By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Engineering mathematics is fundamental in land surveying, where precise calculations ensure accurate measurements and reliable data for mapping and construction projects. Mastery of algebraic laws, especially the law of indices, enables surveyors to manipulate and simplify expressions involving powers and roots efficiently. This chapter focuses on applying the law of indices to solve problems commonly encountered in the field, such as calculating areas, volumes, and scaling factors for land parcels.
The law of indices governs the rules for manipulating powers of numbers or variables, which is essential in simplifying algebraic expressions in surveying computations. Understanding these rules allows surveyors to handle exponential calculations that arise in coordinate transformations, distance scaling, and volume computations with confidence.
When multiplying two powers that have the same base, the exponents add together. This law simplifies expressions where repeated multiplication of the same quantity occurs, such as calculating compounded scale factors or repeated measurements.
The multiplication law is stated as:
$$ a^{m} \times a^{n} = a^{m+n} $$
Example 1:
Calculate the product \(5^{3} \times 5^{2}\).
Given: \(a = 5, m=3, n=2\)$$a^{m} \times a^{n} = a^{m+n}$$$$5^{3} \times 5^{2} = 5^{3+2}$$$$= 5^{5}$$$$= 5 \times 5 \times 5 \times 5 \times 5 = 3125$$ Answer: 3125
Example 2:
Simplify \(x^{4} \times x^{7}\) where \(x = 2\).
Given: \(x=2, m=4, n=7\)$$x^{m} \times x^{n} = x^{m+n}$$$$2^{4} \times 2^{7} = 2^{11}$$$$= 2^{11} = 2048$$ Answer: 2048
Example 3:
A land parcel's scale factor is \(1.1^{3} \times 1.1^{5}\). Calculate the combined scale factor.
Given: \(a=1.1, m=3, n=5\)$$a^{m} \times a^{n} = a^{m+n}$$$$1.1^{3} \times 1.1^{5} = 1.1^{8}$$
Calculate \(1.1^{8}\):$$1.1^{8} = 1.1 \times 1.1 \times 1.1 \times 1.1 \times 1.1 \times 1.1 \times 1.1 \times 1.1$$
Using a calculator: \(1.1^{8} \approx 2.1436\)
Answer: 2.1436
Example 4:
Simplify \( (3y)^{2} \times (3y)^{3} \).
Given: base \(3y\), exponents \(2\) and \(3\)$$(3y)^{2} \times (3y)^{3} = (3y)^{2+3} = (3y)^{5}$$ Calculate \( (3y)^{5} \) for \(y=2\):$$(3 \times 2)^{5} = 6^{5} = 6 \times 6 \times 6 \times 6 \times 6 = 7776$$ Answer: 7776
Example 5:
Calculate the product \( (0.5)^{4} \times (0.5)^{6} \).
Given: \(a=0.5, m=4, n=6\)$$a^{m} \times a^{n} = a^{m+n}$$$$0.5^{4} \times 0.5^{6} = 0.5^{10}$$ Calculate \(0.5^{10}\):$$0.5^{10} = (0.5)^{5} \times (0.5)^{5} = (0.03125) \times (0.03125) = 0.00097656$$ Answer: 0.00097656
Dividing powers with the same base involves subtracting the exponents. This law is frequently used when calculating ratios of distances or areas in land survey computations.
The division law is expressed as:
$$ \frac{a^{m}}{a^{n}} = a^{m-n} $$
Example 1:
Calculate \( \frac{7^{5}}{7^{2}} \).
Given: \(a=7, m=5, n=2\)$$\frac{a^{m}}{a^{n}} = a^{m-n}$$$$\frac{7^{5}}{7^{2}} = 7^{5-2} = 7^{3}$$$$7^{3} = 7 \times 7 \times 7 = 343$$ Answer: 343
Example 2:
Simplify \( \frac{x^{10}}{x^{6}} \) where \(x=3\).
Given: \(x=3, m=10, n=6\)$$x^{m-n} = x^{10-6} = x^{4}$$$$3^{4} = 3 \times 3 \times 3 \times 3 = 81$$ Answer: 81
Example 3:
Calculate \( \frac{(2.5)^{8}}{(2.5)^{3}} \).
Given: \(a=2.5, m=8, n=3\)$$a^{m-n} = 2.5^{8-3} = 2.5^{5}$$ Calculate \(2.5^{5}\):$$2.5 \times 2.5 \times 2.5 \times 2.5 \times 2.5 = 97.65625$$ Answer: 97.65625
Example 4:
Simplify \( \frac{(4y)^{7}}{(4y)^{4}} \) when \(y=1.5\).
Given: base \(4y\), exponents \(7\) and \(4\)$$(4y)^{7-4} = (4y)^{3}$$ Calculate for \(y=1.5\):$$(4 \times 1.5)^{3} = 6^{3} = 6 \times 6 \times 6 = 216$$ Answer: 216
Example 5:
Calculate \( \frac{(0.8)^{6}}{(0.8)^{6}} \).
Given: \(a=0.8, m=6, n=6\)$$a^{m-n} = 0.8^{6-6} = 0.8^{0}$$
Any number to the zero power is 1:
Answer: 1
Raising a power to another power involves multiplying the exponents. This law is useful in land survey when dealing with compounded scaling or repeated transformations.
The multiple powers law is:
$$ (a^{m})^{n} = a^{m \times n} $$
Example 1:
Calculate \( (3^{2})^{4} \).
Given: \(a=3, m=2, n=4\)$$(a^{m})^{n} = a^{m \times n}$$$$(3^{2})^{4} = 3^{2 \times 4} = 3^{8}$$ Calculate \(3^{8}\):$$3^{8} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 6561$$ Answer: 6561
Example 2:
Simplify \( (x^{5})^{3} \) for \(x=2\).
Given: \(x=2, m=5, n=3\)$$x^{m \times n} = 2^{5 \times 3} = 2^{15}$$ Calculate \(2^{15}\):$$2^{15} = 32768$$ Answer: 32768
Example 3:
Calculate \( ((1.2)^{3})^{2} \).
Given: \(a=1.2, m=3, n=2\)$$(a^{m})^{n} = a^{m \times n} = 1.2^{3 \times 2} = 1.2^{6}$$ Calculate \(1.2^{6}\) using a calculator:$$1.2^{6} = 2.985984$$ Answer: 2.985984
Example 4:
Simplify \( ((5y)^{2})^{3} \) when \(y=1\).
Given: \(a=5y, m=2, n=3\)$$(5y)^{2 \times 3} = (5y)^{6}$$ Calculate for \(y=1\):$$(5 \times 1)^{6} = 5^{6} = 15625$$ Answer: 15625
Example 5:
Calculate \( ((0.7)^{4})^{2} \).
Given: \(a=0.7, m=4, n=2\)$$0.7^{4 \times 2} = 0.7^{8}$$ Calculate \(0.7^{8}\):$$0.7^{8} = 0.05764801$$ Answer: 0.05764801
Any nonzero number raised to the zero power equals one. This principle is critical when simplifying survey calculations where variables may cancel out, leaving zero exponents.
The zero index law states:
$$ a^{0} = 1 \quad \text{for } a \neq 0 $$
The law follows from the division law. Dividing any number by itself gives 1, and by the division law it also gives an index of zero:
$$ \frac{a^{m}}{a^{m}} = a^{m-m} = a^{0}, \quad \text{and} \quad \frac{a^{m}}{a^{m}} = 1, \quad \text{so} \quad a^{0} = 1 $$
Example 1:
Calculate \(5^{0}\).
Given: \(a=5\)$$a^{0} = 1$$ Answer: 1
Example 2:
Simplify \( (x^{3} \div x^{3}) \).
Given: \(x\) any nonzero number$$\frac{x^{3}}{x^{3}} = x^{3-3} = x^{0} = 1$$ Answer: 1
Example 3:
Calculate \( (10y)^{0} \) when \(y = 100\).
Given: base \(10y\)$$(10y)^{0} = 1$$ Answer: 1
Example 4:
Simplify \( \frac{(2.5)^{4}}{(2.5)^{4}} \).
Given: \(a=2.5\)$$\frac{2.5^{4}}{2.5^{4}} = 2.5^{4-4} = 2.5^{0} = 1$$ Answer: 1
Example 5:
Calculate \( (0.9)^{0} \).
Given: \(a=0.9\)$$0.9^{0} = 1$$ Answer: 1
Any number raised to the power of one is equal to the number itself:
$$ a^{1} = a $$
This also follows from the division law. Dividing a power by the power one lower leaves a single factor of the base:
$$ \frac{a^{m}}{a^{m-1}} = a^{m-(m-1)} = a^{1}, \quad \text{and} \quad \frac{a \times a \times \cdots \times a \; (m \text{ times})}{a \times a \times \cdots \times a \; (m-1 \text{ times})} = a, \quad \text{so} \quad a^{1} = a $$
The index of one is usually left unwritten, so \(x\) means \(x^{1}\). Remembering it prevents a common error when applying the other laws: in \(x \times x^{3}\), the first \(x\) carries an index of 1, so the product is \(x^{4}\), not \(x^{3}\).
Example 6:
Calculate \(7^{1}\).
Given: \(a=7\)$$a^{1} = a$$$$7^{1} = 7$$ Answer: 7
Example 7:
Simplify \(x \times x^{3}\) where \(x=2\).
Given: \(x = x^{1}\), so \(m=1, n=3\)$$x^{1} \times x^{3} = x^{1+3} = x^{4}$$$$2^{4} = 2 \times 2 \times 2 \times 2 = 16$$ Answer: 16
Example 8:
Simplify \( \frac{(1.05)^{5}}{(1.05)^{4}} \).
Given: \(a=1.05, m=5, n=4\)$$1.05^{5-4} = 1.05^{1} = 1.05$$ Answer: 1.05
A negative index indicates the reciprocal of the base raised to the positive exponent. This rule helps surveyors invert ratios or convert measurements in inverse proportion.
The negative index law is:
$$ a^{-n} = \frac{1}{a^{n}} \quad \text{for } a \neq 0 $$
Example 1:
Calculate \(2^{-3}\).
Given: \(a=2, n=3\)$$a^{-n} = \frac{1}{a^{n}} = \frac{1}{2^{3}} = \frac{1}{8} = 0.125$$ Answer: 0.125
Example 2:
Simplify \(x^{-4}\) for \(x=5\).
Given: \(x=5, n=4\)$$x^{-4} = \frac{1}{x^{4}} = \frac{1}{5^{4}} = \frac{1}{625} = 0.0016$$ Answer: 0.0016
Example 3:
Calculate \( (3y)^{-2} \) when \(y=2\).
Given: \(a=3y=6, n=2\)$$(3y)^{-2} = \frac{1}{(3y)^{2}} = \frac{1}{6^{2}} = \frac{1}{36} = 0.0278$$ Answer: 0.0278
Example 4:
Simplify \( \frac{1}{(4.5)^{3}} \).
Given: \(a=4.5, n=3\)$$\frac{1}{4.5^{3}} = 4.5^{-3}$$ Calculate \(4.5^{3}\):$$4.5 \times 4.5 \times 4.5 = 91.125$$$$\frac{1}{91.125} = 0.01097$$ Answer: 0.01097
Example 5:
Calculate \( (0.7)^{-4} \).
Given: \(a=0.7, n=4\)$$0.7^{-4} = \frac{1}{0.7^{4}}$$ Calculate \(0.7^{4}\):$$0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401$$$$\frac{1}{0.2401} = 4.165$$ Answer: 4.165
Fractional indices represent roots of numbers. For example, an exponent of \(\frac{1}{2}\) corresponds to the square root. This concept is vital when surveyors calculate distances, areas, and volumes involving roots.
The fractional index rule is:
$$ a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m} $$
Example 1:
Calculate \(16^{\frac{1}{2}}\).
Given: \(a=16, m=1, n=2\)$$a^{\frac{1}{2}} = \sqrt{16} = 4$$ Answer: 4
Example 2:
Calculate \(27^{\frac{1}{3}}\).
Given: \(a=27, m=1, n=3\)$$a^{\frac{1}{3}} = \sqrt[3]{27} = 3$$ Answer: 3
Example 3:
Calculate \(81^{\frac{3}{4}}\).
Given: \(a=81, m=3, n=4\)$$a^{\frac{3}{4}} = (\sqrt[4]{81})^{3}$$ Calculate \(\sqrt[4]{81}\):$$\sqrt[4]{81} = 3$$ Then:$$3^{3} = 27$$ Answer: 27
Example 4:
Calculate \(125^{\frac{2}{3}}\).
Given: \(a=125, m=2, n=3\)$$a^{\frac{2}{3}} = (\sqrt[3]{125})^{2}$$ Calculate \(\sqrt[3]{125}\):$$\sqrt[3]{125} = 5$$ Then:$$5^{2} = 25$$ Answer: 25
Example 5:
Calculate \(64^{\frac{5}{6}}\).
Given: \(a=64, m=5, n=6\)$$a^{\frac{5}{6}} = (\sqrt[6]{64})^{5}$$ Calculate \(\sqrt[6]{64}\):$$64 = 2^{6}$$$$\sqrt[6]{64} = 2$$ Then:$$2^{5} = 32$$ Answer: 32
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Create a free accountThis chapter covered essential algebraic techniques starting with the law of indices, including multiplication, division, powers of powers, zero, negative, and fractional indices. Understanding these laws is critical for simplifying expressions involving exponents in engineering calculations. The chapter then explored the laws of logarithms, which provide powerful tools for handling multiplication, division, and powers within logarithmic expressions. Next, methods for solving simultaneous equations were introduced, enabling the determination of unknown variables in systems of linear equations commonly encountered in engineering problems. Finally, the chapter addressed quadratic equations, detailing techniques for finding roots using factorization and the quadratic formula. Mastery of these topics equips students with fundamental skills necessary for complex problem-solving in civil engineering contexts.
Simplify \(4^3 \times 4^2\). (2 marks)
Calculate \(\frac{9^5}{9^2}\). (2 marks)
Type: Individual
| Tools & Equipment | Materials |
|---|---|
| Scientific Calculator | Writing Materials (foolscap paper) |
| Pencil and Rubber | Mathematics Reference Sheet (laws of indices summary) |
| S/N | Item | Quantity |
|---|---|---|
| 1 | Scientific Calculator | 1 Pc per Candidate |
| 2 | Writing Materials (foolscap paper) | 2 Sheets per Candidate |
| 3 | Pencil and Rubber | 1 Set per Candidate |
| 4 | Mathematics Reference Sheet (laws of indices summary) | 1 Sheet per Candidate |
| Items to be Evaluated | Marks Available | Marks Obtained | Comments |
|---|---|---|---|
| TASK 1: Use of PPE and Preparation | |||
| Identified and used PPE appropriately (e.g., wearing apron/dust coat, closed shoes) (Award 1 mark for correct PPE use, 0 if not) | 1 | ||
| Prepared working area and materials before starting calculations (Award 1 mark for neat and organized preparation, 0 if not) | 1 | ||
| Sub-Total | 2 | ||
| TASK 2: Application of Multiplication Law of Indices | |||
| Correctly applied multiplication law of indices in expression (a) and (b) (Award 1.5 marks each for correct application in each expression) | 3 | ||
| Showed clear and logical working steps (Award 1 mark for clear presentation of work, 0 if unclear) | 1 | ||
| Sub-Total | 4 | ||
| TASK 3: Application of Division Law of Indices | |||
| Correctly applied division law of indices in expression (a) and (c) (Award 1.5 marks each for correct application in each expression) | 3 | ||
| Showed clear and logical working steps (Award 1 mark for clear presentation of work, 0 if unclear) | 1 | ||
| Sub-Total | 4 | ||
| TASK 4: Final Simplification and Presentation | |||
| Correctly simplified all expressions to final answers (Award 1.5 marks for each correct final simplified answer for expressions a, b, c, d) | 6 | ||
| Presented answers clearly and legibly (Award 2 marks for neat and legible final answers, 0 if not) | 2 | ||
| Sub-Total | 8 | ||
| PRODUCT CHECKLIST | |||
| Final simplified answers are mathematically correct: (a) 3^3, (b) x^4, (c) 2^7, (d) a^9 (Award 2.5 marks for each correct simplified expression) | 10 | ||
| Sub-Total | 10 | ||
| GRAND TOTAL | 28 | ||
Type: Individual
| Tools & Equipment | Materials |
|---|---|
| Scientific Calculator | Writing materials (foolscaps) |
| Pencil and Rubber |
| S/N | Item | Quantity |
|---|---|---|
| 1 | Scientific Calculator | 1 Pc per Candidate |
| 2 | Writing materials (foolscaps) | 2 Sheets per Candidate |
| 3 | Pencil and Rubber | 1 Set per Candidate |
| Items to be Evaluated | Marks Available | Marks Obtained | Comments |
|---|---|---|---|
| TASK 1: Use of Personal Protective Equipment | |||
| Identified and used PPE appropriately (e.g. dust coat or apron, closed shoes) (Award 1 mark for correct PPE use, 0 if not) | 1 | ||
| Sub-Total | 1 | ||
| TASK 2: Selection of Tools and Materials | |||
| Selected suitable site to perform the simplification task (quiet, well-lit workspace) (Award 1 mark for correct site selection) | 1 | ||
| Correctly identified and arranged tools and materials (calculator, pencil, paper) (Award 1 mark for correct tools and materials selection) | 1 | ||
| Sub-Total | 2 | ||
| TASK 3: Interpretation of Expressions | |||
| Correctly interpreted the algebraic expressions involving powers and zero index (Award 2 marks for accurate interpretation) | 2 | ||
| Marked key terms and indices correctly on the working sheets (Award 2 marks for correct marking and identification) | 2 | ||
| Sub-Total | 4 | ||
| TASK 4: Application of Laws | |||
| Applied multiple powers law correctly in simplification (Award 3 marks for correct application) | 3 | ||
| Applied zero index law correctly in simplification (Award 3 marks for correct application) | 3 | ||
| Sub-Total | 6 | ||
| TASK 5: Presentation of Work | |||
| Presented calculations neatly and logically on paper (Award 2 marks for neatness and clarity) | 2 | ||
| Used correct mathematical notation and symbols throughout (Award 2 marks for correct notation) | 2 | ||
| Sub-Total | 4 | ||
| PRODUCT CHECKLIST | |||
| Simplified expressions are mathematically correct and complete (Award up to 10 marks based on correctness and completeness) | 10 | ||
| Final answers demonstrate correct application of multiple powers and zero index laws (Award up to 7 marks for correct law application in final answers) | 7 | ||
| Sub-Total | 17 | ||
| GRAND TOTAL | 34 | ||
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