2.03 ExpLog - Exponential Finding Rule

2 Exponential Functions 2.03 FINDING THE RULE Exponential Functions 𝑥 𝑓 𝑥 = 𝑎(𝑐) + 𝑘 Case 1: k = 0 1) Make 2 rules using y = a(c)x 8 = a(c)3 1 = a(c)0 2) Elimination by division 8 = a(c)3 1 a(c)0 8 = c3 c=2 3) Substitute to solve “a” 1 = a(2)0 a=1 y = (2)x Case 2: k has given value 1) Make 2 rules using y = a(c)x + k 6246 = a(c)5 - 4 46 = a(c)2 - 4 2) Elimination by division after isolating exponent term 6250 = a(c)5 50 a(c)2 125 = c3 c=5 3) Substitute to solve “a” 6250 = a(5)5 a = 6250/3125 a=2 y = 2(5)x - 4 • Visions 1 • P174: 5ace, 6b Homework 1) -479 = a(c)-5 + 7 61/9 = a(c)2 +7 2) -486 = a(c)-5 -2/9 a(c)2 2187 = c-7 c = 1/3 3) 61/9 = a(1/3)2 + 7 -2/9 = a(1/3)2 -2/9 = a(1/9) a = -2 y = -2(1/3)x + 7 Given a table x y -4 -131 -3 -35 -2 -11 -1 -5 0 -7/2 1 -25/8 +96 +24 +6 +3/2 +3/8 +1/4 +1/4 +1/4 +1/4 1) Check x-values go up by 1 each step 2) Calculate Δy for each step 3) Calculate Δy2/Δy1 for each step This is your base. c = ¼ 4) Make two rules and solve system -5 = a(1/4)-1 + k -7/2 = a(1/4)0 + k -5 = a(4) + k -7/2 = a + k -5 – 4a = k -7/2 – a = k -5 – 4a = -7/2 – a -3a = 3/2 a = -1/2 -> -7/2 – (-1/2) = k k = -3 𝑦 = −12(14)𝑥 −3 EX 1 A bacteria culture doubles every day. Initially the culture contains 8 bacteria. Find the rule. Initial value = a when k = 0 “Rate” = c c=2 -> a = 8 𝑓 𝑥 = 8(2)𝑥 EX 2 If the same culture doubles three times per day, find the rule. Exponent represents # of times rate is applied, so two options: 1) Let x = 1/3 of a day or 2) Let x = # of days 𝑓 𝑥 = 8(2)𝑥 𝑓 𝑥 = 8(2)3𝑥 Test: How many bacteria exist in 5 days? 1) In 5 days, x = 15 or 2) x = 5 𝑓 𝑥 = 8(2)15 𝑓 𝑥 = 8(2)3(5) 𝑓 𝑥 = 8(32768) 𝑓 𝑥 = 8(32768) 𝑓 𝑥 = 262144 𝑓 𝑥 = 262144 If culture doubles every three days, 𝑓 𝑥 = 8(2)𝑥/3 In Words: EX 3 The population of Exponentville varies according to an exponential function. On Jan 1st 1975, it was 130,000. On Jan 1st 1985, it was 260,000. No mention of constant -> k = 0 Let x = yrs since 1975 (0,130000) & (10,260000) 260000 = a(c)10 130000 = a(c)0 2 = c10 c = 10√2 c = (2)1/10 -> a = 130000 1 10 𝑃 𝑥 = 130000 2 𝑃 𝑥 = 𝑥 𝑥 130000(2)10 PERCENTAGE RATES c = (1 ± r) + if growth, - if decay. r = percentage as decimal (ie 8% = 0.08) EX 5: A $35000 car loses 15% of its value every year. Represent the value of the car as a function of time (exponential). No constant value mentioned: k = 0 Value decreases: decay -> c = (1-0.15) c = 0.85 Value of the car is 𝑓 𝑥 = 35000(0.85)𝑥 Also a = 35000 PERCENTAGE RATES EX 6: Re-write the rule if same car is worth $300 as scrap metal. Scrap metal value does not decrease with time: k = 300 Therefore initial value is a + k = 35000 a = 34700 Value of the car is 𝑓 𝑥 = 34700(0.85)𝑥 + 300 PERCENTAGE RATES EX 8: A municipality is planning a tax increase of 6% a year starting in 2019. Mr. Black paid $1225 in taxes in 2018 and he wants to find a rule to calculate his annual tax amount as a function of x years. k=0 a = 1225 c = (1+0.06) = 1.06 𝑓 𝑥 = 1225(1.06)𝑥 COMPOUND INTEREST C 𝑥 = 𝐶𝑜 (1 + 𝑟 𝑛𝑥 ) 𝑛 C(x) = capital after “x” years Co = amount invested r = annual interest rate (decimal) n = number of times per year ‘rate’ is applied x = number of years Ex: terms: semi-annually -> n = 2 monthly -> n = 12 every 5 years -> n = 1/5 EX 9: $1000 is at a fixed annual interest rate of 6.0%, compounded every four months. What is the value of the investment after the five years? Co = 1000 r = 0.06 x = 5 years Compounded every four months -> n = 3 0.06 3(5) C = 1000(1 + ) 3 𝐶 = 1000(1 + 0.02)15 𝐶 = 1000 1.345868338 C = $1345.87 EX 10: Three years ago, Greg invested $1000 at a fixed annual interest rate compounded every six months. His investment is currently valued at $1400. What is the annual rate? Co = 1000 C(x) = 1400 x = 3 years Compounded every six months -> n = 2 𝑟 2(3) 1400 = 1000(1 + ) 2 𝑟 6 1.4 = (1 + ) 2 𝑟 6 1.4 = 1 + 2 𝑟 1.057680926 = 1 + 2 0.115361852 = 𝑟 r = 11.54% per year • Visions 1 • P175: 10, 13bcd, 16, 17 Homework