By the end of this chapter, you will be able to:
Mastering these skills will help you tackle real-world technical challenges and make you a valuable asset in your trade.
Compiled by: Stephen Waweru
Applied mathematics is fundamental in agricultural engineering for solving practical problems involving growth rates, resource allocation, and machinery performance. Algebra, particularly indices algebra, is essential when dealing with powers and exponential relationships common in calculations such as crop yield estimation, soil nutrient decay, and equipment efficiency. Mastery of indices laws enables agricultural engineers to simplify complex expressions and perform accurate computations required in design and analysis. This chapter focuses on applying the rules of indices to solve relevant agricultural engineering problems efficiently.
An index (plural indices, also called a power or an exponent) shows how many times a number is multiplied by itself. In 5^3 the base is 5 and the index is 3, so 5^3 = 5 × 5 × 5 = 125. Indices are used in agricultural engineering to write areas and volumes (m², m³), to describe growth and decay (a fungus on stored grain, a falling water level) and to scale the size of machine parts. Example: a mould on stored maize that doubles in amount every day grows by a factor of 2^5 = 32 in five days.
This topic covers powers of zero, negative powers, fractional powers and the laws that let several powers of the same base be combined into one.
Any non-zero number raised to the power zero equals one. The rule follows from dividing powers of the same base: a^3 ÷ a^3 = 1, and by subtraction of indices a^3 ÷ a^3 = a^(3 − 3) = a^0, so a^0 = 1. It matters when a growth or decay factor is applied for zero periods, because the factor then leaves the starting value unchanged. Example: the volume of water in a farm pond is modelled as 800 × (0.98)^t cubic metres, where t is the number of weeks of seepage loss. At the start t = 0, so the factor (0.98)^0 = 1 and the volume is 800 m³.
a^0 = 1, where a ≠ 0
The expression 0^0 is not defined, so the base must never be zero.
Example 1: A tractor-hire cost model multiplies the hourly rate by 1.08^n, where n is the number of years since purchase. Find the factor in the year of purchase, n = 0.
Given: a = 1.08, n = 0
Formula: a^0 = 1
Substitution: 1.08^0
Result: 1
Answer: the factor is 1, so the rate is unchanged in the year of purchase
Example 2: Simplify (3/5)^0, a factor in a spray-tank mixing formula.
Given: a = 3/5
Formula: a^0 = 1
Substitution: (3/5)^0
Result: 1
Answer: 1
Example 3: Evaluate 6 × 5^0 + 2^0.
Given: the power zero applies only to the base next to it, so 5^0 and 2^0 are worked out first
Formula: a^0 = 1
Substitution: 6 × 1 + 1
Working: 6 × 1 = 6, then 6 + 1 = 7
Answer: 7
A negative power is the reciprocal of the same base raised to the positive power. It is used for rates and inverse relationships. Example: a grain dryer takes 5 hours to dry 1 tonne, so it dries 5^(−1) = 1 ÷ 5 = 0.2 tonne per hour.
a^(−n) = 1 ÷ a^n, where a ≠ 0
When the base is a fraction, a negative power turns the fraction upside down: (a/b)^(−n) = (b/a)^n.
A negative index means a reciprocal, not a negative number. So 2^(−3) = 1 ÷ 8 = 0.125, not −8.
Example 1: A riffle divider halves a grain sample, and the sample is passed through it four times. Find the fraction of the original sample that remains, 2^(−4).
Given: a = 2, n = 4
Formula: a^(−n) = 1 ÷ a^n
Substitution: 2^(−4) = 1 ÷ 2^4
Working: 2^4 = 16, so 1 ÷ 16
Result: 0.0625
Answer: 0.0625 of the original sample, which is 6.25%
Example 2: Simplify (3/4)^(−2).
Given: a = 3, b = 4, n = 2
Formula: (a/b)^(−n) = (b/a)^n
Substitution: (3/4)^(−2) = (4/3)^2
Working: (4/3)^2 = 16/9
Result: 16 ÷ 9 = 1.78 (to 2 decimal places)
Answer: 16/9, which is 1.78
Example 3: Evaluate 10^(−3), the factor for converting millimetres to metres.
Given: a = 10, n = 3
Formula: a^(−n) = 1 ÷ a^n
Substitution: 10^(−3) = 1 ÷ 10^3
Working: 10^3 = 1 000
Result: 1 ÷ 1 000 = 0.001
Answer: 0.001, so 1 mm = 0.001 m
A fractional power is a root combined with a power: the denominator of the fraction is the root and the numerator is the power. Fractional powers are used to work back from an area or a volume to a length. Example: a square seedbed of area 144 m² has side 144^(1/2) = 12 m.
a^(m/n) = ⁿ√(a^m) = (ⁿ√a)^m
It is usually easier to take the root first and then apply the power, because the numbers stay small. A negative fractional power is the reciprocal of the positive one.
Example 1: A cube-shaped water tank holds 64 m³. Find its edge length, 64^(1/3).
Given: a = 64, m = 1, n = 3
Formula: a^(m/n) = ⁿ√(a^m)
Substitution: 64^(1/3) = ³√64
Result: 4
Answer: 4 m
Example 2: Evaluate 8^(2/3).
Given: a = 8, m = 2, n = 3
Formula: a^(m/n) = (ⁿ√a)^m
Substitution: 8^(2/3) = (³√8)^2
Working: ³√8 = 2, so 2^2
Result: 4
Answer: 4
Example 3: Evaluate 16^(−3/4).
Given: a = 16, m = 3, n = 4, with a negative sign
Formula: a^(−m/n) = 1 ÷ (ⁿ√a)^m
Substitution: 16^(−3/4) = 1 ÷ (⁴√16)^3
Working: ⁴√16 = 2, and 2^3 = 8
Result: 1 ÷ 8 = 0.125
Answer: 0.125
The laws of indices let a long expression with powers be reduced to a single power. The first two laws need powers of the same base; the others apply to a power of a quotient, a power of a power and a power of a product. The four sub-sections below are organised as follows:
When powers of the same base are multiplied, the indices are added. This is the product law, and it saves repeated multiplication when a quantity grows in stages. Example: a bacterial count in a livestock water trough grows by a factor of 3^2 = 9 in the first week and by 3^3 = 27 in the second week, so the combined factor is 3^(2 + 3) = 3^5 = 243, which is 9 × 27.
a^m × a^n = a^(m + n)
The rule fails when the bases differ, so 2^3 × 3^2 cannot be written as one power.
Three worked examples follow.
Example 1: Simplify 2^5 × 2^3.
Given: a = 2, m = 5, n = 3
Formula: a^m × a^n = a^(m + n)
Substitution: 2^5 × 2^3 = 2^(5 + 3) = 2^8
Result: 256
Answer: 256
Example 2: Simplify x^4 × x^6.
Given: a = x, m = 4, n = 6
Formula: a^m × a^n = a^(m + n)
Substitution: x^4 × x^6 = x^(4 + 6)
Answer: x^10
Example 3: Simplify 4^3 × 4^(−1).
Given: a = 4, m = 3, n = −1
Formula: a^m × a^n = a^(m + n)
Substitution: 4^3 × 4^(−1) = 4^(3 + (−1)) = 4^2
Check: 64 × 0.25 = 16
Answer: 16
When powers of the same base are divided, the index of the divisor is subtracted from the index of the dividend. This is the quotient law, and it gives a negative index when the divisor has the larger power. Example: a storage bin holds 10^5 cm³ and a sample cup holds 10^2 cm³, so the bin holds 10^(5 − 2) = 10^3 = 1 000 times as much as the cup.
a^m ÷ a^n = a^(m − n), where a ≠ 0
Three worked examples follow.
Example 1: Simplify 6^8 ÷ 6^5.
Given: a = 6, m = 8, n = 5
Formula: a^m ÷ a^n = a^(m − n)
Substitution: 6^8 ÷ 6^5 = 6^(8 − 5) = 6^3
Result: 216
Answer: 216
Example 2: Simplify 3^2 ÷ 3^5.
Given: a = 3, m = 2, n = 5
Formula: a^m ÷ a^n = a^(m − n)
Substitution: 3^2 ÷ 3^5 = 3^(2 − 5) = 3^(−3)
Working: 3^(−3) = 1 ÷ 3^3 = 1 ÷ 27
Result: 0.037 (to 3 decimal places)
Answer: 1/27, which is 0.037
Example 3: Simplify (12a^6) ÷ (4a^2).
Given: the numbers are divided as numbers, and the powers of a by the quotient law
Working: 12 ÷ 4 = 3, and a^6 ÷ a^2 = a^(6 − 2) = a^4
Answer: 3a^4
This section covers a power applied to a quotient: the index is shared between the numerator and the denominator, so a ratio can be handled one part at a time. Dividing two powers of the same base is covered under Subtraction. Example: when every length of a pump housing is scaled up by the ratio 3/2, every area is scaled by (3/2)^2 = 3^2 ÷ 2^2 = 9 ÷ 4 = 2.25 and every volume by (3/2)^3 = 27 ÷ 8 = 3.375.
(a/b)^n = a^n ÷ b^n, where b ≠ 0
Three worked examples follow.
Example 1: 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: 4^3 = 64 and 5^3 = 125
Result: 64 ÷ 125 = 0.512
Answer: 0.512
Example 2: The ratio of the side of a large irrigation plot to the side of a small test plot is 10/4. Find the ratio of their areas, (10/4)^2.
Given: a = 10, b = 4, n = 2
Formula: (a/b)^n = a^n ÷ b^n
Substitution: (10/4)^2 = 10^2 ÷ 4^2
Working: 100 ÷ 16
Result: 6.25
Answer: the large plot has 6.25 times the area of the test plot
Example 3: Simplify (2x/y)^3.
Given: a = 2x, b = y, n = 3
Formula: (a/b)^n = a^n ÷ b^n
Substitution: (2x/y)^3 = (2x)^3 ÷ y^3
Working: (2x)^3 = 2^3 × x^3 = 8x^3
Answer: 8x^3 ÷ y^3
Multiplication of indices covers a power raised to another power, where the indices are multiplied, and a product raised to a power, where each factor takes the power. These rules are used when a growth factor repeats or when a quantity with several dimensions is scaled. Example: a pest that multiplies by 3^2 = 9 every fortnight multiplies by (3^2)^3 = 3^(2 × 3) = 3^6 = 729 in three fortnights, which is the same as 9^3.
Power of a power: (a^m)^n = a^(m × n)
Power of a product: (ab)^n = a^n × b^n
Three worked examples follow.
Example 1: Simplify (5^2)^3.
Given: a = 5, m = 2, n = 3
Formula: (a^m)^n = a^(m × n)
Substitution: (5^2)^3 = 5^(2 × 3) = 5^6
Result: 15 625
Answer: 15 625
Example 2: A sample cube of soil has edge 2 × 5 = 10 cm. Find its volume in cm³ using (2 × 5)^3.
Given: a = 2, b = 5, n = 3
Formula: (ab)^n = a^n × b^n
Substitution: (2 × 5)^3 = 2^3 × 5^3
Working: 2^3 = 8 and 5^3 = 125, so 8 × 125
Result: 1 000
Answer: 1 000 cm³
Example 3: Simplify (3x^2)^3.
Given: a = 3, b = x^2, n = 3
Formula: (ab)^n = a^n × b^n and (a^m)^n = a^(m × n)
Substitution: (3x^2)^3 = 3^3 × (x^2)^3
Working: 3^3 = 27 and (x^2)^3 = x^6
Answer: 27x^6
Many mistakes with indices come from using a law where it does not apply. Checking the base and the operation before choosing a law prevents them. Example: a technician who adds 2^3 + 2^4 as 2^7 gets 128 instead of the true 24.
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter covered the application of algebra starting with indices algebra, explaining the concepts of power zero, negative powers, fractional powers, and the fundamental laws of indices including addition, subtraction, division, and multiplication of powers. The BODMAS rule was introduced to clarify the order of operations in algebraic expressions. Roots were explored in detail, covering square roots, cube roots, and general nth roots. The chapter then examined logarithms, focusing on the product, quotient, and power laws that simplify logarithmic calculations. Practical use of the scientific calculator was demonstrated, highlighting functions such as power on/off, mode settings for degree, radian, gradient, standard deviation, as well as clearing, saving, and using the shift key. Techniques for solving simultaneous equations with up to three variables were explained through elimination, substitution, reduction, and graphical methods. Finally, quadratic equations were addressed using factorization, the quadratic formula, completing the square, and graphical representation to find their roots and analyze their behavior.
Simplify \(x^0\) if \(x = 5\). (1 mark)
Evaluate \(3^{-2}\) and interpret its meaning in the context of fertilizer dosage per hectare. (2 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.