By the end of this chapter, you will be able to: - perform calculations involving indices correctly using the right mathematical concepts - represent linear equations accurately in mathematical form - use a scientific calculator properly to solve mathematical problems following the manufacturer’s manual - solve simultaneous equations correctly by applying the correct mathematical rules - solve simple algebraic equations accurately using the right concepts - form simple algebraic equations correctly based on mathematical principles
Mastering these algebra skills will help you solve real problems efficiently and confidently in your technical trade.
Compiled by: Rose Apiyo
An index (plural indices), also called a power or exponent, shows how many times a number is multiplied by itself, so 2^3 means 2 × 2 × 2 = 8. A logarithm is the reverse operation: it answers the question "to what power must the base be raised to give this number?", so log₂ 8 = 3 because 2^3 = 8. Indices make very large and very small numbers easy to write and calculate, while logarithms allow us to solve equations where the unknown is in the power, such as finding how long savings take to double. Example: the ICT technician at a secondary school in Kericho uses indices daily, since computer storage is counted in powers of 2: 1 kilobyte is 2^10 = 1,024 bytes, and the laboratory's 1 terabyte server holds 2^40 bytes.
The laws of indices apply only when the bases are the same.
| Law | Rule | Example |
|---|---|---|
| Multiplication | a^m × a^n = a^(m+n) | 2^3 × 2^4 = 2^7 = 128 |
| Division | a^m ÷ a^n = a^(m−n) | 5^6 ÷ 5^4 = 5^2 = 25 |
| Power of a power | (a^m)^n = a^(m×n) | (3^2)^3 = 3^6 = 729 |
| Power of a product | (ab)^n = a^n × b^n | (2x)^3 = 8x^3 |
| Zero index | a^0 = 1 (a ≠ 0) | 7^0 = 1 |
| Negative index | a^(−n) = 1 ÷ a^n | 2^(−3) = 1/8 |
| Fractional index (root) | a^(1/n) = nth root of a | 25^(1/2) = √25 = 5 |
| Fractional index (root and power) | a^(m/n) = (nth root of a)^m | 27^(2/3) = 3^2 = 9 |
Simplify (2x^3 y^2)^2 × x^(−4)
(2x^3 y^2)^2 = 2^2 × x^(3×2) × y^(2×2)
(2x^3 y^2)^2 = 4x^6 y^4
4x^6 y^4 × x^(−4) = 4x^(6+(−4)) y^4
Answer = 4x^2 y^4
Evaluate 16^(−1/2)
16^(−1/2) = 1 ÷ 16^(1/2)
16^(−1/2) = 1 ÷ √16
16^(−1/2) = 1 ÷ 4
Answer = 1/4
Evaluate 8^(2/3) × 4^(−1)
8^(2/3) = (cube root of 8)^2 = 2^2 = 4
4^(−1) = 1/4
8^(2/3) × 4^(−1) = 4 × 1/4
Answer = 1
Standard form (scientific notation) uses indices of 10 to write very large or very small numbers as A × 10^n, where A is between 1 and 10 and n is a whole number. It is used for quantities such as electrical resistance, material strengths, distances and populations.
When the unknown is in the index and both sides can be written with the same base, the indices can be equated.
Solve 2^(x+1) = 32
32 = 2^5
2^(x+1) = 2^5
x + 1 = 5
x = 4
A logarithm turns a power statement into a log statement: if a^x = N, then log_a N = x. The base a must be positive and not equal to 1, and N must be positive. Logarithms were traditionally used with printed tables to multiply and divide large numbers by adding and subtracting, and today they are used, through calculators, to solve exponential equations and in formulas for sound levels, acidity and growth.
| Index form | Logarithm form |
|---|---|
| 10^3 = 1,000 | log₁₀ 1,000 = 3 |
| 2^5 = 32 | log₂ 32 = 5 |
| 5^(−2) = 1/25 | log₅ (1/25) = −2 |
| a^0 = 1 | log_a 1 = 0 |
| a^1 = a | log_a a = 1 |
| Law | Rule | Example |
|---|---|---|
| Product law | log (M × N) = log M + log N | log 2 + log 5 = log 10 = 1 |
| Quotient law | log (M ÷ N) = log M − log N | log 50 − log 5 = log 10 = 1 |
| Power law | log (M^n) = n log M | 3 log 2 = log 8 |
| Log of the base | log_a a = 1 | log 10 = 1 |
| Log of one | log_a 1 = 0 | log 1 = 0 |
| Change of base | log_a M = log_b M ÷ log_b a | log₂ 20 = log 20 ÷ log 2 |
Evaluate 3 log 2 + log 125 without a calculator.
3 log 2 = log 2^3 = log 8
log 8 + log 125 = log (8 × 125)
log 8 + log 125 = log 1,000
Answer = 3
Evaluate log₂ 20 correct to 4 decimal places.
log₂ 20 = log 20 ÷ log 2
log₂ 20 = 1.30103 ÷ 0.30103
Answer = 4.3219
| Name | Specification | Use |
|---|---|---|
| Scientific calculator | With x^y (or ^), √, x^(1/y), log, ln, 10^x, e^x and ×10^x (EXP) keys | Evaluating powers, roots, logarithms and standard form numbers |
| log key | Gives logarithm to base 10 | Finding common logarithms, such as log 20 = 1.30103 |
| 10^x key (shift log) | Gives antilogarithm to base 10 | Converting a logarithm back to a number |
| ln and e^x keys | Natural logarithm and its inverse | Growth and decay problems in science and engineering |
| ×10^x (EXP) key | Enters standard form numbers | Entering values such as 4.5 × 10^(−5) without errors |
| Four-figure mathematical tables | Printed tables of logarithms, antilogarithms, squares and roots | Calculating where calculators are not allowed or available |
When the two sides cannot be written with the same base, take logarithms of both sides and use the power law to bring the unknown down from the index.
Solve 3^x = 20
x log 3 = log 20
x = log 20 ÷ log 3
x = 1.30103 ÷ 0.47712
x = 2.727 (to 3 decimal places)
Check: 3^2.727 ≈ 20.0
A = P × (1 + r ÷ 100)^n, where A is the amount, P the principal, r the annual interest rate in % and n the number of years.
Example: a member deposits Ksh 100,000 in a fixed savings account at a SACCO in Meru paying 10% interest compounded annually. How long will it take for the deposit to double to Ksh 200,000?
200,000 = 100,000 × (1.10)^n
(1.10)^n = 2
n log 1.10 = log 2
n = log 2 ÷ log 1.10
n = 0.30103 ÷ 0.04139
n = 7.27 years
Since interest is added once a year, the deposit first exceeds Ksh 200,000 after 8 years.
Value after n years = Original value × (1 − r ÷ 100)^n
Example: a county referral hospital in Kisumu buys an ambulance for Ksh 6,000,000, which loses 15% of its value each year. Find its value after 4 years.
Value = Ksh 6,000,000 × (1 − 15 ÷ 100)^4
Value = Ksh 6,000,000 × 0.85^4
Value = Ksh 6,000,000 × 0.52201
Value = Ksh 3,132,038 (to the nearest shilling)
Future population = Present population × (1 + r ÷ 100)^n
Example: a county planning office in Kajiado projects a town's population of 400,000 growing at 2.5% per year over 5 years, to plan water supply.
Future population = 400,000 × (1.025)^5
Future population = 400,000 × 1.13141
Future population = 452,563 people
pH = −log [H⁺], where [H⁺] is the hydrogen ion concentration in mol/L.
Example: borehole water tested at a hotel in Malindi has [H⁺] = 3.2 × 10^(−6) mol/L.
pH = −log (3.2 × 10^(−6))
pH = −(log 3.2 + log 10^(−6))
pH = −(0.50515 − 6)
pH = 5.49
The water is slightly acidic (below pH 7) and needs treatment before use in boilers and kitchens.
Sound level (dB) = 10 × log (I ÷ I₀), where I is the sound intensity in W/m² and I₀ = 1 × 10^(−12) W/m² is the threshold of hearing.
Example: a plate compactor on a building contractor's site in Thika produces a sound intensity of 1 × 10^(−2) W/m².
Sound level = 10 × log (10^(−2) ÷ 10^(−12))
Sound level = 10 × log 10^10
Sound level = 10 × 10
Sound level = 100 dB
At this level, operators must wear ear protection.
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Create a free accountThis chapter covered essential concepts in basic algebra starting with indices and logarithms, focusing on their properties and how to manipulate them for solving mathematical expressions. It then explored linear equations, demonstrating methods to find unknown variables in straightforward algebraic statements. The use of a calculator was introduced as a practical tool to efficiently solve algebraic problems, enhancing accuracy and speed. Techniques for solving simultaneous equations were presented, enabling the determination of multiple unknowns through systems of equations. The chapter also addressed simple algebraic equations, emphasizing the steps to isolate variables and find solutions. Finally, the various forms of simple algebraic equations were discussed, helping students recognize and work with different equation structures commonly encountered in civil engineering calculations.
The monthly electricity consumption \(E\) (in kWh) at a county referral hospital is modelled by the formula \(E = 2^x\), where \(x\) is the number of days since the hospital upgraded to energy-saving bulbs. If on day 5, find the electricity consumption. (4 marks)
Solve for \(x\) in the equation \(3x - 7 = 2x + 8\). (4 marks)
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