Algebra forms the foundation of mathematical problem solving used across all professional fields in Kenya, including healthcare, agriculture, business, and hospitality. It enables the formulation and solution of equations that describe real-world problems such as budgeting, resource allocation, and operational efficiency. Mastery of algebraic principles such as indices and logarithms is essential for practical computations and data analysis in diverse workplaces, while the use of calculators streamlines complex calculations. This chapter focuses on applying algebra through indices, logarithms, and calculator usage with worked examples drawn from various Kenyan sectors.

1.1 Importance of algebra in engineering

Algebra is the language of quantitative reasoning and problem solving in all technical and professional disciplines. In Kenya’s dynamic workplaces, algebraic methods facilitate decision making, optimization, and performance evaluation. The subtopics below explore key algebraic concepts, indices, logarithms, and calculator use, that underpin practical computations.

1.1.1 Indices

Indices, or powers, represent repeated multiplication of a base number and are essential for expressing large or small quantities efficiently. The basic rules of indices allow simplification of expressions and solving exponential equations commonly encountered in calculations of growth rates, material strength, and financial projections.

The fundamental law of indices is:

$$a^{m} \times a^{n} = a^{m+n}$$

Other important rules include:

• Power of a power: \( (a^{m})^{n} = a^{m \times n} \)
• Power of a product: \( (ab)^{n} = a^{n} b^{n} \)
• Zero power: \( a^{0} = 1 \) (for \(a \neq 0\))
• Negative power: \( a^{-n} = \frac{1}{a^{n}} \)

Worked Examples

Example 1: A county hospital's patient population triples every 2 years. If the current population is 1000, find the population after 6 years.

Given: Initial population \(P_0 = 1000\), growth factor per 2 years \(= 3\), time \(t = 6\) years.

Since population triples every 2 years, after \(t\) years:

$$P = P_0 \times 3^{\frac{t}{2}}$$

$$P = 1000 \times 3^{\frac{6}{2}}$$

$$P = 1000 \times 3^{3}$$

$$P = 1000 \times 27$$

Answer: 27,000 patients

Example 2: A retail business calculates compound interest using the formula \( A = P (1 + r)^n \). If Ksh 50,000 is invested at 5% annual interest compounded yearly, find the amount after 3 years.

Given: \(P = 50,000\), \(r = 0.05\), \(n = 3\).

$$A = 50,000 \times (1 + 0.05)^3$$

$$A = 50,000 \times (1.05)^3$$

Calculate \(1.05^3\):

$$1.05^3 = 1.05 \times 1.05 \times 1.05 = 1.157625$$

$$A = 50,000 \times 1.157625$$

$$A = 57,881.25$$

Answer: Ksh 57,881.25

Example 3: A farm applies fertilizer at a rate that doubles every month. If the initial amount is 20 kg, what is the amount after 5 months?

Given: Initial amount \(F_0 = 20\) kg, doubling factor \(2\), time \(t = 5\) months.

$$F = F_0 \times 2^{t}$$

$$F = 20 \times 2^{5}$$

Calculate \(2^{5}\):

$$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$$

$$F = 20 \times 32$$

Answer: 640 kg

Example 4: A cooperative’s revenue decreases by a factor of \(10^{-0.5}\) annually due to market conditions. If current revenue is Ksh 2,000,000, find revenue after 2 years.

Given: Revenue \(R_0 = 2,000,000\), factor \(10^{-0.5}\) per year, time \(t = 2\) years.

$$R = R_0 \times (10^{-0.5})^{t}$$

$$R = 2,000,000 \times 10^{-0.5 \times 2}$$

$$R = 2,000,000 \times 10^{-1}$$

Calculate \(10^{-1} = \frac{1}{10} = 0.1\):

$$R = 2,000,000 \times 0.1$$

Answer: Ksh 200,000

Example 5: A university’s research budget grows annually by a factor of \(1.08^{n}\) where \(n\) is years. If the initial budget is Ksh 10 million, find the budget after 7 years.

Given: Budget \(B_0 = 10,000,000\), growth factor \(1.08\), time \(n = 7\).

$$B = B_0 \times 1.08^{7}$$

Calculate \(1.08^{7}\):

Stepwise:

$$1.08^{2} = 1.08 \times 1.08 = 1.1664$$

$$1.08^{4} = (1.08^{2})^{2} = 1.1664 \times 1.1664 = 1.3605$$

$$1.08^{7} = 1.08^{4} \times 1.08^{3} = 1.3605 \times (1.08 \times 1.08 \times 1.08)$$

Calculate \(1.08^{3}\):

$$1.08^{3} = 1.08 \times 1.08 \times 1.08 = 1.2597$$

Multiply:

$$1.3605 \times 1.2597 = 1.7139$$

Calculate final budget:

$$B = 10,000,000 \times 1.7139$$

$$B = 17,139,000$$

Answer: Ksh 17,139,000

1.1.2 Logarithms

Logarithms are the inverse operations of indices and are used to solve equations where the unknown is an exponent. They are crucial in fields such as environmental science for measuring pollutant concentrations, in finance for calculating compound interest periods, and in ICT for data processing algorithms.

The logarithm base \(b\) of a number \(x\) is:

$$\log_b x = y \quad \text{if and only if} \quad b^y = x$$

Common logarithms use base 10, denoted \(\log x\), and natural logarithms use base \(e\), denoted \(\ln x\).

Key properties include:

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

Worked Examples

Example 1: A SACCO’s loan portfolio doubles every 5 years. Find the compound growth rate \(r\) per year using logarithms.

Given: Doubling time \(T = 5\) years, growth factor after 5 years = 2.

The formula for compound growth:

$$2 = (1 + r)^5$$

Take logarithm base 10 on both sides:

$$\log 2 = \log (1 + r)^5$$

Use log power rule:

$$\log 2 = 5 \log (1 + r)$$

Solve for \(\log (1 + r)\):

$$\log (1 + r) = \frac{\log 2}{5}$$

Calculate \(\log 2 = 0.3010\):

$$\log (1 + r) = \frac{0.3010}{5} = 0.0602$$

Find \(1 + r\):

$$1 + r = 10^{0.0602}$$

Calculate \(10^{0.0602}\):

$$1 + r = 1.1487$$

Calculate \(r\):

$$r = 1.1487 - 1 = 0.1487 = 14.87\%$$

Answer: 14.87% annual growth rate

Example 2: A hotel’s energy consumption decreases to 40% of its original value after 3 years. Find the annual decay rate \(r\).

Given: Final fraction \(= 0.4\), time \(t = 3\) years.

Formula:

$$0.4 = (1 - r)^3$$

Take natural logarithm:

$$\ln 0.4 = 3 \ln (1 - r)$$

Calculate \(\ln 0.4 = -0.9163\):

$$\ln (1 - r) = \frac{-0.9163}{3} = -0.3054$$

Find \(1 - r\):

$$1 - r = e^{-0.3054} = 0.7367$$

Calculate \(r\):

$$r = 1 - 0.7367 = 0.2633 = 26.33\%$$

Answer: 26.33% annual decay rate

Example 3: A county government wants to find how long it takes for a pollutant concentration to reduce to 10% of the initial level if decay follows:

$$C = C_0 e^{-kt}$$

Given \(C/C_0 = 0.1\), decay constant \(k = 0.5\) per year.

Solve for \(t\):

$$0.1 = e^{-0.5 t}$$

Take natural logarithm:

$$\ln 0.1 = -0.5 t$$

Calculate \(\ln 0.1 = -2.3026\):

$$-2.3026 = -0.5 t$$

Solve for \(t\):

$$t = \frac{2.3026}{0.5} = 4.6052 \text{ years}$$

Answer: 4.61 years

Example 4: A bank calculates that the time \(t\) to double an investment at 8% annual interest is given by:

$$2 = (1 + 0.08)^t$$

Find \(t\) using logarithms.

Take log base 10:

$$\log 2 = t \log 1.08$$

Calculate \(\log 2 = 0.3010\), \(\log 1.08 = 0.0334\):

$$t = \frac{0.3010}{0.0334} = 9.01 \text{ years}$$

Answer: 9.01 years

Example 5: A retail firm’s sales follow the equation:

$$S = 5000 \times 3^{x}$$

If sales are 135,000 units, find \(x\).

Given: \(S = 135,000\)

Solve:

$$135,000 = 5000 \times 3^{x}$$

Divide both sides:

$$\frac{135,000}{5000} = 3^{x}$$

Calculate:

$$27 = 3^{x}$$

Take logarithm base 3:

$$\log_3 27 = x$$

Since \(27 = 3^3\):

$$x = 3$$

Answer: \(x = 3\)

1.1.3 Use of calculator

Calculators are essential tools for performing algebraic calculations efficiently and accurately in professional settings. Modern scientific calculators can handle indices, logarithms, roots, and trigonometric functions, reducing computational errors and saving time. Proficiency in calculator use enables professionals to solve complex problems encountered in financial analysis, environmental monitoring, and data management.

Key Functions for Algebra

• Exponentiation: Use the \(x^y\) or \(y^x\) key to compute powers.
• Logarithms: Use \(\log\) for base 10 and \(\ln\) for natural logarithms.
• Roots: Use the \(\sqrt{}\) key or \(x^{1/y}\) for nth roots.
• Memory: Store and recall intermediate results with \(M+\), \(MR\).
• Parentheses: Use brackets to control operation order.

Worked Examples

Example 1: Calculate \(5^{3}\) using a scientific calculator.

Step 1: Press 5

Step 2: Press \(x^y\) key

Step 3: Press 3

Step 4: Press equals (=)

Display shows 125

Answer: 125

Example 2: Calculate \(\log 250\).

Step 1: Press \(\log\) key

Step 2: Enter 250

Step 3: Press equals (=)

Display shows approximately 2.3979

Answer: 2.3979

Example 3: Calculate \(\ln 0.05\).

Step 1: Press \(\ln\) key

Step 2: Enter 0.05

Step 3: Press equals (=)

Display shows approximately -2.9957

Answer: -2.9957

Example 4: Calculate the cube root of 27.

Step 1: Enter 27

Step 2: Press \(x^{1/y}\) key

Step 3: Enter 3 (for cube root)

Step 4: Press equals (=)

Display shows 3

Answer: 3

Example 5: Solve \(3^{x} = 81\) for \(x\) using logarithms.

Step 1: Take \(\log\) on both sides:

$$\log 3^{x} = \log 81$$

Step 2: Use power rule:

$$x \log 3 = \log 81$$

Step 3: Calculate \(\log 3 = 0.4771\), \(\log 81 = 1.9085\)

Step 4: Calculate \(x\):

$$x = \frac{1.9085}{0.4771} = 4$$

Answer: \(x = 4\)

Chapter Summary

This chapter explored the fundamental role of algebra in engineering, emphasizing its importance in solving practical problems. It began with a detailed study of indices, including the definitions of base and index, the laws governing indices, and how to solve indicial equations. The chapter then covered logarithms, explaining their laws, methods for solving logarithmic equations, and techniques for converting between different bases, along with the effective use of calculators for these calculations. Linear equations were introduced with their standard form, followed by solution methods such as graphical representation, substitution, and elimination, highlighting their applications in engineering contexts. Quadratic equations were examined through their definition, standard form, and solution techniques including factoring, completing the square, and the quadratic formula, with examples drawn from real-world engineering problems. The chapter also discussed various types of functions including linear, quadratic, polynomial, rational, and exponential, outlining their characteristics and properties, and demonstrating how to plot and interpret their graphs on the Cartesian coordinate system. Finally, it addressed the use of algebraic graphs in solving engineering issues and introduced combinations and Pascal’s Triangle as tools for calculating binomial coefficients.