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4.10 Essential Skills 3

Topics

  • Base-10 Logarithms
  • Calculations Using Common Logarithms

Essential Skills 1 in Chapter 1 "Introduction to Chemistry", Section 1.9 "Essential Skills 1" and Essential Skills 2 in Chapter 3 "Chemical Reactions", Section 3.7 "Essential Skills 2" described some fundamental mathematical operations used for solving problems in chemistry. This section introduces you to base-10 logarithms, a topic with which you must be familiar to do the Questions and Problems at the end of Chapter 4 "Reactions in Aqueous Solution". We will return to the subject of logarithms in Essential Skills 6 in Chapter 11 "Liquids", Section 11.9 "Essential Skills 6".

Base-10 (Common) Logarithms

Essential Skills 1 introduced exponential notation, in which a base number is multiplied by itself the number of times indicated in the exponent. The number 103, for example, is the base 10 multiplied by itself three times (10 × 10 × 10 = 1000). Now suppose that we do not know what the exponent is—that we are given only a base of 10 and the final number. If our answer is 1000, the problem can be expressed as

10a = 1000

We can determine the value of a by using an operation called the base-10 logarithm, or common logarithm, abbreviated as log, that represents the power to which 10 is raised to give the number to the right of the equals sign. This relationship is stated as log 10a = a. In this case, the logarithm is 3 because 103 = 1000:

log 103 = 3 log 1000 = 3

Now suppose you are asked to find a when the final number is 659. The problem can be solved as follows (remember that any operation applied to one side of an equality must also be applied to the other side):

10a = 659 log 10a = log 659 a = log 659

If you enter 659 into your calculator and press the “log” key, you get 2.819, which means that a = 2.819 and 102.819 = 659. Conversely, if you enter the value 2.819 into your calculator and press the “10x” key, you get 659.

You can decide whether your answer is reasonable by comparing it with the results you get when a = 2 and a = 3:

a = 2: 102 = 100 a = 2.819: 102.819 = 659 a = 3: 103 = 1000

Because the number 659 is between 100 and 1000, a must be between 2 and 3, which is indeed the case.

Table 4.5 "Relationships in Base-10 Logarithms" lists some base-10 logarithms, their numerical values, and their exponential forms.

Table 4.5 Relationships in Base-10 Logarithms

Numerical Value Exponential Form Logarithm (a)
1000 103 3
100 102 2
10 101 1
1 100 0
0.1 10−1 −1
0.01 10−2 −2
0.001 10−3 −3

Base-10 logarithms may also be expressed as log10, in which the base is indicated as a subscript. We can write log 10a = a in either of two ways:

log 10a = a log10 = (10a) = a

The second equation explicitly indicates that we are solving for the base-10 logarithm of 10a.

The number of significant figures in a logarithmic value is the same as the number of digits after the decimal point in its logarithm, so log 62.2, a number with three significant figures, is 1.794, with three significant figures after the decimal point; that is, 101.794 = 62.2, not 62.23. Skill Builder ES1 provides practice converting a value to its exponential form and then calculating its logarithm.

Skill Builder ES1

Express each number as a power of 10 and then find the common logarithm.

  1. 10,000
  2. 0.00001
  3. 10.01
  4. 2.87
  5. 0.134

Solution

  1. 10,000 = 1 × 104; log 1 × 104 = 4.0
  2. 0.00001 = 1 × 10−5; log 1 × 10−5 = −5.0
  3. 10.01 = 1.001 × 10; log 10.01 = 1.0004 (enter 10.01 into your calculator and press the “log” key); 101.0004 = 10.01
  4. 2.87 = 2.87 × 100; log 2.87 = 0.458 (enter 2.87 into your calculator and press the “log” key); 100.458 = 2.87
  5. 0.134 = 1.34 × 10−1; log 0.134 = −0.873 (enter 0.134 into your calculator and press the “log” key); 10−0.873 = 0.134

Skill Builder ES2

Convert each base-10 logarithm to its numerical value.

  1. 3
  2. −2.0
  3. 1.62
  4. −0.23
  5. −4.872

Solution

  1. 103
  2. 10−2
  3. 101.62 = 42
  4. 10−0.23 = 0.59
  5. 10−4.872 = 1.34 × 10−5

Calculations Using Common Logarithms

Because logarithms are exponents, the properties of exponents that you learned in Essential Skills 1 apply to logarithms as well, which are summarized in Table 4.6 "Properties of Logarithms". The logarithm of (4.08 × 20.67), for example, can be computed as follows:

log(4.08 × 20.67) = log 4.08 + log 20.67 = 0.611 + 1.3153 = 1.926

We can be sure that this answer is correct by checking that 101.926 is equal to 4.08 × 20.67, and it is.

In an alternative approach, we multiply the two values before computing the logarithm:

4.08 × 20.67 = 84.3 log 84.3 = 1.926

We could also have expressed 84.3 as a power of 10 and then calculated the logarithm:

log 84.3 = log(8.43 × 10) = log 8.43 + log 10 = 0.926 + 1 = 1.926

As you can see, there may be more than one way to correctly solve a problem.

We can use the properties of exponentials and logarithms to show that the logarithm of the inverse of a number (1/B) is the negative logarithm of that number (−log B):

log ( 1 B ) = log  B

If we use the formula for division given Table 4.6 "Properties of Logarithms" and recognize that log 1 = 0, then the logarithm of 1/B is

log ( 1 B ) =  log 1 log  B =  0 log  B = log  B

Table 4.6 Properties of Logarithms

Operation Exponential Form Logarithm
multiplication (10a)(10b) = 10a + b log(ab) = log a + log b
division ( 10 a 10 b ) = 10 a     b log ( a b ) = log   a log  b

Skill Builder ES3

Convert each number to exponential form and then calculate the logarithm (assume all trailing zeros on whole numbers are not significant).

  1. 100 × 1000
  2. 0.100 ÷ 100
  3. 1000 × 0.010
  4. 200 × 3000
  5. 20.5 ÷ 0.026

Solution

  1. 100 × 1000 = (1 × 102)(1 × 103)

    log[(1 × 102)(1 × 103)] = 2.0 + 3.0 = 5.0

    Alternatively, (1 × 102)(1 × 103) = 1 × 102 + 3 = 1 × 105

    log(1 × 105) = 5.0

  2. 0.100 ÷ 100 = (1.00 × 10−1) ÷ (1 × 102)

    log[(1.00 × 10−1) ÷ (1 × 102)] = 1 × 10−1−2 = 1 × 10−3

    Alternatively, (1.00 × 10−1) ÷ (1 × 102) = 1 × 10[(−1) − 2] = 1 × 10−3

    log(1 × 10−3) = −3.0

  3. 1000 × 0.010 = (1 × 103)(1.0 × 10−2)

    log[(1 × 103)(1 × 10−2)] = 3.0 + (−2.0) = 1.0

    Alternatively, (1 × 103)(1.0 × 10−2) = 1 × 10[3 + (−2)] = 1 × 101

    log(1 × 101) = 1.0

  4. 200 × 3000 = (2 × 102)(3 × 103)

    log[(2 × 102)(3 × 103)] = log(2 × 102) + log(3 × 103)

    = (log 2 + log 102) + (log 3 + log 103)

    = 0.30 + 2 + 0.48 + 3 = 5.8

    Alternatively, (2 × 102)(3 × 103) = 6 × 102 + 3 = 6 × 105

    log(6 × 105) = log 6 + log 105 = 0.78 + 5 = 5.8

  5. 20.5 ÷ 0.026 = (2.05 × 10) ÷ (2.6 × 10−2)

    log[(2.05 × 10) ÷ (2.6 × 10−2)] = (log 2.05 + log 10) − (log 2.6 + log 10−2)

    = (0.3118 + 1) − [0.415 + (−2)]

    = 1.3118 + 1.585 = 2.90

    Alternatively, (2.05 × 10) ÷ (2.6 × 10−2) = 0.788 × 10[1 − (−2)] = 0.788 × 103

    log(0.79 × 103) = log 0.79 + log 103 = −0.102 + 3 = 2.90

Skill Builder ES4

Convert each number to exponential form and then calculate its logarithm (assume all trailing zeros on whole numbers are not significant).

  1. 10 × 100,000
  2. 1000 ÷ 0.10
  3. 25,000 × 150
  4. 658 ÷ 17

Solution

  1. (1 × 10)(1 × 105); logarithm = 6.0
  2. (1 × 103) ÷ (1.0 × 10−1); logarithm = 4.00
  3. (2.5 × 104)(1.50 × 102); logarithm = 6.57
  4. (6.58 × 102) ÷ (1.7 × 10); logarithm = 1.59