Questions and Answers #10 Modeling

Questions and Answers Sheet 10 Modeling Question #1 Area of triangle Given: Base b=(y+1) units and Height h=2y units. Answer: Calculation: The area of a triangle is half the product of the height and base. 1 𝐴𝑟𝑒𝑎 = × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 2 1 Plug base=(y+1) and height =2y in the formula 𝐴𝑟𝑒𝑎 = × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 1 2 𝐴𝑟𝑒𝑎 = × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 2 1 = ⋅ (𝑦 + 1) ⋅ (2𝑦) = 2 2𝑦(𝑦+1) 2 = 𝑦(𝑦 + 1) = (𝑦 2 + 𝑦) Distributing y over (y+1) = (𝑦 2 + 𝑦) 𝑢𝑛𝑖𝑡 2 Thus, 𝐴𝑟𝑒𝑎 = (𝑦 2 + 𝑦)𝑢𝑛𝑖𝑡 2 Conclusion: = (𝑦 2 + 𝑦)𝑢𝑛𝑖𝑡 2 Question #2 The expression 5 × 103 Answer: Calculation: We have been given the expression 5 × 103 . The exponent on any number states that the number will get multiplied to itself as many times as the exponent. For example in the expression 𝑥 4 , the exponent ''4'' indicates that the base ''x'' will get multiplied to itself four times. Using this concept we can write 103 as 103 = 10 × 10 × 10 = 1000 Therefore, the given expression becomes 5 × 103 = 5 × 1000 =5000 Thus, we have 5 × 103 = 5000 Conclusion: Using the property of exponent, we have 5 × 103 = 5000 Question #3 The opposite of a polynomial Given: −3𝑧 8 Answer: Calculations: Opposite of a polynomial is additive inverse of the polynomial. Thus, addition of the given and the opposite polynomial is zero. (−1)(−3𝑧 8 ) On simplify we get: = (−1 × −3)𝑧 8 = (3)𝑧 8 = 3𝑧 8 Thus, opposite of polynomial −3𝑧 8 is 3𝑧 8 Conclusion: Opposite of polynomial −3𝑧 8 is 3𝑧 8 Question #4 A single expression for the logarithmic expression log 1 0𝑥 5 − log 5 𝑥 Answer: Formula used: 𝑚 Quotient rule of logarithm is given by log 𝑎 𝑚 − log 𝑎 𝑛 = log 𝑎 ( ) Quotient rule of exponent is given by 𝑥𝑎 𝑥𝑏 𝑛 =𝑥 𝑎−𝑏 Calculation: 5 We have been given the expression log 1 0𝑥 − log 5 𝑥 and we need to write it in single logarithmic expression. For this we will do below steps: log 1 0𝑥 5 − log 5 𝑥 Given expression = log ( 10𝑥 5 𝑚 ) Quotient rule: log 𝑎 𝑚 − log 𝑎 𝑛 = log 𝑎 ( ) 5𝑥 4) = log(2𝑥 Exponent property 𝑥𝑎 𝑥𝑏 𝑛 = 𝑥 𝑎−𝑏 Thus, the required single logarithmic expression is given by log(2𝑥 4 ) Conclusion: Using the properties of logarithms, the single logarithmic expression is given by log(2𝑥 4 ) Question #5 The solution to approximate the logarithm to the nearest thousandth using the change base formula 3 log 7 (105). Answer: Formula used: Change of Base Formula: A formula that allows you to rewrite a logarithm in terms of logs written with another base. This is especially helpful when using a calculator to evaluate a log to any base other than 10 or e. Assume that x, and b are all positive. Also assume that 𝑎 ≠ 1, 𝑏 ≠ 1 . Change of base formula: log 𝑎 𝑥 = log𝑏 𝑥 log𝑏 𝑎 (𝑎 ≠ 1, 𝑏 ≠ 1) Calculation: Description 3log 7 (105) 3log10 (105) 3(2.02119) Step 1: Apply change of base formula for given logarithm. 3log 7 (105) = = 7.174973 log10 (7) 0.84510 Step 2: Write the algorithmic value to nearest thousand Conclusion: The solution for the logarithm to nearest thousandth is 3 log 7 (105) = 7.175 Question #6 The equation 𝑦 = 𝑏𝑥 𝑎 can be written as log 𝑦 = log 𝑏 + 𝑎 log 𝑥 3log 7 (105) = 7.175 Answer: Formula used: Logarithmic product rule: log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦 Logarithmic power rule: log 𝑎 𝑥 𝑛 = 𝑛 log 𝑎 𝑥 Calculation: We have been given that 𝑦 = 𝑏𝑥 𝑎 Take log both sides, we get log 𝑦 = log(𝑏𝑥 𝑎 ) Apply the product rule log 𝑎 𝑥 𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦 log 𝑦 = log 𝑏 + 𝑎 log 𝑥 Therefore, 𝑦 = 𝑏𝑥 𝑎 can be written as log 𝑦 = log 𝑏 + 𝑎 log 𝑥 Conclusion: Product rule and Power rule of logarithm are used. 𝑦 = 𝑏𝑥 𝑎 Can be written as log 𝑦 = log 𝑏 + 𝑎 log 𝑥. Question #7 The value of the variable by solving the exponential equation. The given exponential equation is 4𝑒 𝑥 = 5 Answer: Calculation: The given exponential equation is 4𝑒 𝑥 = 5,solve it as follows: 4𝑒 𝑥 = 5 Divide both the sides by 4 𝑒𝑥 = 5 4 Divide 5 by 4 to get 𝑒 𝑥 = 1.25 Take natural logarithm on both the sides to get: ln(𝑒 𝑥 ) = ln(1.25) By the inverse property of the natural log arithm we know: ln(𝑒 𝑥 ) = 𝑥. So, we get 𝑥 = ln(1.25) Find the value of ln(1.25) using calculator. x=0.223 Conclusion: The given exponential equation has been solved to get x=0.223 . Question #8 The solution of equation log(2 − 𝑥) = 0.5 Answer: Formula used: The property of logarithm: log 𝑏 (𝑥) = 𝑦 is equivalent to 𝑏 𝑦 = 𝑥 Calculation: log(2 − 𝑥) = 0.5 Using the property of logarithm: log 𝑏 (𝑥) = 𝑦 is equivalent to 𝑏 𝑦 = 𝑥 ⇒ 2 − 𝑥 = 100.5 ⇒ −𝑥 = 100.5 − 2 ⇒ −𝑥 = 3.1623 − 2 ⇒ −𝑥 = 1.16 ⇒ 𝑥 = −1.16 Conclusion: Using the properties of logarithm, we solved the given equation to get x=-1.16 . Question #9 The value of D and a in T(t)=T(0)+Dat. To estimate the temperature of pan after 10 minutes. The time taken by pan to reach 400 F. Answer: Given: T(t)=T(0)+Dat and T0 is 320 F. Boiling water with a temperature of 2120 and it cools to 700 in 30 minutes. Here D is the difference between room temperature and water temperature. Calculation: a) To determine the value of D and a: We have T(t)=T(0)+Dat ----(1). Here T0=320F ----(2) Water temperature is 2120F Hence D=212-32=180 , Next, D=180, T0=32 and t=30 minutes and t=12 hour. Substitute the above values in the equation (1), T(12)=32+180a12 Here T(12)=70 . On substituting the value of T(12) in the above we get, 70 = 32 + 180𝑎1270 − 32 = 180𝑎12180𝑎12 = 38𝑎12 = 38180𝑎12 = 0.211(𝑎12)2 = (0.211)2𝑎 = 0.044. Hence the values of D, T0 and a are 180, 32 and 0.044 respectively. b) To estimate the temperature of pan after 10 minutes. Here 10 minutes =16 hour. So, t=16 . Substitute the value of t, a, D and t0 in the equation (1) we get, 𝑇(12) = 32 + 180(0.044)16𝑇(12) = 32 + 180(0.594)𝑇(12) = 32 + 106.95𝑇(12) = 138.95𝑇(12) ≅ 1390𝐹 Hence the value of T(t) after 10 minutes is 1390F. c) The time taken by pan to reach 400 F. Given: T(t)=40, T0=32, D=180 and a=0.044 . Substitute the above values in equation (1) we get, 40 = 32 + 180(0.044)𝑡40 − 32 = 180(0.044)𝑡180(0.044)𝑡 = 8(0.044)𝑡 = 8180(0.044)𝑡 = 0.044 Applying log on both sides, log(0.044) 𝑡 = log 0. 044𝑡 log 0. 044 = log 0. 044𝑡 = 1 Hence the time taken by pan to reach 400 F is 1 hour. Conclusion: The value of D is 180, a is 0.044 and T0 is 320 F. The temperature of the pan after 10 minutes is 1390 F. The time taken by pan to reach 400 F is 1 hour. Question #10 The value y when x=40 and the value of x when y=1.5 Answer: Given: 𝑦 = 2 − log(100 − 𝑥) 0.37 given the number of years y required for x percent of the sp ⇒ s to die, where 0 ≤ 𝑥 ≤ 95 . Galculation: a) To determine the value of y when x=40 : We have 𝑦 = 2 − log(100 − 𝑥) 0.37 ----(1). Here x=40 . Substitute the value of x in equation (1) we get, 𝑦 = 2 − log(100 − 𝑥) 0.37𝑦 = 2 − log(100 − 40) 0.37𝑦 = 2 − log 6 00.37𝑦 = 2 − 1.7780.37𝑦 = 0.2220.37𝑦 = 0.6 Hence the values of y when x=40 is 0.6 b) To determine the value of x when y=1.5 . We have 𝑦 = 2 − log(100 − 𝑥) 0.37----(1) Here y=1.5 . Substitute the value of y in equation (1) we get, 𝑦 = 2 − log(100 − 𝑥) 0.371.5 = 2 − log(100 − 𝑥) 0.371.5 × 0.37 = 2 − log(100 − 𝑥) 0.555 = 2 − log(100 − 𝑥) 0.555 − 2 = − log(100 − 𝑥) log(100 − 𝑥) = 1.445100 − 𝑥 = 101.445100 − 𝑥 = 27.86𝑥 = 100 − 27.86𝑥 = 72.14 Hence the value of x when y=1.5 is 72.14. Conclusion: Thus, the value of y when x=40 is 0.6 and the value of x when y=1.5 is 72.14.