By the end of this chapter, you will be able to:
Mastering these algebra skills will help you solve real-world problems efficiently and confidently in your trade.
Welding professionals in Kenya rely heavily on algebraic principles to solve practical problems related to material properties, measurements, and welding parameters. Mastery of algebraic operations such as indices enables accurate calculations essential for determining quantities like heat input, stress factors, and welding consumables. This chapter focuses on applying indices algebra, which is foundational for manipulating expressions involving powers and roots frequently encountered in welding engineering calculations.
Indices are powers or exponents, and working with them simplifies repeated multiplication or division. In a welding workshop, powers appear in area, volume and scaling formulas, for example when finding the size of a square plate or a cubic tank. Understanding zero, negative and fractional powers, and the laws governing indices, is crucial for accurate and efficient calculations.
Any non-zero number raised to the power zero equals one. This follows from the pattern of dividing powers: 5³ ÷ 5³ = 1, and subtracting the indices gives 5^(3−3) = 5⁰, so 5⁰ must equal 1. The rule simplifies expressions where a variable becomes raised to zero during manipulation of a formula. The expression 0⁰ is not defined, so the base must not be zero.
The governing formula is:
a^0 = 1, where a ≠ 0
Example 1: Calculate 5^0.
Given: base a = 5
Formula: a^0 = 1
Substitution: 5^0 = 1
Answer: 1
Example 2: Determine (12.5)^0.
Given: a = 12.5
Formula: a^0 = 1
Substitution: (12.5)^0 = 1
Answer: 1
Example 3: Evaluate x^0, where x is a welding current in amperes and x ≠ 0.
Given: x ≠ 0
Formula: x^0 = 1
Answer: 1
Example 4: Find the value of (0.75)^0.
Given: a = 0.75
Formula: a^0 = 1
Substitution: (0.75)^0 = 1
Answer: 1
Example 5: Calculate (100)^0.
Given: a = 100
Formula: a^0 = 1
Substitution: (100)^0 = 1
Answer: 1
Negative powers represent the reciprocal of the base raised to the corresponding positive power. The negative sign in the index does not make the answer negative: it moves the base to the other side of the fraction line. This is important in calculations involving inverse proportionality, such as intensity falling with the square of distance, and in converting small units, for example 10^(−3) m = 1 mm.
The governing formula is:
a^(−n) = 1 ÷ a^n, where a ≠ 0 and n > 0
Example 1: Calculate 4^(−2).
Given: a = 4, n = 2
Formula: a^(−n) = 1 ÷ a^n
Substitution: 4^(−2) = 1 ÷ 4^2
Working: 1 ÷ 16
Answer: 1/16 = 0.0625
Example 2: Evaluate 10^(−3) m, the size in metres of a 1 mm root gap.
Given: a = 10, n = 3
Formula: a^(−n) = 1 ÷ a^n
Substitution: 10^(−3) = 1 ÷ 10^3
Working: 1 ÷ 1000
Answer: 0.001 m
Example 3: Find 2.5^(−1).
Given: a = 2.5, n = 1
Formula: a^(−n) = 1 ÷ a^n
Substitution: 2.5^(−1) = 1 ÷ 2.5
Answer: 0.4
Example 4: Determine 7^(−3).
Given: a = 7, n = 3
Formula: a^(−n) = 1 ÷ a^n
Substitution: 7^(−3) = 1 ÷ 7^3
Working: 1 ÷ 343
Answer: 0.002915 (to 4 significant figures)
Example 5: Calculate (0.5)^(−2).
Given: a = 0.5, n = 2
Formula: a^(−n) = 1 ÷ a^n
Substitution: 0.5^(−2) = 1 ÷ (0.5)^2
Working: 1 ÷ 0.25
Answer: 4
Fractional powers correspond to roots of numbers or variables. The power 1/2 represents the square root, 1/3 the cube root, and in general the denominator of the fraction gives the root while the numerator gives the power. For example, a fabricator who knows the volume of a cubic tank can find its edge length using the power 1/3. Fractional powers also appear in scaling formulas.
The governing formula is:
a^(m/n) = (n-th root of a^m) = (n-th root of a)^m, where a > 0
Example 1: Calculate 16^(1/2), the side in metres of a square steel plate of area 16 m².
Given: a = 16, m = 1, n = 2
Formula: a^(1/2) = square root of a
Substitution: 16^(1/2) = √16
Answer: 4 m
Example 2: Evaluate 27^(1/3), the edge in metres of a cubic water tank of volume 27 m³.
Given: a = 27, m = 1, n = 3
Formula: a^(1/3) = cube root of a
Substitution: 27^(1/3) = ∛27
Answer: 3 m
Example 3: Find 81^(3/4).
Given: a = 81, m = 3, n = 4
Formula: a^(m/n) = (n-th root of a)^m
Substitution: 81^(3/4) = (4th root of 81)^3
Working: 4th root of 81 = 3, since 3^4 = 81
Working: 3^3 = 27
Answer: 27
Example 4: Calculate 125^(2/3).
Given: a = 125, m = 2, n = 3
Formula: a^(m/n) = (n-th root of a)^m
Substitution: 125^(2/3) = (∛125)^2
Working: ∛125 = 5
Working: 5^2 = 25
Answer: 25
Example 5: Evaluate 64^(5/6).
Given: a = 64, m = 5, n = 6
Formula: a^(m/n) = (n-th root of a)^m
Substitution: 64^(5/6) = (6th root of 64)^5
Working: 6th root of 64 = 2, since 2^6 = 64
Working: 2^5 = 32
Answer: 32
Many practical expressions contain a negative and a fractional index together. The method is to deal with the negative sign first by taking the reciprocal, then evaluate the root, and finally apply the power. The order of the root and the power does not change the answer, but taking the root first keeps the numbers small.
Example 1: Evaluate 8^(−2/3).
Given: a = 8, m = 2, n = 3, with a negative index
Formula: a^(−m/n) = 1 ÷ (n-th root of a)^m
Substitution: 8^(−2/3) = 1 ÷ (∛8)^2
Working: ∛8 = 2, so (∛8)^2 = 4
Working: 1 ÷ 4
Answer: 1/4 = 0.25
Example 2: Evaluate 16^(−3/4).
Given: a = 16, m = 3, n = 4, with a negative index
Formula: a^(−m/n) = 1 ÷ (n-th root of a)^m
Substitution: 16^(−3/4) = 1 ÷ (4th root of 16)^3
Working: 4th root of 16 = 2, since 2^4 = 16, so 2^3 = 8
Working: 1 ÷ 8
Answer: 1/8 = 0.125
The laws of indices allow simplification and manipulation of expressions involving powers. They apply only when the bases are the same (for adding and subtracting indices) or when a power is applied to a product, a quotient or another power. Calculations for plate areas, tank volumes and scaling often require combining or breaking down powers in this way. Each law is treated below under the operation it performs on the indices.
When two powers with the same base are multiplied, the indices are added. This works because a^m means m copies of a multiplied together, and joining n more copies gives m + n copies altogether. For example, 2³ × 2⁴ = (2×2×2) × (2×2×2×2) = 2⁷. Numerical coefficients in front of the variables are multiplied normally.
The governing formula is:
a^m × a^n = a^(m+n)
Worked examples:
Example 1: Simplify 2^3 × 2^4.
Given: a = 2, m = 3, n = 4
Formula: a^m × a^n = a^(m+n)
Substitution: 2^3 × 2^4 = 2^(3+4)
Working: 2^7
Answer: 128
Example 2: Simplify x^5 × x^2.
Given: base x, m = 5, n = 2
Formula: a^m × a^n = a^(m+n)
Substitution: x^5 × x^2 = x^(5+2)
Answer: x^7
Example 3: Evaluate 3^(−2) × 3^5, where one index is negative.
Given: a = 3, m = −2, n = 5
Formula: a^m × a^n = a^(m+n)
Substitution: 3^(−2) × 3^5 = 3^(−2+5)
Working: 3^3
Answer: 27
Example 4: A tank holds 10^2 litres and a depot keeps 10^3 such tanks. Find the total volume.
Given: a = 10, m = 2, n = 3
Formula: a^m × a^n = a^(m+n)
Substitution: 10^2 × 10^3 = 10^(2+3)
Working: 10^5
Answer: 100 000 litres
Example 5: Simplify 2a^3 × 4a^2.
Given: coefficients 2 and 4, base a, m = 3, n = 2
Working: multiply the coefficients: 2 × 4 = 8
Working: add the indices: a^(3+2) = a^5
Answer: 8a^5
When two powers with the same base are divided, the index of the divisor is subtracted from the index of the dividend. The result may be zero or negative, in which case the zero-power and negative-power rules complete the simplification. For example, 5⁶ ÷ 5² = 5⁴, while 2³ ÷ 2⁷ = 2^(−4) = 1/16. Coefficients are divided in the usual way.
The governing formula is:
a^m ÷ a^n = a^(m−n)
Worked examples:
Example 1: Evaluate 5^6 ÷ 5^2.
Given: a = 5, m = 6, n = 2
Formula: a^m ÷ a^n = a^(m−n)
Substitution: 5^6 ÷ 5^2 = 5^(6−2)
Working: 5^4
Answer: 625
Example 2: Simplify y^9 ÷ y^4.
Given: base y, m = 9, n = 4
Formula: a^m ÷ a^n = a^(m−n)
Substitution: y^9 ÷ y^4 = y^(9−4)
Answer: y^5
Example 3: Evaluate 2^3 ÷ 2^7, where the divisor has the larger index.
Given: a = 2, m = 3, n = 7
Formula: a^m ÷ a^n = a^(m−n)
Substitution: 2^3 ÷ 2^7 = 2^(3−7)
Working: 2^(−4) = 1 ÷ 2^4
Answer: 1/16 = 0.0625
Example 4: Evaluate 7^4 ÷ 7^(−1), where the divisor has a negative index.
Given: a = 7, m = 4, n = −1
Formula: a^m ÷ a^n = a^(m−n)
Substitution: 7^4 ÷ 7^(−1) = 7^(4−(−1))
Working: 7^5
Answer: 16 807
Example 5: Simplify 12x^6 ÷ 3x^2.
Given: coefficients 12 and 3, base x, m = 6, n = 2
Working: divide the coefficients: 12 ÷ 3 = 4
Working: subtract the indices: x^(6−2) = x^4
Answer: 4x^4
When a fraction is raised to a power, both the numerator and the denominator are raised to that power. This is the power of a quotient. A negative index on a fraction turns the fraction upside down before the power is applied. For example, (6/2)⁴ can be worked as 6⁴ ÷ 2⁴ = 81, or more quickly as 3⁴ = 81.
The governing formulas are:
(a ÷ b)^n = a^n ÷ b^n, where b ≠ 0
(a ÷ b)^(−n) = (b ÷ a)^n
Worked examples:
Example 1: Evaluate (6 ÷ 2)^4.
Given: a = 6, b = 2, n = 4
Formula: (a ÷ b)^n = a^n ÷ b^n
Substitution: (6 ÷ 2)^4 = 6^4 ÷ 2^4
Working: 1296 ÷ 16
Answer: 81
Example 2: Evaluate (2 ÷ 3)^3.
Given: a = 2, b = 3, n = 3
Formula: (a ÷ b)^n = a^n ÷ b^n
Substitution: (2 ÷ 3)^3 = 2^3 ÷ 3^3
Answer: 8/27
Example 3: Expand (3x ÷ y)^2.
Given: numerator 3x, denominator y, n = 2
Formula: (a ÷ b)^n = a^n ÷ b^n
Substitution: (3x)^2 ÷ y^2
Working: 9x^2 ÷ y^2
Answer: 9x² / y²
Example 4: Evaluate (3 ÷ 4)^(−2), a negative index on a fraction.
Given: a = 3, b = 4, n = 2
Formula: (a ÷ b)^(−n) = (b ÷ a)^n
Substitution: (3 ÷ 4)^(−2) = (4 ÷ 3)^2
Working: 16 ÷ 9
Answer: 16/9 = 1.778 (to 4 significant figures)
Example 5: A fabricator compares two similar round tanks whose diameters are in the ratio 10 ÷ 5. Their volumes are in the ratio of the cube of the diameter ratio. Evaluate (10 ÷ 5)^3.
Given: a = 10, b = 5, n = 3
Formula: (a ÷ b)^n = a^n ÷ b^n
Substitution: (10 ÷ 5)^3 = 10^3 ÷ 5^3
Working: 1000 ÷ 125
Answer: 8
When a power is raised to another power, the indices are multiplied. When a product is raised to a power, the power is applied to each factor and the indices of each factor are multiplied by that power. For example, (3²)⁴ means four groups of 3², so 3^(2×4) = 3⁸. These two laws are used to remove brackets from expressions.
The governing formulas are:
(a^m)^n = a^(m×n)
(a × b)^n = a^n × b^n
Worked examples:
Example 1: Calculate (3^2)^4.
Given: a = 3, m = 2, n = 4
Formula: (a^m)^n = a^(m×n)
Substitution: (3^2)^4 = 3^(2×4)
Working: 3^8
Answer: 6561
Example 2: Simplify (x^3)^5.
Given: base x, m = 3, n = 5
Formula: (a^m)^n = a^(m×n)
Substitution: x^(3×5)
Answer: x^15
Example 3: Evaluate (2^(−3))^2.
Given: a = 2, m = −3, n = 2
Formula: (a^m)^n = a^(m×n)
Substitution: 2^((−3)×2)
Working: 2^(−6) = 1 ÷ 64
Answer: 1/64 = 0.015625
Example 4: Simplify (4 × 5)^3.
Given: a = 4, b = 5, n = 3
Formula: (a × b)^n = a^n × b^n
Substitution: (4 × 5)^3 = 4^3 × 5^3
Working: 64 × 125
Answer: 8000
Example 5: Expand (2x^2)^3.
Given: coefficient 2, base x with index 2, n = 3
Formula: (a × b)^n = a^n × b^n
Working: 2^3 × (x^2)^3 = 8 × x^(2×3)
Answer: 8x^6
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Create a free accountThis chapter covered the application of algebra starting with indices algebra, including the concepts of power zero, negative and fractional powers, as well as the fundamental laws of indices such as addition, subtraction, division, and multiplication. The importance of the BODMAS rule was emphasized for correctly evaluating algebraic expressions. The chapter then explored roots, focusing on square roots, cube roots, and nth roots, followed by logarithms with their key laws: product, quotient, and power. Practical use of the scientific calculator was explained, detailing functions like power on/off, mode settings for degree, radian, gradient, standard deviation, and operations such as clear, save, and shift. Methods for solving simultaneous equations involving up to three equations were demonstrated through elimination, substitution, reduction, and graphical techniques. Finally, quadratic equations were addressed using factorization, the quadratic formula, completing the square, and graphical solutions to provide comprehensive problem-solving tools.
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