Electronics engineering relies heavily on mathematical tools to analyze, design, and troubleshoot complex circuits and systems. Mastery of algebraic operations, particularly indices and logarithms, is essential for understanding signal processing, amplifier gains, frequency responses, and decibel calculations. This chapter develops foundational skills in indices and logarithms, enabling electronics engineers to perform calculations critical in areas such as transistor biasing, filter design, and communication systems.

2.1 Indices and logarithms

Indices and logarithms simplify calculations involving exponential relationships common in electronics. Understanding their properties and applications allows engineers to manipulate equations involving signal attenuation, power ratios, and frequency scaling efficiently.

2.1.1 Indices (powers) and base

Indices describe repeated multiplication of a base number. In electronics, they represent quantities like gain expressed as powers of 10, or signal decay modeled exponentially.

The general form of an index expression is

$$a^n$$

where \(a\) is the base and \(n\) is the index or power.

Worked Examples

Example 1: Calculate \(2^3\) representing a voltage gain factor.

Given: base \(a=2\), power \(n=3\)

$$a^n = 2^3$$

$$= 2 \times 2 \times 2$$

$$= 8$$

Answer: 8

Example 2: Find the value of \(5^0\) representing normalized signal level.

Given: \(a=5\), \(n=0\)

$$a^n = 5^0$$

Any number raised to zero equals 1:

$$= 1$$

Answer: 1

Example 3: Determine \(10^4\), which might represent frequency scaling in Hz.

Given: \(a=10\), \(n=4\)

$$a^n = 10^4$$

$$= 10 \times 10 \times 10 \times 10$$

$$= 10000$$

Answer: 10000

Example 4: Calculate \(3^{-2}\), used in attenuation factors.

Given: \(a=3\), \(n=-2\)

$$a^n = 3^{-2} = \frac{1}{3^2}$$

$$= \frac{1}{9}$$

Answer: \frac{1}{9}

Example 5: Evaluate \(4^{\frac{1}{2}}\), representing square root of power.

Given: \(a=4\), \(n=\frac{1}{2}\)

$$a^n = 4^{\frac{1}{2}} = \sqrt{4}$$

$$= 2$$

Answer: 2

2.1.2 Laws of indices (including positive integers)

The laws of indices govern how powers combine in multiplication, division, and exponentiation. These laws simplify complex algebraic expressions in electronics calculations.

Key laws for positive integers:

• Product of powers with same base: \(a^m \times a^n = a^{m+n}\)
• Quotient of powers with same base: \(\frac{a^m}{a^n} = a^{m-n}\) (for \(m > n\))
• Power of a power: \((a^m)^n = a^{mn}\)
• Power of a product: \((ab)^n = a^n b^n\)

Worked Examples

Example 1: Simplify \(2^3 \times 2^4\), representing combined amplifier gains.

Given: \(a=2\), \(m=3\), \(n=4\)

$$a^m \times a^n = 2^3 \times 2^4 = 2^{3+4}$$

$$= 2^7$$

$$= 128$$

Answer: 128

Example 2: Simplify \(\frac{5^6}{5^2}\), representing signal power ratio.

Given: \(a=5\), \(m=6\), \(n=2\)

$$\frac{a^m}{a^n} = \frac{5^6}{5^2} = 5^{6-2}$$

$$= 5^4$$

$$= 625$$

Answer: 625

Example 3: Simplify \((3^2)^4\), used in calculating energy stored in capacitors.

Given: \(a=3\), \(m=2\), \(n=4\)

$$(a^m)^n = (3^2)^4 = 3^{2 \times 4}$$

$$= 3^8$$

$$= 6561$$

Answer: 6561

Example 4: Simplify \((2 \times 5)^3\), representing combined signal components.

Given: \(a=2\), \(b=5\), \(n=3\)

$$(ab)^n = (2 \times 5)^3 = 2^3 \times 5^3$$

$$= 8 \times 125$$

$$= 1000$$

Answer: 1000

Example 5: Calculate \(10^2 \times 10^5\), for frequency multiplication.

Given: \(a=10\), \(m=2\), \(n=5\)

$$10^2 \times 10^5 = 10^{2+5} = 10^7$$

$$= 10,000,000$$

Answer: 10,000,000

2.1.3 Laws of indices (including negative integers)

Negative indices represent reciprocals and are essential in electronics to express attenuation or inverse proportionality.

Key laws:

• \(a^{-n} = \frac{1}{a^n}\)
• \(a^m \times a^{-n} = a^{m-n}\)
• \(\frac{a^{-m}}{a^{-n}} = a^{n-m}\)

Worked Examples

Example 1: Simplify \(4^{-3}\), representing signal attenuation.

Given: \(a=4\), \(n=3\)

$$4^{-3} = \frac{1}{4^3}$$

$$= \frac{1}{64}$$

Answer: \frac{1}{64}

Example 2: Simplify \(5^4 \times 5^{-2}\), calculating net gain.

Given: \(a=5\), \(m=4\), \(n=2\)

$$5^4 \times 5^{-2} = 5^{4-2} = 5^2$$

$$= 25$$

Answer: 25

Example 3: Simplify \(\frac{2^{-3}}{2^{-5}}\), for ratio of inverse powers.

Given: \(a=2\), \(m=3\), \(n=5\)

$$\frac{2^{-3}}{2^{-5}} = 2^{-3 - (-5)} = 2^{2}$$

$$= 4$$

Answer: 4

Example 4: Calculate \((\frac{1}{3})^{-2}\), representing squared reciprocal.

Given: \(a=\frac{1}{3}\), \(n=-2\)

$$(\frac{1}{3})^{-2} = 3^{2}$$

$$= 9$$

Answer: 9

Example 5: Simplify \(\frac{10^{-4}}{10^{-6}}\), for frequency scaling.

Given: \(a=10\), \(m=4\), \(n=6\)

$$\frac{10^{-4}}{10^{-6}} = 10^{-4 - (-6)} = 10^{2}$$

$$= 100$$

Answer: 100

2.1.4 Laws of indices (including fractional indices)

Fractional indices represent roots and are used in electronics to calculate quantities like RMS voltage or frequency roots.

Key laws:

• \(a^{\frac{m}{n}} = \sqrt[n]{a^m}\)
• \(a^{\frac{1}{n}} = \sqrt[n]{a}\)
• Multiplication and division laws apply as for integers.

Worked Examples

Example 1: Calculate \(16^{\frac{1}{2}}\), representing RMS voltage.

Given: \(a=16\), \(n=2\)

$$16^{\frac{1}{2}} = \sqrt{16}$$

$$= 4$$

Answer: 4

Example 2: Calculate \(27^{\frac{2}{3}}\), used in power calculations.

Given: \(a=27\), \(m=2\), \(n=3\)

$$27^{\frac{2}{3}} = (\sqrt[3]{27})^2$$

$$= (3)^2$$

$$= 9$$

Answer: 9

Example 3: Simplify \(8^{\frac{4}{3}}\), representing frequency scaling.

Given: \(a=8\), \(m=4\), \(n=3\)

$$8^{\frac{4}{3}} = (\sqrt[3]{8})^4$$

$$= (2)^4$$

$$= 16$$

Answer: 16

Example 4: Calculate \((81^{\frac{3}{4}})\), used in signal power.

Given: \(a=81\), \(m=3\), \(n=4\)

$$81^{\frac{3}{4}} = (\sqrt[4]{81})^3$$

$$= (3)^3$$

$$= 27$$

Answer: 27

Example 5: Find \(100^{\frac{1}{2}}\), representing square root of power.

Given: \(a=100\), \(n=2\)

$$100^{\frac{1}{2}} = \sqrt{100}$$

$$= 10$$

Answer: 10

2.1.5 Powers of 10 and common logarithms

Powers of 10 simplify expressing very large or small values common in electronics, such as frequencies and signal powers. Common logarithms use base 10 and convert multiplication/division into addition/subtraction.

The common logarithm of \(x\) is

$$\log_{10} x$$

and satisfies

$$10^{\log_{10} x} = x$$

Worked Examples

Example 1: Calculate \(\log_{10} 1000\), representing gain in decibels.

Given: \(x=1000\)

$$\log_{10} 1000 = \log_{10} (10^3) = 3$$

Answer: 3

Example 2: Calculate \(\log_{10} 0.01\), representing attenuation.

Given: \(x=0.01\)

$$\log_{10} 0.01 = \log_{10} (10^{-2}) = -2$$

Answer: -2

Example 3: Find \(10^{4}\), representing frequency in Hz.

Given: exponent 4

$$10^{4} = 10000$$

Answer: 10000

Example 4: Calculate \(\log_{10} 50\) approximately.

Given: \(x=50\)

Using calculator or table:

$$\log_{10} 50 \approx 1.69897$$

Answer: 1.69897

Example 5: Calculate \(10^{-3}\), representing millivolt scale.

Given: exponent \(-3\)

$$10^{-3} = 0.001$$

Answer: 0.001

2.1.6 Common logarithms

Common logarithms convert multiplication and division of large numbers into simpler addition and subtraction, easing calculations in electronics such as signal gain or attenuation.

The logarithm of a number \(x\) is the exponent to which 10 must be raised:

$$\log_{10} x = y \iff 10^y = x$$

Worked Examples

Example 1: Calculate \(\log_{10} 200\).

Given: \(x=200\)

Using calculator:

$$\log_{10} 200 \approx 2.30103$$

Answer: 2.30103

Example 2: Calculate \(\log_{10} 0.005\).

Given: \(x=0.005\)

$$\log_{10} 0.005 = \log_{10} (5 \times 10^{-3}) = \log_{10} 5 + \log_{10} 10^{-3}$$

$$= 0.69897 - 3 = -2.30103$$

Answer: -2.30103

Example 3: Calculate \(\log_{10} 1\).

Given: \(x=1\)

$$\log_{10} 1 = 0$$

Answer: 0

Example 4: Calculate \(\log_{10} 100000\).

Given: \(x=10^5\)

$$\log_{10} 10^5 = 5$$

Answer: 5

Example 5: Calculate \(\log_{10} 7.5\).

Given: \(x=7.5\)

Using calculator:

$$\log_{10} 7.5 \approx 0.87506$$

Answer: 0.87506

2.1.7 Characteristics of common logarithms

The characteristic of a common logarithm is the integer part of the logarithm, representing the order of magnitude of the number.

If \(\log_{10} x = c + m\), where \(c\) is the characteristic and \(m\) the mantissa.

For \(x > 1\), \(c\) is positive; for \(0 < x < 1\), \(c\) is negative.

Worked Examples

Example 1: Find the characteristic of \(\log_{10} 150\).

Given: \(\log_{10} 150 \approx 2.176\)

Characteristic \(c = 2\)

Answer: 2

Example 2: Find the characteristic of \(\log_{10} 0.025\).

Given: \(\log_{10} 0.025 \approx -1.602\)

Characteristic \(c = -2\) (since \(x < 1\), characteristic is one less than the integer part)

Answer: -2

Example 3: Find the characteristic of \(\log_{10} 1\).

Given: \(\log_{10} 1 = 0\)

Characteristic \(c = 0\)

Answer: 0

Example 4: Find the characteristic of \(\log_{10} 999\).

Given: \(\log_{10} 999 \approx 2.9996\)

Characteristic \(c = 2\)

Answer: 2

Example 5: Find the characteristic of \(\log_{10} 0.5\).

Given: \(\log_{10} 0.5 \approx -0.3010\)

Characteristic \(c = -1\)

Answer: -1

2.1.8 Mantissa of common logarithms

The mantissa is the decimal part of the logarithm and is always positive. It determines the precise value within the order of magnitude.

If \(\log_{10} x = c + m\), mantissa \(m\) is \(0 \leq m < 1\).

Worked Examples

Example 1: Find mantissa of \(\log_{10} 150\).

Given: \(\log_{10} 150 \approx 2.176\)

Mantissa \(m = 0.176\)

Answer: 0.176

Example 2: Find mantissa of \(\log_{10} 0.025\).

Given: \(\log_{10} 0.025 \approx -1.602\)

Mantissa \(m = 0.398\) (since characteristic is -2, mantissa = 1 - 0.602)

Answer: 0.398

Example 3: Find mantissa of \(\log_{10} 999\).

Given: \(\log_{10} 999 \approx 2.9996\)

Mantissa \(m = 0.9996\)

Answer: 0.9996

Example 4: Find mantissa of \(\log_{10} 7.5\).

Given: \(\log_{10} 7.5 \approx 0.87506\)

Mantissa \(m = 0.87506\)

Answer: 0.87506

Example 5: Find mantissa of \(\log_{10} 0.5\).

Given: \(\log_{10} 0.5 \approx -0.3010\)

Mantissa \(m = 0.6990\)

Answer: 0.6990

2.1.9 Logarithm tables

Before calculators, logarithm tables were essential for electronics engineers to perform fast multiplications and divisions involving large numbers.

Tables list mantissas for numbers, allowing lookup of logarithms and antilogarithms.

Worked Examples

Example 1: Use logarithm tables to multiply 25 by 40.

Step 1: Find \(\log_{10} 25\), approximate mantissa = 0.39794

Step 2: Find \(\log_{10} 40\), mantissa = 0.60206

Step 3: Add mantissas: 0.39794 + 0.60206 = 1.00000

Step 4: Characteristic adjustment: 1.00000 corresponds to \(10^1 = 10\)

Result: \(25 \times 40 = 10^{1.00000} = 10^{1} = 10 \times 10^0 = 1000\)

Answer: 1000

Example 2: Use tables to divide 500 by 25.

Step 1: \(\log_{10} 500\) mantissa ~0.69897, characteristic 2

Step 2: \(\log_{10} 25\) mantissa ~0.39794, characteristic 1

Step 3: Subtract mantissas: 0.69897 - 0.39794 = 0.30103

Step 4: Subtract characteristics: 2 - 1 = 1

Step 5: Result \(\log_{10} = 1 + 0.30103 = 1.30103\)

Step 6: Antilog of 0.30103 is 2

Result: \(500 / 25 = 10^{1.30103} = 10^1 \times 2 = 20\)

Answer: 20

Example 3: Use tables to find \(\sqrt{1000}\).

Step 1: \(\log_{10} 1000 = 3\)

Step 2: Divide by 2 for square root: \(\frac{3}{2} = 1.5\)

Step 3: Antilog of 1.5 is \(10^{1.5} = 31.62\)

Answer: \sqrt{1000} = 31.62

2.1.10 Application of common logarithms in multiplication

Logarithms convert multiplication into addition, simplifying electronics calculations of power and voltage gains.

Given \(A \times B\),

$$\log_{10} (A \times B) = \log_{10} A + \log_{10} B$$

Worked Examples

Example 1: Calculate \(200 \times 300\) using logarithms.

Given: \(A=200\), \(B=300\)

Step 1: \(\log_{10} 200 \approx 2.30103\)

Step 2: \(\log_{10} 300 \approx 2.47712\)

Step 3: Sum: \(2.30103 + 2.47712 = 4.77815\)

Step 4: Antilog: \(10^{4.77815} \approx 60000\)

Answer: 60000

Example 2: Calculate \(0.05 \times 0.2\).

Given: \(A=0.05\), \(B=0.2\)

Step 1: \(\log_{10} 0.05 = \log_{10} (5 \times 10^{-2}) = 0.69897 - 2 = -1.30103\)

Step 2: \(\log_{10} 0.2 = \log_{10} (2 \times 10^{-1}) = 0.30103 - 1 = -0.69897\)

Step 3: Sum: \(-1.30103 + (-0.69897) = -2.0000\)

Step 4: Antilog: \(10^{-2} = 0.01\)

Answer: 0.01

Example 3: Calculate \(1500 \times 0.4\).

Given: \(A=1500\), \(B=0.4\)

Step 1: \(\log_{10} 1500 = \log_{10} (1.5 \times 10^{3}) = 0.17609 + 3 = 3.17609\)

Step 2: \(\log_{10} 0.4 = \log_{10} (4 \times 10^{-1}) = 0.60206 - 1 = -0.39794\)

Step 3: Sum: \(3.17609 + (-0.39794) = 2.77815\)

Step 4: Antilog: \(10^{2.77815} \approx 600\)

Answer: 600

2.1.11 Application of common logarithms in division

Division becomes subtraction of logarithms, useful for calculating ratios like voltage division or power loss.

Given \(\frac{A}{B}\),

$$\log_{10} (\frac{A}{B}) = \log_{10} A - \log_{10} B$$

Worked Examples

Example 1: Calculate \(\frac{500}{25}\).

Given: \(A=500\), \(B=25\)

Step 1: \(\log_{10} 500 = 2.69897\)

Step 2: \(\log_{10} 25 = 1.39794\)

Step 3: Subtract: \(2.69897 - 1.39794 = 1.30103\)

Step 4: Antilog: \(10^{1.30103} = 20\)

Answer: 20

Example 2: Calculate \(\frac{0.02}{0.005}\).

Given: \(A=0.02\), \(B=0.005\)

Step 1: \(\log_{10} 0.02 = \log_{10} (2 \times 10^{-2}) = 0.30103 - 2 = -1.69897\)

Step 2: \(\log_{10} 0.005 = \log_{10} (5 \times 10^{-3}) = 0.69897 - 3 = -2.30103\)

Step 3: Subtract: \(-1.69897 - (-2.30103) = 0.60206\)

Step 4: Antilog: \(10^{0.60206} = 4\)

Answer: 4

Example 3: Calculate \(\frac{1000}{400}\).

Given: \(A=1000\), \(B=400\)

Step 1: \(\log_{10} 1000 = 3\)

Step 2: \(\log_{10} 400 = \log_{10} (4 \times 10^{2}) = 0.60206 + 2 = 2.60206\)

Step 3: Subtract: \(3 - 2.60206 = 0.39794\)

Step 4: Antilog: \(10^{0.39794} = 2.5\)

Answer: 2.5

2.1.12 Application of common logarithms in finding roots

Roots are found by dividing the logarithm by the root index, simplifying calculations of square roots or cube roots in electronics.

For nth root:

$$\sqrt[n]{A} = A^{\frac{1}{n}}$$

and

$$\log_{10} \sqrt[n]{A} = \frac{1}{n} \log_{10} A$$

Worked Examples

Example 1: Find \(\sqrt{400}\).

Given: \(A=400\), \(n=2\)

Step 1: \(\log_{10} 400 = \log_{10} (4 \times 10^{2}) = 0.60206 + 2 = 2.60206\)

Step 2: Divide by 2: \(\frac{2.60206}{2} = 1.30103\)

Step 3: Antilog: \(10^{1.30103} = 20\)

Answer: 20

Example 2: Find \(\sqrt[3]{1000}\).

Given: \(A=1000\), \(n=3\)

Step 1: \(\log_{10} 1000 = 3\)

Step 2: Divide by 3: \(\frac{3}{3} = 1\)

Step 3: Antilog: \(10^{1} = 10\)

Answer: 10

Example 3: Find \(\sqrt{0.01}\).

Given: \(A=0.01\), \(n=2\)

Step 1: \(\log_{10} 0.01 = \log_{10} (10^{-2}) = -2\)

Step 2: Divide by 2: \(\frac{-2}{2} = -1\)

Step 3: Antilog: \(10^{-1} = 0.1\)

Answer: 0.1

Example 4: Find \(\sqrt[4]{81}\).

Given: \(A=81\), \(n=4\)

Step 1: \(\log_{10} 81 = \log_{10} (8.1 \times 10^{1}) = 0.90849 + 1 = 1.90849\)

Step 2: Divide by 4: \(\frac{1.90849}{4} = 0.47712\)

Step 3: Antilog: \(10^{0.47712} \approx 3\)

Answer: 3

Example 5: Find \(\sqrt[5]{100000}\).

Given: \(A=100000\), \(n=5\)

Step 1: \(\log_{10} 100000 = 5\)

Step 2: Divide by 5: \(\frac{5}{5} = 1\)

Step 3: Antilog: \(10^{1} = 10\)

Answer: 10

Chapter Summary

This chapter covered the fundamental concepts of indices and logarithms, starting with the definitions of indices and their bases, followed by the laws governing positive, negative, and fractional indices. It explained powers of 10 and common logarithms, detailing the characteristics and mantissa, and demonstrated how logarithm tables are used in calculations such as multiplication, division, and root extraction. The section on algebra introduced algebraic expressions including fractions, focusing on simplification, factorization by grouping, removal of brackets, and substitution for evaluation, with applications to real-life problem solving. The chapter then explored simultaneous equations, presenting methods for solving them through elimination, substitution, and graphical techniques, along with practical problem-solving examples. Finally, quadratic expressions and equations were examined, including expansion, the three key quadratic identities, their use in factorization, solving quadratic equations by factoring, and the formation and solution of quadratic equations from given conditions. This comprehensive treatment equips students with essential algebraic tools for engineering mathematics.