Logarithm Properties

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Logarithm Properties · Grade 11 · Worksheet 1

1. log₃(27) + log₃(9) - log₃(81) = ? Answer: 1 Solution: log₃(27) + log₃(9) - log₃(81) Rewrite each term as a power of 3 27 = 3³ → log₃(27) = log₃(3³) 9 = 3² → log₃(9) = log₃(3²) 81 = 3⁴ → log₃(81) = log₃(3⁴) log₃(3³) + log₃(3²) - log₃(3⁴) The rule is: logₐ(aᵇ) = b log₃(3³) = 3 log₃(3²) = 2 log₃(3⁴) = 4 3 + 2 - 4 3 + 2 = 5 5 - 4 = 1 Final answer: 1
   Full step-by-step solution

   Let's solve step by step. We have: log₃(27) + log₃(9) - log₃(81) --- **Step 1: Rewrite each term as a power of 3** 27 = 3³ → log₃(27) = log₃(3³) 9 = 3² → log₃(9) = log₃(3²) 81 = 3⁴ → log₃(81) = log₃(3⁴) So the expression becomes: log₃(3³) + log₃(3²) - log₃(3⁴) --- **Step 2: Apply the logarithm power rule** The rule is: logₐ(aᵇ) = b So: log₃(3³) = 3 log₃(3²) = 2 log₃(3⁴) = 4 Now the expression is: 3 + 2 - 4 --- **Step 3: Simplify the arithmetic** 3 + 2 = 5 5 - 4 = 1 --- **Final answer:** 1

2. A right triangle is drawn on a coordinate plane with vertices at (0,0), (6,0), and (6,8). A circle is inscribed in this triangle, tangent to all three sides. What is the exact area of this inscribed circle? Answer: 4π Solution: The inradius of any triangle can be found by dividing twice the area by the perimeter. For right triangles specifically, there's also a relationship between the inradius and the lengths of the legs and hypotenuse.
   Full step-by-step solution

   The inradius of any triangle can be found by dividing twice the area by the perimeter. For right triangles specifically, there's also a relationship between the inradius and the lengths of the legs and hypotenuse. Understanding these geometric relationships helps solve problems involving circles inscribed in polygons.

3. Olivia is an environmental scientist studying the acidity of rainwater in two different regions. The pH of a solution is defined by the formula pH = -log[H⁺], where [H⁺] is the hydrogen ion concentration in moles per liter. In Region A, the hydrogen ion concentration is 5.0 × 10⁻⁵ M. In Region B, the hydrogen ion concentration is 2.0 × 10⁻⁶ M. Using properties of logarithms, determine how many times more acidic the rainwater in Region A is compared to Region B. Answer: 25 Solution: The ratio of hydrogen ion concentrations (how many times more acidic) is [H⁺]ₐ / [H⁺]₆ = (5.0 × 10⁻⁵) / (2.0 × 10⁻⁶). Simplify the ratio: (5.0 / 2.0) × (10⁻⁵ / 10⁻⁶) = 2.5 × 10⁻⁵⁻⁽⁻⁶⁾ = 2.5 × 10¹ = 25.
   Full step-by-step solution

   Step 1: The ratio of hydrogen ion concentrations (how many times more acidic) is [H⁺]ₐ / [H⁺]₆ = (5.0 × 10⁻⁵) / (2.0 × 10⁻⁶). Step 2: Simplify the ratio: (5.0 / 2.0) × (10⁻⁵ / 10⁻⁶) = 2.5 × 10⁻⁵⁻⁽⁻⁶⁾ = 2.5 × 10¹ = 25. Step 3: Therefore, the rainwater in Region A is 25 times more acidic than in Region B. The answer is 25.

4. log₂(8x) + log₂(x/2) = 6 Answer: x = 4 Solution: Use the product rule: log₂(8x) + log₂(x/2) = log₂((8x)(x/2)) = log₂(4x²). The equation becomes log₂(4x²) = 6. Rewrite in exponential form: 2⁶ = 4x².
   Full step-by-step solution

   Step 1: Use the product rule: log₂(8x) + log₂(x/2) = log₂((8x)(x/2)) = log₂(4x²). Step 2: The equation becomes log₂(4x²) = 6. Step 3: Rewrite in exponential form: 2⁶ = 4x². Step 4: Compute 2⁶ = 64, so 64 = 4x². Step 5: Divide both sides by 4: 16 = x². Step 6: Take the square root: x = 4 (since x > 0 for the logarithm to be defined). The answer is x = 4.

5. log₇(343) + log₉(81) - log₅(125) = ? Answer: 2 Solution: Evaluate log₇(343) 7^x = 343 7^3 = 343 So log₇(343) = 3 Evaluate log₉(81) 9^x = 81 9^2 = 81 So log₉(81) = 2 Evaluate log₅(125) 5^x = 125 5^3 = 125 So log₅(125) = 3 Substitute the values into the original expression 3 + 2 - 3 = 5 - 3 = 2 The answer is 2.
   Full step-by-step solution

   Step 1: Evaluate log₇(343) 7^x = 343 7^3 = 343 So log₇(343) = 3 Step 2: Evaluate log₉(81) 9^x = 81 9^2 = 81 So log₉(81) = 2 Step 3: Evaluate log₅(125) 5^x = 125 5^3 = 125 So log₅(125) = 3 Step 4: Substitute the values into the original expression 3 + 2 - 3 = 5 - 3 = 2 The answer is 2.

6. Ava is an environmental scientist studying the pH levels of rainwater in a polluted region. She measures the hydrogen ion concentration [H⁺] of a sample to be 6.31 × 10⁻⁶ moles per liter. The pH is defined by the formula pH = -log[H⁺]. Using the properties of logarithms, determine the pH of the rainwater sample. Round your final answer to one decimal place. Answer: 5.2 Solution: Write the pH formula: pH = -log[H⁺] Substitute the given concentration: pH = -log(6.31 × 10⁻⁶) Apply the product property of logarithms: log(6.31 × 10⁻⁶) = log(6.31) + log(10⁻⁶) Evaluate log(10⁻⁶) = -6 (since log base 10 of 10 to any power is that power) Use a calculator to find log(6.31) ≈…
   Full step-by-step solution

   Step 1: Write the pH formula: pH = -log[H⁺] Step 2: Substitute the given concentration: pH = -log(6.31 × 10⁻⁶) Step 3: Apply the product property of logarithms: log(6.31 × 10⁻⁶) = log(6.31) + log(10⁻⁶) Step 4: Evaluate log(10⁻⁶) = -6 (since log base 10 of 10 to any power is that power) Step 5: Use a calculator to find log(6.31) ≈ 0.8000 Step 6: So, log(6.31 × 10⁻⁶) = 0.8000 + (-6) = -5.2000 Step 7: Now apply the negative sign: pH = -(-5.2000) = 5.2000 Step 8: Round to one decimal place: pH = 5.2 The answer is 5.2.

7. Liam is studying the decay of a radioactive isotope in his chemistry lab. The amount of the substance remaining after t years is given by the function A(t) = 500 × e^(-0.025t). He wants to determine how many years it will take for exactly half of the original sample to remain. Using the properties of logarithms, find the half-life of this isotope. Answer: 27.73 Solution: A(t) = 500 × e^(-0.025t) The initial amount is 500. Half of the initial amount is 500/2 = 250.
   Full step-by-step solution

   We are given the decay function: A(t) = 500 × e^(-0.025t) The initial amount is 500. Half of the initial amount is 500/2 = 250. We set A(t) = 250 and solve for t: 250 = 500 × e^(-0.025t) Step 1: Divide both sides by 500: 250/500 = e^(-0.025t) 1/2 = e^(-0.025t) Step 2: Take the natural logarithm of both sides: ln(1/2) = ln(e^(-0.025t)) Step 3: Use the property ln(e^x) = x: ln(1/2) = -0.025t Step 4: Rewrite ln(1/2) as ln(1) - ln(2): 0 - ln(2) = -0.025t -ln(2) = -0.025t Step 5: Multiply both sides by -1: ln(2) = 0.025t Step 6: Solve for t: t = ln(2) / 0.025 Step 7: Compute ln(2) ≈ 0.693147: t ≈ 0.693147 / 0.025 Step 8: Perform the division: 0.693147 / 0.025 = 693.147 / 25 = 27.72588 Step 9: Round to two decimal places: t ≈ 27.73 years Thus, the half-life of the isotope is 27.73 years.

8. log₄(64x) + log₄(x/4) = 4 Answer: x = 4 Solution: Use the product property: log₄(64x) + log₄(x/4) = log₄[(64x)(x/4)] Simplify inside the logarithm: (64x)(x/4) = 64x²/4 = 16x² So the equation becomes: log₄(16x²) = 4 Rewrite in exponential form: 4⁴ = 16x² Calculate 4⁴ = 256, so 256 = 16x² Divide both sides by 16: x² = 16 Take the positive square…
   Full step-by-step solution

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