By the end of this chapter, you will be able to:
Mastering these skills will help you confidently tackle engineering problems and make precise calculations essential in your trade.
Electrical engineers frequently use algebraic concepts such as indices and logarithms to analyze circuits, signal processing, and system behaviours. Mastery of these mathematical tools enables precise calculation of power, voltage ratios, and exponential growth or decay in electrical components. This chapter develops your skills in applying algebraic laws and logarithmic operations essential for solving practical electrical engineering problems in Kenya.
Indices and logarithms form the foundation for manipulating exponential expressions and solving equations involving powers, which are common in electrical engineering calculations such as decibel conversions and capacitor charging. This section covers the principles of indices, the laws governing their operations, and the use of common logarithms in practical contexts.
Indices represent repeated multiplication of a base number and are used to express large or small values compactly. The general form is \(a^n\), where \(a\) is the base and \(n\) is the index or power, which can be a positive integer, negative integer, or fractional number.
$$a^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ times}}$$
Example 1: Calculate \(5^3\) representing the power rating increase of an electrical transformer.
Given: Base \(a=5\), power \(n=3\)
$$a^n = 5^3$$
$$= 5 \times 5 \times 5$$
$$= 125$$
Answer: 125
Example 2: Evaluate \(2^{-2}\) representing a voltage ratio expressed as a power.
Given: \(a=2\), \(n=-2\)
$$a^n = 2^{-2}$$
$$= \frac{1}{2^2}$$
$$= \frac{1}{4} = 0.25$$
Answer: 0.25
Example 3: Find \(9^{\frac{1}{2}}\) representing the square root of resistance in ohms.
Given: \(a=9\), \(n=\frac{1}{2}\)
$$a^n = 9^{\frac{1}{2}}$$
$$= \sqrt{9}$$
$$= 3$$
Answer: 3
The laws of indices simplify calculations involving powers with positive integers, critical for analyzing electrical quantities such as power and energy.
Key laws:
Example 1: Simplify \(3^2 \times 3^4\) to calculate combined gain in a circuit.
Given: \(a=3, m=2, n=4\)
$$a^m \times a^n = 3^{2+4} = 3^6$$
$$= 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729$$
Answer: 729
Example 2: Simplify \(\frac{10^5}{10^3}\) to find voltage ratio.
Given: \(a=10, m=5, n=3\)
$$\frac{a^m}{a^n} = 10^{5-3} = 10^2 = 100$$
Answer: 100
Example 3: Simplify \((2^3)^4\) to determine power factor exponent.
Given: \(a=2, m=3, n=4\)
$$(a^m)^n = 2^{3 \times 4} = 2^{12}$$
$$= 4096$$
Answer: 4096
Negative indices represent reciprocal powers, important when calculating attenuation or inverse relationships in electrical circuits.
Key law:
$$a^{-n} = \frac{1}{a^n}, \quad a eq 0$$
Example 1: Calculate \(4^{-2}\) representing inverse power.
Given: \(a=4, n=2\)
$$4^{-2} = \frac{1}{4^2} = \frac{1}{16} = 0.0625$$
Answer: 0.0625
Example 2: Simplify \(\frac{5^{-3}}{5^{-1}}\) representing voltage attenuation ratio.
Given: \(a=5\)
$$\frac{5^{-3}}{5^{-1}} = 5^{-3 - (-1)} = 5^{-2} = \frac{1}{25} = 0.04$$
Answer: 0.04
Example 3: Simplify \((3^{-2})^3\) representing power loss exponent.
Given: \(a=3, m=-2, n=3\)
$$(3^{-2})^3 = 3^{-2 \times 3} = 3^{-6} = \frac{1}{3^6} = \frac{1}{729} \approx 0.00137$$
Answer: 0.00137
Fractional indices denote roots, essential for calculating RMS values and impedance in AC circuits.
Key law:
$$a^{\frac{m}{n}} = \sqrt[n]{a^m}$$
Example 1: Calculate \(16^{\frac{1}{2}}\) representing RMS voltage.
Given: \(a=16, m=1, n=2\)
$$16^{\frac{1}{2}} = \sqrt{16} = 4$$
Answer: 4
Example 2: Find \(27^{\frac{2}{3}}\) representing impedance magnitude.
Given: \(a=27, m=2, n=3\)
$$27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$$
Answer: 9
Example 3: Evaluate \(81^{\frac{3}{4}}\) for power calculations.
Given: \(a=81, m=3, n=4\)
$$81^{\frac{3}{4}} = (\sqrt[4]{81})^3 = (3)^3 = 27$$
Answer: 27
Powers of 10 express magnitudes in engineering notation, while common logarithms (base 10) convert multiplicative relationships into additive ones, simplifying calculations in signal processing and decibel computations.
The power of 10 is:
$$10^n$$
The common logarithm is:
$$\log_{10} x$$
Example 1: Calculate \(10^4\) representing a large current multiplier.
Given: \(n=4\)
$$10^4 = 10,000$$
Answer: 10,000
Example 2: Find \(\log_{10} 1000\) representing decibel conversion.
Given: \(x=1000\)
$$\log_{10} 1000 = 3$$
Answer: 3
Example 3: Calculate \(10^{-3}\) representing micro-scale voltage.
Given: \(n=-3\)
$$10^{-3} = 0.001$$
Answer: 0.001
Common logarithms transform multiplication, division, and powers into simpler addition, subtraction, and multiplication operations, facilitating complex electrical calculations.
Example 1: Calculate \(\log_{10} 50\).
Given: \(x=50\)
Use logarithm tables or calculators:
$$\log_{10} 50 \approx 1.69897$$
Answer: 1.69897
Example 2: Find \(\log_{10} 0.02\).
Given: \(x=0.02\)
$$\log_{10} 0.02 = \log_{10} (2 \times 10^{-2}) = \log_{10} 2 + \log_{10} 10^{-2} = 0.30103 - 2 = -1.69897$$
Answer: -1.69897
Example 3: Calculate \(\log_{10} 1,000,000\).
Given: \(x=1,000,000\)
$$\log_{10} 1,000,000 = \log_{10} (10^6) = 6$$
Answer: 6
The characteristic is the integer part of a common logarithm representing the order of magnitude of the number.
Example 1: Find the characteristic of \(\log_{10} 450\).
Given: \(x=450\), \(\log_{10} 450 \approx 2.65321\)
Characteristic = 2
Answer: 2
Example 2: Find the characteristic of \(\log_{10} 0.0075\).
Given: \(x=0.0075\), \(\log_{10} 0.0075 \approx -2.12494\)
Characteristic = -3 (since the logarithm is negative, characteristic is one less than integer part)
Answer: -3
Example 3: Find the characteristic of \(\log_{10} 0.1\).
Given: \(x=0.1\), \(\log_{10} 0.1 = -1\)
Characteristic = -1
Answer: -1
The mantissa is the decimal part of a common logarithm representing the significant digits of the number.
Example 1: Find mantissa of \(\log_{10} 450\).
Given: \(\log_{10} 450 \approx 2.65321\)
Mantissa = 0.65321
Answer: 0.65321
Example 2: Find mantissa of \(\log_{10} 0.0075\).
Given: \(\log_{10} 0.0075 \approx -2.12494\)
Mantissa = 0.87506 (Mantissa = 1 - decimal part of logarithm when characteristic is negative)
Answer: 0.87506
Example 3: Find mantissa of \(\log_{10} 0.1\).
Given: \(\log_{10} 0.1 = -1.00000\)
Mantissa = 0.00000
Answer: 0.00000
Logarithm tables provide mantissa values for numbers, enabling manual calculation of logarithms before calculators, still useful in understanding logarithmic calculations.
Example 1: Use logarithm table to find \(\log_{10} 3.16\).
Given: \(x=3.16\)
From table, mantissa ≈ 0.49969
Characteristic = 0 (since 3.16 between 1 and 10)
$$\log_{10} 3.16 = 0 + 0.49969 = 0.49969$$
Answer: 0.49969
Example 2: Find \(\log_{10} 7.21\).
Given: \(x=7.21\)
Mantissa ≈ 0.85733
Characteristic = 0
$$\log_{10} 7.21 = 0 + 0.85733 = 0.85733$$
Answer: 0.85733
Example 3: Find \(\log_{10} 0.45\).
Given: \(x=0.45 = 4.5 \times 10^{-1}\)
From table, mantissa for 4.5 ≈ 0.65321
Characteristic = -1
$$\log_{10} 0.45 = -1 + 0.65321 = -0.34679$$
Answer: -0.34679
Common logarithms convert multiplication into addition, simplifying calculations in electrical engineering such as power gain.
Example 1: Calculate \(23 \times 45\) using logarithms.
Given: \(a=23, b=45\)
$$\log_{10} 23 = 1.36173$$
$$\log_{10} 45 = 1.65321$$
Add:
$$1.36173 + 1.65321 = 3.01494$$
Find antilog:
$$10^{3.01494} \approx 1033.8$$
Answer: 1033.8
Example 2: Multiply \(12.5 \times 7.8\).
Given:
$$\log_{10} 12.5 = 1.09691$$
$$\log_{10} 7.8 = 0.89209$$
Sum:
$$1.09691 + 0.89209 = 1.98900$$
Antilog:
$$10^{1.98900} \approx 97.5$$
Answer: 97.5
Example 3: Multiply \(0.35 \times 0.65\).
Given:
$$\log_{10} 0.35 = -0.45593$$
$$\log_{10} 0.65 = -0.18752$$
Sum:
$$-0.45593 + (-0.18752) = -0.64345$$
Antilog:
$$10^{-0.64345} \approx 0.227$$
Answer: 0.227
Division becomes subtraction of logarithms, useful in voltage and current ratio calculations.
Example 1: Calculate \(\frac{150}{25}\) using logarithms.
Given:
$$\log_{10} 150 = 2.17609$$
$$\log_{10} 25 = 1.39794$$
Subtract:
$$2.17609 - 1.39794 = 0.77815$$
Antilog:
$$10^{0.77815} \approx 6.0$$
Answer: 6.0
Example 2: Calculate \(\frac{0.75}{0.25}\).
Given:
$$\log_{10} 0.75 = -0.12494$$
$$\log_{10} 0.25 = -0.60206$$
Subtract:
$$-0.12494 - (-0.60206) = 0.47712$$
Antilog:
$$10^{0.47712} \approx 3.0$$
Answer: 3.0
Example 3: Find \(\frac{5000}{200}\).
Given:
$$\log_{10} 5000 = 3.69897$$
$$\log_{10} 200 = 2.30103$$
Subtract:
$$3.69897 - 2.30103 = 1.39794$$
Antilog:
$$10^{1.39794} \approx 25$$
Answer: 25
Roots can be found by dividing the logarithm of the number by the root index.
Formula:
$$\sqrt[n]{a} = 10^{\frac{\log_{10} a}{n}}$$
Example 1: Find \(\sqrt{2500}\).
Given: \(a=2500\), \(n=2\)
Calculate:
$$\log_{10} 2500 = 3.39794$$
Divide:
$$\frac{3.39794}{2} = 1.69897$$
Antilog:
$$10^{1.69897} = 50$$
Answer: 50
Example 2: Find \(\sqrt[3]{1000}\).
Given: \(a=1000\), \(n=3\)
Calculate:
$$\log_{10} 1000 = 3$$
Divide:
$$\frac{3}{3} = 1$$
Antilog:
$$10^{1} = 10$$
Answer: 10
Example 3: Calculate \(\sqrt[4]{81}\).
Given: \(a=81\), \(n=4\)
Calculate:
$$\log_{10} 81 = 1.90849$$
Divide:
$$\frac{1.90849}{4} = 0.47712$$
Antilog:
$$10^{0.47712} = 3$$
Answer: 3
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Create a free accountThis chapter covered fundamental algebraic concepts essential for engineering mathematics. It began with indices and logarithms, explaining the laws of indices for positive, negative, and fractional powers, followed by the properties of powers of 10 and the use of common logarithms including their characteristics and mantissa. The use of logarithm tables was introduced to simplify multiplication, division, and root extraction problems. The chapter then progressed to algebraic expressions, focusing on simplification, factorization by grouping, removal of brackets, and substitution for evaluation and problem solving. It further explored simultaneous equations and demonstrated methods to solve them by elimination, substitution, and graphing, including applications to real-life problems. Quadratic expressions and equations were examined through expansion, the three key quadratic identities, factorization techniques, and solving quadratic equations by factoring. Finally, the formation and solution of quadratic equations in practical contexts were discussed to consolidate the understanding of algebraic problem solving.
Calculate \(2^5 \times 2^3\) using the laws of indices. (2 marks)
Simplify \(\frac{5^4}{5^2}\) and give the answer as a power of 5. (2 marks)
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