Fractions #16 - Questions and Answers

Question 1 Use the properties of logarithms, given that ln ⁡ ( 2 ) ≈ 0.6931 ln(2)≈0.6931 and ln ⁡ ( 3 ) ≈ 1.0986 ln(3)≈1.0986 to approximate the logarithms. Use a calculator to confirm your approximations. (a) ln ⁡ ( 0.75 ) ≈ ln(0.75)≈ (b) ln ⁡ ( 288 ) ≈ ln(288)≈ (c) ln ⁡ ( 3 × 24 ) ≈ ln( 3×24 ​ )≈ (d) ln ⁡ ( 1 6 ) ≈ ln( 6 1 ​ )≈ Answer 1 The handwritten notes in the image provide the following steps and results for each logarithmic approximation: (a) ln ⁡ ( 0.75 ) = ln ⁡ ( 3 4 ) = ln ⁡ ( 3 ) − ln ⁡ ( 4 ) = ln ⁡ ( 3 ) − 2 ln ⁡ ( 2 ) = 1.0986 − 2 ( 0.6931 ) = 0.224 ln(0.75)=ln( 4 3 ​ )=ln(3)−ln(4)=ln(3)−2ln(2)=1.0986−2(0.6931)=0.224 (b) ln ⁡ ( 288 ) = ln ⁡ ( 2 5 ⋅ 3 2 ) = 5 ln ⁡ ( 2 ) + 2 ln ⁡ ( 3 ) = 5 ( 0.6931 ) + 2 ( 1.0986 ) = 3.4655 + 2.1972 = 5.6627 ln(288)=ln(2 5 ⋅3 2 )=5ln(2)+2ln(3)=5(0.6931)+2(1.0986)=3.4655+2.1972=5.6627 (c) ln ⁡ ( 3 ⋅ 24 ) = ln ⁡ ( 72 ) = 1 2 ln ⁡ ( 72 ) = 1 2 [ ln ⁡ ( 2 3 ⋅ 3 2 ) ] = 1 2 [ 3 ln ⁡ ( 2 ) + 2 ln ⁡ ( 3 ) ] = 1 2 [ 3 ( 0.6931 ) + 2 ( 1.0986 ) ] = 1.753 ln( 3⋅24 ​ )=ln( 72 ​ )= 2 1 ​ ln(72)= 2 1 ​ [ln(2 3 ⋅3 2 )]= 2 1 ​ [3ln(2)+2ln(3)]= 2 1 ​ [3(0.6931)+2(1.0986)]=1.753 (d) ln ⁡ ( 1 6 ) = ln ⁡ ( 1 ) − ln ⁡ ( 6 ) = 0 − [ ln ⁡ ( 2 ) + ln ⁡ ( 3 ) ] = − ( 0.6931 + 1.0986 ) = − 1.7917 ln( 6 1 ​ )=ln(1)−ln(6)=0−[ln(2)+ln(3)]=−(0.6931+1.0986)=−1.7917 Question 2 Set up the form for the partial fraction decomposition. Do not solve for A A, B B, C C, and so on. Answer 1 Step 1: Background Concept Partial fraction decomposition is the process of converting a simplified rational function into initial individual separate rational functions. Step 2: Obtain the partial fractions The given simplified rational polynomial is: x − 14 x 2 − 3 x x 2 −3x x−14 ​ The given rational function can be decomposed into partial fractions as: x − 14 x ( x − 3 ) = A x + B x − 3 x(x−3) x−14 ​ = x A ​ + x−3 B ​ Question 3 Write an informal explanation of partial fraction decomposition. Answer 1 Step 1: Partial fraction decomposition is the process of starting with the simplified answer and taking it back apart, "decomposing" the final expression into its initial polynomial fractions. For example: 2 x + 1 x + 1 = 2 ( x + 1 ) + x x ( x + 1 ) = 2 x + 2 + x x ( x + 1 ) = 3 x + 2 x ( x + 1 ) x 2 ​ + x+1 1 ​ = x(x+1) 2(x+1)+x ​ = x(x+1) 2x+2+x ​ = x(x+1) 3x+2 ​ Step 2: To decompose a fraction, you first factor the denominator. Let’s work backwards from the example above. The denominator is x ( x + 1 ) x(x+1). Step 3: Then you write the fractions with one of the factors for each of the denominators. Assign variables A A and B B for these unknown values: A x + B x + 1 x A ​ + x+1 B ​ Step 4: Then set this sum equal to the simplified result: 3 x + 2 x ( x + 1 ) = A x + B x + 1 x(x+1) 3x+2 ​ = x A ​ + x+1 B ​ Question 4 Perform the indicated operations and express the answers in simplified form: x + 3 10 + 2 ( x + 1 ) 15 − x − 2 18 10 x+3 ​ + 15 2(x+1) ​ − 18 x−2 ​ Answer 1 Step 1: Given expression is: x + 3 10 + 2 ( x + 1 ) 15 − x − 2 18 10 x+3 ​ + 15 2(x+1) ​ − 18 x−2 ​ Step 2: The objective is to express the answers in simplified form. First, find the least common multiplier (LCM) of 10, 15, and 18. So, 90 is the least common multiplier of 10, 15, 18. Step 3: Adjust the fractions based on the LCM: ( x + 3 ) ⋅ 9 90 + 2 ( x + 1 ) ⋅ 6 90 − ( x − 2 ) ⋅ 5 90 90 (x+3)⋅9 ​ + 90 2(x+1)⋅6 ​ − 90 (x−2)⋅5 ​ Step 4: Since the denominators are equal, combine the fractions: ( x + 3 ) ⋅ 9 + 2 ( x + 1 ) ⋅ 6 − ( x − 2 ) ⋅ 5 90 = 15 x + 43 90 90 (x+3)⋅9+2(x+1)⋅6−(x−2)⋅5 ​ = 90 15x+43 ​ Question 5 Find the LCD of each set of fractions: 1 5 , 3 14 , 4 35 5 1 ​ , 14 3 ​ , 35 4 ​ Answer 1 Step 1: LCD is the least common denominator of the fractions. To calculate the LCD, we need to find the LCM of the denominators of the fractions after simplification. Step 2: Find the LCM of 5, 14, and 35: 5 = 5 14 = 7 × 2 7×2 35 = 7 × 5 7×5 LCM of 5, 14, and 35 = Product of highest powers of all the prime factors of the numbers: LCM ( 5 , 14 , 35 ) = 7 × 5 × 2 = 70 LCM(5,14,35)=7×5×2=70 Thus, the LCD is 70.