Electronics engineering technicians frequently encounter calculations involving powers and logarithms when analyzing circuits, signal strengths, and electronic measurements. Mastery of indices and logarithms enables efficient manipulation of exponential relationships, essential in understanding amplifier gains, decibel scales, and filter responses. This chapter develops the algebraic tools needed to handle such expressions accurately and confidently in practical electronics engineering tasks.

2.1 Indices and logarithms

2.1.1 Indices (powers) and base

Indices express repeated multiplication of a base number. In electronics, they appear in formulas for capacitive reactance, signal attenuation, and transistor current gain calculations. 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 \(3^{4}\) representing the voltage gain raised to the fourth power.

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

$$ a^{n} = 3^{4} $$

$$ 3 \times 3 \times 3 \times 3 = 81 $$

Answer: 81

Example 2: Find the value of \(5^{0}\) representing a normalized gain.

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

$$ a^{n} = 5^{0} $$

By definition, any non-zero number raised to zero is 1.

Answer: 1

Example 3: Compute \(2^{-3}\) representing an attenuation factor.

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

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

$$ = \frac{1}{2 \times 2 \times 2} = \frac{1}{8} $$

Answer: \frac{1}{8}

Example 4: Evaluate \(9^{\frac{1}{2}}\) representing a root in frequency response.

Given: \(a = 9\), \(n = \frac{1}{2}\)

$$ a^{n} = 9^{\frac{1}{2}} = \sqrt{9} $$

$$ = 3 $$

Answer: 3

Example 5: Calculate \(4^{\frac{3}{2}}\) used in power calculations.

Given: \(a = 4\), \(n = \frac{3}{2}\)

$$ a^{n} = 4^{\frac{3}{2}} = (4^{\frac{1}{2}})^{3} = (\sqrt{4})^{3} = 2^{3} $$

$$ = 8 $$

Answer: 8

2.1.2 Laws of indices (positive integers)

The laws of indices simplify expressions with powers and are fundamental in electronics calculations involving exponential signals and power relationships. The main laws for positive integers are:

• Product law: \(a^{m} \times a^{n} = a^{m+n}\)
• Quotient law: \(\frac{a^{m}}{a^{n}} = a^{m-n}\)
• Power of a power: \((a^{m})^{n} = a^{m \times n}\)
• Power of a product: \((ab)^{n} = a^{n} b^{n}\)
• Power of a quotient: \((\frac{a}{b})^{n} = \frac{a^{n}}{b^{n}}\)

Worked Examples

Example 1: Simplify \(2^{3} \times 2^{4}\) representing combined signal gains.

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

$$ a^{m} \times a^{n} = 2^{3} \times 2^{4} = 2^{3+4} = 2^{7} $$

$$ 2^{7} = 128 $$

Answer: 128

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

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

$$ \frac{a^{m}}{a^{n}} = \frac{5^{6}}{5^{2}} = 5^{6-2} = 5^{4} $$

$$ 5^{4} = 625 $$

Answer: 625

Example 3: Simplify \((3^{2})^{4}\) representing power of power in signal strength.

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

$$ (a^{m})^{n} = 3^{2 \times 4} = 3^{8} $$

$$ 3^{8} = 6561 $$

Answer: 6561

Example 4: Simplify \((2 \times 5)^{3}\) representing combined effects.

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

$$ (ab)^{n} = 2^{3} \times 5^{3} = 8 \times 125 $$

$$ = 1000 $$

Answer: 1000

Example 5: Simplify \((\frac{6}{2})^{2}\) representing voltage ratio squared.

Given: \(a=6\), \(b=2\), \(n=2\)

$$ (\frac{a}{b})^{n} = \frac{6^{2}}{2^{2}} = \frac{36}{4} $$

$$ = 9 $$

Answer: 9

2.1.3 Laws of indices (negative integers)

Negative indices represent reciprocals, often used in electronics when dealing with attenuation or inverse relationships.

• Negative index law: \(a^{-n} = \frac{1}{a^{n}}\)
• Other laws apply as for positive indices, but with negative powers.

Worked Examples

Example 1: Simplify \(4^{-2}\) representing inverse square law in signal strength.

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

$$ a^{-n} = \frac{1}{4^{2}} = \frac{1}{16} $$

Answer: \frac{1}{16}

Example 2: Simplify \(2^{3} \times 2^{-5}\) representing gain and attenuation.

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

$$ a^{m} \times a^{n} = 2^{3} \times 2^{-5} = 2^{3-5} = 2^{-2} $$

$$ = \frac{1}{2^{2}} = \frac{1}{4} $$

Answer: \frac{1}{4}

Example 3: Simplify \((3^{-1})^{4}\) representing inverse gain raised to a power.

Given: \(a=3\), \(n=1\), power=4

$$ (a^{-n})^{m} = 3^{-1 \times 4} = 3^{-4} = \frac{1}{3^{4}} = \frac{1}{81} $$

Answer: \frac{1}{81}

Example 4: Simplify \(\frac{5^{-3}}{5^{-1}}\) representing ratio of inverse powers.

Given: \(a=5\), \(m=-3\), \(n=-1\)

$$ \frac{a^{m}}{a^{n}} = 5^{-3 - (-1)} = 5^{-3 + 1} = 5^{-2} = \frac{1}{25} $$

Answer: \frac{1}{25}

Example 5: Simplify \((\frac{2}{3})^{-2}\) representing inverse ratio squared.

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

$$ (\frac{a}{b})^{-n} = (\frac{3}{2})^{2} = \frac{3^{2}}{2^{2}} = \frac{9}{4} $$

Answer: \frac{9}{4}

2.1.4 Laws of indices (fractional indices)

Fractional indices represent roots, common in electronics for frequency and power calculations.

• Fractional index law: \(a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m}\)

Worked Examples

Example 1: Calculate \(16^{\frac{1}{2}}\) representing square root of power.

Given: \(a=16\), \(m=1\), \(n=2\)

$$ a^{\frac{1}{2}} = \sqrt{16} = 4 $$

Answer: 4

Example 2: Calculate \(27^{\frac{1}{3}}\) representing cube root of signal amplitude.

Given: \(a=27\), \(m=1\), \(n=3\)

$$ a^{\frac{1}{3}} = \sqrt[3]{27} = 3 $$

Answer: 3

Example 3: Calculate \(8^{\frac{2}{3}}\) representing square of cube root.

Given: \(a=8\), \(m=2\), \(n=3\)

$$ a^{\frac{2}{3}} = (\sqrt[3]{8})^{2} = 2^{2} = 4 $$

Answer: 4

Example 4: Calculate \(81^{\frac{3}{4}}\) representing cube of fourth root.

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

$$ a^{\frac{3}{4}} = (\sqrt[4]{81})^{3} = 3^{3} = 27 $$

Answer: 27

Example 5: Calculate \(32^{\frac{4}{5}}\) representing fourth power of fifth root.

Given: \(a=32\), \(m=4\), \(n=5\)

$$ a^{\frac{4}{5}} = (\sqrt[5]{32})^{4} = 2^{4} = 16 $$

Answer: 16

2.1.5 Powers of 10 and common logarithms

Powers of 10 are fundamental in electronics for expressing quantities in scientific notation and decibels. Common logarithms (base 10) help transform multiplicative relationships into additive ones.

The power of 10 is written as:

$$ 10^{n} $$

where \(n\) is an integer or fraction.

The common logarithm of a number \(x\) is:

$$ \log_{10} x = y \quad \text{if and only if} \quad 10^{y} = x $$

Worked Examples

Example 1: Calculate \(10^{3}\) representing voltage gain scaling.

Given: \(n = 3\)

$$ 10^{3} = 1000 $$

Answer: 1000

Example 2: Calculate \(\log_{10} 1000\) representing logarithmic gain.

Given: \(x = 1000\)

$$ \log_{10} 1000 = y $$

Since \(10^{3} = 1000\), \(y = 3\)

Answer: 3

Example 3: Calculate \(10^{-2}\) representing signal attenuation.

Given: \(n = -2\)

$$ 10^{-2} = \frac{1}{10^{2}} = \frac{1}{100} = 0.01 $$

Answer: 0.01

Example 4: Calculate \(\log_{10} 0.01\) representing logarithm of attenuation.

Given: \(x = 0.01\)

$$ 10^{y} = 0.01 = 10^{-2} $$

So,

$$ y = -2 $$

Answer: -2

Example 5: Calculate \(10^{0.5}\) representing square root of 10.

Given: \(n = 0.5\)

$$ 10^{0.5} = \sqrt{10} \approx 3.162 $$

Answer: 3.162

2.1.6 Common logarithms

Common logarithms convert multiplication and division into addition and subtraction, simplifying electronics calculations such as decibel computations and signal processing.

The main properties are:

• \(\log_{10} (xy) = \log_{10} x + \log_{10} y\)
• \(\log_{10} (\frac{x}{y}) = \log_{10} x - \log_{10} y\)
• \(\log_{10} (x^{n}) = n \log_{10} x\)

Worked Examples

Example 1: Calculate \(\log_{10} (100 \times 1000)\).

Given: \(x=100\), \(y=1000\)

$$ \log_{10} (xy) = \log_{10} 100 + \log_{10} 1000 $$

$$ = 2 + 3 = 5 $$

Answer: 5

Example 2: Calculate \(\log_{10} (\frac{1000}{10})\).

Given: \(x=1000\), \(y=10\)

$$ \log_{10} (\frac{x}{y}) = \log_{10} 1000 - \log_{10} 10 = 3 - 1 = 2 $$

Answer: 2

Example 3: Calculate \(\log_{10} (100^{3})\).

Given: \(x=100\), \(n=3\)

$$ \log_{10} (x^{n}) = n \log_{10} x = 3 \times 2 = 6 $$

Answer: 6

Example 4: Calculate \(\log_{10} (0.01 \times 100)\).

Given: \(x=0.01\), \(y=100\)

$$ \log_{10} (xy) = \log_{10} 0.01 + \log_{10} 100 = -2 + 2 = 0 $$

Answer: 0

Example 5: Calculate \(\log_{10} (\frac{0.1}{0.01})\).

Given: \(x=0.1\), \(y=0.01\)

$$ \log_{10} (\frac{x}{y}) = \log_{10} 0.1 - \log_{10} 0.01 = -1 - (-2) = 1 $$

Answer: 1

2.1.7 Characteristics

The characteristic of a logarithm is the integer part of the logarithm value and indicates the order of magnitude.

For a number \(x\):

$$ \log_{10} x = \text{Characteristic} + \text{Mantissa} $$

where the characteristic is the integer part.

Worked Examples

Example 1: Find the characteristic of \(\log_{10} 500\).

Given: \(x=500\)

Since \(10^{2} = 100\) and \(10^{3} = 1000\), \(\log_{10} 500\) is between 2 and 3.

Approximate \(\log_{10} 500 \approx 2.69897\)

Characteristic = 2

Answer: 2

Example 2: Find the characteristic of \(\log_{10} 0.05\).

Given: \(x=0.05\)

Since \(10^{-2} = 0.01\) and \(10^{-1} = 0.1\), \(\log_{10} 0.05\) is between -2 and -1.

Approximate \(\log_{10} 0.05 \approx -1.30103\)

Characteristic = -2 (note for numbers less than 1, characteristic is one less than integer part)

Answer: -2

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

Given: \(x=1\)

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

Characteristic = 0

Answer: 0

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

Given: \(x=7.5\)

Since \(10^{0} = 1\), \(10^{1} = 10\), \(\log_{10} 7.5 \approx 0.87506\)

Characteristic = 0

Answer: 0

Example 5: Find the characteristic of \(\log_{10} 0.003\).

Given: \(x=0.003\)

Since \(10^{-3} = 0.001\), \(10^{-2} = 0.01\), \(\log_{10} 0.003 \approx -2.52288\)

Characteristic = -3

Answer: -3

2.1.8 Mantissa

The mantissa is the fractional part of the logarithm and is always positive. It represents the decimal part after the characteristic.

If:

$$ \log_{10} x = \text{Characteristic} + \text{Mantissa} $$

then Mantissa \(= \log_{10} x - \text{Characteristic}\).

Worked Examples

Example 1: Find mantissa of \(\log_{10} 500\).

Given: \(\log_{10} 500 \approx 2.69897\), characteristic = 2

$$ \text{Mantissa} = 2.69897 - 2 = 0.69897 $$

Answer: 0.69897

Example 2: Find mantissa of \(\log_{10} 0.05\).

Given: \(\log_{10} 0.05 \approx -1.30103\), characteristic = -2

$$ \text{Mantissa} = -1.30103 - (-2) = 0.69897 $$

Answer: 0.69897

Example 3: Find mantissa of \(\log_{10} 7.5\).

Given: \(\log_{10} 7.5 \approx 0.87506\), characteristic = 0

$$ \text{Mantissa} = 0.87506 - 0 = 0.87506 $$

Answer: 0.87506

Example 4: Find mantissa of \(\log_{10} 1\).

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

$$ \text{Mantissa} = 0 - 0 = 0 $$

Answer: 0

Example 5: Find mantissa of \(\log_{10} 0.003\).

Given: \(\log_{10} 0.003 \approx -2.52288\), characteristic = -3

$$ \text{Mantissa} = -2.52288 - (-3) = 0.47712 $$

Answer: 0.47712

2.1.9 Logarithm tables

Before calculators, logarithm tables provided mantissa values for different numbers to simplify multiplication, division, and root extraction. Technicians can still benefit from understanding table use for quick approximations.

Steps to use a logarithm table:

1. Identify the number to find the log for.
2. Separate the characteristic and mantissa.
3. Use the table to find the mantissa value corresponding to the significant digits.
4. Combine characteristic and mantissa for the full logarithm.
5. Use anti-log tables for inverse operations.

Worked Examples

Example 1: Use a logarithm table to find \(\log_{10} 234\).

Given: Number = 234

Step 1: Characteristic = 2 (since 234 is between 100 and 1000)

Step 2: Find mantissa for 234 from table (approximate mantissa = 0.3692)

Step 3:

$$ \log_{10} 234 = 2 + 0.3692 = 2.3692 $$

Answer: 2.3692

Example 2: Use a logarithm table to find \(\log_{10} 0.056\).

Given: Number = 0.056

Step 1: Characteristic = -2 (since 0.056 is between 0.01 and 0.1)

Step 2: Mantissa for 560 (ignoring leading zeros) from table = 0.7482

Step 3:

$$ \log_{10} 0.056 = -2 + 0.7482 = -1.2518 $$

Note: For numbers less than 1, characteristic is one less than negative integer part.

Answer: -1.2518

Example 3: Use logarithm table to find \(\log_{10} 7.12\).

Given: Number = 7.12

Step 1: Characteristic = 0

Step 2: Mantissa for 712 from table = 0.8520

Step 3:

$$ \log_{10} 7.12 = 0 + 0.8520 = 0.8520 $$

Answer: 0.8520

Example 4: Use logarithm table to find \(\log_{10} 1000\).

Given: Number = 1000

Step 1: Characteristic = 3

Step 2: Mantissa for 1000 is 0.0000 (exact power of 10)

Step 3:

$$ \log_{10} 1000 = 3 + 0 = 3 $$

Answer: 3

Example 5: Use logarithm table to find \(\log_{10} 0.0091\).

Given: Number = 0.0091

Step 1: Characteristic = -3

Step 2: Mantissa for 910 from table = 0.9590

Step 3:

$$ \log_{10} 0.0091 = -3 + 0.9590 = -2.0410 $$

Answer: -2.0410

2.1.10 Application of common logarithms in multiplication, division and finding

Common logarithms simplify multiplication, division, and root extraction by converting these operations into addition, subtraction, and multiplication respectively.

• Multiplication: \(\log_{10} (xy) = \log_{10} x + \log_{10} y\)
• Division: \(\log_{10} (\frac{x}{y}) = \log_{10} x - \log_{10} y\)
• Roots: \(\log_{10} \sqrt[n]{x} = \frac{1}{n} \log_{10} x\)

Worked Examples

Example 1: Multiply \(25 \times 400\) using logarithms.

Given: \(x=25\), \(y=400\)

Step 1: Find \(\log_{10} 25 \approx 1.39794\)

Step 2: Find \(\log_{10} 400 \approx 2.60206\)

Step 3: Add logs

$$ 1.39794 + 2.60206 = 4.0000 $$

Step 4: Find antilog of 4.0000

$$ 10^{4.0000} = 10000 $$

Answer: 10000

Example 2: Divide \(5000 \div 20\) using logarithms.

Given: \(x=5000\), \(y=20\)

Step 1: \(\log_{10} 5000 \approx 3.69897\)

Step 2: \(\log_{10} 20 \approx 1.30103\)

Step 3: Subtract logs

$$ 3.69897 - 1.30103 = 2.39794 $$

Step 4: Antilog

$$ 10^{2.39794} \approx 250 $$

Answer: 250

Example 3: Find \(\sqrt{2500}\) using logarithms.

Given: \(x=2500\), \(n=2\)

Step 1: \(\log_{10} 2500 \approx 3.39794\)

Step 2: Divide log by 2

$$ \frac{3.39794}{2} = 1.69897 $$

Step 3: Antilog

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

Answer: 50

Example 4: Multiply \(0.03 \times 0.2\) using logarithms.

Given: \(x=0.03\), \(y=0.2\)

Step 1: \(\log_{10} 0.03 \approx -1.52288\)

Step 2: \(\log_{10} 0.2 \approx -0.69897\)

Step 3: Add logs

$$ -1.52288 + (-0.69897) = -2.22185 $$

Step 4: Antilog

$$ 10^{-2.22185} \approx 0.006 $$

Answer: 0.006

Example 5: Find cube root of \(125000\) using logarithms.

Given: \(x=125000\), \(n=3\)

Step 1: \(\log_{10} 125000 \approx 5.09691\)

Step 2: Divide log by 3

$$ \frac{5.09691}{3} = 1.69897 $$

Step 3: Antilog

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

Answer: 50

Chapter Summary

This chapter covered the fundamental concepts of indices and logarithms, starting with the definition of indices and their bases, followed by the laws governing positive, negative, and fractional indices. It explained powers of 10 and common logarithms, including the understanding of characteristics, mantissa, and the use of logarithm tables. The practical application of common logarithms was demonstrated in multiplication, division, and root extraction. The chapter then explored algebraic expressions, including algebraic fractions, and methods for simplification, factorization by grouping, removal of brackets, and substitution for evaluation, emphasizing problem solving in real-life contexts. It introduced simultaneous equations and detailed three methods of solution: elimination, substitution, and graphical methods, with examples from practical situations. Finally, quadratic expressions and equations were examined, covering expansion, the three quadratic identities, factorization of quadratic expressions, and solving quadratic equations by factorization, including the formation and solution of quadratic equations from given problems.