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== 5.4#h(0.6em)Logarithmic Scales


    
      
=== Introduction


      Because logarithmic functions grow very slowly, they 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.


      #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([Animal], [Shrew], [Cat], [Wolf], [Horse], [Elephant], [Whale]),
  [Mass, kg], [#math.equation(block: false, alt: "0.004")[$0.004$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "5400")[$5400$]], [#math.equation(block: false, alt: "70 , 000")[$70 , 000$]],
))


      Imagine trying to scale the #math.equation(block: false, alt: "x")[$x$]-axis to show all of these values. If we set tick marks at intervals of #math.equation(block: false, alt: "10 , 000")[$10 , 000$] 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.


      #figure(figph[number line with masses of mammals], alt: "number line with masses of mammals", caption: none)


      On the other hand, we can plot the mass of the cat if we set tick marks at intervals of #math.equation(block: false, alt: "1")[$1$] 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


      #figure(figph[number line with masses of mammals], alt: "number line with masses of mammals", caption: none)


      To get around this problem, we'll compute the the log of each mass, and use the logs on a new scale. The table below shows the base 10 log of each animal's mass, rounded to #math.equation(block: false, alt: "2")[$2$] decimal places.


      #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#emph[Animal]], [Shrew], [Cat], [Wolf], [Horse], [Elephant], [Whale]),
  [Mass, kg], [#math.equation(block: false, alt: "0.004")[$0.004$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "5400")[$5400$]], [#math.equation(block: false, alt: "70 , 000")[$70 , 000$]],
  [Log (mass)], [#math.equation(block: false, alt: "minus 2.40")[$− 2.40$]], [#math.equation(block: false, alt: "0.60")[$0.60$]], [#math.equation(block: false, alt: "1.90")[$1.90$]], [#math.equation(block: false, alt: "2.48")[$2.48$]], [#math.equation(block: false, alt: "3.73")[$3.73$]], [#math.equation(block: false, alt: "4.85")[$4.85$]],
))


      The logs of the masses range from #math.equation(block: false, alt: "minus 2.40")[$− 2.40$] to #math.equation(block: false, alt: "4.85")[$4.85$]. We can easily plot these values on a single scale, as shown below.


      #figure(figph[mammal masses plotted on log scale], alt: "mammal masses plotted on log scale", caption: none)


      We'd need to keep in mind that we are plotting the logs of the animals' masses, and not the actual masses. However, remember that a logarithm is really an exponent! For example, the mass of the horse is 300 kg, and


      #math.equation(block: true, alt: "since log sub 10 ⁡ open parenthesis 300 close parenthesis equals 2.48 , then 10 to the power 2.48 equals 300")[$"since" " " " " " " log_(10) ( 300 ) = 2.48 , " " " " " " "then" " " " " " " 10^(2.48) = 300$]
      So instead of plotting the logs from the table, we will plot powers of 10 that give the actual masses of the animals, like this:


      #figure(figph[mammal masses plotted on log scale], alt: "mammal masses plotted on log scale", caption: none)


      Compare this new scale to the previous one. It looks almost the same, except that the number line is labeled with powers of 10. Even though we computed the log of each mass, we still plotted the actual mass of each animal, in its form as a power of 10. It is the scale on the number line that has changed.


      A scale labeled with powers of 10 is called a #strong[logarithmic scale], or log scale. The powers of 10 on a log scale are evenly spaced, so that the actual values at the tick marks look like this.


      #figure(figph[logscale with integer exponents], alt: "logscale with integer exponents", caption: none)


      We can see right away that the increments between tick marks on a log scale are not equal, as they are on a usual linear scale. The increments get larger as we move from left to right on the scale. However, when we are plotting powers of 10 we use the exponents to place the data points on the scale.


      For example, you can check that the mass of the horse, at #math.equation(block: false, alt: "10 to the power 2.48 equals 300")[$10^(2.48) = 300$] kg, is plotted about half-way between #math.equation(block: false, alt: "10 squared equals 100")[$10^(2) = 100$] and #math.equation(block: false, alt: "10 cubed equals 1000")[$10^(3) = 1000$] on the log scale, because 2.48 is about half-way between 2 and 3. Similarly, the mass of the cat, at #math.equation(block: false, alt: "10 to the power 0.60 equals 4")[$10^(0.60) = 4$] kg, is plotted between #math.equation(block: false, alt: "10 to the power 0 equals 1")[$10^(0) = 1$] and #math.equation(block: false, alt: "10 to the power 1 equals 10")[$10^(1) = 10$] on the log scale.


      #examplebox("Example 1")[][
Plot the values on a log scale.


          #figure(table(
  columns: 7, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.0007")[$0.0007$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "1600")[$1600$]], [#math.equation(block: false, alt: "72 , 000")[$72 , 000$]], [#math.equation(block: false, alt: "4 times 10 to the power 8")[$4 × 10^(8)$]],
))


        
        #solutionbox[
We first compute the base #math.equation(block: false, alt: "10")[$10$] logarithm of each number.


          #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.0007")[$0.0007$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "1600")[$1600$]], [#math.equation(block: false, alt: "72 , 000")[$72 , 000$]], [#math.equation(block: false, alt: "4 times 10 to the power 8")[$4 × 10^(8)$]]),
  [#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [#math.equation(block: false, alt: "minus 3.15")[$− 3.15$]], [#math.equation(block: false, alt: "minus 0.70")[$− 0.70$]], [#math.equation(block: false, alt: "0.54")[$0.54$]], [#math.equation(block: false, alt: "3.20")[$3.20$]], [#math.equation(block: false, alt: "4.86")[$4.86$]], [#math.equation(block: false, alt: "8.60")[$8.60$]],
))


          Thus, for example, we see that #math.equation(block: false, alt: "0.0007 equals 10 to the power minus 3.15")[$0.0007 = 10^(− 3.15)$]. Then we use the logs to plot each number as a power of 10, estimating its position between integer powers of 10. For example, we plot the first value, #math.equation(block: false, alt: "10 to the power minus 3.15")[$10^(− 3.15)$], closer to #math.equation(block: false, alt: "10 to the power minus 3")[$10^(− 3)$] than to #math.equation(block: false, alt: "10 to the power minus 4")[$10^(− 4)$]. The finished plot is shown below.


          #figure(figph[points plotted on log scale], alt: "points plotted on log scale", caption: none)
]
]


      
        A value of 5682.7 would be plotted between which two integers on a log scale?


        \_\_\_\_\_


        #solutionbox[
3 and 4
]


      
      
        A value of 5682.7 would be plotted between which two integers on a log scale?


        + 5682 and 5683
+ 5000 and 6000
+ 5 and 6
+ 3 and 4


      
      
        #figure(figph[points on logscale], alt: "points on logscale", caption: none)


        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.


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_]),
  [#math.equation(block: false, alt: "x")[$x$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_],
))


        #solutionbox[
#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 2.5")[$− 2.5$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "4.25")[$4.25$]]),
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.0001")[$0.0001$]], [#math.equation(block: false, alt: "0.00316")[$0.00316$]], [#math.equation(block: false, alt: "31.6")[$31.6$]], [#math.equation(block: false, alt: "17 , 782.8")[$17 , 782.8$]],
))
]


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


        #figure(figph[points on logscale], alt: "points on logscale", caption: none)


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$]),
  [#math.equation(block: false, alt: "x")[$x$]], [], [], [], [],
))


        #solutionbox[
#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 2.5")[$− 2.5$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "4.25")[$4.25$]]),
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.0001")[$0.0001$]], [#math.equation(block: false, alt: "0.00316")[$0.00316$]], [#math.equation(block: false, alt: "31.6")[$31.6$]], [#math.equation(block: false, alt: "17 , 782.8")[$17 , 782.8$]],
))
]


      
      
        What is a log scale used for?


        \_\_\_\_\_


        #solutionbox[
To plot data that covers a wide range of values.
]


      
      
        What is a log scale used for?


        + To find the logarithm of a number.
+ To plot data that covers a wide range of values.
+ To highlight the curvature in the graph of an exponential function.
+ To convert logarithms to base 10.


        #solutionbox[
To plot data that covers a wide range of values.
]


      
    
    
      
=== Using Log Scales


      By now, you have noticed that the values represented by points on a log scale increase rapidly as we move to the right along the scale. Also notice that #math.equation(block: false, alt: "10 to the power 0 equals 1")[$10^(0) = 1$], so the "middle" of a log scale represents #math.equation(block: false, alt: "1")[$1$] (not zero, as on a linear scale).


      Points to the left of #math.equation(block: false, alt: "10 to the power 0")[$10^(0)$] represent fractions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "1")[$1$], because powers of #math.equation(block: false, alt: "10")[$10$] with negative exponents are numbers less than #math.equation(block: false, alt: "1")[$1$]. Their values decrease toward #math.equation(block: false, alt: "0")[$0$] as we move to the left, but they never become negative.


      We cannot plot negative numbers or zero on a log scale, because the log of a negative number or zero is undefined.


      #examplebox("Example 2")[][
The figure shows a timeline for life on Earth, in units of Mya (million years ago).


          #figure(figph[timeline of events on a logscale], alt: "timeline of events on a logscale", caption: none)


          Approximately how long ago did each of the following events occur?


          + Formation of Earth
+ Dinosaurs became extinct
+ The last ice age
+ The Crusades


        
        #solutionbox[
+ We read from the timeline that the Earth was formed between #math.equation(block: false, alt: "10 cubed")[$10^(3)$] and #math.equation(block: false, alt: "10 to the power 4")[$10^(4)$], or between #math.equation(block: false, alt: "1000")[$1000$] and #math.equation(block: false, alt: "10 , 000")[$10 , 000$] million years ago. We estimate that Earth formed #math.equation(block: false, alt: "5000")[$5000$] million years ago.
+ The extinction of the dinosaurs is plotted between #math.equation(block: false, alt: "10 to the power 1")[$10^(1)$] and #math.equation(block: false, alt: "10 squared")[$10^(2)$], or between #math.equation(block: false, alt: "10")[$10$] and #math.equation(block: false, alt: "100")[$100$] million years ago. Because the point is closer to #math.equation(block: false, alt: "10 squared")[$10^(2)$], we estimate their extinction at #math.equation(block: false, alt: "70")[$70$] million years ago.
+ The last ice age is plotted just after #math.equation(block: false, alt: "10 to the power minus 2")[$10^(− 2)$], or #math.equation(block: false, alt: "0.01")[$0.01$] million years ago. One-hundredth of a million is #math.equation(block: false, alt: "10 , 000")[$10 , 000$], so we estimate that the ice age occurred a little more than #math.equation(block: false, alt: "10 , 000")[$10 , 000$] years ago.
+ The Crusades occurred about #math.equation(block: false, alt: "10 to the power minus 3")[$10^(− 3)$], or about #math.equation(block: false, alt: "0.001")[$0.001$] million years ago. One-thousandth of a million is #math.equation(block: false, alt: "1000")[$1000$], so the Crusades occurred about #math.equation(block: false, alt: "1000")[$1000$] years ago, or about #math.equation(block: false, alt: "1000")[$1000$] A.D.
]
]


      
        Plot the following dollar values on a log scale.


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Postage stamp], [#math.equation(block: false, alt: "0.47")[$0.47$]]),
  [Notebook computer], [#math.equation(block: false, alt: "679")[$679$]],
  [One year at Harvard], [#math.equation(block: false, alt: "88 , 600")[$88 , 600$]],
  [2016 Lamborghini], [#math.equation(block: false, alt: "530 , 075")[$530 , 075$]],
  [Kobe Bryant salary], [#math.equation(block: false, alt: "25 , 000 , 000")[$25 , 000 , 000$]],
  [Bill Gates financial worth], [#math.equation(block: false, alt: "79 , 400 , 000 , 000")[$79 , 400 , 000 , 000$]],
  [U.S. National debt], [#math.equation(block: false, alt: "19 , 341 , 810 , 000")[$19 , 341 , 810 , 000$]],
))


        #solutionbox[
A graph is below.
]


        #figure(figph[logscale], alt: "logscale", caption: none)


      
      
        Plot the following dollar values on a log scale.


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Postage stamp], [#math.equation(block: false, alt: "0.47")[$0.47$]]),
  [Notebook computer], [#math.equation(block: false, alt: "679")[$679$]],
  [One year at Harvard], [#math.equation(block: false, alt: "88 , 600")[$88 , 600$]],
  [2016 Lamborghini], [#math.equation(block: false, alt: "530 , 075")[$530 , 075$]],
  [Kobe Bryant salary], [#math.equation(block: false, alt: "25 , 000 , 000")[$25 , 000 , 000$]],
  [Bill Gates financial worth], [#math.equation(block: false, alt: "79 , 400 , 000 , 000")[$79 , 400 , 000 , 000$]],
  [U.S. National debt], [#math.equation(block: false, alt: "19 , 341 , 810 , 000 , 000")[$19 , 341 , 810 , 000 , 000$]],
))


        #solutionbox[
#figure(figph[logscale], alt: "logscale", caption: none)
]


      
      
        How does a log scale differ from a linear scale?


        \_\_\_\_\_


      
      
        How does a log scale differ from a linear scale?


      
    
    
      
=== Equal Increments on 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 integer values, as well as powers of 10. The difference between #math.equation(block: false, alt: "10 to the power 1")[$10^(1)$] and #math.equation(block: false, alt: "10 to the power 0")[$10^(0)$] is #math.equation(block: false, alt: "10 minus 1 equals 9")[$10 − 1 = 9$], but the difference between #math.equation(block: false, alt: "10 squared")[$10^(2)$] and #math.equation(block: false, alt: "10 to the power 1")[$10^(1)$] is #math.equation(block: false, alt: "100 minus 10 equals 90")[$100 − 10 = 90$].


      #figure(figph[logscale with integer exponents], alt: "logscale with integer exponents", caption: none)


      If we include tick marks for intermediate values on the log scale, they look like this.


      #figure(figph[logscale with decimal exponents], alt: "logscale with decimal exponents", caption: none)


      Once again, the difference between, say, #math.equation(block: false, alt: "10 to the power 0.1")[$10^(0.1)$] and #math.equation(block: false, alt: "10 to the power 0.2")[$10^(0.2)$] is not the same as the difference between #math.equation(block: false, alt: "10 to the power 0.2")[$10^(0.2)$] and #math.equation(block: false, alt: "10 to the power 0.3")[$10^(0.3)$]. The decimal values of the powers #math.equation(block: false, alt: "10 to the power 0.1")[$10^(0.1)$] through #math.equation(block: false, alt: "10 to the power 0.9")[$10^(0.9)$], rounded to two places, are shown below.


      #figure(figph[logscale showing values of powers], alt: "logscale showing values of powers", caption: none)


      As we move from left to right on this scale, we #emph[multiply] the value at the previous tick mark by #math.equation(block: false, alt: "10 to the power 0.1")[$10^(0.1)$], or about #math.equation(block: false, alt: "1.258")[$1.258$]. For example,


      #math.equation(block: true, alt: "10 to the power 0.2, equals 1.258 times 10 to the power 0.1 equals 1.585; 10 to the power 0.3, equals 1.585 times 10 to the power 0.1 equals 1.995")[$10^(0.2) & = 1.258 × 10^(0.1) = 1.585 \ 10^(0.3) & = 1.585 × 10^(0.1) = 1.995$]
      and so on. Moving up by equal increments on a log scale does not add equal amounts to the values plotted; it #emph[multiplies] the values by equal #emph[factors].


      
        Which statement is false?


        \_\_\_\_\_


        #solutionbox[
"Equal increments on a log scale correspond to equal differences in value" is a false statement.
]


      
      
        Which statement is false?


        + We use log scales to graph a variable that has a wide range of values.
+ On a log scale, we actually plot exponents.
+ Values less than one appear as negative numbers on a log scale.
+ Equal increments on a log scale correspond to equal differences in value.


        #solutionbox[
"Equal increments on a log scale correspond to equal differences in value" is a false statement.
]


      
      #examplebox("Example 3")[][
What number is halfway between #math.equation(block: false, alt: "10")[$10$] and #math.equation(block: false, alt: "100")[$100$] on a log scale?


        
        #solutionbox[
On a log scale, the number #math.equation(block: false, alt: "10 to the power 1.5")[$10^(1.5)$] is halfway between #math.equation(block: false, alt: "10 to the power 1")[$10^(1)$] and #math.equation(block: false, alt: "10 squared")[$10^(2)$], as shown below.


          #figure(figph[log scale showing point halfway between 10 and 100], alt: "log scale showing point halfway between 10 and 100", caption: none)


          Now, #math.equation(block: false, alt: "10 to the power 1.5 equals 10 the square root of 10")[$10^(1.5) = 10 sqrt(10)$], or approximately #math.equation(block: false, alt: "31.62")[$31.62$]. Note how equal increments of #math.equation(block: false, alt: "0.5")[$0.5$] on the log scale correspond to equal factors of #math.equation(block: false, alt: "10 to the power 0.5")[$10^(0.5)$] in the values plotted:


          #math.equation(block: true, alt: "10 times 3.162, equals 31.62 and 31.62 times 3.162 equals 100; 10 to the power 1 times 10 to the power 0.5, equals 101.5 and 101.5 times 10 to the power 0.5 equals 10 squared")[$10 × 3.162 & = 31.62 " " " " " " " " " and " " " " " " " " " 31.62 × 3.162 = 100 \ 10^(1) × 10^(0.5) & = 101.5 " " " " " " " " " and " " " " " " " " " 101.5 × 10^(0.5) = 10^(2)$]
]
]


      
        What number is halfway between #math.equation(block: false, alt: "10 to the power 1.5")[$10^(1.5)$] and #math.equation(block: false, alt: "10 squared")[$10^(2)$] on a log scale?


        Answer: \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "56.23")[$56.23$]
]


      
      
        What number is halfway between #math.equation(block: false, alt: "10 to the power 1.5")[$10^(1.5)$] and #math.equation(block: false, alt: "10 squared")[$10^(2)$] on a log scale?


        #solutionbox[
#math.equation(block: true, alt: "56.23")[$56.23$]
]


      
      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.


      #examplebox("Example 4")[][
Plot the integer values 2 through 9 and 20 through 90 on a log scale.


        
        #solutionbox[
We compute the logarithm of each integer value.


          #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]]),
  [#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [#math.equation(block: false, alt: "0.301")[$0.301$]], [#math.equation(block: false, alt: "0.477")[$0.477$]], [#math.equation(block: false, alt: "0.602")[$0.602$]], [#math.equation(block: false, alt: "0.699")[$0.699$]], [#math.equation(block: false, alt: "0.778")[$0.778$]], [#math.equation(block: false, alt: "0.845")[$0.845$]], [#math.equation(block: false, alt: "0.903")[$0.903$]], [#math.equation(block: false, alt: "0.954")[$0.954$]],
))


          #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]]),
  [#math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$]], [#math.equation(block: false, alt: "1.301")[$1.301$]], [#math.equation(block: false, alt: "1.477")[$1.477$]], [#math.equation(block: false, alt: "1.602")[$1.602$]], [#math.equation(block: false, alt: "1.699")[$1.699$]], [#math.equation(block: false, alt: "1.778")[$1.778$]], [#math.equation(block: false, alt: "1.845")[$1.845$]], [#math.equation(block: false, alt: "1.903")[$1.903$]], [#math.equation(block: false, alt: "1.954")[$1.954$]],
))


          We plot on a log scale, as shown below.


          #figure(figph[log scale showing integer points], alt: "log scale showing integer points", caption: none)
]
]


      On the log scale in Example, notice how the integer values are spaced: They get closer together as they approach the next power of #math.equation(block: false, alt: "10")[$10$]. You will often see log scales labeled not with powers of #math.equation(block: false, alt: "10")[$10$], but with integer values, like this:


      #figure(figph[log scale showing tic marks at integer points], alt: "log scale showing tic marks at integer points", caption: none)


      In fact, #strong[log-log graph paper] scales both axes with logarithmic scales.


      
        #figure(figph[mouse-to-elephant curve on log-log graph], alt: "mouse-to-elephant curve on log-log graph", caption: none)


        The opening page of Power Functions 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.


        #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Animal], [Mouse], [Dog], [Sheep]),
  [Mass (kg)], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_],
  [Metabolic rate (kcal/day)], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_],
))


        #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([Animal], [Cow], [Elephant]),
  [Mass (kg)], [\_\_\_\_\_], [\_\_\_\_\_],
  [Metabolic rate (kcal/day)], [\_\_\_\_\_], [\_\_\_\_\_],
))


        Do not use commas. For example, use "10000" rather than "10,000".


        #solutionbox[
#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([Animal], [Mouse], [Dog], [Sheep], [Cow], [Elephant]),
  [Mass (kg)], [#math.equation(block: false, alt: "0.02")[$0.02$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "4000")[$4000$]],
  [Metabolic rate (kcal/day)], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "1500")[$1500$]], [#math.equation(block: false, alt: "6000")[$6000$]], [#math.equation(block: false, alt: "50 , 000")[$50 , 000$]],
))
]


      
      
        The opening page of Power Functions shows the "mouse-to-elephant" curve, a graph of the metabolic rate of mammals as a function of their mass. Here it is again.


        #figure(figph[Kleiber mouse-to-elephant-curve], alt: "Kleiber mouse-to-elephant-curve", caption: none)


        (The elephant does not appear on that graph, because its mass is too big.) The figure below shows the same function, graphed on log-log paper.


        #figure(figph[mouse-to-elephant curve on log-log graph], alt: "mouse-to-elephant curve on log-log graph", caption: none)


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


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([Animal], [Mouse], [Dog], [Sheep], [Cow], [Elephant]),
  [Mass (kg)], [], [], [], [], [],
  [Metabolic rate (kcal/day)], [], [], [], [], [],
))


        #solutionbox[
#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([Animal], [Mouse], [Dog], [Sheep], [Cow], [Elephant]),
  [Mass (kg)], [#math.equation(block: false, alt: "0.02")[$0.02$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "4000")[$4000$]],
  [Metabolic rate (kcal/day)], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "1500")[$1500$]], [#math.equation(block: false, alt: "6000")[$6000$]], [#math.equation(block: false, alt: "50 , 000")[$50 , 000$]],
))
]


      
      
        If #math.equation(block: false, alt: "B equals 100 A")[$B = 100 A$], the difference between #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$] on a log scale is 2 units. Use the properties of logarithms to explain why this is true.


        \_\_\_\_\_


      
      
        If #math.equation(block: false, alt: "B equals 100 A")[$B = 100 A$], the difference between #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$] on a log scale is 2 units. Use the properties of logarithms to explain why this is true.


      
    
    
      
=== Acidity and the pH Scale


      You may have already encountered log scales in some everyday applications. A simple example is the #strong[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 #math.equation(block: false, alt: "open bracket H to the power plus close bracket")[$[ H^(+) ]$]. The pH value is defined by the formula


      #math.equation(block: true, alt: "pH equals minus log sub 10 ⁡ open parenthesis open bracket H to the power plus close bracket close parenthesis")[$"pH" = − log_(10) ( [ H^(+) ] )$]
      Values for pH fall between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "14")[$14$], with #math.equation(block: false, alt: "7")[$7$] 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.


      #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([Substance], [pH], [#math.equation(block: false, alt: "open bracket H to the power plus close bracket")[$[ H^(+) ]$]]),
  [Battery acid], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0.1")[$0.1$]],
  [Lemon juice], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "0.01")[$0.01$]],
  [Vinegar], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "0.001")[$0.001$]],
  [Milk], [#math.equation(block: false, alt: "6.4")[$6.4$]], [#math.equation(block: false, alt: "10 to the power minus 6.4")[$10^(− 6.4)$]],
  [Baking soda], [#math.equation(block: false, alt: "8.4")[$8.4$]], [#math.equation(block: false, alt: "10 to the power minus 8.4")[$10^(− 8.4)$]],
  [Milk of magnesia], [#math.equation(block: false, alt: "10.5")[$10.5$]], [#math.equation(block: false, alt: "10 to the power minus 10.5")[$10^(− 10.5)$]],
  [Lye], [#math.equation(block: false, alt: "13")[$13$]], [#math.equation(block: false, alt: "10 to the power minus 13")[$10^(− 13)$]],
))


      #examplebox("Example 5")[][
+ Calculate the pH of a solution with a hydrogen ion concentration of #math.equation(block: false, alt: "3.98 times 10 to the power minus 5")[$3.98 × 10^(− 5)$].
+ The water in a swimming pool should be maintained at a pH of #math.equation(block: false, alt: "7.5")[$7.5$]. What is the hydrogen ion concentration of the water?


        
        #solutionbox[
+ We use a calculator to evaluate the pH formula with #math.equation(block: false, alt: "open bracket H to the power plus close bracket equals 3.98 times 10 to the power minus 5")[$[ H^(+) ] = 3.98 × 10^(− 5)$].
			#math.equation(block: true, alt: "pH equals minus log sub 10 ⁡ open parenthesis 3.98 times 10 to the power minus 5 close parenthesis approximately equals 4.4")[$"pH" = − log_(10) ( 3.98 × 10^(− 5) ) ≈ 4.4$]
+ We solve the equation
			#math.equation(block: true, alt: "7.5 equals minus log sub 10 ⁡ open parenthesis open bracket H to the power plus close bracket close parenthesis")[$7.5 = − log_(10) ( [ H^(+) ] )$]
		for #math.equation(block: false, alt: "open bracket H to the power plus close bracket")[$[ H^(+) ]$]. First, we write
			#math.equation(block: true, alt: "minus 7.5 equals log sub 10 ⁡ open parenthesis open bracket H to the power plus close bracket close parenthesis")[$− 7.5 = log_(10) ( [ H^(+) ] )$]
		Then we convert the equation to exponential form to get
			#math.equation(block: true, alt: "open bracket H to the power plus close bracket equals 10 to the power minus 7.5 approximately equals 3.2 times 10 to the power minus 8")[$[ H^(+) ] = 10^(− 7.5) ≈ 3.2 × 10^(− 8)$]
		The hydrogen ion concentration of the water is #math.equation(block: false, alt: "3.2 times 10 to the power minus 8")[$3.2 × 10^(− 8)$].
]
]


      
        The pH of the water in a tide pool is #math.equation(block: false, alt: "8.3")[$8.3$]. What is the hydrogen ion concentration of the water?


        Answer: \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "10 to the power minus 8.3 approximately equals 5.01 times 10 to the power minus 9")[$10^(− 8.3) ≈ 5.01 × 10^(− 9)$]
]


      
      
        The pH of the water in a tide pool is #math.equation(block: false, alt: "8.3")[$8.3$]. What is the hydrogen ion concentration of the water?


        #solutionbox[
#math.equation(block: true, alt: "10 to the power minus 8.3 approximately equals 5.01 times 10 to the power minus 9")[$10^(− 8.3) ≈ 5.01 × 10^(− 9)$]
]


      
      A decrease of #math.equation(block: false, alt: "1")[$1$] on the pH scale corresponds to an increase in acidity by a factor of #math.equation(block: false, alt: "10")[$10$]. Thus, lemon juice is #math.equation(block: false, alt: "10")[$10$] times more acidic than vinegar, and battery acid is #math.equation(block: false, alt: "100")[$100$] times more acidic than vinegar.


    
    
      
=== Decibels


      The #strong[decibel scale], used to measure the loudness or intensity of a sound, is another example of a logarithmic scale. The loudness of a sound is measured in decibels, #math.equation(block: false, alt: "D ,")[$D ,$] by


      #math.equation(block: true, alt: "D equals 10 log sub 10 ⁡ open parenthesis the fraction I over 10 to the power minus 12 close parenthesis")[$D = 10 log_(10) ( frac(I, 10^(− 12)) )$]
      where #math.equation(block: false, alt: "I")[$I$] 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.


      #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([Sound], [Intensity (watts/m#math.equation(block: false, alt: "2")[$2$])], [Decibels]),
  [Whisper], [#math.equation(block: false, alt: "10 to the power minus 10")[$10^(− 10)$]], [#math.equation(block: false, alt: "20")[$20$]],
  [Background music], [#math.equation(block: false, alt: "10 to the power minus 8")[$10^(− 8)$]], [#math.equation(block: false, alt: "40")[$40$]],
  [Loud conversation], [#math.equation(block: false, alt: "10 to the power minus 6")[$10^(− 6)$]], [#math.equation(block: false, alt: "60")[$60$]],
  [Heavy traffic], [#math.equation(block: false, alt: "10 to the power minus 4")[$10^(− 4)$]], [#math.equation(block: false, alt: "80")[$80$]],
  [Jet airplane], [#math.equation(block: false, alt: "10 to the power minus 2")[$10^(− 2)$]], [#math.equation(block: false, alt: "100")[$100$]],
  [Thunder], [#math.equation(block: false, alt: "10 to the power minus 1")[$10^(− 1)$]], [#math.equation(block: false, alt: "110")[$110$]],
))


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


      #math.equation(block: true, alt: "the fraction Intensity of thunder over Intensity of a whisper equals the fraction 10 to the power minus 1 over 10 to the power minus 10 equals 10 to the power 9")[$frac("Intensity of thunder", "Intensity of a whisper") = frac(10^(− 1), 10^(− 10)) = 10^(9)$]
      Thunder is #math.equation(block: false, alt: "10 to the power 9")[$10^(9)$], 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.


      #examplebox("Example 6")[][
+ Normal breathing generates about #math.equation(block: false, alt: "10 to the power minus 11")[$10^(− 11)$] watts per square meter at a distance of #math.equation(block: false, alt: "3")[$3$] feet. Find the number of decibels for a breath #math.equation(block: false, alt: "3")[$3$] feet away.
+ Normal conversation registers at about #math.equation(block: false, alt: "40")[$40$] decibels. How many times more intense than breathing is normal conversation?


        
        #solutionbox[
+ We evaluate the decibel formula with #math.equation(block: false, alt: "I equals 10 to the power minus 11")[$I = 10^(− 11)$] to find
			#math.equation(block: true, alt: "D, equals 10 log sub 10 ⁡ open parenthesis the fraction 10 to the power minus 11 over 10 to the power minus 12 close parenthesis equals 10 log sub 10 ⁡ open parenthesis 10 to the power 1 close parenthesis; equals 10 open parenthesis 1 close parenthesis equals 10 decibels")[$D & = 10 log_(10) ( frac(10^(− 11), 10^(− 12)) ) = 10 log_(10) ( 10^(1) ) \  & = 10 ( 1 ) = 10 " decibels"$]
+ We let #math.equation(block: false, alt: "I sub b")[$I_(b)$] stand for the sound intensity of breathing, and #math.equation(block: false, alt: "I sub c")[$I_(c)$] stand for the intensity of normal conversation. We are looking for the ratio #math.equation(block: false, alt: "I sub c / I sub b")[$I_(c) / I_(b)$]. From part (a), we know that
			#math.equation(block: true, alt: "I sub w equals 10 to the power minus 11")[$I_(w) = 10^(− 11)$]
	and from the formula for decibels, we have
			#math.equation(block: true, alt: "40 equals 10 log sub 10 ⁡ open parenthesis the fraction I sub c over 10 to the power minus 12 close parenthesis")[$40 = 10 log_(10) ( frac(I_(c), 10^(− 12)) )$]
	which we can solve for #math.equation(block: false, alt: "I sub c")[$I_(c)$]. Dividing both sides of the equation by #math.equation(block: false, alt: "10")[$10$] and rewriting in exponential form, we have
			#math.equation(block: true, alt: "the fraction I sub c over 10 to the power minus 12, equals 10 to the power 4, Multiply both sides by 10 to the power minus 12.; I sub c, equals 10 to the power 4 open parenthesis 10 to the power minus 12 close parenthesis equals 10 to the power minus 8")[$display(frac(I_(c), 10^(− 12))) & = 10^(4) &  & "Multiply both sides by " 10^(− 12) . \ I_(c) & = 10^(4) ( 10^(− 12) ) = 10^(− 8)$]
	Finally, we compute the ratio #math.equation(block: false, alt: "the fraction I sub c over I sub b")[$display(frac(I_(c), I_(b)))$]:
			#math.equation(block: true, alt: "the fraction I sub c over I sub b equals the fraction 10 to the power minus 8 over 10 to the power minus 11 equals 10 cubed")[$frac(I_(c), I_(b)) = frac(10^(− 8), 10^(− 11)) = 10^(3)$]
	Normal conversation is #math.equation(block: false, alt: "1000")[$1000$] times more intense than breathing.
]
]


      
        The noise of city traffic registers at about #math.equation(block: false, alt: "70")[$70$] decibels.


        + What is the intensity of traffic noise, in watts per square meter? #linebreak() Answer: #math.equation(block: false, alt: "I equals")[$I =$]\_\_\_\_\_  watts/m#math.equation(block: false, alt: "2")[$2$]
+ How many times more intense is traffic noise than conversation? #linebreak() Answer: \_\_\_\_\_ times


        #solutionbox[
+ #math.equation(block: false, alt: "I equals 10 to the power minus 5")[$I = 10^(− 5)$] watts/m#math.equation(block: false, alt: "2")[$2$]
+ #math.equation(block: false, alt: "1000")[$1000$]
]


      
      
        The noise of city traffic registers at about #math.equation(block: false, alt: "70")[$70$] decibels.


        + What is the intensity of traffic noise, in watts per square meter?
+ How many times more intense is traffic noise than conversation?


        #solutionbox[
+ #math.equation(block: false, alt: "I equals 10 to the power minus 5")[$I = 10^(− 5)$] watts/m#math.equation(block: false, alt: "2")[$2$]
+ #math.equation(block: false, alt: "1000")[$1000$]
]


      
      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Both the decibel model and the Richter scale in the next example use expressions of the form #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction a over b close parenthesis")[$log ( display(frac(a, b)) )$]. Be careful to follow the order of operations when using these models. We must compute the quotient #math.equation(block: false, alt: "the fraction a over b")[$display(frac(a, b))$] before taking a logarithm. In particular, recall that  #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction a over b close parenthesis")[$log ( display(frac(a, b)) )$]  is #emph[not] equivalent to #math.equation(block: false, alt: "the fraction log ⁡ open parenthesis a close parenthesis over log ⁡ open parenthesis b close parenthesis")[$display(frac(log ( a ), log ( b )))$].
]


    
    
      
=== The Richter Scale


      One method for measuring the magnitude of an earthquake compares the amplitude #math.equation(block: false, alt: "A")[$A$] of its seismographic trace with the amplitude #math.equation(block: false, alt: "A sub 0")[$A_(0)$] of the smallest detectable earthquake. The log of their ratio is the Richter magnitude, #math.equation(block: false, alt: "M")[$M$]. Thus,


      #math.equation(block: true, alt: "M equals log sub 10 ⁡ open parenthesis the fraction A over A sub 0 close parenthesis")[$M = log_(10) ( frac(A, A_(0)) )$]
      #examplebox("Example 7")[][
+ 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?
+ How many times more powerful than the Northridge quake was the San Francisco earthquake of 1989, which registered 7.1 on the Richter scale?


        
        #solutionbox[
+ The amplitude #math.equation(block: false, alt: "A")[$A$] of the Northridge quake is given by
            #math.equation(block: true, alt: "6.9 equals log sub 10 ⁡ open parenthesis the fraction A over A sub 0 close parenthesis")[$6.9 = log_(10) ( display(frac(A, A_(0))) )$]
    and by rewriting in exponential form we find
            #math.equation(block: true, alt: "A equals 10 to the power 6.9 A sub 0")[$A = 10^(6.9) A_(0)$]
    An earthquake 100 times as powerful would have amplitude
        #math.equation(block: true, alt: "100 A equals 100 times 10 to the power 6.9 A sub 0 equals 10 to the power 8.9 A sub 0")[$100 A = 100 ⋅ 10^(6.9) A_(0) = 10^(8.9) A_(0)$]
    Thus, the magnitude of the more powerful quake is
            #math.equation(block: true, alt: "M, equals log sub 10 ⁡ open parenthesis the fraction 10 to the power 8.9 A sub 0 over A sub 0 close parenthesis; equals log sub 10 ⁡ open parenthesis 10 to the power 8.9 close parenthesis equals 8.9")[$M & = log_(10) ( display(frac(10^(8.9) A_(0), A_(0))) ) \  & = log_(10) ( 10^(8.9) ) = 8.9$]
+ In part (a) we used the Richter formula to find that the amplitude of the Northridge quake was 
        #math.equation(block: true, alt: "A equals 10 to the power 6.9 A sub 0")[$A = 10^(6.9) A_(0)$]
    Similarly, the amplitude of the San Francisco quake was
        #math.equation(block: true, alt: "A equals 10 to the power 7.1 A sub 0")[$A = 10^(7.1) A_(0)$]
    So the ratio of their amplitudes is
        #math.equation(block: true, alt: "the fraction 10 to the power 7.1 A sub 0 over 10 to the power 6.9 A sub 0 equals 10 to the power 0.2")[$frac(10^(7.1) A_(0), 10^(6.9) A_(0)) = 10^(0.2)$]
    The San Francisco earthquake was #math.equation(block: false, alt: "10 to the power 0.2")[$10^(0.2)$], 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?


        Answer: \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "10 to the power .5 approximately equals 3.16")[$10^(.5) ≈ 3.16$]
]


      
      
        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?


        #solutionbox[
#math.equation(block: true, alt: "10 to the power .5 approximately equals 3.16")[$10^(.5) ≈ 3.16$]
]


      
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
An earthquake #math.equation(block: false, alt: "100")[$100$], or #math.equation(block: false, alt: "10 squared")[$10^(2)$], times as strong is only two units greater in magnitude on the Richter scale. In general, a difference of #math.equation(block: false, alt: "K")[$K$] units on the Richter scale (or any logarithmic scale) corresponds to a factor of #math.equation(block: false, alt: "10 to the power K")[$10^(K)$] units in the intensity of the quake.
]


      
        How much stronger is magnitude 4 earthquake than a magnitude 2 earthquake?


        \_\_\_\_\_


        #solutionbox[
100 times as strong.
]


      
      
        How much stronger is magnitude 4 earthquake than a magnitude 2 earthquake?


        + Twice as strong.
+ Four times as strong.
+ 16 times as strong.
+ 100 times as strong.


      
      #examplebox("Example 8")[][
On a log scale, the weights of two animals differ by #math.equation(block: false, alt: "1.6")[$1.6$] units. What is the ratio of their actual weights?


        
        #solutionbox[
A difference of #math.equation(block: false, alt: "1.6")[$1.6$] on a log scale corresponds to a factor of #math.equation(block: false, alt: "10 to the power 1.6")[$10^(1.6)$] in the actual weights. Thus, the heavier animal is #math.equation(block: false, alt: "10 to the power 1.6")[$10^(1.6)$], or #math.equation(block: false, alt: "39.8")[$39.8$] times as heavy as the lighter animal.
]
]


      
        Two points, labeled #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], differ by #math.equation(block: false, alt: "2.5")[$2.5$] units on a log scale. What is the ratio of their decimal values?


        Answer: \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "10 to the power 2.5 approximately equals 316.2")[$10^(2.5) ≈ 316.2$]
]


      
      
        Two points, labeled #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], differ by #math.equation(block: false, alt: "2.5")[$2.5$] units on a log scale. What is the ratio of their decimal values?


        #solutionbox[
#math.equation(block: true, alt: "10 to the power 2.5 approximately equals 316.2")[$10^(2.5) ≈ 316.2$]
]


      
      
        Explain what negative values on a log scale mean.


        \_\_\_\_\_


      
      
        Explain what negative values on a log scale mean.


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Log scale
- Log-log paper


      
      
        
==== CONCEPTS


        + A #strong[log scale] is useful for plotting values that vary greatly in magnitude. We plot the log of the variable instead of the variable itself.
+ A log scale is a #strong[multiplicative scale]: Each increment of equal length on the scale indicates that the value is multiplied by an equal amount.
+ The pH value of a substance is defined by the formula
			#math.equation(block: true, alt: "pH equals minus log sub 10 ⁡ open parenthesis open bracket H to the power plus close bracket close parenthesis")[$"pH" = − log_(10) ( [ H^(+) ] )$]
			where #math.equation(block: false, alt: "open bracket H to the power plus close bracket")[$[ H^(+) ]$] denotes the concentration of hydrogen ions in the substance.
+ The loudness of a sound is measured in decibels, #math.equation(block: false, alt: "D")[$D$], by
			#math.equation(block: true, alt: "D equals 10 log sub 10 ⁡ open parenthesis the fraction I over 10 to the power minus 12 close parenthesis")[$D = 10 log_(10) ( frac(I, 10^(− 12)) )$] 	where #math.equation(block: false, alt: "I")[$I$] is the intensity of its sound waves (in watts per square meter).
+ The Richter magnitude, #math.equation(block: false, alt: "M")[$M$], of an earthquake is given by
    		#math.equation(block: true, alt: "M equals log sub 10 ⁡ open parenthesis the fraction A over A sub 0 close parenthesis")[$M = log_(10) ( frac(A, A_(0)) )$]where #math.equation(block: false, alt: "A")[$A$] is the amplitude of its seismographic trace and #math.equation(block: false, alt: "A sub 0")[$A_(0)$] is the amplitude of the smallest detectable earthquake.
+ A #emph[difference] of #math.equation(block: false, alt: "K")[$K$] units on a logarithmic scale corresponds to a #emph[factor] of #math.equation(block: false, alt: "10 to the power K")[$10^(K)$] units in the value of the variable.


      
      
        
==== STUDY QUESTIONS


        + What numbers are used to label the axis on a log scale?
+ What does it mean to say that a log scale is a multiplicative scale?
+ Delbert says that #math.equation(block: false, alt: "80")[$80$] decibels is twice as loud as #math.equation(block: false, alt: "40")[$40$] decibels. Is he correct? Why or why not?
+ Which is farther on a log scale, the distance between #math.equation(block: false, alt: "5")[$5$] and #math.equation(block: false, alt: "15")[$15$], or the distance between #math.equation(block: false, alt: "0.5")[$0.5$] and #math.equation(block: false, alt: "1.5")[$1.5$]?


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Plot values on a log scale: \#1–4, 9 and 10
+ Read values from a log scale: \#5–8, 11–14, 19 and 20
+ Compare values on a log scale: \#15–18
+ Use log scales in applications: \#21–40


      
    
    
      
=== Homework 5.4


      
        + The log scale is labeled with powers of #math.equation(block: false, alt: "10")[$10$]. Finish labeling the tick marks in the figure with their corresponding decimal values.#figure(figph[log scale with exponents shown], alt: "log scale with exponents shown", caption: none)
+ The log scale is labeled with integer values. Label the tick marks in the figure with the corresponding powers of #math.equation(block: false, alt: "10")[$10$].#figure(figph[log scale with exponents shown], alt: "log scale with exponents shown", caption: none)


        
          + #figure(figph[log scale], alt: "log scale", caption: none)
+ #figure(figph[log scale], alt: "log scale", caption: none)


        
      
      
        + The log scale is labeled with powers of #math.equation(block: false, alt: "10")[$10$]. Finish labeling the tick marks in the figure with their corresponding decimal values.#figure(figph[log scale with exponents shown], alt: "log scale with exponents shown", caption: none)
+ The log scale is labeled with decimal values. Label the tick marks in the figure with the corresponding powers of #math.equation(block: false, alt: "10")[$10$].#figure(figph[log scale with exponents shown], alt: "log scale with exponents shown", caption: none)


      
      
        Plot the values on a log scale.


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.075")[$0.075$]], [#math.equation(block: false, alt: "1.3")[$1.3$]], [#math.equation(block: false, alt: "4200")[$4200$]], [#math.equation(block: false, alt: "87 , 000")[$87 , 000$]], [#math.equation(block: false, alt: "6.5 times 10 to the power 7")[$6.5 × 10^(7)$]],
))


        
          #figure(figph[logscale], alt: "logscale", caption: none)


        
      
      
        Plot the values on a log scale.


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "4 times 10 to the power minus 4")[$4 × 10^(− 4)$]], [#math.equation(block: false, alt: "0.008")[$0.008$]], [#math.equation(block: false, alt: "0.9")[$0.9$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "90")[$90$]],
))


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


        #figure(figph[logscale], alt: "logscale", caption: none)


        
          #math.equation(block: true, alt: "1.58")[$1.58$], #math.equation(block: true, alt: "6.31")[$6.31$], #math.equation(block: true, alt: "15.8")[$15.8$], #math.equation(block: true, alt: "63.1")[$63.1$]


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


        #figure(figph[logscale], alt: "logscale", caption: none)


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


        #figure(figph[logscale], alt: "logscale", caption: none)


        
          #math.equation(block: true, alt: "1")[$1$], #math.equation(block: true, alt: "80")[$80$], #math.equation(block: true, alt: "330")[$330$], #math.equation(block: true, alt: "1600")[$1600$], #math.equation(block: true, alt: "7000")[$7000$], #math.equation(block: true, alt: "4 times 10 to the power 7")[$4 × 10^(7)$]


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


        #figure(figph[logscale], alt: "logscale", caption: none)


      
      
        Plot the values of #math.equation(block: false, alt: "open bracket H to the power plus close bracket")[$[ H^(+) ]$] in the section "Acidity and the pH Scale" on a log scale.


        
          #figure(figph[pH on log scale], alt: "pH on log scale", caption: none)


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


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


        #math.equation(block: true, alt: "m equals 4.83 minus 2.5 log ⁡ L")[$m = 4.83 − 2.5 log L$]
        where #math.equation(block: false, alt: "L")[$L$] is the luminosity of the star, measured in solar units. Calculate the magnitude of the stars whose luminosities are given in the figure.


        #figure(figph[star magnitudes on log scale], alt: "star magnitudes on log scale", caption: none)


        
          Proxima Centauri: #math.equation(block: false, alt: "15.5")[$15.5$]; Barnard: #math.equation(block: false, alt: "13.2")[$13.2$]; Sirius: #math.equation(block: false, alt: "1.4")[$1.4$]; Vega: #math.equation(block: false, alt: "0.6")[$0.6$]; Arcturus: #math.equation(block: false, alt: "minus 0.4")[$− 0.4$]; Antares: #math.equation(block: false, alt: "minus 4.7")[$− 4.7$]; Betelgeuse: #math.equation(block: false, alt: "minus 7.2")[$− 7.2$]


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


        #figure(figph[radiation wavelength on log scale], alt: "radiation wavelength on log scale", caption: none)


      
      
        The #strong[risk magnitude] of an event is defined by #math.equation(block: false, alt: "R equals 10 plus log ⁡ p")[$R = 10 + log p$], where #math.equation(block: false, alt: "p")[$p$] is the probability of the event occurring. Calculate the probability of each event.


        + The sun will rise tomorrow, #math.equation(block: false, alt: "R equals 10")[$R = 10$].
+ The next child born in Arizona will be a boy, #math.equation(block: false, alt: "R equals 9.7")[$R = 9.7$].
+ A major hurricane will strike North Carolina this year, #math.equation(block: false, alt: "R equals 9.1")[$R = 9.1$].
+ A 100-meter asteroid will collide with Earth this year, #math.equation(block: false, alt: "R equals 8.0")[$R = 8.0$].
+ You will be involved in an automobile accident during a 10-mile trip, #math.equation(block: false, alt: "R equals 5.9")[$R = 5.9$].
+ A comet will collide with Earth this year, #math.equation(block: false, alt: "R equals 3.5")[$R = 3.5$].
+ You will die in an automobile accident on a 1000-mile trip, #math.equation(block: false, alt: "R equals 2.3")[$R = 2.3$]
+ You will die in a plane crash on a 1000-mile trip, #math.equation(block: false, alt: "R equals 0.9")[$R = 0.9$].


        
          + #math.equation(block: false, alt: "1")[$1$]
+ #math.equation(block: false, alt: "0.5012")[$0.5012$]
+ #math.equation(block: false, alt: "0.1259")[$0.1259$]
+ #math.equation(block: false, alt: "0.01")[$0.01$]
+ #math.equation(block: false, alt: "0.000079")[$0.000079$]
+ #math.equation(block: false, alt: "3.2 times 10 to the power minus 7")[$3.2 × 10^(− 7)$]
+ #math.equation(block: false, alt: "2 times 10 to the power minus 8")[$2 × 10^(− 8)$]
+ #math.equation(block: false, alt: "8 times 10 to the power minus 10")[$8 × 10^(− 10)$]


        
      
      
        Have you ever wondered why time seems to pass more quickly as we grow older? One theory suggests that the human mind judges the length of a long period of time by comparing it with its current age. For example, a year is #math.equation(block: false, alt: "20 %")[$20 upright(%)$] of a #math.equation(block: false, alt: "5")[$5$]-year-old's lifetime, but only #math.equation(block: false, alt: "5 %")[$5 upright(%)$] of a #math.equation(block: false, alt: "20")[$20$]-year-old's, so a year feels longer to a #math.equation(block: false, alt: "5")[$5$]-year-old. Thus, psychological time follows a log scale, like the one shown in the figure.


        #figure(figph[ages on log scale], alt: "ages on log scale", caption: none)


        + Label the tick marks with their base #math.equation(block: false, alt: "10")[$10$] logarithms, rounded to #math.equation(block: false, alt: "3")[$3$] decimal places. What do you notice about the values?
+ By computing their logs, locate #math.equation(block: false, alt: "18")[$18$] and #math.equation(block: false, alt: "22")[$22$] on the scale
+ Four years of college seems like a long time to an #math.equation(block: false, alt: "18")[$18$]-year-old. What length of time feels the same to a #math.equation(block: false, alt: "40")[$40$]-year-old?
+ How long will the rest of your life feel? Let #math.equation(block: false, alt: "A")[$A$] be your current age, and let #math.equation(block: false, alt: "L")[$L$] be the age to which you think you will live. Compute the difference of their logs. Now move backward on the log scale an equal distance from your current age. What is the age at that spot? Call that age #math.equation(block: false, alt: "B")[$B$]. The rest of your life will feel the same as your life from age #math.equation(block: false, alt: "B")[$B$] until now.
+ Compute #math.equation(block: false, alt: "B")[$B$] using a proportion instead of logs.


      
      
        + What number is halfway between #math.equation(block: false, alt: "10 to the power 1.5")[$10^(1.5)$] and #math.equation(block: false, alt: "10 squared")[$10^(2)$] on a log scale?
+ What number is halfway between #math.equation(block: false, alt: "20")[$20$] and #math.equation(block: false, alt: "30")[$30$] on a log scale?


        
          + #math.equation(block: false, alt: "10 to the power 1.75 approximately equals 56.2341")[$10^(1.75) ≈ 56.2341$]
+ #math.equation(block: false, alt: "10 to the power open parenthesis log ⁡ 600 close parenthesis / 2 approximately equals 24.4949")[$10^(( log 600 ) / 2) ≈ 24.4949$]


        
      
      
        + What number is halfway between #math.equation(block: false, alt: "10 to the power 3.0")[$10^(3.0)$] and #math.equation(block: false, alt: "10 to the power 3.5")[$10^(3.5)$] on a log scale?
+ What number is halfway between #math.equation(block: false, alt: "500")[$500$] and #math.equation(block: false, alt: "600")[$600$] on a log scale?


      
      
        The distances to two stars are separated by #math.equation(block: false, alt: "3.4")[$3.4$] units on a log scale. What is the ratio of their distances?


        
          #math.equation(block: true, alt: "10 to the power 3.4 approximately equals 2512")[$10^(3.4) ≈ 2512$]


        
      
      
        The populations of two cities are separated by #math.equation(block: false, alt: "2.8")[$2.8$] units on a log scale. What is the ratio of their populations?


      
      
        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)


        #figure(figph[probabilty of discovery vs diameter on log-log], alt: "probabilty of discovery vs diameter on log-log", caption: none)


        
          A: #math.equation(block: false, alt: "a approximately equals 45")[$a ≈ 45$], #math.equation(block: false, alt: "p approximately equals 7.4 %")[$p ≈ 7.4 upright(%)$]; B: #math.equation(block: false, alt: "a approximately equals 400")[$a ≈ 400$], #math.equation(block: false, alt: "p approximately equals 15 %")[$p ≈ 15 upright(%)$]; C: #math.equation(block: false, alt: "a approximately equals 6000")[$a ≈ 6000$], #math.equation(block: false, alt: "p approximately equals 50 %")[$p ≈ 50 upright(%)$]; D: #math.equation(block: false, alt: "a approximately equals 13000")[$a ≈ 13000$], #math.equation(block: false, alt: "p approximately equals 45 %")[$p ≈ 45 upright(%)$]


        
      
      
        The #strong[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 #math.equation(block: false, alt: "1")[$1$] through #math.equation(block: false, alt: "4")[$4$]. (Source: Leopold, Wolman, and Miller)


        #figure(figph[stream drainage vs order on semi-log], alt: "stream drainage vs order on semi-log", caption: none)


      
      In Problems 21–40, use the appropriate formulas for logarithmic models.


      
        The hydrogen ion concentration of vinegar is about #math.equation(block: false, alt: "6.3 times 10 to the power minus 4")[$6.3 × 10^(− 4)$]. Calculate the pH of vinegar.


        
          #math.equation(block: true, alt: "3.2")[$3.2$]


        
      
      
        The hydrogen ion concentration of spinach is about #math.equation(block: false, alt: "3.2 times 10 to the power minus 6")[$3.2 × 10^(− 6)$]. Calculate the pH of spinach.


      
      
        The pH of lime juice is #math.equation(block: false, alt: "1.9")[$1.9$]. Calculate its hydrogen ion concentration.


        
          #math.equation(block: true, alt: "0.0126")[$0.0126$]


        
      
      
        The pH of ammonia is #math.equation(block: false, alt: "9.8")[$9.8$]. Calculate its hydrogen ion concentration.


      
      
        A lawn mower generates a noise of intensity #math.equation(block: false, alt: "10 to the power minus 2")[$10^(− 2)$] watts per square meter. Find the decibel level of the sound of a
lawn mower.


        
          #math.equation(block: true, alt: "100")[$100$]


        
      
      
        A jet airplane generates #math.equation(block: false, alt: "100")[$100$] watts per square meter at a distance of #math.equation(block: false, alt: "100")[$100$] feet. Find the decibel level for a jet airplane.


      
      
        The loudest sound emitted by any living source is made by the blue whale. Its whistles have been measured at #math.equation(block: false, alt: "188")[$188$] decibels and are detectable #math.equation(block: false, alt: "500")[$500$] miles away. Find the intensity of the blue whale's whistle in watts per square meter.


        
          #math.equation(block: false, alt: "6 , 309 , 573")[$6 , 309 , 573$] watts per square meter


        
      
      
        The loudest sound created in a laboratory registered at #math.equation(block: false, alt: "210")[$210$] decibels. The energy from such a sound is sufficient to bore holes in solid material. Find the intensity of a #math.equation(block: false, alt: "210")[$210$]-decibel sound.


      
      
        At a concert by The Who in 1976, the sound level #math.equation(block: false, alt: "50")[$50$] meters from the stage registered #math.equation(block: false, alt: "120")[$120$] decibels. How many times more intense was this than a #math.equation(block: false, alt: "90")[$90$]-decibel sound (the threshold of pain for the human ear)?


        
          #math.equation(block: true, alt: "1000")[$1000$]


        
      
      
        The loudest scientifically measured shouting by a human being registered #math.equation(block: false, alt: "123.2")[$123.2$] decibels. How many times more intense was this than normal conversation at #math.equation(block: false, alt: "40")[$40$] decibels?


      
      
        The pH of normal rain is #math.equation(block: false, alt: "5.6")[$5.6$]. Some areas of Ontario have experienced acid rain with a pH of #math.equation(block: false, alt: "4.5")[$4.5$]. How many times more acidic is acid rain than normal rain?


        
          #math.equation(block: true, alt: "12.6")[$12.6$]


        
      
      
        The pH of normal hair is about #math.equation(block: false, alt: "5")[$5$], the average pH of shampoo is #math.equation(block: false, alt: "8")[$8$], and #math.equation(block: false, alt: "4")[$4$] for conditioner. Compare the acidity of normal hair, shampoo, and conditioner.


      
      
        How much more acidic is milk than baking soda? (Refer to the table in this section.)


        
          #math.equation(block: true, alt: "100")[$100$]


        
      
      
        Compare the acidity of lye and milk of magnesia. (Refer to the table in this section.)


      
      
        In 1964, an earthquake in Alaska measured #math.equation(block: false, alt: "8.4")[$8.4$] on the Richter scale. An earthquake measuring #math.equation(block: false, alt: "4.0")[$4.0$] is considered small and causes little damage. How many times stronger was the Alaska quake than one measuring #math.equation(block: false, alt: "4.0")[$4.0$]?


        
          #math.equation(block: true, alt: "approximately equals 25 , 000")[$≈ 25 , 000$]


        
      
      
        On April 30, 1986, an earthquake in Mexico City measured #math.equation(block: false, alt: "7.0")[$7.0$] on the Richter scale. On September 21, a second earthquake, this one measuring #math.equation(block: false, alt: "8.1")[$8.1$], hit Mexico City. How many times stronger was the September quake than the one in April?


      
      
        A small earthquake measured #math.equation(block: false, alt: "4.2")[$4.2$] on the Richter scale. What is the magnitude of an earthquake three times as strong?


        
          #math.equation(block: true, alt: "4.7")[$4.7$]


        
      
      
        Earthquakes measuring #math.equation(block: false, alt: "3.0")[$3.0$] on the Richter scale often go unnoticed. What is the magnitude of a quake #math.equation(block: false, alt: "200")[$200$] times as strong as a #math.equation(block: false, alt: "3.0")[$3.0$] quake?


      
      
        The sound of rainfall registers at #math.equation(block: false, alt: "50")[$50$] decibels. What is the decibel level of a sound twice as loud?


        
          #math.equation(block: true, alt: "53")[$53$]


        
      
      
        The magnitude, #math.equation(block: false, alt: "m")[$m$], of a star is a function of its luminosity, #math.equation(block: false, alt: "L")[$L$], given by


        #math.equation(block: true, alt: "m equals 4.83 minus 2.5 log ⁡ open parenthesis L close parenthesis")[$m = 4.83 − 2.5 log ( L )$]
        If one star is #math.equation(block: false, alt: "10")[$10$] times as luminous as another star,  is the difference in their magnitudes?