Laws of Indices and Indicial Equations for SS 1

Laws of Indices

Performance Objectives

By the end of the lesson, students should be able to:

1. generate the laws of indices;
2. use the laws of indices to solve indices problems;
3. solve simple indicial equations;
4. apply laws of indices to solve practical daily problems.

Introduction to Indices

An index is the mathematical representation of the number of times a base number is multiplied by itself.

The plural of index is indices. In the expression $a^n$, the number $a$ is called the base, while the superscript $n$ is called the index, exponent, or power. This notation is a concise way of writing repeated multiplications of the same number.

For example, the expression $5 \times 5 \times 5 \times 5$ can be written in index form as $5^4$. Here, $5$ is the base and $4$ is the index.

Observe the repeated multiplication chart below to see how index form relates to expanded form and evaluated values.

Generating the Laws of Indices

The laws of indices are fundamental rules used to simplify expressions involving powers. These laws can be generated by expanding the index terms and observing the patterns that emerge.

1. The Multiplication Law

To generate this law, consider the product of $a^3$ and $a^2$:

$$a^3 \times a^2 = (a \times a \times a) \times (a \times a)$$

By counting the total number of times the base $a$ is multiplied, the expression becomes:

$$a^3 \times a^2 = a \times a \times a \times a \times a = a^5$$

Since $5 = 3 + 2$, the relationship can be written as:

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

This pattern leads to the general multiplication law:

$$a^x \times a^y = a^{x+y}$$

2. The Division Law

To generate this law, consider the division of $a^5$ by $a^2$:

$$a^5 \div a^2 = \frac{a^5}{a^2} = \frac{a \times a \times a \times a \times a}{a \times a}$$

By cancelling out the common factors of $a$ in both the numerator and the denominator, the expression simplifies to:

$$a^5 \div a^2 = a \times a \times a = a^3$$

Since $3 = 5 – 2$, the relationship can be written as:

$$a^5 \div a^2 = a^{5-2}$$

This pattern leads to the general division law:

$$a^x \div a^y = a^{x-y}$$

3. The Power Law

To generate this law, consider the expression $(a^3)^2$:

$$(a^3)^2 = a^3 \times a^3$$

Expanding each term gives:

$$(a^3)^2 = (a \times a \times a) \times (a \times a \times a) = a^6$$

Since $6 = 3 \times 2$, the relationship can be written as:

$$(a^3)^2 = a^{3 \times 2}$$

This pattern leads to the general power law:

$$(a^x)^y = a^{xy}$$

4. The Zero Index Law

To generate this law, apply the division law to the expression $a^x \div a^x$:

$$a^x \div a^x = a^{x-x} = a^0$$

Mathematically, any non-zero quantity divided by itself is equal to $1$:

$$\frac{a^x}{a^x} = 1$$

Equating the two results leads to the zero index law:

$$a^0 = 1 \quad (\text{where } a \neq 0)$$

5. The Negative Index Law

To generate this law, apply the division law to the expression $a^0 \div a^x$:

$$a^0 \div a^x = a^{0-x} = a^{-x}$$

Using the zero index law, substitute $1$ for $a^0$ in the division fraction:

$$a^0 \div a^x = \frac{a^0}{a^x} = \frac{1}{a^x}$$

Equating the two results leads to the negative index law:

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

6. The Fractional Index Law

To generate this law, let $y = a^{\frac{1}{2}}$. Squaring both sides of the equation and applying the power law gives:

$$y^2 = \left(a^{\frac{1}{2}}\right)^2 = a^{\frac{1}{2} \times 2} = a^1 = a$$

Taking the square root of both sides gives:

$$y = \sqrt{a}$$

Therefore, $a^{\frac{1}{2}} = \sqrt{a}$. In general, for any positive integer $n$:

$$a^{\frac{1}{n}} = \sqrt[n]{a}$$

For a general fraction $\frac{m}{n}$, the law is expressed as:

$$a^{\frac{m}{n}} = \left(a^{\frac{1}{n}}\right)^m = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$$

The diagram below provides a summary of these generated laws for quick reference.

Solving Problems Using the Laws of Indices

Simplifying algebraic expressions requires the systematic application of one or more laws of indices. The following worked examples demonstrate these procedures.

Worked Example 1

Simplify the expression:

$$\frac{27^{2/3} \times 64^{-1/3}}{81^{1/4}}$$

Solution

First, express each base number as a power of its prime factors:

The prime factor forms are:

1. $27 = 3^3$
2. $64 = 2^6$
3. $81 = 3^4$

Substitute these prime factor forms back into the original expression:

$$\frac{(3^3)^{2/3} \times (2^6)^{-1/3}}{(3^4)^{1/4}}$$

Apply the power law, $(a^x)^y = a^{xy}$, to simplify the exponents:

$$\frac{3^{3 \times \frac{2}{3}} \times 2^{6 \times -\frac{1}{3}}}{3^{4 \times \frac{1}{4}}}$$
$$\frac{3^2 \times 2^{-2}}{3^1}$$

Apply the negative index law, $2^{-2} = \frac{1}{2^2}$, to the numerator:

$$\frac{3^2 \times \frac{1}{4}}{3}$$
$$\frac{\frac{9}{4}}{3} = \frac{9}{4} \times \frac{1}{3}$$
$$\frac{3}{4}$$

Worked Example 2

Simplify the algebraic expression:

$$\frac{12x^4y^5 \times 3x^2y^{-2}}{9x^5y^2}$$

Solution

First, group the numerical coefficients and the variables with like bases together:

$$\frac{(12 \times 3) \cdot (x^4 \times x^2) \cdot (y^5 \times y^{-2})}{9x^5y^2}$$
$$\frac{36 \cdot x^{4+2} \cdot y^{5+(-2)}}{9x^5y^2}$$
$$\frac{36x^6y^3}{9x^5y^2}$$

Divide the numerical coefficients and apply the division law to the variables:

$$\left(\frac{36}{9}\right) \cdot x^{6-5} \cdot y^{3-2}$$
$$4 \cdot x^1 \cdot y^1 = 4xy$$

Simple Indicial Equations

An indicial equation is an equation in which the unknown variable appears as an index or exponent.

To solve simple indicial equations, the terms on both sides of the equation must be expressed in the same base. Once the bases are equal, the exponents can be equated using the principle:

$$\text{If } a^x = a^y, \text{ then } x = y \quad (\text{where } a > 0, a \neq 1)$$

Worked Example 1

Solve the equation:

$$2^{x+3} = 32$$

Solution

Express $32$ as a power of base $2$:

$$32 = 2^5$$

Substitute this back into the equation:

$$2^{x+3} = 2^5$$

Since the bases are equal, equate the exponents:

$$x + 3 = 5$$

Subtract $3$ from both sides of the equation:

$$x = 5 – 3$$
$$x = 2$$

Worked Example 2

Solve the equation:

$$9^{2x-1} = 27^{x+1}$$

Solution

Express both bases, $9$ and $27$, as powers of a common prime base, which is $3$:

The base conversions are:

1. $9 = 3^2$
2. $27 = 3^3$

Substitute these values into the equation:

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

Apply the power law to expand the exponents on both sides:

$$3^{2(2x-1)} = 3^{3(x+1)}$$
$$3^{4x-2} = 3^{3x+3}$$

Since the bases are equal, equate the exponents:

$$4x – 2 = 3x + 3$$

Collect like terms by subtracting $3x$ from both sides and adding $2$ to both sides:

$$4x – 3x = 3 + 2$$
$$x = 5$$

Practical Applications of Indices

Indices are used in daily life to model scenarios involving rapid growth or decay. Examples include population growth, bacterial reproduction, financial compound interest, and radioactive decay.

Worked Example 1 (Bacterial Growth)

A certain species of bacteria doubles in population every hour. The population $P$ after $t$ hours is modeled by the formula:

$$P = P_0 \times 2^t$$

where $P_0$ is the initial population. If a laboratory experiment starts with $150$ bacteria, calculate the population after $6$ hours.

Solution

Identify the given values from the problem:

The parameters are:

1. Initial population, $P_0 = 150$
2. Time, $t = 6$ hours

Substitute these values into the growth formula:

$$P = 150 \times 2^6$$

Evaluate the index term $2^6$:

$$2^6 = 64$$

Calculate the final population:

$$P = 150 \times 64$$
$$P = 9,600$$

The population of the bacteria after $6$ hours is $9,600$.

Worked Example 2 (Radioactive Decay)

A radioactive substance decays such that its remaining mass $M$ (in grams) after $t$ years is given by the formula:

$$M = 160 \times 2^{-0.5t}$$

Calculate the remaining mass of the substance after $6$ years.

Solution

Identify the given values from the problem:

The parameters are:

1. Time, $t = 6$ years

Substitute $t = 6$ into the decay formula:

$$M = 160 \times 2^{-0.5 \times 6}$$
$$M = 160 \times 2^{-3}$$

Apply the negative index law to evaluate $2^{-3}$:

$$2^{-3} = \frac{1}{2^3} = \frac{1}{8}$$

Calculate the remaining mass:

$$M = 160 \times \frac{1}{8}$$
$$M = 20 \text{ grams}$$

The remaining mass of the substance after $6$ years is $20$ grams.

Practice Exercises

Multiple Choice Questions

Choose the correct option for each of the following questions:

1. What is the value of $32^{3/5}$?
   1. $4$
   2. $8$
   3. $16$
   4. $64$
2. Simplify the expression $\frac{a^7 \times a^{-3}}{a^2}$.
   1. $a^2$
   2. $a^3$
   3. $a^6$
   4. $a^8$
3. Solve the indicial equation $3^{2x-1} = 27$.
   1. $x = 1$
   2. $x = 2$
   3. $x = 3$
   4. $x = 4$
4. A population of insects triples every week according to the formula $I = I_0 \times 3^w$. If the initial population is $50$, what is the population after $3$ weeks?
   1. $150$
   2. $450$
   3. $1,350$
   4. $4,050$

Theory Questions

Solve the following problems, showing all steps clearly:

1. Simplify the expression:
   $$\frac{16^{3/4} \times 8^{-2/3}}{4^{1/2}}$$
2. Solve the indicial equation:
   $$25^{2x-3} = 125^{x+1}$$
3. The value of a piece of agricultural machinery depreciates according to the formula:
   $$V = 800,000 \times 2^{-0.2t}$$
   where $V$ is the value in Naira and $t$ is the time in years. Calculate the value of the machinery after $10$ years.