Math 3.5 Worksheet Solution Key

3.5 Day 2 WS Name: Hour: Find an exponential function having the given values. a. f(0) = 5, f(3) = 40 y = ab^x 40 = 5b³ 8 = b³ b = 2 y = 5(2)^x b. f(3) = 16, f(6) = 128 y = ab^x 16 = ab³ 128 = ab⁶ b² = 8 b = 2 16 = a(2)³ a = 2 y = 2(2)^x 2. A colony of bacteria decays so that the population t days from now is given by f(t) = 1000(1/2)^(t/4). a. What is the amount present when t = 0? 1000 bacteria b. How much will be present in 4 days? 1000(1/2)^(4/4) = 500 bacteria c. What is the half-life? 4 days 3. The table shows the amount A(t) in grams of a radioactive element present after t days. Suppose that A(t) decays exponentially.   | t (days) | 0 | 2 | 4 | 6 | 8 | 10 | |---|---|---|---|---|---|---| | A(t) | 320 | 226 | 160 | 115 | 80 | 57 | a. What is the half-life of the element? 4 days b. About how much will be present after 16 days? 320(1/2)^(16/4) = 20 g c. Find an equation for A(t). A(t) = 320(1/2)^(t/4) 5. An amount A₀ of a radioactive iodine has a half-life of 8.1 days. In terms of A₀, how much is present after 5 days? A₀(1/2)^(5/8.1) 6. In a research experiment, a population of fruit flies is increasing according to the law of exponential growth. After 2 days, there are 125   flies, and after 4 days, there are 350 flies. How many flies will be there after 6 days? Round down to the nearest whole number.   (2, 125) (4, 350) 350 = ab⁴ 125 = ab² b² = 14/5 b = √(14/5) 125 = a(√(14/5))² a = 625/14 y = (625/14)(√(14/5))^x N = (625/14)(√(14/5))^6 ≈ 925 flies 7. The population, P (in thousands) of Las Vegas, Nevada can be modeled by P = 258.0e^(kt) where t is the year, with t = 0 corresponding to the year 1990. In 2000, the population was 478,000. a. Find the value of k for the model. Round your result to four decimal places.   478,000 = 258,000e^(10k) 478,000/258,000 = e^(10k) ln(478,000/258,000) = 10k k = ln(478,000/258,000)/10 k ≈ 0.0617 b. Use your model to predict the population in 2017. P = 258e^(0.0617 * 27) P ≈ 1,349,710 8. The population of the world in 2000 was 6.1 billion, and the estimated relative growth rate was 1.4% per year. If the population continues to grow at this   rate, when will it reach 122   billion? Use y = ae^(bt) 122 = 6.1e^(0.014t) 122/6.1 = e^(0.014t) ln(122/6.1) = 0.014t t ≈ 213.9 years 9. A culture starts with 10,000 bacteria and the number doubles every 40 minutes. a. In the half-life formula, use a 2 instead of 1/2 to find a function that models the number of bacteria at time t. A = 10,000(2)^(t/40) b. Find the number of bacteria after 1 hour. A = 10,000(2)^(60/40) ≈ 28,284 bacteria c. After how many minutes will there be 50,000 bacteria? 50,000 = 10,000(2)^(t/40) 5 = 2^(t/40) log(5) = (t/40)log(2) t ≈ 92.88 minutes 10. A culture starts with 8000 bacteria. After 1 hour, the count is 10,000. a. Find a function that models the number of bacteria   after t hours. Use y = ae^(bt) 10,000 = 8000e^b 10,000/8,000 = e^b ln(10,000/8,000) = b b ≈ 0.1508 b. Find the number of bacteria after 2 hours. y = 8000e^(0.1508 * 2) y ≈ 11,627 bacteria c. After how many hours will the number of bacteria double? 16,000 = 8000e^(0.1508t) 2 = e^(0.1508t) ln(2) = 0.1508t t ≈ 4.60 hours