Logarithm Properties

lessonbunny.com

Answer Key & Explanations

Logarithm Properties · Grade 11 · Worksheet 3

1. log₃(81x) + log₃(x/9) = 6 Answer: x = 27 Solution: The equation becomes log₃(9x²) = 6. Rewrite in exponential form: 3⁶ = 9x². Compute 3⁶ = 729, so 729 = 9x².
   Full step-by-step solution

   Step 1: Apply the product property: log₃(81x) + log₃(x/9) = log₃(81x * x/9) = log₃(81x²/9) = log₃(9x²). Step 2: The equation becomes log₃(9x²) = 6. Step 3: Rewrite in exponential form: 3⁶ = 9x². Step 4: Compute 3⁶ = 729, so 729 = 9x². Step 5: Divide both sides by 9: x² = 81. Step 6: Take the square root: x = 9 or x = -9. Since the domain of log₃(x/9) requires x/9 > 0, x must be positive. Thus x = 9. The answer is x = 9.

2. log₂(32x) + log₂(x/2) = 7 Answer: x = 4 Solution: The equation becomes log₂(16x²) = 7. Rewrite in exponential form: 2⁷ = 16x². Compute 2⁷ = 128, so 128 = 16x².
   Full step-by-step solution

   Step 1: Apply the product rule: log₂(32x) + log₂(x/2) = log₂((32x)(x/2)) = log₂(16x²). Step 2: The equation becomes log₂(16x²) = 7. Step 3: Rewrite in exponential form: 2⁷ = 16x². Step 4: Compute 2⁷ = 128, so 128 = 16x². Step 5: Divide both sides by 16: x² = 8. Step 6: Take the square root: x = √8 = 2√2. Step 7: Since the original equation has log₂(x/2), x must be positive, so x = 2√2. The answer is x = 2√2.

3. log₄(64x) + log₄(x/16) = 4 Answer: x = 8 Solution: Step 1: Apply the product property: log₄(64x) + log₄(x/16) = log₄[(64x)(x/16)] Step 2: Simplify inside the logarithm: (64x)(x/16) = (64/16)x² = 4x² Step 3: So the equation becomes: log₄(4x²) = 4 Step 4: Rewrite in exponential form: 4⁴ = 4x² Step 5: Calculate 4⁴ = 256, so 256 = 4x² Step 6: Divide…
   Full step-by-step solution

   Step 1: Apply the product property: log₄(64x) + log₄(x/16) = log₄[(64x)(x/16)] Step 2: Simplify inside the logarithm: (64x)(x/16) = (64/16)x² = 4x² Step 3: So the equation becomes: log₄(4x²) = 4 Step 4: Rewrite in exponential form: 4⁴ = 4x² Step 5: Calculate 4⁴ = 256, so 256 = 4x² Step 6: Divide both sides by 4: 64 = x² Step 7: Take the square root: x = 8 (since x > 0 for the logarithm to be defined) The answer is x = 8.

4. log₃(x) + log₃(x - 8) = 2 Answer: 9 Solution: Rewrite in exponential form: 3² = x(x - 8). Simplify: 9 = x² - 8x. Rearrange to standard form: x² - 8x - 9 = 0.
   Full step-by-step solution

   Step 1: Apply the product property: log₃(x) + log₃(x - 8) = log₃[x(x - 8)] = 2. Step 2: Rewrite in exponential form: 3² = x(x - 8). Step 3: Simplify: 9 = x² - 8x. Step 4: Rearrange to standard form: x² - 8x - 9 = 0. Step 5: Factor the quadratic: (x - 9)(x + 1) = 0. Step 6: Solve: x = 9 or x = -1. Step 7: Check for extraneous solutions: For x = 9, log₃(9) and log₃(1) are defined (both arguments positive). For x = -1, log₃(-1) is undefined. So x = -1 is extraneous. The answer is 9.

5. Hana is studying the pH levels of soil samples from her family's farm. The pH of a solution is defined as pH = -log[H⁺], where [H⁺] is the hydrogen ion concentration in moles per liter. One soil sample has a hydrogen ion concentration of 3.2 × 10⁻⁶ M, and another sample from a different field has a concentration of 1.6 × 10⁻⁵ M. Using properties of logarithms, how many times more acidic is the second sample compared to the first? (Note: A higher [H⁺] means more acidic.) Answer: 5 Solution: Let [H⁺]₁ = 3.2 × 10⁻⁶ and [H⁺]₂ = 1.6 × 10⁻⁵. The ratio of acidity is [H⁺]₂ / [H⁺]₁ = (1.6 × 10⁻⁵) / (3.2 × 10⁻⁶). Simplify the ratio: 1.6 / 3.2 = 0.5, and 10⁻⁵ / 10⁻⁶ = 10¹ = 10.
   Full step-by-step solution

   Step 1: Let [H⁺]₁ = 3.2 × 10⁻⁶ and [H⁺]₂ = 1.6 × 10⁻⁵. Step 2: The ratio of acidity is [H⁺]₂ / [H⁺]₁ = (1.6 × 10⁻⁵) / (3.2 × 10⁻⁶). Step 3: Simplify the ratio: 1.6 / 3.2 = 0.5, and 10⁻⁵ / 10⁻⁶ = 10¹ = 10. So the ratio = 0.5 × 10 = 5. Step 4: Therefore, the second sample is 5 times more acidic than the first. The answer is 5.

6. log₂(16) + log₂(4) - log₂(8) = ? Answer: 3 Solution: Since 2^3 = 8, log₂(8) = 3 The answer is 3.
   Full step-by-step solution

   Step 1: Apply the product rule to the first two terms: log₂(16) + log₂(4) = log₂(16 × 4) = log₂(64) Step 2: Now apply the quotient rule: log₂(64) - log₂(8) = log₂(64 ÷ 8) = log₂(8) Step 3: Evaluate log₂(8). Since 2^3 = 8, log₂(8) = 3 The answer is 3.

7. log₅(25x) + log₅(x/5) = 5 Answer: x = 25 Solution: Use the product property: log₅(25x) + log₅(x/5) = log₅(25x * x/5) = log₅(25x²/5) = log₅(5x²) The equation becomes log₅(5x²) = 5 Rewrite in exponential form: 5⁵ = 5x² 5⁵ = 3125, so 3125 = 5x² Divide both sides by 5: x² = 625 Take the square root: x = 25 (since x > 0 for the original logarithms to…
   Full step-by-step solution

   Step 1: Use the product property: log₅(25x) + log₅(x/5) = log₅(25x * x/5) = log₅(25x²/5) = log₅(5x²) Step 2: The equation becomes log₅(5x²) = 5 Step 3: Rewrite in exponential form: 5⁵ = 5x² Step 4: 5⁵ = 3125, so 3125 = 5x² Step 5: Divide both sides by 5: x² = 625 Step 6: Take the square root: x = 25 (since x > 0 for the original logarithms to be defined) The answer is x = 25.

8. A marine biologist is studying the acidity of ocean water samples. She measures the pH of a sample using the formula pH = -log[H⁺], where [H⁺] is the hydrogen ion concentration in moles per liter. If one sample has a hydrogen ion concentration of 2.5 × 10⁻⁸ M, while another sample has a concentration of 4.0 × 10⁻⁹ M, how many times more acidic is the first sample compared to the second? Answer: 6.25 Solution: Calculate the pH of the first sample: pH₁ = -log(2.5 × 10⁻⁸) = -[log(2.5) + log(10⁻⁸)] = -[0.3979 - 8] = 7.6021 Calculate the pH of the second sample: pH₂ = -log(4.0 × 10⁻⁹) = -[log(4.0) + log(10⁻⁹)] = -[0.6021 - 9] = 8.3979 The ratio of hydrogen ion concentrations is [H⁺]₁/[H⁺]₂ = (2.5 ×…
   Full step-by-step solution

   Step 1: Calculate the pH of the first sample: pH₁ = -log(2.5 × 10⁻⁸) = -[log(2.5) + log(10⁻⁸)] = -[0.3979 - 8] = 7.6021 Step 2: Calculate the pH of the second sample: pH₂ = -log(4.0 × 10⁻⁹) = -[log(4.0) + log(10⁻⁹)] = -[0.6021 - 9] = 8.3979 Step 3: The ratio of hydrogen ion concentrations is [H⁺]₁/[H⁺]₂ = (2.5 × 10⁻⁸)/(4.0 × 10⁻⁹) = 6.25 Step 4: Since acidity is directly proportional to hydrogen ion concentration, the first sample is 6.25 times more acidic than the second sample. The answer is 6.25.