Questions and Answers #7 Modeling

Questions and Answers Sheet 7 Modeling Question #1 To write and solve the system of linear equations to find the price of 1 cup of dried fruit and 1 cup of almonds. Given: A small bag of trail mix contains 3 cups of driedd fruit and 4 cups of almonds. A large bag contains 1 4 cups of dried fruit and 6 cups of almonds. 2 Answer: Calculation: Let the cost of 1 cup of dried fruit be x And let the cost of 1 cup of almonds be y According to the question, 3x+4y=6 1 9 4 𝑥 + 6𝑦 = 9 ⇒ 𝑥 + 6𝑦 = 9 2 2 ⇒ 3𝑥 + 4𝑦 = 6 … … … . . (𝑖) ⇒ 9𝑥 + 12𝑦 = 18 … … … … . (𝑖𝑖) Now, divide equation (ii) by 3 ⇒ 3𝑥 + 4𝑦 = 6 … … … (𝑖𝑖𝑖) Now, subtract equation (i) from equation (iii), ⇒ (3𝑥 + 4𝑦) − (3𝑥 + 4𝑦) = 6 − 6 ⇒ 3𝑥 + 4𝑦 − 3𝑥 − 4𝑦 = 0 ⇒0=0 Therefore, there will be an infinite number of solutions that satisfy both the equations. Conclusion: Therefore, the price of 1 cup of dried fruit and 1 cup of almonds has infinite set of solutions. Question #2 25−3/2 by hand using properties of exponents. Answer: Calculation: 25−3/2 is in the form of 𝑎−𝑚 . Hence, using the second formula 𝑎−𝑚 = 1 𝑎𝑚 we rewrite 25−3/2 as 25−3/2 = 1 In order to evaluate as 253/2 = 25 1 ( )3 2 253/2 1 253/2 3/2 , first we evaluate 25 which is in the denominator. For this we rewrite 253/2 . Now using the first formula 𝑎𝑚𝑛 = (𝑎𝑚 )𝑛 = (𝑎𝑛 )𝑚 , we rewrite 25 1 2 ( )3 as 25 1 2 ( )3 1 3 ) = (25{2 . We know that 1 2 25 = √25 = 5. 1 ( )3 Hence, 253/2 = 25 2 = 53 = 125 As the denominator is evaluated, we can now substitute this in 25−3/2 = Therefore, our original expression to be evaluated, 25 −3/2 = 1 253/2 = 1 125 1 253/2 to complete the calculation. i.e., 25−3/2 = Conclusion: Therefore, using the properties of exponent, 𝑎𝑚𝑛 = (𝑎𝑚 )𝑛 = (𝑎𝑛 )𝑚 and 𝑎−𝑚 = 25 −3/2 = 1 125 without using calculator. 1 𝑎𝑚 1 125 we easily found out Question #3 (0.5−2 )2 by hand using properties of exponents. Answer: To evaluate (0.5−2 )2 , note that 0.5 is Thus, (0.5−2 )2 = [( 1−2 2 1 −2 2 𝑎𝑚 𝑏 𝑏𝑚 ( ) = 2 1 −2 as ( ) 2 2 . )] . 1 −2 Now consider ( ) . Were write ( ) 𝑎 𝑚 1 2 =( 1−2 2−2 )= using the property of exponents, 1 2−2 By another property of exponents, 𝑎−𝑚 = 1 −2 ( ) 2 =( 1−2 2−2 )= 1 2−2 1 , 1 𝑎𝑚 2−2 becomes 2−(−2) = 22 = 4, hence, =4 2 1 −2 Therefore, (0.5−2 )2 = [( ) ] = 42 = 16 2 Conclusion: Therefore, (0.5−2 )2 = 16 and is evaluated using properties of exponents without using calculator. Question #4 The solution of the given equation. 𝑥 + 13 = 2𝑥 − 13 Answer: Key concepts used: We will cute both sides to remove radicals and solve the equation for x. Calculation: The given equation: 𝑥 + 13 = 2𝑥 − 13 . Taking cube of both sides, we get (𝑥 + 13)3 = (2𝑥 − 13)3[∵ (𝑎3)3 = 𝑎] ⇒ 𝑥 + 1 = 2𝑥 − 1 ⇒ 𝑥 = 2. Thus, the possible solution is x=2 . Let us now verify this solution. To check the solution. To check the solution, we need to substitute in the given equation. For 𝑥 = 22 + 13 = 2(2) − 13 ⇒ 2 + 13 = 4 − 13 ⇒ 33 = 33 (True) So, for x=2 , the given equation is verified. Thus, the required solution is x=2. Conclusion: We cubed the given equation. Then we got a linear equation. Finally, we solved this linear equation to get the values. Substituting the value in the given equation, we could identify the final solution which satisfied the given equation. Question #5 The average rate of change of S. Given: Distance function 𝑆(𝑡) = 96𝑡, 𝑡1 = 12, 𝑡2 = 92 Answer: Formula used: Average rate = 𝑓(𝑏) − 𝑓(𝑎)𝑏 − 𝑎 Calculation: We are given 𝑆(𝑡) = 96𝑡, 𝑡1 = 12, 𝑡2 = 92 , we will plug these in the formula. Average rate of change = 𝑓(𝑏) − 𝑓(𝑎)𝑏 − 𝑎 We get Average rate of change of 𝑆(𝑡) = 𝑆(92) − 𝑆(12)92 − 12 Average rate of change of 𝑆(𝑡) = (96)(92) − (96)(12)82 Average rate of change of 𝑆(𝑡) = (432) − (48)4 Average change of rate of 𝑆(𝑡) = 20.784 − 6.9284 Average rate of change of 𝑆(𝑡) = 3.5 Conclusion: The average rate of change of S is 3.5 Question #6 The solution of the given equation. 5(1.3) 𝑥 + 4 = 104 Answer: Concept used: The property of natural logarithmic functions. Calculation: 51.3 + 4 = 104 ⇒ 51.3 = 104 − 4 ⇒ 51.3 = 100 100 ⇒ 1.3𝑥 = 5 ⇒ 1.3𝑥 = 20 Taking natural logarithm of both sides: ⇒ ln(1.3𝑥 ) = ln 2 0 ⇒ 𝑥 ln 1. 3 = ln 2 0[∵ ln 𝑎𝑚 = 𝑚 ln 𝑎] ⇒𝑥= ⇒𝑥= ln 20 ln 1.3 2.995732273 0.262364264 (Using a scientific calculator) ⇒ 𝑥 = 11.418217643 ⇒ 𝑥 = 11.418 (Rounded to three decimal places) Conclusion: The equation 5(1.3) 𝑥 + 4 = 104 has the solution x=11.418 . Question #7 The solution of the given equation. Given: log 3 𝑥 = 4 Answer: Concept used: The property of natural logarithmic functions. Calculation: log 3 𝑥 = 4 ⇒ 𝑥 = 34 ⇒ 𝑥 = 81 Conclusion: The equation log 3 𝑥 = 4 has the solution x=81 . Question #8 Solve the logarithmic equation. The logarithm equation is ln(5 − 𝑥) − ln(5 + 𝑥) = − ln 9 Answer: Formula used: log 𝑏 (𝑚𝑛 ) = 𝑛 log 𝑏 𝑚 m log b ( ) = log b m − log b n n Calculation: The expression can be further simplified by using the exponent property of the logarithm logb (mn ) = n logb m ln(5 − 𝑥) − ln(5 + 𝑥) = − ln 9 ln(5 − 𝑥) − ln(5 + 𝑥) = ln 9−1 1 ln(5 − 𝑥) − ln(5 + 𝑥) = ln 9 𝑚 The further expression can be simplified using the division property of the logarithm. log 𝑏 ( ) = log 𝑏 𝑚 − 𝑛 log 𝑏 𝑛 ln ( 5−𝑥 5+𝑥 ) = ln 1 9 If the equation contains only two logarithms on opposite sides of the equal sign with the same base then the problem can be solved by simply dropping the logarithm. 5−𝑥 1 = 5+𝑥 9 9(5 − 𝑥) = (5 + 𝑥) 45 − 9𝑥 = 5 + 𝑥 10𝑥 = 40 𝑥=4 Conclusion: By using the properties of logarithm, the solution of logarithmic equation is 4. Question #9 To Find: Determine the following values a) (𝑓 + 𝑔)(2) b) (𝑓 − 𝑔)(12) c) (𝑓𝑔)(4) d) (𝑓𝑔)(0) Answer: Given: The functions 𝑓(𝑥) = 2𝑥 + 1 and 𝑔(𝑥) = 1𝑥 We can define addition of two functions as follows (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) We can define subtraction of two functions as follows (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) We can define multiplication of two functions as follows (𝑓𝑔)(𝑥) = 𝑓(𝑥) × 𝑔(𝑥) We can define division of two functions as follows (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) Calculation: a) Since (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) Substituting f(x) and g(x) we get, (𝑓 + 𝑔)(𝑥) = 2𝑥 + 1 + 1𝑥 Substituting 𝑥 = 2 we get, (𝑓 + 𝑔)(2) = 2(2) + 1 + 12 (𝑓 + 𝑔)(2) = 5 + 12 (𝑓 + 𝑔)(2) = 112 b) Since (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) Substituting f(x) and g(x) we get, (𝑓 − 𝑔)(𝑥) = 2𝑥 + 1 − 1𝑥 Substituting 𝑥 = 12 we get, (𝑓 − 𝑔)(12) = 2(12) + 1 − 112 (𝑓 − 𝑔)(12) = 1 + 1 − 2 (𝑓 − 𝑔)(12) = 0 c) Since (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) Substituting f(x) and g(x) we get, (𝑓𝑔)(𝑥) = (2𝑥 + 1)(1𝑥) Substituting 𝑥 = 4 we get, (𝑓𝑔)(4) = (2 × 4 + 1)(14) (𝑓𝑔)(4) = (94) d) Since (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) Substituting f(X) and g(x) we get, (𝑓𝑔)(𝑥) = (2𝑥 + 1)(1𝑥) (𝑓𝑔)(𝑥) = 𝑥(2𝑥 + 1) Substituting x=0 we get, (𝑓𝑔)(0) = 0(2 × 0 + 1) (fg)(0)=0 Conclusion: The value of the functions are as below a) (𝑓 + 𝑔)(2) = 112 b) (𝑓 − 𝑔)(12) = 0 c) (𝑓𝑔)(4) = (94) d) (𝑓𝑔)(0) = 0 Question #10 To Evaluate: The value of the following. a) g(-3) b) g(b) c) g(x3) d) g(2x-3) Answer: Given: 𝑔(𝑥) = 2𝑥 + 1 Calculation: a) We have given 𝑔(𝑥) = 2𝑥 + 1 On Substituting 𝑥 = −3 we get, 𝑔(−3) = 2(−3) + 1 𝑔(−3) = −6 + 1 𝑔(−3) = −5 b) We have given 𝑔(𝑥) = 2𝑥 + 1 On Substituting 𝑏 = 𝑥 we get, 𝑔(𝑏) = 2(𝑏) + 1 𝑔(𝑏) = 2𝑏 + 1 c) We have given 𝑔(𝑥) = 2𝑥 + 1 On replacing x by x3 we get, 𝑔(𝑥3) = 2(𝑥3) + 1 𝑔(𝑥3) = 2𝑥3 + 1 d) We have given g(x)=2x+1 On replacing x by 2x-3 we get, 𝑔(2𝑥 − 3) = 2(2𝑥 − 3) + 1 𝑔(2𝑥 − 3) = 4𝑥 − 6 + 1 𝑔(2𝑥 − 3) = 4𝑥 − 5 Conclusion: a) The value of 𝑔(−3) is -5 b) The value of g(b) is 2𝑏 + 1 c) The value of 𝑔(𝑥3) is 2𝑥3 + 1 d) The value of 𝑔(2𝑥 − 3) is 4𝑥 − 5