Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Logarithm
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Logarithm
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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 logarithms with a positive base in calculations Change the base of a logarithm
Definition of Logarithm: A logarithm is the power to which a base must be raised to produce a given number. If $b^y = x$, then $y = \log_b x$. Here, 'b' is the base ($b > 0, b \neq 1$), 'y' is the logarithm (or index), and 'x' is the number. For instance, since $2^3 = 8$, it means $3 = \log_2 8$. Similarly, $10^2 = 100$ means $2 = \log_{10} 100$. Relationship between Index Form and Logarithmic Form: It is crucial for students to seamlessly convert between these two forms: Index Form: $b^y = x$ Logarithmic Form: $y = \log_b x$ Example 2.1: Convert the following to logarithmic form: a) $3^4 = 81$ b) $5^{-2} = \frac{1}{25}$ Solution: a) From $3^4 = 81$, the base $b=3$, the power $y=4$, and the number $x=81$. So, $4 = \log_3 81$. b) From $5^{-2} = \frac{1}{25}$, the base $b=5$, the power $y=-2$, and the number $x=\frac{1}{25}$. So, $-2 = \log_5 \frac{1}{25}$. Example 2.2: Convert the following to index form: a) $\log_{10} 1000 = 3$ b) $\log_4 2 = \frac{1}{2}$ Solution: a) From $\log_{10} 1000 = 3$, the base $b=10$, the logarithm $y=3$, and the number $x=1000$. So, $10^3 = 1000$. b) From $\log_4 2 = \frac{1}{2}$, the base $b=4$, the logarithm $y=\frac{1}{2}$, and the number $x=2$. So, $4^{1/2} = 2$. --- Laws of Logarithms (with positive base): These laws are derived directly from the laws of indices. For $M, N > 0$, $b > 0$, $b \neq 1$, and $p$ being any real number: Law 1: Product Law $\log_b (MN) = \log_b M + \log_b N$ Explanation: When multiplying numbers, their logarithms (to the same base) are added. This simplifies multiplication to addition. Example 2.3: Simplify $\log_3 (9 \times 27)$.
Solution: $\log_3 (9 \times 27) = \log_3 9 + \log_3 27 = 2 + 3 = 5$. (Since $3^2=9$ and $3^3=27$).
Law 2: Quotient Law $\log_b (\frac{M}{N}) = \log_b M - \log_b N$ Explanation: When dividing numbers, the logarithm of the divisor is subtracted from the logarithm of the dividend. This simplifies division to subtraction. Example 2.4: Simplify $\log_2 (\frac{32}{8})$.
Solution: $\log_2 (\frac{32}{8}) = \log_2 32 - \log_2 8 = 5 - 3 = 2$. (Since $2^5=32$ and $2^3=8$).
Law 3: Power Law $\log_b (M^p) = p \log_b M$ Explanation: The logarithm of a number raised to a power is the power multiplied by the logarithm of the number. This simplifies exponentiation to multiplication. Example 2.5: Simplify $\log_5 (25^3)$.
Solution: $\log_5 (25^3) = 3 \log_5 25 = 3 \times 2 = 6$. (Since $5^2=25$).
Law 4: Identity Law $\log_b b = 1$ Explanation: Any positive number (except 1) raised to the power of 1 equals itself. Example 2.6: $\log_{10} 10 = 1$, $\log_7 7 = 1$.
Law 5: Zero Law $\log_b 1 = 0$ Explanation: Any positive number (except 1) raised to the power of 0 equals
1. Example 2.7: $\log_{10} 1 = 0$, $\log_e 1 = 0$.
Derived Laws (from the main three): $\log_b (\frac{1}{M}) = -\log_b M$ (Using power law: $\log_b (M^{-1}) = -1 \log_b M = -\log_b M$) $\log_{b^k} M = \frac{1}{k} \log_b M$ (Less common but useful for simplifying bases) If $\log_b x = \log_b y$, then $x=y$. (Useful for solving logarithmic equations). --- Change of Base of a Logarithm: Sometimes, a logarithm needs to be expressed in a different base, often to convert it to base 10 or base 'e' so that a calculator can be used.
Formula for Change of Base: $\log_b x = \frac{\log_c x}{\log_c b}$ where $c$ is any convenient new base ($c > 0, c \neq 1$). Typically, $c=10$ (common logarithm, often written as $\log x$) or $c=e$ (natural logarithm, written as $\ln x$).
Explanation: This formula allows conversion of a logarithm from an arbitrary base $b$ to a new base $c$. Let $y = \log_b x$. By definition, $b^y = x$. Taking the logarithm to base $c$ on both sides: $\log_c (b^y) = \log_c x$ Using the power law of logarithms: $y \log_c b = \log_c x$ $y = \frac{\log_c x}{\log_c b}$ Since $y = \log_b x$, we have $\log_b x = \frac{\log_c x}{\log_c b}$.
Special Case: If we $c=10$ (common logarithm, often written as $\log x$) or $c=e$ (natural logarithm, written as $\ln x$).
Explanation: This formula allows conversion of a logarithm from an arbitrary base $b$ to a new base $c$. Let $y = \log_b x$. By definition, $b^y = x$. Taking the logarithm to base $c$ on both sides: $\log_c (b^y) = \log_c x$ Using the power law of logarithms: $y \log_c b = \log_c x$ $y = \frac{\log_c x}{\log_c b}$ Since $y = \log_b x$, we have $\log_b x = \frac{\log_c x}{\log_c b}$.
Special Case: If we choose the new base $c=x$: $\log_b x = \frac{\log_x x}{\log_x b} = \frac{1}{\log_x b}$. This means $\log_b x = \frac{1}{\log_x b}$. This is a useful identity. Example 2.8: Convert $\log_3 7$ to base 10 and evaluate.
Solution: Using the change of base formula with $c=10$: $\log_3 7 = \frac{\log_{10} 7}{\log_{10} 3}$ Using a calculator: $\log_{10} 7 \approx 0.8451$ $\log_{10} 3 \approx 0.4771$ $\log_3 7 \approx \frac{0.8451}{0.4771} \approx 1.7713$ (to 4 decimal places). Example 2.9: Evaluate $\log_4 64$ using change of base to base
2. Solution: $\log_4 64 = \frac{\log_2 64}{\log_2 4}$ Since $2^6=64$ and $2^2=4$: $\log_2 64 = 6$ $\log_2 4 = 2$ Therefore, $\log_4 64 = \frac{6}{2} = 3$.
Self-check: Since $4^3 = 64$, the answer is correct.
Teacher Activities: Introduction & Review (10 mins): Begin by reviewing the concept of indices and their relationship to logarithms. Ask students to recall the definition of a logarithm. Briefly explain the importance of logarithms in simplifying complex calculations and their relevance in various scientific and economic fields (e.g., pH scales, population growth). Present the performance objectives for the lesson. Explanation of Laws of Logarithms (20 mins): Introduce each law of logarithm (Product, Quotient, Power, Identity, Zero) sequentially. For each law, provide a clear explanation of its derivation (briefly connecting to index laws) and illustrate with 2-3 worked examples on the board, ensuring step-by-step clarity. Encourage students to ask questions and provide simple examples for them to try immediately after each law is explained. Emphasise the conditions ($M, N > 0$, $b > 0, b \neq 1$).
Explanation of Change of Base (15 mins): Introduce the necessity of changing the base of a logarithm, particularly for calculator use (e.g., using common logarithms, base 10).
Derive the change of base formula: $\log_b x = \frac{\log_c x}{\log_c b}$. Work through 2-3 examples on the board, demonstrating how to convert to base 10 and evaluate using a calculator, and also how to convert to a simpler base (e.g., $\log_8 16$ to base 2).
Guided Practice Facilitation (15 mins): Divide the class into small groups (3-4 students). Distribute prepared practice questions covering both the laws of logarithms and change of base. Circulate among groups, providing support, clarification, and monitoring progress. Encourage peer teaching and discussion within groups.
Plenary Session & Conclusion (10 mins): Invite selected groups to present their solutions to the guided practice questions on the board. Facilitate a class discussion to correct errors and reinforce key concepts. Summarise the main learning points of the lesson.
Student Activities: Active Listening & Note-Taking: Students will listen attentively to explanations, take detailed notes, and ask clarifying questions.
Individual Practice: Students will attempt simple examples provided by the teacher immediately after each concept is introduced.
Group Work: In small groups, students will collaborate to solve the guided practice problems, discussing strategies and solutions.
Presentation: Selected students/groups will present their solutions to the class, explaining their reasoning.
Q&A: Students will participate in the class discussion, asking questions and providing answers.
Objective: To reinforce the application of logarithm laws and change of base.
Question 1 (Laws of Logarithms): Simplify the following expression: $\log_2 8 + \log_2 6 - \log_2 3$.
Solution: Using the Product Law and Quotient Law: $\log_2 8 + \log_2 6 - \log_2 3$ $= \log_2 (8 \times 6) - \log_2 3$ (Applying Product Law) $= \log_2 48 - \log_2 3$ $= \log_2 (\frac{48}{3})$ (Applying Quotient Law) $= \log_2 16$ Since $2^4 = 16$, then $\log_2 16 = 4$.
Final Answer: 4
Commentary: This question tests the combined application of the product and quotient laws. Students should apply them sequentially and then evaluate the resulting simple logarithm.
Question 2 (Laws of Logarithms): Evaluate $\frac{2 \log_3 81}{3 \log_3 9}$.
Solution: First, evaluate the individual logarithms: $\log_3 81$: Since $3^4 = 81$, $\log_3 81 = 4$. $\log_3 9$: Since $3^2 = 9$, $\log_3 9 = 2$. Substitute these values into the expression: $\frac{2 \times 4}{3 \times 2}$ $= \frac{8}{6}$ $= \frac{4}{3}$ Final Answer: $\frac{4}{3}$
Commentary: This problem requires students to evaluate basic logarithms and then perform arithmetic operations. It's important to simplify the individual log terms first.
Question 3 (Laws of Logarithms): Express $2 \log_{10} x + 3 \log_{10} y - \log_{10} z$ as a single logarithm.
Solution: Using the Power Law: $2 \log_{10} x = \log_{10} (x^2)$ $3 \log_{10} y = \log_{10} (y^3)$ The expression becomes: $\log_{10} (x^2) + \log_{10} (y^3) - \log_{10} z$ Using the Product Law: $= \log_{10} (x^2 y^3) - \log_{10} z$ Using the Quotient Law: $= \log_{10} \left(\frac{x^2 y^3}{z}\right)$ Final Answer: $\log_{10} \left(\frac{x^2 y^3}{z}\right)$
Commentary: This question requires students to reverse the application of the laws of logarithms to combine multiple terms into a single logarithm. The order of operations (power first, then product/quotient) is crucial.
Question 4 (Change of Base): Evaluate $\log_5 18$ correct to three significant figures, using common logarithms (base 10).
Solution: Using the change of base formula $\log_b x = \frac{\log_{10} x}{\log_{10} b}$: $\log_5 18 = \frac{\log_{10} 18}{\log_{10} 5}$ Using a calculator: $\log_{10} 18 \approx 1.25527$ $\log_{10} 5 \approx 0.69897$ $\log_5 18 \approx \frac{1.25527}{0.69897} \approx 1.79590$ Rounding to three significant figures: $1.79590 \approx 1.80$ Final Answer: $1.80$
Commentary: This problem directly tests the application of the change of base formula to a practical calculation using a calculator. Emphasis should be placed on correct rounding. Question 5 (Change of Base with Simplification): Evaluate $\log_8 16$ without using a calculator.
Solution: Using the change of base formula, we can convert to a common base that is a factor of both 8 and 16, which is base 2. $\log_8 16 = \frac{\log_2 16}{\log_2 8}$ Evaluate the logarithms in base 2: $\log_2 16$: Since $2^4 = 16$, $\log_2 16 = 4$. $\log_2 8$: Since $2^3 = 8$, $\log_2 8 = 3$.
Substitute these values: $\log_8 16 = \frac{4}{3}$ Final Answer: $\frac{4}{3}$
Commentary: This question requires strategic selection of the new base (base 2 in this case) to simplify calculations without a calculator, demonstrating a deeper understanding of the change of base concept.
Population Growth Modelling in Nigeria: Logarithms are used to model exponential growth or decay. For instance, if Nigeria's population grows at a certain annual rate, logarithms can help to determine the doubling time (the number of years it takes for the population to double). This is crucial for national planning, resource allocation, and infrastructure development, helping policy makers in Nigeria understand future demographic challenges and opportunities.
Example: If Nigeria's population is growing at 2.6% per annum, the doubling time $t$ can be estimated using the formula $2 = (1 + 0.026)^t$. Taking $\log_{10}$ on both sides, $\log_{10} 2 = t \log_{10} 1.026$, allowing for calculation of $t$.
Financial Calculations in Nigerian Banks: Logarithms are essential in finance for calculating compound interest, loan repayments, and investment growth over time. For individuals saving or borrowing money in Nigerian banks, logarithms help determine how long it will take for an investment to reach a certain value, or to calculate the effective interest rate.
Example: A customer invests ₦50,000 in a fixed deposit account at a Nigerian bank yielding 8% interest compounded annually. Logarithms can be used to find out how many years it will take for the investment to grow to ₦100,000. $100,000 = 50,000(1 + 0.08)^t \implies 2 = (1.08)^t \implies t = \log_{1.08} 2$. This requires a change of base for calculation. Environmental Science (pH Measurement for Agriculture): In agriculture, particularly in Nigeria where soil quality is vital for crop yield, the pH scale is logarithmic. pH measures the acidity or alkalinity of soil, which affects nutrient availability for crops like maize, cassava, or cocoa. Understanding logarithms helps agronomists interpret pH values (where a change of 1 unit on the pH scale represents a tenfold change in acidity/alkalinity) and advise farmers on appropriate soil amendments.
Example: If a soil sample has a pH of 5 and another has a pH of 7, the pH 5 soil is 100 times more acidic than the pH 7 soil. This is because pH is defined as $-\log_{10} [H^+]$.