Chemical engineering involves quantitative analysis to design, optimize, and control chemical processes. Algebra is fundamental for manipulating equations describing reaction rates, mass balances, and thermodynamic properties. This chapter applies algebraic methods including indices, logarithms, simultaneous equations, and progressions to solve typical chemical engineering problems encountered in Kenyan industries such as pharmaceuticals, petrochemicals, and water treatment.

1.1 Base and Index

Algebraic indices (exponents) simplify expressions involving repeated multiplication, a common occurrence in kinetics and process calculations. Understanding laws of indices enables manipulation of powers of variables representing concentrations, volumes, or rate constants.

1.1.1 Laws of Indices

The laws of indices govern operations on powers of the same base, essential for simplifying algebraic expressions in chemical engineering formulas. The primary laws are:

• \(a^m \times a^n = a^{m+n}\)
• \(\frac{a^m}{a^n} = a^{m-n}\)
• \((a^m)^n = a^{m \times n}\)
• \(a^{\frac{m}{n}} = \sqrt[n]{a^m}\)
• \(a^{-n} = \frac{1}{a^n}\)
• \(a^0 = 1\) for \(a \neq 0\)

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

Worked Examples

Example 1: In a reaction, the concentration changes proportional to \(t^3 \times t^2\). Find the simplified power of \(t\).

Given: \(t^3 \times t^2\)

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

$$t^3 \times t^2 = t^{3+2}$$

$$t^5$$

Answer: \(t^5\)

Example 2: Simplify \(\frac{(5x)^4}{(5x)^2}\) where \(x\) is concentration.

Given: \(\frac{(5x)^4}{(5x)^2}\)

$$\frac{a^m}{a^n} = a^{m-n}$$

$$(5x)^4 \div (5x)^2 = (5x)^{4-2}$$

$$(5x)^2 = 25x^2$$

Answer: \(25x^2\)

Example 3: Calculate \((2x^3)^2\) for volume \(x\).

Given: \((2x^3)^2\)

$$(a^m)^n = a^{m \times n}$$

$$(2)^2 \times (x^3)^2 = 4x^{6}$$

Answer: \(4x^6\)

Example 4: Express \(27^{\frac{2}{3}}\) as a simplified number (used in reaction rate constants).

Given: \(27^{\frac{2}{3}}\)

$$a^{\frac{m}{n}} = \sqrt[n]{a^m}$$

$$27^{\frac{2}{3}} = (\sqrt[3]{27})^2$$

$$= (3)^2 = 9$$

Answer: 9

Example 5: Simplify \(x^{-3}\) where \(x\) is pressure.

Given: \(x^{-3}\)

$$a^{-n} = \frac{1}{a^n}$$

$$x^{-3} = \frac{1}{x^3}$$

Answer: \(\frac{1}{x^3}\)

1.1.2 Indicial Equations

Indicial equations involve variables with exponents that require solving for unknown powers, common in rate laws and empirical formulae.

Equation form: \(a^x = b\), find \(x\).

$$a^x = b$$

Worked Examples

Example 1: Solve \(3^x = 81\) for the reaction order \(x\).

Given: \(3^x = 81\)

Express 81 as power of 3:

$$81 = 3^4$$

$$3^x = 3^4$$

$$x = 4$$

Answer: \(x=4\)

Example 2: Find \(x\) if \(2^{3x} = 64\), where \(x\) is a time exponent.

Given: \(2^{3x} = 64\)

Express 64 as power of 2:

$$64 = 2^6$$

So,

$$2^{3x} = 2^6$$

$$3x = 6$$

$$x = \frac{6}{3} = 2$$

Answer: \(x=2\)

Example 3: Solve \(5^{2x+1} = 625\) for \(x\), representing concentration exponent.

Given: \(5^{2x+1} = 625\)

Express 625 as power of 5:

$$625 = 5^4$$

Equate powers:

$$2x + 1 = 4$$

$$2x = 3$$

$$x = \frac{3}{2} = 1.5$$

Answer: \(x=1.5\)

Example 4: Find \(x\) if \(9^{x-1} = 27\) for a process variable.

Given: \(9^{x-1} = 27\)

Express in base 3:

$$9 = 3^2, \quad 27 = 3^3$$

So,

$$(3^2)^{x-1} = 3^3$$

$$3^{2(x-1)} = 3^3$$

$$2(x-1) = 3$$

$$2x - 2 = 3$$

$$2x = 5$$

$$x = \frac{5}{2} = 2.5$$

Answer: \(x=2.5\)

Example 5: Solve \(4^{x+2} = 32\) for \(x\).

Given: \(4^{x+2} = 32\)

Express in base 2:

$$4 = 2^2, \quad 32 = 2^5$$

So,

$$(2^2)^{x+2} = 2^5$$

$$2^{2(x+2)} = 2^5$$

$$2x + 4 = 5$$

$$2x = 1$$

$$x = \frac{1}{2} = 0.5$$

Answer: \(x=0.5\)

1.1.3 Laws of Logarithm

Logarithms transform multiplicative relationships into additive ones, simplifying complex chemical engineering calculations such as pH, reaction rates, and kinetics.

Key laws:

• \(\log(ab) = \log a + \log b\)
• \(\log(\frac{a}{b}) = \log a - \log b\)
• \(\log a^n = n \log a\)
• \(\log 1 = 0\)
• \(\log a = c \implies a = 10^c\)

$$\log(ab) = \log a + \log b$$

Worked Examples

Example 1: Calculate \(\log(100 \times 1000)\).

Given: \(\log(100 \times 1000)\)

$$\log(ab) = \log a + \log b$$

$$\log(100) + \log(1000)$$

$$2 + 3 = 5$$

Answer: 5

Example 2: Find \(\log(\frac{1000}{10})\).

Given: \(\log(\frac{1000}{10})\)

$$\log(1000) - \log(10)$$

$$3 - 1 = 2$$

Answer: 2

Example 3: Evaluate \(\log(10^4)\).

Given: \(\log(10^4)\)

$$n \log 10 = 4 \times 1 = 4$$

Answer: 4

Example 4: Calculate \(\log 1\).

Given: \(\log 1\)

Answer: 0

Example 5: If \(\log a = 2\), find \(a\).

Given: \(\log a = 2\)

$$a = 10^2 = 100$$

Answer: 100

1.1.4 Logarithmic Equations

Solving logarithmic equations is vital in determining unknown variables in chemical kinetics and equilibrium constants.

Typical form: \(\log f(x) = c\)

Worked Examples

Example 1: Solve \(\log x = 3\).

Given: \(\log x = 3\)

$$x = 10^3 = 1000$$

Answer: \(x=1000\)

Example 2: Solve \(\log (2x) = 2\).

Given: \(\log (2x) = 2\)

$$2x = 10^2 = 100$$

$$x = \frac{100}{2} = 50$$

Answer: \(x=50\)

Example 3: Solve \(\log (x-1) - \log (x+1) = 0\).

Given: \(\log (x-1) - \log (x+1) = 0\)

Use \(\log a - \log b = \log \frac{a}{b}\):

$$\log \frac{x-1}{x+1} = 0$$

$$\frac{x-1}{x+1} = 10^0 = 1$$

$$x-1 = x+1$$

$$-1 = 1$$ (No solution from this)

Check domain: \(x > 1\)

Since no equality, no solution.

Answer: No solution

Example 4: Solve \(2 \log x = 4\).

Given: \(2 \log x = 4\)

$$\log x = 2$$

$$x = 10^2 = 100$$

Answer: \(x=100\)

Example 5: Solve \(\log (x^2) = 3\).

Given: \(\log (x^2) = 3\)

Use \(n \log a = \log a^n\):

$$2 \log x = 3$$

$$\log x = \frac{3}{2} = 1.5$$

$$x = 10^{1.5} = 31.62$$

Answer: \(x=31.62\)

Chapter Summary

This chapter covered the fundamentals of algebra beginning with the concepts of base and index and the laws governing indices, which are essential for simplifying expressions involving powers. It progressed to indicial equations, demonstrating how to solve equations where variables appear as exponents. The chapter then explored the laws of logarithms, providing methods to simplify and solve logarithmic equations, including techniques for converting between different logarithmic bases. Students learned how to effectively use scientific calculators to handle complex calculations involving indices and logarithms. The chapter also addressed the solution of simultaneous linear equations involving three unknowns, applying systematic algebraic methods. Further, it examined equations that can be reduced to quadratic form, showing how to transform and solve them efficiently. Finally, the chapter applied algebraic principles to solve problems involving arithmetic progressions and geometric progressions, highlighting their practical significance in engineering contexts.