Math Exponential Growth and Decay Solution

Honors Pre-Calculus 3.5 D2 Worksheet Name: Exponential Growth and Decay Exponential Growth: y = ab^x for b > 1 y = a(1 + r)^t y = ae^(rt) for population growth Exponential Decay: y = ab^x for 0 < b < 1 y = a(1 - r)^t y = ae^(-rt) for population decay Compound Interest: A = P(1 + r/n)^(nt) Compound Interest Continuously: A = Pe^(rt) Half-Life: A = a(1/2)^(t/k) *k is the interval of the half-life n = 1 annual n = 2 semiannual n = 4 quarterly n = 12 monthly n = 52 weekly n = 365 daily 1. Find an exponential function having the given values. a. f(0) = 3, f(1) = 15 y = ab^x 15 = 3b^1 b = 5 y = 3(5)^x b. f(0) = 5, f(3) = 40 40 = 5b^3 8 = b^3 b = 2 y = 5(2)^x c. f(0) = 64, f(2) = 4 4 = 64b^2 1/16 = b^2 b = 1/4 y = 64(1/4)^x d. f(0) = 80, f(4) = 5 5 = 80b^4 1/16 = b^4 b = 1/2 y = 80(1/2)^x e. f(-2) = 10, f(4) = 40 10 = ab^-2 40 = ab^4 a = 10b^2 40 = 10b^2 * b^4 4 = b^6 b = 2^1/3 y = 10(2^(1/3))^x f. f(3) = 16, f(6) = 128 128 = ab^6 16 = ab^3 a = 16/b^3 128 = 16/b^3 * b^6 8 = b^3 b = 2 y = 2(2)^x g. f(2) = 8, f(5) = 20 20 = ab^5 8 = ab^3 a = 8/b^3 20 = 8/b^3 * b^5 5/2 = b^2 b = sqrt(5/2) y = 8/((sqrt(5/2))^3) * (sqrt(5/2))^x h. f(4) = 24, f(6) = 80 80 = ab^6 24 = ab^4 a = 24/b^4 80 = 24/b^4 * b^6 10/3 = b^2 b = sqrt(10/3) y = 24/(sqrt(10/3))^4 * (sqrt(10/3))^x 2. Suppose the population of a nation grows at 3% per year. If the population was 30,000,000 people in 1990, what will be the population, to the nearest million, in the year 2000? Y = ae^(rt) Y = 30,000,000e^(0.03*10) ≈ 40,495,764.23 3. A colony of bacteria decays so that the population t days from now is given by n(t) = 1000(1/2)^(t/4). a. What is the amount present when t = 0? 1000(1/2)^(0/4) = 1000 b. How much will be present in 4 days? 1000(1/2)^(4/4) = 500 c. What is the half-life? 4 days 4. The half-life of a radioactive isotope is 4 days. If 3.2 kg are present now, how much will be present after: a. 4 days? 3.2(1/2)^(4/4) = 1.6 kg b. 8 days? 3.2(1/2)^(8/4) = 0.8 kg c. 20 days? 3.2(1/2)^(20/4) = 0.1 kg d. t days? 3.2(1/2)^(t/4) 5. The half-life of radium is about 1600 years. If 1 kg is present now, how much will be present after: a. 3200 years? 1(1/2)^(3200/1600) = 0.25 kg b. 16000 years? 1(1/2)^(16000/1600) = 0.000977 kg c. 800 years? 1(1/2)^(800/1600) ≈ 0.707 kg d. t years? (1/2)^(t/1600) 6. 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) | A(t) (grams) | |---|---| | 0 | 320 | | 2 | 226 | | 4 | 160 | | 6 | 113 | | 8 | 80 | | 10 | 57 | a. What is the half-life of the element? 4 days; it takes 4 days for 320g to become 160g b. About how much will be present after 16 days? 320(1/2)^(16/4) = 20g c. Find an equation for A(t). A(t) = 320(1/2)^(t/4) 7. When a certain medicine enters the bloodstream, it gradually dilutes, decreasing exponentially with a half-life of 3 days. The initial amount of the medicine   in the bloodstream is A₀ milliliters. What will the amount be 30 days later? A = A₀(1/2)^(30/3) = A₀(1/2)^10 ≈ 0.098% A₀ 8. A radioactive iodine has a half-life of 8.1 days. In terms of A₀, how much is present after 5 days? A = A₀(1/2)^(5/8.1) ≈ 0.652A₀ or 65.2% A₀ 9. A bacteria colony triples in population every 6 days. The population is P₀ bacteria. What will the population P(t) be t days later? P = P₀(3)^(t/6) 10. In a research experiment, a population of fruit flies is increasing according to the law of exponential growth. After 2 days there are 100 flies, and after 4 days there are 300 flies. How many flies will there be after   5 days? Round down to the nearest whole number. (2, 100) (4, 300) (5, ?) 300 = ab⁴ 100 = ab² 3 = b² b = √3 100 = a√3² a = 100/3 y = (100/3)√3^5 ≈ 519.62 or 519 flies