adrian was calculating the pitch of a sound and mistakenly transposed the frequency f of the sound with the reference frequency f0.f0. The correct answer is 40 cents. What was her answer?

  5. If the A-string of a violin is tuned to 880 Hz with an error of ±10±10 cents, what is the frequency range of the A-string?

  6. Show that the formula for star brightness can be written as b1b2=100.4(m2−m1).b1b2=100.4(m2−m1).

  7. A star of magnitude m is 176 times brighter than a star of magnitude 3.42. Find m.

  8. Star luminosity. The luminosity L of a star is the power output at a star’s surface (measured in watts). The star’s apparent brightness b is the wattage received on Earth’s surface. The apparent magnitude m of a star depends on its absolute magnitude M (related to L) and its distance D in parsecs (1 parsec≈3.26 light years)(1 parsec≈3.26 light years) by the formula

    m−M=2.5 log Lb=5 log D10.
    m−M=2.5 log Lb=5 log D10.

    1. Given m and M for a star, show that its distance D from Earth is D=101+(m−M)/5D=101+(m−M)/5 parsecs.

    2. For the star Alpha Centauri, m=0m=0 and M=4.39.M=4.39. How far is it from Earth?

    3. Compare the luminosity and brightness of Alpha Centauri.

Getting Ready for the Next Section

In Exercises 85–91, find the value of a variable that satisfies the given condition.

  1. Find x if 3x+2y=73x+2y=7 and y=12.y=12.

  2. Find x if −2x+9y=5−2x+9y=5 and y=3.y=3.

  3. Find y if −4x+7y=7−4x+7y=7 and x=0.x=0.

  4. Find y if 23x−14y=−523x−14y=−5 and x=1.x=1.

  5. Find the slope of the line with equation 10x−2y=28.10x−2y=28.

  6. Find the slope of the line with equation 15x+5y=2.15x+5y=2.

  7. Find an equation of the line through the point (5, −1)(5, −1) parallel to the line with equation 6x−3y=7.6x−3y=7.

  8. Find an equation of the line through the point (3, 3) parallel to the line with equation x+2y=1.x+2y=1.

  9. Find values for a and b so that the equation ax+by=3ax+by=3 has the same graph as the equation 2x−4y=12.2x−4y=12.

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