Theory of Logarithms.

Theory of Logarithms. Grade 10 Mathematics: Logarithms of Numbers Subtopics Navigator Introduction Meaning of Logarithms Simplifying Logarithmic Expressions Operations with Logarithms Applications Cumulative Exercises Conclusion Lesson Objectives Understand what logarithms are and identify them in mathematical expressions. Simplify logarithmic expressions using appropriate techniques. Perform basic operations with logarithms. Apply logarithms to solve real-world problems. Lesson Introduction Logarithms are an important mathematical concept that helps us solve exponential equations and work with very large or very small numbers. For example, the logarithm base 2 of 32 is 5, written as [latex]log_2 32 = 5[/latex] because [latex]2^5 = 32[/latex]. Logarithms are important in mathematics because they allow us to transform multiplicative relationships into additive ones, making complex calculations more manageable. Meaning of Logarithms A logarithm answers the question: "To what exponent must we raise a base to get a certain number?" The general form is: [latex]log_b a = c[/latex] which means [latex]b^c = a[/latex], where: b is the base (b > 0, b ≠ 1) a is the argument (a > 0) c is the logarithm or exponent Common logarithms have base 10 ([latex]log_{10}[/latex]) and natural logarithms have base e ([latex]ln[/latex]). Example 1: Evaluate [latex]log_3 81[/latex] Solution: We need to find the exponent to which 3 must be raised to get 81. Since [latex]3^4 = 81[/latex], then [latex]log_3 81 = 4[/latex]. Meaning of Logarithms Evaluate [latex]log_2 16[/latex] Evaluate [latex]log_5 125[/latex] Evaluate [latex]log_{10} 1000[/latex] Logarithmic Expressions We can simplify logarithmic expressions using several important properties: [latex]log_b 1 = 0[/latex] (since [latex]b^0 = 1[/latex]) [latex]log_b b = 1[/latex] (since [latex]b^1 = b[/latex]) [latex]log_b b^x = x[/latex] [latex]b^{log_b x} = x[/latex] Example 2: Simplify [latex]log_7 1 + log_5 5 - log_3 9[/latex] Solution: [latex]log_7 1 = 0[/latex] (property 1) [latex]log_5 5 = 1[/latex] (property 2) [latex]log_3 9 = log_3 3^2 = 2[/latex] (property 3) So the expression becomes: 0 + 1 - 2 = -1 Simplifying Logarithmic Expressions Simplify [latex]log_4 1 + log_2 8[/latex] Simplify [latex]3^{log_3 7}[/latex] Simplify [latex]log_6 36 - log_{10} 1[/latex] Operations with Logarithms Logarithms follow specific rules that make operations easier: Product Rule: [latex]log_b (mn) = log_b m + log_b n[/latex] Quotient Rule: [latex]log_b left(frac{m}{n}right) = log_b m - log_b n[/latex] Power Rule: [latex]log_b (m^n) = n log_b m[/latex] Change of Base: [latex]log_b a = frac{log_c a}{log_c b}[/latex] Example 3: Expand [latex]log_2 left(frac{8x^3}{y}right)[/latex] Solution: Using the quotient rule: [latex]log_2 (8x^3) - log_2 y[/latex] Using the product rule on the first term: [latex]log_2 8 + log_2 x^3 - log_2 y[/latex] Using the power rule: [latex]log_2 2^3 + 3log_2 x - log_2 y[/latex] Simplifying: [latex]3 + 3log_2 x - log_2 y[/latex] Operations with Logarithms Expand [latex]log_5 (25x^2y)[/latex] Condense [latex]2log_3 a + frac{1}{2}log_3 b - 3log_3 c[/latex] Evaluate [latex]log_2 15[/latex] using common logarithms Applications Logarithms have many real-world applications: Earthquake measurement - The Richter scale uses logarithms to measure earthquake intensity Sound intensity - Decibels use a logarithmic scale pH scale - Acidity and alkalinity are measured using logarithms Compound interest - Logarithms help calculate investment growth over time Example 4: The magnitude of an earthquake is given by [latex]M = log_{10} left(frac{I}{I_0}right)[/latex], where I is the intensity of the earthquake and I₀ is a reference intensity. If an earthquake has an intensity 100,000 times greater than I₀, what is its magnitude? Solution: [latex]M = log_{10} (100,000) = log_{10} (10^5) = 5[/latex] The earthquake has a magnitude of 5 on the Richter scale. Exercises (Applications) If a sound is 1000 times more intense than the threshold of hearing, what is its decibel level? (dB = 10·log₁₀(I/I₀)) How long will it take for an investment to double at 5% annual interest compounded annually? (Use the rule of 72 or logarithms) The pH of a solution is given by pH = -log₁₀[H⁺]. If [H⁺] = 0.001, what is the pH? Cumulative Exercises Evaluate [latex]log_4 64[/latex] Simplify [latex]log_2 32 + log_3 1 - log_5 25[/latex] Expand [latex]log_7 left(frac{49x^4}{y^3}right)[/latex] Condense [latex]3log_2 a - 2log_2 b + frac{1}{2}log_2 c[/latex] Solve for x: [latex]log_3 x = 4[/latex] Solve for x: [latex]log_x 16 = 2[/latex] Evaluate [latex]log_5 20[/latex] using common logarithms If [latex]log_{10} 2 ≈ 0.3010[/latex] and [latex]log_{10} 3 ≈ 0.4771[/latex], find [latex]log_{10} 6[/latex] How many times more intense is a 60 dB sound than a 30 dB sound? An investment grows according to A = P(1 + r)ⁿ. How long will it take for $1000 to grow to $2000 at 7% annual interest? Show/Hide Answers Problem: Evaluate [latex]log_4 64[/latex] Step 1: Recognize that 64 = 4³ Step 2: Therefore, [latex]log_4 64 = log_4 4^3 = 3[/latex] Answer: 3 Problem: Simplify [latex]log_2 32 + log_3 1 - log_5 25[/latex] Step 1: [latex]log_2 32 = log_2 2^5 = 5[/latex] Step 2: [latex]log_3 1 = 0[/latex] Step 3: [latex]log_5 25 = log_5 5^2 = 2[/latex] Step 4: 5 + 0 - 2 = 3 Answer: 3 Problem: Expand [latex]log_7 left(frac{49x^4}{y^3}right)[/latex] Step 1: Use quotient rule: [latex]log_7 (49x^4) - log_7 y^3[/latex] Step 2: Use product rule: [latex]log_7 49 + log_7 x^4 - log_7 y^3[/latex] Step 3: Use power rule: [latex]log_7 7^2 + 4log_7 x - 3log_7 y[/latex] Step 4: Simplify: [latex]2 + 4log_7 x - 3log_7 y[/latex] Answer: [latex]2 + 4log_7 x - 3log_7 y[/latex] Problem: Condense [latex]3log_2 a - 2log_2 b + frac{1}{2}log_2 c[/latex] Step 1: Apply power rule in reverse: [latex]log_2 a^3 - log_2 b^2 + log_2 c^{1/2}[/latex] Step 2: Combine using product and quotient rules: [latex]log_2 left(frac{a^3 sqrt{c}}{b^2}right)[/latex] Answer: [latex]log_2 left(frac{a^3 sqrt{c}}{b^2}right)[/latex] Problem: Solve for x: [latex]log_3 x = 4[/latex] Step 1: Convert to exponential form: [latex]x = 3^4[/latex] Step 2: Calculate: [latex]x = 81[/latex] Answer: 81 Problem: Solve for x: [latex]log_x 16 = 2[/latex] Step 1: Convert to exponential form: [latex]x^2 = 16[/latex] Step 2: Solve for x: [latex]x = sqrt{16} = 4[/latex] (we take the positive root since base must be positive) Answer: 4 Problem: Evaluate [latex]log_5 20[/latex] using common logarithms Step 1: Use change of base formula: [latex]log_5 20 = frac{log_{10} 20}{log_{10} 5}[/latex] Step 2: Calculate: [latex]frac{log_{10} (2 times 10)}{log_{10} (10/2)} = frac{log_{10} 2 + 1}{1 - log_{10} 2}[/latex] Step 3: Using [latex]log_{10} 2 ≈ 0.3010[/latex]: [latex]frac{0.3010 + 1}{1 - 0.3010} = frac{1.3010}{0.6990} ≈ 1.861[/latex] Answer: Approximately 1.861 Problem: If [latex]log_{10} 2 ≈ 0.3010[/latex] and [latex]log_{10} 3 ≈ 0.4771[/latex], find [latex]log_{10} 6[/latex] Step 1: Note that 6 = 2 × 3 Step 2: Use product rule: [latex]log_{10} 6 = log_{10} 2 + log_{10} 3[/latex] Step 3: Calculate: 0.3010 + 0.4771 = 0.7781 Answer: 0.7781 Problem: How many times more intense is a 60 dB sound than a 30 dB sound? Step 1: Use the formula dB = 10·log₁₀(I/I₀) Step 2: For 60 dB: 60 = 10·log₁₀(I₁/I₀) ⇒ log₁₀(I₁/I₀) = 6 ⇒ I₁/I₀ = 10⁶ Step 3: For 30 dB: 30 = 10·log₁₀(I₂/I₀) ⇒ log₁₀(I₂/I₀) = 3 ⇒ I₂/I₀ = 10³ Step 4: Ratio: I₁/I₂ = (10⁶)/(10³) = 10³ = 1000 Answer: 1000 times more intense Problem: How long will it take for $1000 to grow to $2000 at 7% annual interest? Step 1: Use the formula A = P(1 + r)ⁿ with A = 2000, P = 1000, r = 0.07 Step 2: 2000 = 1000(1.07)ⁿ ⇒ 2 = (1.07)ⁿ Step 3: Take logarithms: log(2) = n·log(1.07) Step 4: Solve for n: n = log(2)/log(1.07) ≈ 0.3010/0.0294 ≈ 10.24 years Answer: Approximately 10.24 years Conclusion/Recap Logarithms are a powerful mathematical tool that helps us work with exponential relationships. We learned how to evaluate logarithms, simplify logarithmic expressions using properties like the product, quotient, and power rules, and apply logarithms to solve real-world problems. 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