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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 1.2.2 Logarithms

Why Logarithms?

In the previous section, indices allowed us to simplify very large or small numbers by expressing them in terms of powers. However, there are situations where we need to reverse this process, rather than finding the result of a power, we need to determine the exponent itself. This is where logarithms come into play.
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A \(logarithm\) is the inverse of an exponent (index). It tells us the power to which a specific base must be raised to produce a given number.
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For example:
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Exponent Form: \(2^3 = 8.\)
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Logarithm Form: \(\log_2 8 = 3.\)
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In simpler terms, a logarithm answers the question: "To what power must the base be raised to produce a certain number?"
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Why Are Logarithms Important?
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Logarithms are crucial for solving equations involving exponents. They simplify computations and are widely used in fields like:
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  • Finance: calculating compound interest or investment growth.
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  • Science and Engineering: understanding radioactive decay, measuring sound intensity, or modeling population growth.
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Subsubsection 1.2.2.1 Introduction to Logarithms

Learner Experience 1.2.6.

Work in groups: Form groups of \(2\) or \(3\) students
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Materials: A Paper/book and a pen
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Instructions:
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Pick a number from the set: \(81, 243, 512\) or \(1000\)
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Break down the chosen number into its prime factors.
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Start by dividing the number by its smallest prime factor repeatedly until only prime factors remain.
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e.g. \(8 = 2 \times 2 \times 2 = 2^3\)
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Express your final result in index form
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Reflect on how the indices relate to logarithms:
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For example, \(2^3 =8\) translates to \(\log_2 8 = 3.\)
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Key Takeaway 1.2.30.

Logarithms are simply another way of expressing indices. They bridge the gap between exponential and logarithmic notation:
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  • Exponential Form: \(a^b = c.\)
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    Logarithmic Form:\(\log_a c = b.\)
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\(e.g\)
\begin{equation*} 8 = 2 \times 2 \times 2 = 2^3 \end{equation*}
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\begin{equation*} 4 = 2 \times 2 = 2^2 \end{equation*}
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\begin{equation*} 2 = 2^1 \end{equation*}
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The powers/ indices \(3, 2, 1\) are the \(logarithms\)
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For \(2^3 = 8\) is written as \(\log_2 8 = 3\)
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And is read as: logarithm of \(8\) to base \(2\) is equal to \(3\)
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The general form is:
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\begin{equation*} a^b = c \iff \log_a c = b \end{equation*}
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\(a^b = c\) represent the Index notation
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\(\log_a c = b\) represent the Logarithmic notation
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Example 1.2.31.

The table below contains numbers in index form and logarithm form. Fill and complete the table.
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Table 1.2.32.
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Exponential Form Logarithmic Form
__ \(\log_2 32 = 5\)
\(10^4 = 10,000\) __
\(3^3 = 27\) __
\(5^2 = 25\) __
__ \(\log_7 7 = 1\)
__ \(\log_4 64 = 3\)
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Example 1.2.33.

What is the logarithmic form of the following exponents:
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  1. \(6^2 = 36\)
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    When \(6^2 = 36\) then, \(\log_6 36 = 2\)
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  2. \(9^3 = 729\)
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    When \(9^3 = 729\) then, \(\log_9 729 = 3\)
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  3. \(4^5 = 1024\)
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    When \(4^5 = 1024\) then, \(\log_4 1024 = 5\)
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Standard form is a way of writing very large or very small numbers in a more manageable format. It is expressed as:
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\begin{equation*} B \times 10^x \end{equation*}
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Where:
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  • \(B\) is a number between \(1\) and \(10\) \((1 \leq B \lt 100)\)
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  • \(x\) is an integer (positive for large numbers and negative for small numbers)
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Learner Experience 1.2.7.

Work in groups: Form groups of \(2\) or \(3\) students
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Instruction:
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  1. Write any five digit number in your book
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  2. Identify the first non-zero digit from the left.
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  3. Place a decimal point after this digit.
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  4. Count how many places the decimal has moved;
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    • If moved left, the exponent is positive.
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    • If moved right, the exponent is negative.
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  5. Express the number in the form \(B \times 10^x\)
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Examples:
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  • The distance from the Earth to the Sun is approximately \(149,600,000 \, km. \)
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    \begin{equation*} 149,600,000 = 1.496 \times 10^8 \, km \end{equation*}
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  • The size of a red blood cell is about \(0.000007\) m
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    \begin{equation*} 0.000007 = 7.0 \times 10^{-6} m \end{equation*}
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Further Activity:
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  • The population of Kenya is approximately \(54,985,698\text{.}\) Express this number in standard form.
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  • The size of a human hair is about \(0.00008\) metres. Express the length in standard form.
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  • The charge of an electron is \(0.00000000000000000016\) coulombs. Convert this to standard form.
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Finding Logarithms of Numbers in Standard Form:
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Example 1.2.34.

Find \(\log (4.5 \times 10^5)\)
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Solution.
\begin{equation*} \log (4.5 \times 10^5) = \log 4.5 \times 10^5 \end{equation*}
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\begin{equation*} \log 4.5 = 0.6535 \end{equation*}
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\begin{equation*} \therefore \log 450000 = 5.6535 \end{equation*}
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In this case \(5\) is the characteristic. The characteristic is the whole number part of a logarithm. It tells us how large or small the number is based on powers of \(10\text{.}\)
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\(0.6535\) is the mantissa. The mantissa is the decimal part of the logarithm, found using logarithm tables. It depends on the significant digits of the number but is always positive.
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Key Takeaway 1.2.35.

  • For numbers greater than \(1\text{,}\) the characteristic is one less than the number of digits before the decimal point.
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  • For numbers less than \(1\text{,}\) the characteristic is negative, often written in bar notation (e.g \(\overline{3}\) instead of \(-3\)).
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Checkpoint 1.2.36. Using Logarithms with Standard Form.

Load the question by clicking the button below.
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Checkpoint 1.2.37. Using Logarithms to Find the Power Part of Standard Form.

Load the question by clicking the button below.
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Checkpoint 1.2.38. Compute Logs Base 10.

Load the question by clicking the button below.
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Checkpoint 1.2.39. Expressing an Exponential Number in Logarithmic Form.

Load the question by clicking the button below.
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Checkpoint 1.2.40. Rewrite Complex Exponential Relationship into Logarithmic Form.

Load the question by clicking the button below.
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Exercises Exercises

1.
What are the logarithmic form of:
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  1. \(\displaystyle 2^6 = 64\)
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  2. \(\displaystyle 5^3 = 125\)
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  3. \(\displaystyle 3^x = 81\)
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  4. \(\displaystyle 8^{\frac {2}{3}} = 4\)
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  5. \(\displaystyle 8^4 = 4096\)
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  6. \(\displaystyle 6^y = 216\)
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  7. \(\displaystyle 9^{\frac {1}{2}} = 3\)
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  8. \(\displaystyle 4^{-2} = \frac {1}{16}\)
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Answer.
  1. \(\displaystyle \log_{2}64 = 6\)
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  2. \(\displaystyle \log_{5}125 = 3\)
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  3. \(\displaystyle \log_{3}81 = x = 4\)
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  4. \(\displaystyle \log_{8}4 = \tfrac{2}{3}\)
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  5. \(\displaystyle \log_{8}4096 = 4\)
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  6. \(\displaystyle \log_{6}216 = y = 3\)
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  7. \(\displaystyle \log_{9}3 = \tfrac{1}{2}\)
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  8. \(\displaystyle \log_{4}\tfrac{1}{16} = -2\)
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2.
Express \(10^4 = 10000\) in logarithmic form.
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Answer.
\(\log_{10}10000 = 4\quad\text{(i.e., }\log 10000 = 4\text{)}\)
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3.
Find the value of \(y\) given that \(\log_y 81 = 4\)
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Answer.
\(y^{4}=81\Rightarrow y=3\)
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5.
Convert \(8^x = 512\) to logarithmic form and solve for \(x\text{.}\)
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Answer.
\(\log_{8}512=x\Rightarrow 8^{x}=512\Rightarrow x=3\)
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Subsubsection 1.2.2.2 Finding Logarithms Using Tables and Calculators

Learner Experience 1.2.8.

Understanding the Table Structure
  1. Examine your logarithm table. What do you notice about its structure?
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  2. Identify the following parts:
    • The first column, titled Integer, which contains the numbers \(10\)–\(99\text{.}\)
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    • The middle columns, titled First Decimal, which contains the digits \(0\)–\(9\)
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    • The right columns, titled Mean Difference, which contains the digits \(0\)–\(9\)
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  3. What do you think each part represents?
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  4. What happens to the numbers in the table as you go down the middle columns (the first decimal columns)? Or across the middle columns (i.e. from \(0\) to \(9\text{?}\))
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  5. What happens to the numbers in the table as you go down the right columns (the mean difference columns)? Or across the right columns (i.e. from \(0\) to \(9\text{?}\))
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Using Logarithm Tables
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In Grade \(8\text{,}\) you learned how to use mathematical tables to find squares, square roots, and reciprocals of numbers. Here we will use the mathematical tables to determine logarithms to base \(10\text{.}\)
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Table 1.2.42. Base 10 Logarithm Table
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First Decimal Mean Difference
Integer 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 8 12 17 21 25 29 33 37
11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 27 30 34
12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 31
13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 29
14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 3 6 9 12 15 18 21 24 27
15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 3 6 8 11 14 17 20 22 25
16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 3 5 8 11 13 16 19 21 24
17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2 5 8 10 12 15 18 20 22
18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 2 5 7 9 12 14 16 19 21
19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2 4 7 9 11 13 16 18 20
20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 6 8 11 13 15 17 19
21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2 4 6 8 10 12 14 16 18
22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 2 4 6 8 10 12 14 15 17
23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2 4 6 7 9 11 13 15 17
24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 2 4 5 7 9 11 12 14 16
25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2 3 5 7 9 10 12 14 15
26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2 3 5 7 8 10 12 13 15
27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 2 3 5 6 8 9 11 13 14
28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2 3 5 6 8 9 11 12 14
29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 1 3 4 6 7 9 10 12 13
30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1 3 4 6 7 9 10 11 13
31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 3 4 6 7 8 10 11 12
32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 1 3 4 5 7 8 9 11 12
33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 1 3 4 5 6 8 9 10 12
34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 1 3 4 5 6 8 9 10 11
35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1 2 4 5 6 7 9 10 11
36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 1 2 4 5 6 7 8 10 11
37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 1 2 3 5 6 7 8 9 10
38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 1 2 3 4 6 7 8 9 10
39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 1 2 3 4 6 7 8 9 10
40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 1 2 3 4 5 6 7 9 10
41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 1 2 3 4 5 6 7 8 9
42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 1 2 3 4 5 6 7 8 9
43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 1 2 3 4 5 6 7 8 9
44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 1 2 3 4 5 6 7 8 9
45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 1 2 3 4 5 6 7 8 9
46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1 2 3 4 5 6 7 7 8
47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 1 2 3 4 5 5 6 7 8
48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 1 2 3 4 4 5 6 7 8
49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 1 2 3 4 4 5 6 7 8
50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1 2 3 3 4 5 6 7 8
51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 2 3 3 4 5 6 7 8
52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 3 4 5 6 7 8
53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 2 2 3 4 5 6 6 7
54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 1 2 2 3 4 5 6 6 7
55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 3 4 5 5 6 7
56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 1 2 2 3 4 5 5 6 7
57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 2 2 3 4 5 5 6 7
58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 1 1 2 3 4 4 5 6 7
59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 1 1 2 3 4 4 5 6 6
60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1 1 2 3 4 4 5 6 6
61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 1 1 2 3 4 4 5 6 6
62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 1 1 2 3 4 4 5 6 6
63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 1 1 2 3 3 4 5 6 6
64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 1 1 2 3 3 4 5 5 6
65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1 1 2 3 3 4 5 5 6
66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 1 1 2 3 3 4 5 5 6
67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 1 2 3 3 4 5 5 6
68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 1 2 3 3 4 4 5 6
69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 1 2 3 3 4 4 5 6
70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1 1 2 2 3 4 4 5 6
71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 1 2 2 3 4 4 5 5
72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 1 1 2 2 3 4 4 5 5
73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 1 2 2 3 4 4 5 5
74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 1 2 2 3 4 4 5 5
75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 1 2 2 3 3 4 5 5
76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 1 1 2 2 3 3 4 5 5
77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 1 1 2 2 3 3 4 4 5
78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 1 1 2 2 3 3 4 4 5
79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 1 1 2 2 3 3 4 4 5
80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1 1 2 2 3 3 4 4 5
81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 1 1 2 2 3 3 4 4 5
82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 1 1 2 2 3 3 4 4 5
83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 1 1 2 2 3 3 4 4 5
84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 1 1 2 2 3 3 4 4 5
85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 2 3 3 4 4 5
86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 0 1 2 2 2 3 4 4 4
87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 0 1 2 2 2 3 4 4 4
88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 0 1 1 2 2 3 3 4 4
89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 0 1 1 2 2 3 3 4 4
90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0 1 1 2 2 3 3 4 4
91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 0 1 1 2 2 3 3 4 4
92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 0 1 1 2 2 3 3 4 4
93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 0 1 1 2 2 3 3 4 4
94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 0 1 1 2 2 3 3 4 4
95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0 1 1 2 2 3 3 4 4
96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 0 1 1 2 2 3 3 4 4
97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 0 1 1 2 2 3 3 4 4
98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 0 1 1 2 2 3 3 4 4
99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 0 1 1 2 2 3 3 4 4

Key Takeaway 1.2.43.

Structure of a Common Logarithm:
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Every common logarithm has two parts:
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  1. Characteristic: The integer part (determined by the position of the decimal point)
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  2. Mantissa: The decimal part (found from the logarithm table)
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Finding the Characteristic:
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Number Type Rule Example
Numbers \(\geq 1\) Characteristic \(=\) (digits before decimal) \(-1\) \(472.8 \rightarrow \text{Char} = 3 - 1 = 2\)
Numbers \(< 1\) Characteristic \(= -\)(zeros after decimal \(+ 1\)) \(0.00534 \rightarrow \text{Char} = -(2 + 1) = -3 \;(\text{written as} \;\bar{3})\)
Finding the Mantissa (5-Step Process):
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  1. Move the decimal so that there are only two digits to the left of it.
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  2. Locate the row corresponding to the first two digits (the integer) in the table.
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  3. Move across to the column for the first decimal digit, and write down the main part of he mantissa where the row and column intersect.
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  4. Use the second decimal digit to find the mean difference column in the table and add the value in the corresponding row to the main part of the mantissa.
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  5. Combine characteristic and mantissa, subtracting if necessary.
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Bar Notation for Negative Characteristics:
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When the characteristic is negative, we write it with a bar over the number:
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  1. \(\displaystyle \bar{3}.7272 = -3 + 0.7272 = -2.2728\)
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  2. The bar only applies to the characteristic, NOT the mantissa
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Remember: The mantissa is ALWAYS positive. Only the characteristic can be negative. Also, always verify your table reading with a calculator when possible.
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Example 1.2.44.

Use the table to find:
  1. \(\displaystyle \log_{10}(14.67)\)
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  2. \(\displaystyle \log_{10}(146.7)\)
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  3. \(\displaystyle \log_{10}(0.01467)\)
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You may find Figure 1.2.45 helpful:
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Figure 1.2.45. Help finding \(\log_{10}(14.67)\)
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Solution.
  1. To use the table, we need to split \(14.67\) into three components.
    • The integer part, left of the decimal, is \(14\)
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    • The first decmial, which is the digit immediately to the right of the decimal: \(6\)
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    • The second decimal, following the first decimal, is \(7\)
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    We can use the table to find the mantissa of \(\log_{10}(14.67)\text{.}\) This is the decimal part of the answer. The steps are as follows:
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    1. As shown in Figure 1.2.45, we use the integer part \((14)\) to find which row of the table we need, which is highlighted in \(\color{purple}{\text{purple}}\text{.}\)
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    2. Next, we use the first decimal \((6)\) to find the column, which is highlighted in \(\color{green}{\text{green}}\text{.}\) Where the row and column overlap (show in the \(\color{red}{\text{red}}\) box) we have the main part of the mantissa: \(1644\text{.}\)
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    3. However, we are taking \(\log_{10}(14.67)\text{,}\) not just \(\log_{10}(14.6)\text{.}\) So we need to adjust our mantissa using the second decimal \((7)\text{.}\) Locate the column for this under the Mean Difference part of the table (coloured in \(\color{blue}{\text{blue}}\)), and find where this intersects with the row from before. This is shown in \(\color{orange}{\text{orange}}\text{,}\) with mean difference \(21\) in our example.
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    4. We add this adjustment to the main part of our mantissa, to get the mantissa as: \(1644+21 = 1665\)
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    Thus, we know that \(\log_{10}(14.67) = X.1665\text{.}\) But we need to find \(X\text{,}\) called the characteristic.
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    There is an easy shortcut:
    \begin{equation*} X = \text{Number of digits before the decimal} - 1 \end{equation*}
    So with \(14.67\text{,}\) we have 2 digits before the decimal (\(14\) has two digits), so \(X = 2 - 1 = 1\)
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    Thus \(\log_{10}(14.67) = 1.665\)
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  2. If we try the same procedure as (a) with \(\log_{10}(146.7)\text{,}\) we run into a problem: there’s no row for \(146\text{!}\) The table only works directly for \(\log_{10}(x)\) when \(10 \le x < 100 \text{.}\)
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    Remembering our logarithm laws from Logarithms of Products and Ratios, we know that:
    \begin{equation*} \log_{10}(146.7) = \log_{10}\left(10 \times 14.67\right) = \log_{10}(10) + \log_{10}(14.67) \end{equation*}
    So we can shift the decimal so there are only two digits to the left of it \((14)\text{,}\) and calculate the mantissa using the same procedure as (a).
    🔗
    You now need to find the mantissa of \(\log_{10}(14.67)\) using the steps in (a). As we’ve just done that, we’ll reuse our answer of \(.1665\)
    🔗
    For the characteristic, the same rule applies:
    \begin{equation*} X = \text{Number of digits before the decimal} - 1 \end{equation*}
    So with our original number \(146.77\text{,}\) we have 3 digits before the decimal (\(146\) has three digits), so \(X = 3 - 1 = 2\)
    🔗
    Why is this still true? Since \(146.7\) has an extra digit left of the decimal than \(14.67\text{,}\) we get a \(\log_{10}(10)\) term from the logarithm rule. And \(\log_{10}(10) = 1\text{,}\) so our characteristic is bigger by \(1\text{,}\) the exact number of extra digits.
    🔗
    Hence combining the characteristic and the mantissa, we have \(\log_{10}(146.7) = 2.1665\)
    🔗
    🔗
🔗
🔗

Example 1.2.46.

Find \(\log (472.8)\) using a logarithm table.
🔗
Solution.
First, we split \(472.8\) into its components. Since the table only works for numbers with two digits before the decimal, we shift the decimal to get \(47.28\text{:}\)
🔗
From the table: row \(47\text{,}\) first decimal column \(2\) gives \(6739\text{.}\) The mean difference for \(8\) is \(7\text{.}\)
🔗
So the mantissa is \(6739 + 7 = 6746\text{,}\) i.e. \(.6746\)
🔗
For the characteristic: \(472.8\) has \(3\) digits before the decimal, so \(X = 3 - 1 = 2\text{.}\)
🔗
Therefore:
\begin{equation*} \log (472.8) = 2.6746 \end{equation*}
🔗
Using a calculator, we can confirm that \(\log (472.8) \approx 2.6747\text{,}\) which closely matches the table value.
🔗
🔗
🔗

Example 1.2.47.

Use logarithm tables to find \(\log 0.00534\)
🔗
Solution.
First, we shift the decimal to get two digits before it: \(53.4\text{.}\)
🔗
From the table: row \(53\text{,}\) first decimal column \(4\) gives \(7275\text{.}\) No mean difference is needed.
🔗
So the mantissa is \(.7275\)
🔗
For the characteristic: \(0.00534\) has \(2\) zeros after the decimal point, so \(X = -(2 + 1) = -3\text{,}\) written as \(\bar{3}\text{.}\)
🔗
Therefore:
\begin{equation*} \log (0.00534) = \bar{3}.7275 \end{equation*}
which, as a negative number, is \(-3 + 0.7275 = -2.2725\text{.}\)
🔗
Using a calculator, we can confirm that \(\log (0.00534) \approx -2.2725\text{,}\) matching the table value.
🔗
🔗
🔗

Checkpoint 1.2.48. Computing Simple Logarithms.

Load the question by clicking the button below.
🔗
🔗

Checkpoint 1.2.49. Determining Common Logarithms Using a Calculator.

Load the question by clicking the button below.
🔗
🔗

Exercises Exercises

1.
Use a logarithm table to find the logarithm of the following numbers:
🔗
  1. \(\displaystyle 0.00893\)
    🔗
    🔗
  2. \(\displaystyle 3140\)
    🔗
    🔗
  3. \(\displaystyle 52.7\)
    🔗
    🔗
  4. \(\displaystyle 0.000978\)
    🔗
    🔗
  5. \(\displaystyle 0.000245\)
    🔗
    🔗
  6. \(\displaystyle 6.42 \times 10^3\)
    🔗
    🔗
  7. \(\displaystyle 78900\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle \log(0.00893)=\bar{3}.9509\quad(\approx -2.0491)\)
    🔗
  2. \(\displaystyle \log(3140)=3.4969\)
    🔗
  3. \(\displaystyle \log(52.7)=1.7218\)
    🔗
  4. \(\displaystyle \log(0.000978)=\bar{4}.9903\quad(\approx -3.0097)\)
    🔗
  5. \(\displaystyle \log(0.000245)=\bar{4}.3892\quad(\approx -3.6108)\)
    🔗
  6. \(\displaystyle \log(6.42\times10^{3})=3.8075\)
    🔗
  7. \(\displaystyle \log(78900)=4.8971\)
    🔗
🔗
🔗
🔗
🔗
🔗

Subsubsection 1.2.2.3 Antilogarithms

The antilogarithm (antilog) of a number is the inverse operation of taking a logarithm. If \(\log x = y\text{,}\) then \(x\) is the antilog of \(y\text{.}\) This means:
🔗
\begin{equation*} x = 10^y \end{equation*}
🔗
To find the antilogarithm of a number, we use logarithm tables or a calculator.
🔗

Finding Antilogarithms Using Logarithm Tables.

Learner Experience 1.2.9.
Work in groups: Form groups of \(2\) or \(3\) students.
🔗
Materials: Logarithm table, paper and pen
🔗
Instructions:
🔗
  1. Rewrite \(\log(100) = 2\) and \(\log(1000) = 3\) in exponential form.
    🔗
  2. If \(\log(x) = 2.6756\text{,}\) write this in exponential form.
    🔗
  3. Separate \(2.6756\) into its characteristic and mantissa.
    🔗
  4. Using the logarithm table, find the antilog of the mantissa.
    🔗
  5. Apply the characteristic as a power of \(10\) to obtain the value of \(x\text{.}\)
    🔗
  6. Verify your result by taking the logarithm of your answer.
    🔗
What role does the characteristic play? What role does the mantissa play?
🔗
🔗
Key Takeaway 1.2.51.
A logarithm can be reversed by raising the base to the given power.
🔗
If:
\begin{equation*} \log(x) = y \end{equation*}
🔗
then:
\begin{equation*} x = 10^y \end{equation*}
🔗
The operation of finding \(x\) from \(y\) is called the antilogarithm.
🔗
When using tables:
🔗
  1. Separate the characteristic (integer part).
    🔗
  2. Use the table to find the antilog of the mantissa (decimal part).
    🔗
  3. Multiply by \(10^{\text{characteristic}}\text{.}\)
    🔗
🔗
The mantissa determines the significant digits, while the characteristic determines the position of the decimal point.
🔗
🔗
Example 1.2.52.
Find the antilogarithm of \(2.6756\) using logarithm tables.
🔗
Solution.
Separate the characteristic and mantissa
🔗
\begin{equation*} 2.6756 = 2 + 0.6756 \end{equation*}
🔗
Find the antilogarithm of \(0.6756\) using tables:
🔗
\begin{equation*} \text{Antilog } (0.6756) = 4.735 \end{equation*}
🔗
Apply the characteristic:
🔗
\begin{equation*} 10^2 \times 4.735 = 473.5 \end{equation*}
🔗
Thus, the antilogarithm of \(2.6756\) is \(473.5\text{.}\)
🔗
🔗
🔗
🔗

Finding Antilogarithms Using a Calculator.

When using a calculator, simply input the given logarithm value and use the inverse logarithm function:
🔗
\begin{equation*} x = 10^y \end{equation*}
🔗
For example, to find \(\text{Antilog } (1.3562)\) on a calculator:
🔗
  • Press 10^x or INV + LOG and enter \(1.3562\text{.}\)
    🔗
    🔗
  • Record the results
    🔗
    🔗
🔗
🔗

Checkpoint 1.2.53.

🔗

Checkpoint 1.2.54.

🔗

Exercises Exercises

1.
Find the antilogarithms of the following logarithmic values using logarithm tables:
🔗
  1. \(\displaystyle \log P = 1.3562\)
    🔗
    🔗
  2. \(\displaystyle \log Q = 3.4821\)
    🔗
    🔗
  3. \(\displaystyle \log R = 0.7294\)
    🔗
    🔗
  4. \(\displaystyle \log S = 4.2187\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle P = \approx 22.7091\)
    🔗
  2. \(\displaystyle Q = \approx 3034.5898\)
    🔗
  3. \(\displaystyle R = \approx 5.3629\)
    🔗
  4. \(\displaystyle S = \approx 16546.2659\)
    🔗
🔗
🔗
🔗
2.
Solve the following logarithms using a calculator:
🔗
  1. \(\displaystyle \log a = 2.1457\)
    🔗
    🔗
  2. \(\displaystyle \log b = 0.8743\)
    🔗
    🔗
  3. \(\displaystyle \log c = 3.5961\)
    🔗
    🔗
  4. \(\displaystyle \log d = 1.9999\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle a = \approx 139.8621\)
    🔗
  2. \(\displaystyle b = \approx 7.4869\)
    🔗
  3. \(\displaystyle c = \approx 3945.4814\)
    🔗
  4. \(\displaystyle d = \approx 99.9770\)
    🔗
🔗
🔗
🔗
🔗
🔗

Subsubsection 1.2.2.4 Logarithms of Products and Ratios

Learner Experience 1.2.10.

Work in groups: Form groups of \(2\) or \(3\) students.
🔗
Materials: Paper, pen, and calculator.
🔗
Instructions:
🔗
  1. Evaluate: \(2^2 \times 2^3\text{.}\)
    🔗
  2. Rewrite the result as a single power of \(2\text{.}\)
    🔗
  3. Find: \(\log_2 (2^2 \times 2^3)\text{.}\)
    🔗
  4. Compare with: \(\log_2 2^2 + \log_2 2^3\text{.}\)
    🔗
What do you notice?
🔗
🔗

Key Takeaway 1.2.55.

Since:
\begin{equation*} 2^2 \times 2^3 = 2^{2+3} = 2^5 \end{equation*}
🔗
Taking logarithm base \(2\text{:}\)
\begin{equation*} \log_2 (2^2 \times 2^3) = \log_2 2^5 = 5 \end{equation*}
🔗
But:
\begin{equation*} \log_2 2^2 + \log_2 2^3 = 2 + 3 = 5 \end{equation*}
🔗
Therefore,
\begin{equation*} \log_a (MN) = \log_a M + \log_a N \end{equation*}
🔗
This is called the Product Law of Logarithms.
🔗
Logarithms convert multiplication into addition.
🔗
🔗

Learner Experience 1.2.11.

Repeat the process for division:
🔗
  1. Evaluate: \(5^4 \div 5^2\text{.}\)
    🔗
  2. Rewrite as a single power of \(5\text{.}\)
    🔗
  3. Find: \(\log_5 (5^4 \div 5^2)\text{.}\)
    🔗
  4. Compare with: \(\log_5 5^4 - \log_5 5^2\text{.}\)
    🔗
🔗

Key Takeaway 1.2.56.

Since:
\begin{equation*} 5^4 \div 5^2 = 5^{4-2} = 5^2 \end{equation*}
🔗
Therefore,
\begin{equation*} \log_a \left(\frac{M}{N}\right) = \log_a M - \log_a N \end{equation*}
🔗
This is called the Quotient Law of Logarithms.
🔗
Logarithms convert division into subtraction.
🔗
🔗

Example 1.2.57.

Express as a sum or difference of logarithms and evaluate:
  1. \(\displaystyle \log_3 (9 \times 27)\)
    🔗
  2. \(\displaystyle \log_5 \left(\frac{125}{25}\right)\)
    🔗
  3. \(\displaystyle \log_2 (16 \times 8)\)
    🔗
🔗
Solution.
a) Using the product law:
\begin{equation*} \log_3 (9 \times 27) = \log_3 9 + \log_3 27 = 2 + 3 = 5 \end{equation*}
🔗
b) Using the quotient law:
\begin{equation*} \log_5 \left(\frac{125}{25}\right) = \log_5 125 - \log_5 25 = 3 - 2 = 1 \end{equation*}
🔗
c) Using the product law:
\begin{equation*} \log_2 (16 \times 8) = \log_2 16 + \log_2 8 = 4 + 3 = 7 \end{equation*}
🔗
🔗
🔗

Checkpoint 1.2.58.

🔗

Checkpoint 1.2.59.

🔗

Exercises Exercises

1.
Express the following as a sum or difference of logarithms and evaluate:
🔗
  1. \(\displaystyle \log_4 (64 \times 16)\)
    🔗
  2. \(\displaystyle \log_4 (64 \times 16)\)
    🔗
  3. \(\displaystyle \log_6 \left(\frac{216}{36}\right)\)
    🔗
  4. \(\displaystyle \log_5 (25 \times 125)\)
    🔗
  5. \(\displaystyle \log_8 \left(\frac{512}{64}\right)\)
    🔗
🔗
Answer.
  1. \(\displaystyle 6\)
    🔗
  2. \(\displaystyle 2\)
    🔗
  3. \(\displaystyle 3\)
    🔗
  4. \(\displaystyle 3\)
    🔗
🔗
🔗
🔗
2.
Given that \(\log_{10} 2 \approx 0.3010\) and \(\log_{10} 3 \approx 0.4771\text{,}\) evaluate:
🔗
  1. \(\displaystyle \log_{10} 6\)
    🔗
  2. \(\displaystyle \log_{10} 1.5\)
    🔗
  3. \(\displaystyle \log_{10} 5\)
    🔗
🔗
Answer.
  1. \(\displaystyle \log_{10}(2 \times 3) = 0.3010 + 0.4771 = 0.7781\)
    🔗
  2. \(\displaystyle \log_{10}(3 \div 2) = 0.4771 - 0.3010 = 0.1761\)
    🔗
  3. \(\displaystyle \log_{10}(10 \div 2) = 1 - 0.3010 = 0.6990\)
    🔗
🔗
🔗
🔗
3.
Write each of the following as a single logarithm:
🔗
  1. \(\displaystyle \log_a 5 + \log_a 4\)
    🔗
  2. \(\displaystyle \log_x 18 - \log_x 3\)
    🔗
  3. \(\displaystyle \log_{10} 8 + \log_{10} 125\)
    🔗
  4. \(\displaystyle \log_b 81 - \log_b 9\)
    🔗
🔗
Answer.
  1. \(\displaystyle \log_a 20\)
    🔗
  2. \(\displaystyle \log_x 6\)
    🔗
  3. \(\displaystyle \log_{10} 1000 = 3\)
    🔗
  4. \(\displaystyle \log_b 9\)
    🔗
🔗
🔗
🔗
🔗
🔗

Subsubsection 1.2.2.5 Using Logarithms to Compute Products and Ratios

Learner Experience 1.2.12. The Calculator Challenge.

Work in groups to explore the following challenge.
🔗
Part A: The Long Way.
Without using a calculator, try to compute:
🔗
  1. \(\displaystyle 236.5 \times 42.8\)
    🔗
    🔗
  2. \(\displaystyle 528.6 \div 24.7\)
    🔗
    🔗
🔗
Part B: Discovering the Pattern.
Use your logarithm tables to find:
🔗
  • \(\displaystyle \log(236.5)\)
    🔗
    🔗
  • \(\displaystyle \log(42.8)\)
    🔗
    🔗
  • What do you notice when you add these two logarithms?
    🔗
    🔗
  • What happens when you find the antilog of the sum?
    🔗
    🔗
🔗
Discussion Questions.
  1. How does adding logarithms relate to multiplication?
    🔗
    🔗
  2. How does subtracting logarithms relate to division?
    🔗
    🔗
  3. What operation on logarithms gives you a power?
    🔗
    🔗
  4. What operation on logarithms gives you a root?
    🔗
    🔗
🔗
🔗

Key Takeaway 1.2.61. Multiplicative Laws of Logarithms.

The main laws used for calculations are:
🔗
  1. \(\log(ab)=\log a+\log b\) — Product Law (multiply by adding logs)
    🔗
    🔗
  2. \(\log(a/b)=\log a-\log b\) — Quotient Law (divide by subtracting logs)
    🔗
    🔗
The general process: find the logs, apply the operation (+, −, ×, ÷ on logs), then find the antilog of the result.
🔗
Misconception: log(a+b) ≠ log a + log b. The laws only apply to multiplication, division, powers and roots.
🔗
🔗

Example 1.2.62. Multiplication using Logarithms.

Find \(236.5 \times 42.8\) using logarithm tables.
🔗
Solution.
Step 1: From tables, \(\log(236.5)=2.3741\) and \(\log(42.8)=1.6318\text{.}\)
🔗
Step 2: Add the logs: \(2.3741+1.6318=4.0059\text{.}\)
🔗
Step 3: Antilog of \(4.0059\) gives \(\approx 10\,160\text{,}\) so \(236.5\times42.8\approx10\,160\text{.}\)
🔗
🔗
🔗

Example 1.2.63. Division using Logarithms.

Find \(528.6 \div 24.7\) using logarithm tables.
🔗
Solution.
Step 1: From tables, \(\log(528.6)=2.7233\) and \(\log(24.7)=1.3927\text{.}\)
🔗
Step 2: Subtract the logs: \(2.7233-1.3927=1.3306\text{.}\)
🔗
Step 3: Antilog of \(1.3306\) gives \(\approx 21.4\text{,}\) so \(528.6\div24.7\approx21.4\text{.}\)
🔗
🔗
🔗

Checkpoint 1.2.64.

🔗

Checkpoint 1.2.65.

🔗

Exercises Exercises

1.
Use logarithm tables to evaluate the following:
🔗
  1. \(\displaystyle 345.6 \times 78.9\)
    🔗
    🔗
  2. \(\displaystyle 6284 \times 92.5\)
    🔗
    🔗
  3. \(\displaystyle 0.0482 \times 53.7\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle \approx 27\,267.84\)
    🔗
  2. \(\displaystyle \approx 581\,270\)
    🔗
  3. \(\displaystyle \approx 2.5883\)
    🔗
🔗
🔗
🔗
2.
Use logarithms to calculate:
🔗
  1. \(\displaystyle 652.3 \div 12.7\)
    🔗
    🔗
  2. \(\displaystyle 0.0854 \div 3.42\)
    🔗
    🔗
  3. \(\displaystyle 4312 \div 58.3\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle \approx 51.3622\)
    🔗
  2. \(\displaystyle \approx 0.024971\)
    🔗
  3. \(\displaystyle \approx 73.9623\)
    🔗
🔗
🔗
🔗
🔗
🔗

Subsubsection 1.2.2.6 Logarithms of Powers and Roots

Logarithms can be used to simplify calculations involving powers and roots by converting exponentiation into multiplication and roots into division.
🔗

Learner Experience 1.2.13.

Material needed:
🔗
🔗
Instructions:
🔗
Use your calculator and math reasoning to solve the following expressions involving logarithms:
🔗
\begin{equation*} \log(10000) \end{equation*}
🔗
Now rewrite \(10000\) as a power of \(10\text{:}\)
🔗
\begin{equation*} \log(10000) = \log(10^4) \end{equation*}
🔗
Apply the logarithmic law:
\begin{equation*} \log(a^b) = b \times \log(a) \end{equation*}
🔗
\begin{equation*} \log(10^4) = 4 \times \log(10) \end{equation*}
🔗
\begin{equation*} = 4 \times 1 = 4 \end{equation*}
🔗
Try the following:
🔗
  • \begin{equation*} \log(100) \end{equation*}
    🔗
    🔗
  • \begin{equation*} \log(1000000) \end{equation*}
    🔗
    🔗
  • \begin{equation*} \log(\sqrt{100}) \end{equation*}
    🔗
    🔗
  • \begin{equation*} \log(\sqrt[3]{1000}) \end{equation*}
    🔗
    🔗
🔗
  • What do you observe when applying logarithms to square and cube roots?
    🔗
    🔗
  • Why does
    \begin{equation*} \log(\sqrt{100}) \end{equation*}
    give half of
    \begin{equation*} \log(100)\text{?} \end{equation*}
    🔗
    🔗
  • What general rule can you form for logarithms and powers?
    🔗
    🔗
🔗
🔗

Key Takeaway 1.2.67.

Logarithms help simplify calculations involving powers and roots.
🔗
For example: \(\log(a^b) = b \times \log(a)\text{,}\) and \(\log(\sqrt{a}) = \frac{1}{2} \times \log(a)\text{.}\)
🔗
🔗

Example 1.2.68.

We can also use logarithms to simplify calculations that don’t directly involve logarithms. For example, computing \(2^{20}\) by multiplying would be lot of work. Approximate \(2^{20}\) using logarithms.
🔗
Solution.
Note that \(\text{Antilog}(\log 2^{20}) = 2^{20}\text{.}\)
🔗
Then \(\log 2^{20} = 20 \times \log 2 \approx 20 \times 0.3010 = 6.020 \approx 6\) using logarithm tables.
🔗
Then \(2^{20} = \text{Antilog}(\log 2^{20}) \approx \text{Antilog}(6) = 10^6 = 1000000\text{.}\)
🔗
(Remark: The exact value is 1048576.)
🔗
🔗
🔗

Example 1.2.69.

Evaluate \((23.5)^4\) using logarithm tables.
🔗
Solution.
Find the logarithm of \(23.5\) from the logarithm table:
🔗
\begin{equation*} \log 23.5 = 1.3711 \end{equation*}
🔗
Multiply by the exponent \(4\text{:}\)
🔗
\begin{equation*} 1.3711 \times 4 = 5.4844 \end{equation*}
🔗
Find the antilogarithm of \(5.4844\text{:}\)
🔗
\begin{equation*} \text{Antilog} (5.4844) \approx 304000 \end{equation*}
🔗
Thus, \((23.5)^4 \approx 304000\text{.}\)
🔗
🔗
🔗

Example 1.2.70.

Evaluate \(\sqrt[3]{524.8}\) using logarithm tables.
🔗
Solution.
Find the logarithm of \(524.8\) from the logarithm table:
🔗
\begin{equation*} \log 524.8 = 2.7200 \end{equation*}
🔗
Divide the logarithm by \(3\) (since it is a cube root):
🔗
\begin{equation*} \frac{2.7200}{3} = 0.9067 \end{equation*}
🔗
Find the antilogarithm of \(0.9067\text{:}\)
🔗
\begin{equation*} \text{Antilog} (0.9067) \approx 8.1 \end{equation*}
🔗
Thus, \(\sqrt[3]{524.8} \approx 8.1\text{.}\)
🔗
🔗
🔗

Checkpoint 1.2.71. Adding Logarithms for Mixed Power and Root.

🔗

Checkpoint 1.2.72. Applying Logarithms to a Power.

🔗

Checkpoint 1.2.73. Applying Logarithms to Roots.

🔗

Exercises Exercises

1.
Compute the following using logarithms:
🔗
  1. \(\displaystyle (78.5)^3\)
    🔗
    🔗
  2. \(\displaystyle (254.6)^4\)
    🔗
    🔗
  3. \(\displaystyle (12.75)^{2.5}\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle \approx 483\,736.6\)
    🔗
  2. \(\displaystyle \approx 4.2018 \times 10^{9}\)
    🔗
  3. \(\displaystyle \approx 580.4642\)
    🔗
🔗
🔗
🔗
2.
Compute the following roots using logarithms:
🔗
  1. \(\displaystyle \sqrt[3]{658.4}\)
    🔗
    🔗
  2. \(\displaystyle \sqrt{82.6}\)
    🔗
    🔗
  3. \(\displaystyle \sqrt[4]{3126}\)
    🔗
    🔗
🔗
Answer.
  1. \(\displaystyle \approx 8.6995\)
    🔗
  2. \(\displaystyle \approx 9.0885\)
    🔗
  3. \(\displaystyle \approx 7.4773\)
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3.
A square field has an area of \(18432\) square metres. Use logarithms to determine the length of one side.
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Answer.
\(\text{Side length} = \sqrt{18432} \approx 135.7645\ \text{m}\)
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4.
Use logarithm tables to evaluate:
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  1. \(\displaystyle \sqrt{ \frac{(4.56 \times 12.3)}{24.7} }\)
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  2. \(\displaystyle \sqrt{ \frac{(2.718 \times 9.81)}{5.432 \times 3.14} }\)
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  3. \(\displaystyle \frac{\sqrt{52.8 \times 24.6}}{\sqrt{31.5}}\)
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  4. \(\displaystyle \sqrt[3]{\frac{(6.75 \times 432)}{0.89}}\)
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  5. \(\displaystyle \sqrt[4]{\frac{(8462 \times 23.7)}{673}}\)
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Answer.
  1. \(\displaystyle \approx 1.5069\)
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  2. \(\displaystyle \approx 1.2503\)
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  3. \(\displaystyle \approx 6.4214\)
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  4. \(\displaystyle \approx 14.8525\)
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  5. \(\displaystyle \approx 4.1548\)
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5.
The volume of a cube is \(79507\) cubic centimetres. Use logarithms to find the length of one side.
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Answer.
\(43\ \text{cm}\)
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Subsubsection 1.2.2.7 Logarithms in Real-Life

Learner Experience 1.2.14. Exploring Logarithms in Different Fields.

At an individual level, explore how logarithms are used in different fields, such as:
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  • Engineering: Signal processing, decibel measurements, circuit analysis
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  • Finance: Compound interest, investment growth, loan calculations
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  • Science: pH scale in chemistry, radioactive decay, population biology
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  • Geology: Richter scale for earthquake measurement
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  • Astronomy: Stellar magnitude and brightness
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  • Medicine: Drug concentration decay in the body
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Individual Research Task:
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  1. Choose one field that interests you.
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  2. Find at least one specific example of how logarithms are used.
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  3. Write down the formula or application.
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  4. Be prepared to share with your group.
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Share your findings with your group and discuss:
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  • What common patterns do you see across different applications?
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  • Why do you think logarithms are useful for these applications?
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  • How do logarithms help us work with very large or very small numbers?
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Key Takeaway 1.2.75. Why Do We Use Logarithms in Real Life?

Logarithms help us:
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  • Compress large ranges of numbers into manageable scales
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  • Solve exponential equations (find unknown exponents)
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  • Model phenomena that grow or decay exponentially
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  • Simplify multiplication and division into addition and subtraction
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Key Real-Life Applications.
Table 1.2.76.
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Application Formula What It Measures
pH Scale (Chemistry) \(pH = -\log [H^+]\) Acidity/alkalinity
Richter Scale (Geology) \(R = \log (I/I_0)\) Earthquake intensity
Decibel Scale (Sound) \(dB = 10\log (I/I_0)\) Sound intensity
Population Growth \(P = P_0 e^{rt}\) Exponential population change
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Important Properties for Applications.
  • \(\log(ab) = \log a + \log b\) - useful for compound calculations
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  • \(\log(a/b) = \log a - \log b\) - useful for ratio comparisons
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  • \(\log(a^n) = n\log a\) - useful for exponential problems
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  • \(\log(\sqrt{a}) = \tfrac12\log a\) - useful for root calculations
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Addressing misconceptions: Remember, the negative sign in the pH formula is part of its definition, not the answer; each unit increase on the Richter scale represents a tenfold increase in intensity.
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Example 1.2.77.

A country’s population grows by 5% each year. So if the initial population is \(P_0\text{,}\) then after \(t\) years the population is \(P=P_0 (1.05)^{t}\text{.}\) How long does it take for the population to double?
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Solution.
For doubling \(2P_0=P_0 e^{rt}\) so \(2=e^{rt}\) giving \(\ln 2 = rt\) and \(t=\frac{\ln2}{0.05}\approx 13.86\text{ years}\text{.}\) (At 5% the doubling time is \(\approx 14\) years; to double in 10 years \(r\approx6.93\%\text{.}\))
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Example 1.2.78. Carbon Dating and Half-Life.

Carbon-14 dating is a method for estimating the age of formerly living things. While alive, organisms take in Carbon-14; after death the Carbon-14 decays and is not replaced, so the remaining fraction indicates how long it has been since death.
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If the amount of carbon at the beginning is \(N_0\text{,}\) after \(t\) years the amount remaining is \(N = N_0\left(\tfrac{1}{2}\right)^{t/T_{1/2}}\text{,}\) where \(T_{1/2}\) is the half-life. For Carbon-14, \(T_{1/2}=5730\) years. In other words, every \(T_{1/2}=5730\) years the amount of Carbon-14 in the sample decreases by one half.
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A sample of human bones contains 12.5% of the amount of Carbon-14 in the bones of a living human being. Estimate the age of the sample.
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Solution.
As 12.5% remain, we have \(N=0.125N_0\text{.}\) Using the decay formula, we get \(0.125N_0=N_0\left(\tfrac{1}{2}\right)^{t/5730}\text{.}\) Dividing by \(N_0\text{,}\) we have \(0.125=\left(\tfrac{1}{2}\right)^{t/5730}\text{.}\)
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Take log with base 2 on both sides: \(\log_2 0.125=\tfrac{t}{5730}\log_2(1/2)\text{,}\) so \(t=5730\dfrac{\log_2 0.125}{\log_2 (1/2)}\text{.}\) Since \(\log_2 0.125=-3\) and \(\log_2 (1/2)=-1\text{,}\) we get \(t=5730\times3=17190\) years.
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Example 1.2.79.

The pH scale in chemistry is based on logarithms. Given that \(pH = -\log [H^+]\text{,}\) determine the pH of a solution where \([H^+] = 3.2\times 10^{-4}\text{.}\)
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Solution.
\(pH = -\log(3.2\times 10^{-4}) = -(\log 3.2 + \log 10^{-4}) = -(0.5051 - 4) = 3.49\) (acidic solution)
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Example 1.2.80.

The Richter scale measures earthquake intensity using \(R = \log(I/I_0)\text{.}\) If an earthquake is 1000 times more intense than the reference, what is its magnitude?
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Solution.
Here \(I/I_0 = 1000 = 10^3\text{,}\) so \(R = \log 10^3 = 3\text{.}\) Magnitude = 3.0.
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Checkpoint 1.2.81.

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Checkpoint 1.2.82.

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Exercises Exercises

1.
Assume the number of algae in Lake Victoria doubles every week.
  1. If today Lake Victoria is one quarter covered in algae, how long does it take until it is fully covered?
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  2. If today Lake Victoria is 1% covered in algae, how long does it take until it is fully covered?
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Answer 1.
2 weeks
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Answer 2.
Approximately 6.64 weeks
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Solution.
  1. If at the beginning X% are covered, after \(n\) weeks the coverage is \(X \cdot 2^n\text{.}\) So we want to find \(n\) such that \(X \cdot 2^n = 100\%\text{,}\) i.e., \(2^n = \frac{100}{X}\text{.}\) If \(X=25\%\text{,}\) then \(2^n = 4\text{,}\) so \(n= \log_2 4 = 2\) weeks.
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    (We can also see this directly: start at 1/4. After 1 week → 1/2; after 2 weeks → 1 (fully covered). So it takes 2 weeks.)
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  2. If \(X=1\%\text{,}\) then \(2^n = 100\text{,}\) so \(n= \log_2 100\text{.}\) Since \(\log_2 100 = \frac{\ln 100}{\ln 2} \approx \frac{4.605}{0.693} \approx 6.64\text{,}\) it takes approximately 6.64 weeks.
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2.
You put 10000 ksh into an account that grows by 6% each year. How many years will it take for the money to double to 20000 ksh if the interest is added once per year?
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Answer.
About 11.9 years (≈ 12 years)
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Solution.
Each year, the new amount \(A\) of money is the old amount \(P + 0.06 P = P(1 + 0.06) = P(1.06)\text{.}\) So after \(t\) years, \(A = P(1.06)^t\text{.}\) Put \(A=20000\text{,}\) \(P=10000\text{.}\) So \(20000 = 10000(1.06)^t\) which simplifies to \(2 = 1.06^t\text{.}\) Take natural logs of both sides: \(\ln 2 = t\ln 1.06\text{,}\) therefore \(t = \dfrac{\ln 2}{\ln 1.06} \approx \dfrac{0.6931}{0.05827} \approx 11.9\ \text{years}.\) In plain terms: it takes about 12 years for the investment to double.
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3.
The sound level in decibels is given by \(L = 10 \,\log_{10}\left(\frac{I}{I_0}\right)\text{,}\) where \(I_0 = 10^{-12}\,\text{W/m}^2\) is a reference intensity. Calculate the decibel level of a sound whose intensity is \(1.0 \times 10^{-3}\,\text{W/m}^2\text{.}\)
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Answer.
90 dB
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Solution.
Compute the ratio: \(I/I_0 = 1.0\times10^{-3}/10^{-12} = 10^{9}.\) Then \(L = 10\log_{10}(10^{9}) = 10\times9 = 90\text{ dB}.\)
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4.
The brightness of a star is measured on a logarithmic magnitude scale where a difference of \(5\) magnitudes corresponds to a brightness ratio of \(100\text{.}\) If one star is \(250\) times brighter than another, what is the difference in magnitudes between them? (Use logarithms.)
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Answer.
Approximately 6.0 magnitudes
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Solution.
If brightness ratio \(R = 100^{\Delta m/5}\text{,}\) then \(250 = 100^{\Delta m/5}\text{.}\) Take base-10 logs: \(\log_{10}250 = (\Delta m/5)\log_{10}100 = 2(\Delta m/5)\text{,}\) so \(\Delta m = \tfrac{5}{2}\log_{10}250\text{.}\) Using \(\log_{10}250\approx 2.39794\text{,}\) \(\Delta m\approx(5/2)\times 2.39794\approx 5.9949\approx 6.0.\)
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