Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Indices
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Indices
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Subject: Further Mathematics
Class: Senior Secondary 1
Term: 3rd Term
Week: 2
Theme: Pure Mathematics
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Use the laws of in dices in calculations Solve problems of in dicial equations
Definition of Indices: An index (plural: indices) or exponent indicates the number of times a base number is multiplied by itself. In the expression $a^n$: $a$ is the base $n$ is the index (or exponent or power) $a^n$ means $a$ multiplied by itself $n$ times ($a \times a \times a \times \dots \times a$ ($n$ times)).
Laws of Indices: The following laws govern operations with indices: Law 1: Multiplication Law Statement: When multiplying numbers with the same base, add their indices.
Formula: $a^m \times a^n = a^{m+n}$ Explanation: This law simplifies expressions where the same base is raised to different powers and then multiplied. For example, $2^3 \times 2^2 = (2 \times 2 \times 2) \times (2 \times 2) = 2^5 = 32$. Worked Example 2.1: Simplify $5^4 \times 5^2 \times 5^{-1}$.
Solution: $5^4 \times 5^2 \times 5^{-1} = 5^{(4+2-1)} = 5^5 = 3125$.
Law 2: Division Law Statement: When dividing numbers with the same base, subtract the index of the denominator from the index of the numerator.
Formula: $a^m \div a^n = a^{m-n}$ (where $a \neq 0$)
Explanation: This law helps in simplifying fractions or division problems involving powers. For example, $3^5 \div 3^2 = (3 \times 3 \times 3 \times 3 \times 3) / (3 \times 3) = 3 \times 3 \times 3 = 3^3 = 27$. Worked Example 2.2: Simplify $7^6 / 7^3$.
Solution: $7^6 / 7^3 = 7^{(6-3)} = 7^3 = 343$.
Law 3: Power Law (Power of a Power)
Statement: When raising a power to another power, multiply the indices.
Formula: $(a^m)^n = a^{mn}$ Explanation: This law indicates how to handle nested powers. For example, $(2^3)^2 = (2 \times 2 \times 2)^2 = (8)^2 = 64$. Alternatively, $(2^3)^2 = 2^{(3 \times 2)} = 2^6 = 64$. Worked Example 2.3: Simplify $(b^3)^{-2}$.
Solution: $(b^3)^{-2} = b^{(3 \times -2)} = b^{-6}$.
Law 4: Zero Index Law Statement: Any non-zero number raised to the power of zero is
1. Formula: $a^0 = 1$ (where $a \neq 0$)
Explanation: This can be derived from the division law: $a^m / a^m = a^{(m-m)} = a^0$. Also, any number divided by itself is
1. Thus, $a^0 = 1$. Worked Example 2.4: Simplify $(15x^2y^3)^0$.
Solution: $(15x^2y^3)^0 = 1$.
Law 5: Negative Index Law Statement: A number raised to a negative index is the reciprocal of the number raised to the positive index.
Formula: $a^{-n} = 1/a^n$ (where $a \neq 0$)
Explanation: This law defines negative powers. For example, $2^{-3} = 1/2^3 = 1/8$. Worked Example 2.5: Express $4^{-2}$ as a fraction.
Solution: $4^{-2} = 1/4^2 = 1/16$.
Law 6: Fractional Index Law Statement: A number raised to a fractional index $1/n$ is the $n$-th root of that number.
Formula: $a^{1/n} = \sqrt[n]{a}$ Explanation: This law connects indices with roots. For example, $8^{1/3} = \sqrt[3]{8} = 2$. Worked Example 2.6: Evaluate $81^{1/4}$.
Solution: $81^{1/4} = \sqrt[4]{81} = 3$ (since $3 \times 3 \times 3 \times 3 = 81$).
Law 7: General Fractional Index Law Statement: A number raised to a fractional index $m/n$ is the $n$-th root of the number, raised to the power of $m$. Alternatively, it is the $n$-th root of the number raised to the power of $m$.
Formula: $a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$ Explanation: This combines the power and fractional index laws. For example, $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. Or $8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$. Worked Example 2.7: Evaluate $27^{2/3}$.
Solution: $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$.
Law 8: Power of a Product/Quotient Statement: The power of a product is the product of the powers. The power of a quotient is the quotient of the powers.
Formula: $(ab)^n = a^n b^n$ and $(a/b)^n = a^n / b^n$ Explanation: This applies the index to each factor within parentheses. For example, $(2x)^3 = 2^3 x^3 = 8x^3$. Worked Example 2.8: Simplify $(3pq^2)^3$. * Solution: $(3pq^2)^3 = 3^3 p^3 (q^2)^3 = 27p^3q^6$.
Indicial Equations: An indicial equation Evaluate $27^{2/3}$.
Solution: $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$.
Law 8: Power of a Product/Quotient Statement: The power of a product is the product of the powers. The power of a quotient is the quotient of the powers.
Formula: $(ab)^n = a^n b^n$ and $(a/b)^n = a^n / b^n$ Explanation: This applies the index to each factor within parentheses. For example, $(2x)^3 = 2^3 x^3 = 8x^3$. Worked Example 2.8: Simplify $(3pq^2)^3$.
Solution: $(3pq^2)^3 = 3^3 p^3 (q^2)^3 = 27p^3q^6$.
Indicial Equations: An indicial equation is an equation where the unknown variable appears as an exponent. To solve indicial equations, the primary strategy is to express both sides of the equation with the same base. Once the bases are the same, the exponents can be equated.
Steps to solve indicial equations:
1. Simplify both sides: Use the laws of indices to simplify each side of the equation as much as possible.
2. Express with the same base: Try to rewrite numbers as powers of a common base. For example, $8$ can be written as $2^3$, $27$ as $3^3$, $64$ as $2^6$ or $4^3$ or $8^2$.
3. Equate the exponents: Once both sides have the same base, set the exponents equal to each other.
4. Solve the resulting equation: Solve the simple linear or quadratic equation obtained from equating the exponents. Worked Example 2.9: Solve $2^x = 32$.
Solution:
1. Recognise that $32$ can be expressed as a power of $2$. $32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$.
2. Rewrite the equation: $2^x = 2^5$.
3. Since the bases are the same, equate the exponents: $x = 5$. Worked Example 2.10: Solve $3^{2x-1} = 27$.
Solution:
1. Express $27$ as a power of $3$: $27 = 3^3$.
2. Rewrite the equation: $3^{2x-1} = 3^3$.
3. Equate the exponents: $2x-1 = 3$.
4. Solve for $x$: $2x = 3+1 \Rightarrow 2x = 4 \Rightarrow x = 2$. Worked Example 2.11: Solve $4^{x+1} = 8^{x-1}$.
Solution:
1. Find a common base for $4$ and $8$. Both can be expressed as powers of $2$. 2. $4 = 2^2$ and $8 = 2^3$.
3. Substitute these into the equation: $(2^2)^{x+1} = (2^3)^{x-1}$.
4. Apply the power law: $2^{2(x+1)} = 2^{3(x-1)}$.
5. Simplify exponents: $2^{2x+2} = 2^{3x-3}$.
6. Equate exponents: $2x+2 = 3x-3$.
7. Solve for $x$: $2+3 = 3x-2x \Rightarrow 5 = x$. So, $x = 5$.
Introduction (10 minutes): Teacher Activity: Begin by reviewing the concept of repeated multiplication (e.g., $2 \times 2 \times 2 = 8$) and introduce the shorthand notation using indices ($2^3$). Ask students for examples from real life where numbers might grow or shrink rapidly (e.g., population growth, interest rates, bacterial growth).
Student Activity: Students recall basic multiplication and powers. They provide examples of large numbers or repeated processes. They observe the initial examples and ask clarifying questions about the notation. Explanation of Laws of Indices (25 minutes): Teacher Activity: Systematically introduce each law of indices, one by one.
For each law: State the law clearly with its formula. Provide a simple numerical example to illustrate the law (e.g., $2^3 \times 2^2$). Guide students through a worked example (similar to those in Section 2) that might involve variables or negative/fractional indices. Emphasise common misconceptions (e.g., confusing $a^m + a^n$ with $a^{m+n}$).
Student Activity: Students actively listen, take notes, and work through the numerical examples alongside the teacher. They ask questions for clarification and attempt to solve the guided worked examples.
Solving Indicial Equations (20 minutes): Teacher Activity: Explain the concept of indicial equations and the core strategy of equating bases. Demonstrate the step-by-step process using Worked Examples 2.9, 2.10, and 2.
1
1. Stress the importance of simplifying expressions using the laws of indices first. Provide a real-world scenario where such equations might be applied (e.g., calculating years for investment to double).
Student Activity: Students observe the problem-solving methodology. They follow the steps demonstrated and note key strategies. They may attempt to anticipate the next step in the solution process.
Guided Practice (15 minutes): Teacher Activity: Distribute a few practice questions (as outlined in Section 4). Circulate around the classroom, providing individual or small-group support. Encourage students to discuss their approaches.
Student Activity: Students work individually or in pairs to solve the guided practice questions. They share their methods and solutions with their peers and seek assistance from the teacher when stuck.
Class Discussion and Recap (5 minutes): Teacher Activity: Facilitate a brief discussion on common challenges encountered during the guided practice. Recap the main laws of indices and the strategy for solving indicial equations. Assign independent practice (Section 5) as homework.
Student Activity: Students share their solutions or challenges. They participate in the recap, reinforcing their understanding of the key concepts. They copy down the homework assignment. The teacher should present these questions and guide students through their solutions, allowing time for individual attempts before revealing the full solution.
Question 1: Simplify the expression: $(3x^2y^{-3})^2 \times (2x^{-1}y^4)$.
Solution: Apply the power law to the first term: $(3x^2y^{-3})^2 = 3^2 (x^2)^2 (y^{-3})^2 = 9x^4y^{-6}$. Now multiply the simplified first term by the second term: $9x^4y^{-6} \times 2x^{-1}y^4$ Group terms with the same base and multiply coefficients: $(9 \times 2) \times (x^4 \times x^{-1}) \times (y^{-6} \times y^4)$ Apply the multiplication law for indices: $18 \times x^{(4-1)} \times y^{(-6+4)}$ $18x^3y^{-2}$ Express with positive indices (optional, but good practice): $18x^3/y^2$
Commentary: This question tests the application of the power law, multiplication law, and negative index law in a single problem, requiring careful step-by-step execution.
Question 2: Evaluate $125^{2/3} + (1/4)^{-2}$.
Solution: Evaluate $125^{2/3}$: $125^{2/3} = (\sqrt[3]{125})^2 = 5^2 = 25$. (Alternatively, $125^{2/3} = \sqrt[3]{125^2} = \sqrt[3]{15625} = 25$) Evaluate $(1/4)^{-2}$: $(1/4)^{-2} = (4/1)^2 = 4^2 = 16$.
Add the results: $25 + 16 = 41$.
Commentary: This question combines the general fractional index law and the negative index law, requiring careful evaluation of each term before addition.
Question 3: Solve for $p$ in the equation: $9^{2p-1} = 27^{p+2}$.
Solution: Identify a common base for $9$ and $27$. Both are powers of $3$. Rewrite $9$ as $3^2$ and $27$ as $3^3$.
Substitute into the equation: $(3^2)^{2p-1} = (3^3)^{p+2}$.
Apply the power law: $3^{2(2p-1)} = 3^{3(p+2)}$.
Simplify the exponents: $3^{4p-2} = 3^{3p+6}$. Equate the exponents since the bases are the same: $4p-2 = 3p+6$. Solve the linear equation for $p$: $4p - 3p = 6 + 2$ $p = 8$.
Commentary: This question requires students to first identify a common base, apply the power law correctly, and then solve a resulting linear equation. It directly addresses the second performance objective.
Compound Interest and Economic Growth: In Nigeria, understanding compound interest is vital for personal finance, investment, and national economic planning. The formula for compound interest, $A = P(1+r)^n$, is an indicial expression where $A$ is the final amount, $P$ is the principal, $r$ is the interest rate, and $n$ is the number of compounding periods. Students can calculate how long it takes for an investment to double at a certain interest rate, or how much a loan will accrue over time, directly linking to the economy and daily financial decisions of Nigerians.
Population Dynamics and Epidemiology: Indices are used to model population growth (e.g., of humans, livestock, or even bacteria causing disease like cholera in a community). If a population doubles every $n$ years, the formula $P_t = P_0 \times 2^{t/n}$ is an indicial model. This can be used to project Nigeria's future population or to understand the spread of diseases, informing public health policies and resource allocation. For instance, calculating how many bacteria are present after a certain number of hours if they double every 20 minutes. Scientific Notation in Engineering and Science: Nigerian scientists, engineers, and healthcare professionals frequently encounter very large or very small numbers (e.g., distances in astronomy, sizes of microorganisms, energy calculations). Scientific notation, which uses powers of 10 ($a \times 10^n$), relies heavily on the laws of indices for operations like multiplication and division. For example, calculating the safe dosage of a drug for a patient based on extremely small quantities or dealing with large energy units in power generation (e.g., from Kainji Dam).