F.LE.A.4: Exponential Growth

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Regens Exam Quesions F.LE.A.4: Exponenial Growh www.jmap.org Name: F.LE.A.4: Exponenial Growh 1 A populaion of rabbis doubles every days according o he formula P = 10(2), where P is he populaion of rabbis on day. Wha is he value of when he populaion is 320? 1) 240 2) 300 3) 6 4) 9 2 The growh of baceria in a dish is modeled by he funcion f() = 2 f() = 32? 1) 8 2) 2 3) 15 4) 16 3. For which value of is 3 Given a saring populaion of 100 baceria, he formula b = 100(2 ) can be used o find he number of baceria, b, afer periods of ime. If each period is 15 minues long, how many minues will i ake for he populaion of baceria o reach 51,200? 5 If ae b = c, where a, b, and c are posiive, hen equals 1) ln c ab 2) ln cb a ln c a 3) b ln c a 4) ln b 6 Susie invess $500 in an accoun ha is compounded coninuously a an annual ineres rae of 5%, according o he formula A = Pe r, where A is he amoun accrued, P is he principal, r is he rae of ineres, and is he ime, in years. Approximaely how many years will i ake for Susie s money o double? 1) 1.4 2) 6.0 3) 13.9 4) 14.7 4 Drew s parens invesed $1,500 in an accoun such ha he value of he invesmen doubles every seven years. The value of he invesmen, V, is deermined by he equaion V = 1500(2), where represens he number of years since he money was deposied. How many years, o he neares enh of a year, will i ake he value of he invesmen o reach $1,000,000? 7 7 Akeem invess $25,000 in an accoun ha pays 4.75% annual ineres compounded coninuously. Using he formula A = Pe r, where A = he amoun in he accoun afer years, P = principal invesed, and r = he annual ineres rae, how many years, o he neares enh, will i ake for Akeem s invesmen o riple? 1) 10.0 2) 14.6 3) 23.1 4) 24.0 1

Regens Exam Quesions F.LE.A.4: Exponenial Growh www.jmap.org 8 Sean invess $10,000 a an annual rae of 5% compounded coninuously, according o he formula A = Pe r, where A is he amoun, P is he principal, e = 2.718, r is he rae of ineres, and is ime, in years. Deermine, o he neares dollar, he amoun of money he will have afer 2 years. Deermine how many years, o he neares year, i will ake for his iniial invesmen o double. Name: 12 The number of baceria presen in a Peri dish can be modeled by he funcion N = 50e 3, where N is he number of baceria presen in he Peri dish afer hours. Using his model, deermine, o he neares hundredh, he number of hours i will ake for N o reach 30,700. 9 Judih pus $5000 ino an invesmen accoun wih ineres compounded coninuously. Which approximae annual rae is needed for he accoun o grow o $9110 afer 30 years? 1) 2% 2) 2.2% 3) 0.02% 4) 0.022% 10 In New York Sae, he minimum wage has grown exponenially. In 1966, he minimum wage was $1.25 an hour and in 2015, i was $8.75. Algebraically deermine he rae of growh o he neares percen. 11 A house purchased 5 years ago for $100,000 was jus sold for $135,000. Assuming exponenial growh, approximae he annual growh rae, o he neares percen. 13 Afer siing ou of he refrigeraor for a while, a urkey a room emperaure (68 F) is placed ino an oven a 8 a.m., when he oven emperaure is 325 F. Newon s Law of Heaing explains ha he emperaure of he urkey will increase proporionally o he difference beween he emperaure of he urkey and he emperaure of he oven, as given by he formula below: T = T a + T 0 T a e k T a = he emperaure surrounding he objec T 0 = he iniial emperaure of he objec = he ime in hours T = he emperaure of he objec afer hours k = decay consan The urkey reaches he emperaure of approximaely 100 F afer 2 hours. Find he value of k, o he neares housandh, and wrie an equaion o deermine he emperaure of he urkey afer hours. Deermine he Fahrenhei emperaure of he urkey, o he neares degree, a 3 p.m. 2

F.LE.A.4: Exponenial Growh Answer Secion 1 ANS: 2 320 = 10(2) 32 = (2) log 32 = log(2) log 32 = log 2 log32 log 2 = 300 = REF: 011205a2 2 ANS: 3... REF: 080502b 3 ANS: 135. REF: 010923b 1

4 ANS: 65.7.... REF: 080729b 5 ANS: 3 e b = c a ln e b = ln c a bln e = ln c a = ln c a b REF: 011813aii 6 ANS: 3 1000 = 500e.05 2 = e.05 ln 2 = ln e.05 ln 2.05 13.9 =.05 ln e.05 REF: 061313a2 2

7 ANS: 3 75000 = 25000e.0475 3 = e.0475 ln 3 = ln e.0475 ln 3.0475 ln e =.0475.0475 23.1 REF: 061117a2 8 ANS: 11052, 14.... REF: 0330b 9 ANS: 1 9110 = 5000e 30r ln 911 30r = ln e 500 ln 911 500 = r 30 r.02 REF: 011810aii 10 ANS: 8.75 = 1.25x 49 4 7 = x 49 49 x = 7 1.04 REF: 081730aii 3

11 ANS: A = Pe r 135000 = 100000e 5r 1.35 = e 5r ln1.35 = ln e 5r ln1.35 = 5r.06 r or 6% REF: 061632aii 12 ANS: 30700 = 50e 3 614 = e 3 ln614 = ln e 3 ln614 = 3 ln e ln614 = 3 2.14 REF: 011333a2 13 ANS: 100 = 325 + (68 325)e 2k 225 = 257e 2k ln 225 257 k = 2 k 0.066 T = 325 257e 0.066 T = 325 257e 0.066(7) 163 REF: fall1513aii 4