Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Logarithm
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Logarithm
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Subject: General Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 2
Theme: Number And Numeration
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show the basiclaws of logarithm revision of the useof logarithm table for calculation
Phase 1: Introduction and Recap (10-15 minutes)
Teacher Activity: Begins by revising the concept of indices, asking students to state the laws of indices and solve simple index problems (e.g., $2^3 \times 2^4$, $3^5 \div 3^2$, $(4^2)^3$). Introduces logarithm as the inverse of indices, providing the definition $b^x = N \iff \log_b N = x$. Gives simple examples like $\log_2 8 = 3$. Briefly explains the importance of logarithms for simplifying calculations, especially with large numbers.
Student Activity: Participate in the revision of indices, answering questions and providing examples. Take notes on the definition of logarithm and its relation to indices.
Phase 2: Explanation of Logarithm Laws (20-25 minutes)
Teacher Activity: Systematically introduces each law of logarithm: Product, Quotient, and Power laws. Provides a brief, intuitive explanation or a quick proof sketch for each law (as outlined in section 2B). Works through 2-3 algebraic examples for each law (e.g., expressing $\log(ab^2/c)$ as sum/difference, or simplifying $\log_2 16 + \log_2 4$).
Highlights the special cases: $\log_b b = 1$ and $\log_b 1 = 0$.
Student Activity: Listen attentively, ask clarifying questions. Copy down the laws and worked examples. Attempt to work out the examples alongside the teacher.
Phase 3: Revision of Logarithm Table Usage (25-30 minutes)
Teacher Activity: Distributes logarithm and antilogarithm tables to all students (if available). If not, uses a projected image or a large blackboard drawing of portions of the tables. Revises the process of finding the characteristic for numbers greater than and less than
1. Gives practice numbers (e.g., 543.2, 7.8, 0.0091) and asks students to state the characteristic. Demonstrates finding the mantissa using the logarithm table for various numbers, including interpolation for mean differences. Explains finding the antilogarithm using the antilog table, emphasizing how the characteristic determines the decimal point position. Works through a simple calculation involving multiplication (e.g., $25.3 \times 1.7$) using the full log table process (log, add, antilog). Guides students through a division problem (e.g., $98.1 \div 0.23$) emphasizing careful handling of negative characteristics during subtraction.
Student Activity: Actively participate in finding characteristics and reading mantissas from tables. Follow the teacher's steps for using log and antilog tables. Practice simple calculations on their own or in pairs using their tables. Ask questions regarding characteristic rules or table readings.
Phase 4: Application and Practice (15-20 minutes)
Teacher Activity: Presents a more complex problem combining multiplication, division, and powers/roots, similar to Example 4 in section 2D. Breaks down the problem into logical steps (taking logs, applying laws, performing arithmetic, finding antilogs). Monitors students as they attempt to solve parts of the problem. Provides individual guidance and correction as needed.
Student Activity: Work in small groups or individually to solve the presented problem using log tables. Collaborate to ensure correct application of laws and table readings. Present their steps and final answers to the class.
Phase 5: Conclusion and Assignment (5 minutes)
Teacher Activity: Summarizes the key concepts: definition of logarithm, the three main laws (product, quotient, power), and the process of using logarithm tables for calculations. Assigns independent practice questions for homework.
Student Activity: Review their notes. Note down the homework assignment.
Materials: Whiteboard/Blackboard and markers/chalk NERDC approved General Mathematics textbook (for reference, though this note is self-contained) Logarithm and Antilogarithm tables (essential for each student if possible, or projected copies) Calculators (optional, for checking answers after manual calculation, not for the main task) The teacher should guide students through these questions, explaining each step and ensuring correct application of laws and table usage.
Question 1: Express $2 \log_a 3 + \log_a 5 - \log_a 15$ as a single logarithm.
Solution: Apply the power law: $2 \log_a 3 = \log_a 3^2 = \log_a 9$. The expression becomes $\log_a 9 + \log_a 5 - \log_a 15$.
Apply the product law: $\log_a 9 + \log_a 5 = \log_a (9 \times 5) = \log_a 45$. The expression becomes $\log_a 45 - \log_a 15$.
Apply the quotient law: $\log_a 45 - \log_a 15 = \log_a (45 / 15) = \log_a 3$.
Final Answer: $\log_a 3$.
Question 2: If $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$, find the value of $\log_{10} 72$.
Solution: Break down 72 into its prime factors: $72 = 8 \times 9 = 2^3 \times 3^2$.
Apply the product law: $\log_{10} 72 = \log_{10} (2^3 \times 3^2) = \log_{10} 2^3 + \log_{10} 3^2$.
Apply the power law: $\log_{10} 2^3 + \log_{10} 3^2 = 3 \log_{10} 2 + 2 \log_{10} 3$.
Substitute the given values: $3(0.3010) + 2(0.4771)$.
Calculate: $0.9030 + 0.9542 = 1.8572$.
Final Answer: $\log_{10} 72 = 1.8572$.
Question 3: Using logarithm tables, evaluate $12.56 \times 7.93 \div 0.482$.
Solution: Let $X = \frac{12.56 \times 7.93}{0.482}$. $\log X = \log 12.56 + \log 7.93 - \log 0.482$. Find $\log 12.56$: Char: $1$. Mantissa of $12$ under $5$ mean diff $6 \implies 0969 + 21 = 0990$. $\log 12.56 = 1.0990$. Find $\log 7.93$: Char: $0$. Mantissa of $79$ under $3 \implies 8993$. $\log 7.93 = 0.8993$. Find $\log 0.482$: Char: $\bar{1}$. Mantissa of $48$ under $2 \implies 6830$. $\log 0.482 = \bar{1}.6830$.
Perform additions and subtractions: $\log X = 1.0990 + 0.8993 - \bar{1}.6830$ $\log X = (1.0990 + 0.8993) - \bar{1}.6830$ $\log X = 1.9983 - \bar{1}.6830$ $\log X = 1.9983 - (-1 + 0.6830)$ $\log X = 1.9983 + 1 - 0.6830$ $\log X = 2.9983 - 0.6830 = 2.3153$. Find antilog $2.3153$: Mantissa: $0.3153$. Antilog of $31$ under $5$ mean diff $3 \implies 2065 + 1 = 2066$.
Characteristic: $2$. So, $2+1=3$ digits before decimal point. $X = 206.6$.
Final Answer: $206.6$.
Question 4: Calculate $\sqrt[4]{0.00782}$ using logarithm tables.
Solution: Let $Y = (0.00782)^{1/4}$. $\log Y = \frac{1}{4} \log 0.00782$. Find $\log 0.00782$: Characteristic: $2$ zeros after decimal $\implies -(2+1) = \bar{3}$. Mantissa of $78$ under $2 \implies 8932$. $\log 0.00782 = \bar{3}.8932$.
Divide by 4: $\log Y = \frac{1}{4} \times \bar{3}.8932$. To divide $\bar{3}.8932$ by 4, adjust the characteristic to be a multiple of 4, keeping the mantissa positive. $\bar{3}.8932 = -3 + 0.8932 = (-4 + 1) + 0.8932 = -4 + 1.8932$.
Now divide by 4: $\frac{1}{4} (-4 + 1.8932) = -1 + 0.4733 = \bar{1}.4733$. Find antilog $\bar{1}.4733$: Mantissa: $0.4733$. Antilog of $47$ under $3$ mean diff $3 \implies 2972 + 2 = 2974$.
Characteristic: $\bar{1}$. So, $1-1=0$ zeros after decimal. $Y = 0.2974$.
Final Answer: $0.2974$.
Differentiation (for diverse learners): Collaborative Learning: Pair stronger students with weaker ones to facilitate peer tutoring, especially during guided practice and table usage. Varied
Examples: Provide a range of examples from basic application of one law to complex multi-step problems, allowing students to tackle questions suitable for their current understanding.
Visual Aids: Utilize charts, diagrams, and projected log tables to cater to visual learners.
Remediation (for struggling learners): Revisit Prerequisites: If students struggle with logarithms, review the basic laws of indices thoroughly, as logarithms are built upon this foundation. Provide extra practice with converting between index and logarithm forms.
Focus on Fundamentals: Break down the lesson into smaller, more manageable parts. For instance, dedicate separate mini-lessons to mastering characteristic calculation, then mantissa lookup, and then antilog determination before combining them.
Simplified Practice: Provide exercises that focus on one specific law at a time. For log table usage, start with numbers that only require direct table lookup (no mean differences) and simple operations (just multiplication).
One-on-One Support: Offer individualised attention to address specific difficulties in characteristic determination (especially negative characteristics) or mantissa interpolation.
Use of Partial Tables: Create simplified logarithm tables with fewer entries or pre-calculated characteristics for numbers to reduce cognitive load.
Extension (for high-achieving learners): Logarithmic Equations: Introduce basic logarithmic equations requiring the application of log laws to solve for an unknown variable (e.g., $\log_2 (x+1) + \log_2 (x-1) = 3$).
Change of Base: Explore the change of base formula and how it allows for calculating logarithms to bases other than 10 using standard log tables or calculators. For instance, calculating $\log_2 5$.
Graphing Logarithmic Functions: Introduce the basic graph of $y = \log_b x$ and discuss its properties (domain, range, asymptote, inverse of exponential function).
Comparison with Calculator: Allow students to use calculators to find logarithms and compare the results with those obtained from tables, discussing the precision and limitations of each method.
Product Law: The logarithm of the product of two numbers is the sum of their individual logarithms to the same base. $\log_b (MN) = \log_b M + \log_b N$ Proof sketch: Let $\log_b M = x$ and $\log_b N = y$. Then $M = b^x$ and $N = b^y$. $MN = b^x \cdot b^y = b^{x+y}$ (from laws of indices). Taking logarithm to base $b$ on both sides: $\log_b (MN) = x+y$ Substituting back $x$ and $y$: $\log_b (MN) = \log_b M + \log_b N$.
Example: $\log_{10} (5 \times 7) = \log_{10} 5 + \log_{10} 7$. $\log_{10} 35 = \log_{10} 5 + \log_{10} 7$.
Quotient Law: The logarithm of the quotient of two numbers is the difference between the logarithm of the numerator and the logarithm of the denominator to the same base. $\log_b (M/N) = \log_b M - \log_b N$ Proof sketch: Let $\log_b M = x$ and $\log_b N = y$. Then $M = b^x$ and $N = b^y$. $M/N = b^x / b^y = b^{x-y}$. Taking logarithm to base $b$ on both sides: $\log_b (M/N) = x-y$ Substituting back $x$ and $y$: $\log_b (M/N) = \log_b M - \log_b N$.
Example: $\log_{10} (100/10) = \log_{10} 100 - \log_{10} 10 = 2 - 1 = 1$. (Since $\log_{10} 10 = 1$)
Power Law: The logarithm of a number raised to a power is the product of the power and the logarithm of the number to the same base. $\log_b (M^p) = p \log_b M$ Proof sketch: Let $\log_b M = x$, so $M = b^x$. Then $M^p = (b^x)^p = b^{xp}$. Taking logarithm to base $b$ on both sides: $\log_b (M^p) = xp$ Substituting back $x$: $\log_b (M^p) = p \log_b M$.
Example: $\log_{10} (10^3) = 3 \log_{10} 10 = 3 \times 1 = 3$.
Special Logarithms: $\log_b b = 1$ (because $b^1 = b$) $\log_b 1 = 0$ (because $b^0 = 1$)
Change of Base Law: (Less emphasis for log tables, but useful for understanding) $\log_b M = \frac{\log_c M}{\log_c b}$ This law allows conversion of logarithms from one base to another.
Example 1: Simplify $\log_3 27 + \log_3 9 - \log_3 81$ Solution: Using the laws of logarithm: $\log_3 27 = 3$ (since $3^3 = 27$) $\log_3 9 = 2$ (since $3^2 = 9$) $\log_3 81 = 4$ (since $3^4 = 81$) So, $3 + 2 - 4 = 1$. Alternatively, using product and quotient laws: $\log_3 27 + \log_3 9 - \log_3 81 = \log_3 (27 \times 9 / 81)$ $= \log_3 (243 / 81)$ $= \log_3 3 = 1$.
Example 2: Express $3 \log x + 2 \log y - \log z$ as a single logarithm.
Solution: Using the power law: $3 \log x = \log x^3$ $2 \log y = \log y^2$ So, the expression becomes $\log x^3 + \log y^2 - \log z$.
Using the product and quotient laws: $\log x^3 + \log y^2 - \log z = \log (x^3 y^2 / z)$.
Example 3: Calculate $24.75 \times 0.816$ using logarithm tables.
Solution: Let $P = 24.75 \times 0.816$ $\log P = \log 24.75 + \log 0.816$ Find $\log 24.75$: Characteristic: $2$ digits before decimal $\implies 2-1 = 1$.
Mantissa: Look up 24 under 7, difference 5. $\log 24.75 = 1.3936 + (\text{mean diff for 5}) = 1.3953$. (Assuming a standard 4-figure table, 24 under 7 is 3927, mean diff 5 is 9, so 3927+9=
3
9
3
6. This depends on the specific table.) Let's assume standard values. $\log 24.75 = 1.3953$ (from table: 24 under 7 is 3927, mean diff for 5 is 9, so 3927+9=
3
9
3
6. I will use $1.3936$) Find $\log 0.816$: Characteristic: $0$ zeros after decimal $\implies -(0+1) = \bar{1}$.
Mantissa: Look up 81 under 6. $\log 0.816 = \bar{1}.9117$ (from table: 81 under 6 is 9117).
Add the logarithms: $\log P = 1.3936 + \bar{1}.9117$ $= (1 + \bar{1}) + (0.3936 + 0.9117)$ $= 0 + 1.3053$ $= 1.3053$ Find antilog 1.3053: Mantissa: $0.3053$. Look up $30$ under $5$, difference $3$. Antilog of $0.3053 \approx 20.19$ (from table: 30 under 5 is 2018, mean diff for 3 is 1, so 2018+1=2019).
Characteristic: $1$. So, $1+1=2$ digits before the decimal. $P = 20.19$ Example 4: Evaluate $\frac{\sqrt[3]{78.9}}{\sqrt{0.56}}$ using logarithm tables.
Solution: Let $P = \frac{(78.9)^{1/3}}{(0.56)^{1/2}}$ $\log P = \log (78.9)^{1/3} - \log (0.56)^{1/2}$ $\log P = \frac{1}{3} \log 78.9 - \frac{1}{2} \log 0.56$ Find $\log 78.9$: Characteristic: $2-1=1$.
Mantissa: Look up $78$ under $9$. $\log 78.9 = 1.8971$ Find $\log 0.56$: Characteristic: $-(0+1)=\bar{1}$.
Mantissa: Look up $56$ under $0$. $\log 0.56 = \bar{1}.7482$ Perform multiplication by powers: $\frac{1}{3} \log 78.9 = \frac{1}{3} \times 1.8971 = 0.6324$ (approx) $\frac{1}{2} \log 0.56 = \frac{1}{2} \times \bar{1}.7482$ Note for negative characteristic: Convert $\bar{1}.7482$ into a form easily divisible by 2. $\bar{1}.7482 = -1 + 0.7482 = (-2 + 1) + 0.7482 = -2 + 1.7482$.
Now divide by 2: $\frac{1}{2} (-2 + 1.7482) = -1 + 0.8741 = \bar{1}.8741$.
Subtract the logarithms: $\log P = 0.6324 - \bar{1}.8741$ $\log P = 0.6324 - (-1 + 0.8741)$ $\log P = 0.6324 + 1 - 0.8741$ $\log P = 1.6324 - 0.8741 = 0.7583$ Find antilog 0.7583: Mantissa: $0.7583$. Look up $75$ under $8$, difference $3$. Antilog of $0.7583 \approx 5.732$ (from table: 75 under 8 is 5728, mean diff for 3 is 4, so 5728+4=5732).
Characteristic: $0$. So, $0+1=1$ digit before the decimal. $P = 5.732$
Financial Calculations in Nigerian Businesses: Logarithms are extensively used in finance to calculate compound interest, growth rates of investments, and loan repayments over time. For example, a small business owner in Lagos might need to calculate the future value of an investment or the repayment schedule for a bank loan. Logarithms can simplify these exponential growth/decay calculations, especially when dealing with long periods or complex rates.
Example: Calculating how many years it would take for a sum of money invested in a Nigerian bank to double at a given annual interest rate.
Formula for compound interest: $A = P(1 + r/n)^{nt}$. Taking logs can help solve for $t$ (time). Scientific Measurements and Environmental Studies: In Nigeria, particularly in environmental agencies or academic research, logarithms are crucial for measuring quantities that vary over a wide range. pH Scale: Used in agriculture and water quality monitoring to measure acidity or alkalinity. For instance, soil pH is vital for crop yield in different Nigerian regions. pH is defined as $-\log_{10}[H^+]$, where $[H^+]$ is the hydrogen ion concentration.
Sound Intensity (Decibels): Used in urban planning and noise pollution control (e.g., around major markets like Onitsha Main Market or transportation hubs). Decibel level is $10 \log_{10}(I/I_0)$, where $I$ is sound intensity.
Earthquake Magnitude (Richter Scale): Used by geologists to measure the energy released by earthquakes, relevant in regions with seismic activity. The Richter scale is logarithmic, comparing the amplitude of seismic waves.
Population Dynamics and Epidemiology: Logarithms can model population growth in Nigerian cities or the spread of diseases (epidemiology). Understanding logarithmic growth helps demographers and public health officials predict future trends and plan interventions. For instance, modeling the growth of Abuja's population or the spread of an infectious disease like cholera in a particular community.