Exponential Decay GEOLOGY The half-life of Sodium-22 is 2.6 years. Determine the value of k and the equation of decay for Sodium-22. If a is the initial amount of the substance, then the amount y that remains after 2.6 years is or 0.5a. Exponential decay formula Replace y with 0.5a and t with 2.6. Divide each side by a. Exponential Decay In 0.5 = In e–2.6k In 0.5 = –2.6k Property of Equality for Logarithmic Functions Inverse Property of Exponents and Logarithms Divide each side by –2.6. 0.2666 ≈ k Use a calculator. Answer: The value of k of Sodium-22 is 0.2666. Thus, the equation for the decay of Sodium-22 is y = ae–0.2666t, where t is given in years. HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. What is the value of k for radioactive iodine? A. k = –0.0866 B. k = –4.1589 C. k = 0.0866 D. k = 4.1589 Carbon Dating GEOLOGY A geologist examining a meteorite estimates that it contains only about 10% as much Sodium-22 as it would have contained when it reached the surface of the Earth. How long ago did the meteorite reach the surface of the Earth? Understand The formula for the decay of Sodium-22 is y = ae–kt. You want to find how long ago the meteorite reached Earth. Plan Let a be the initial amount of Sodium-22 in the meteorite. The amount y that remains after t years is 10% of a or 0.10a. Carbon Dating Solve y = ae–0.2666t 0.1a = ae–0.2666t 0.1 = e–0.2666t Formula for the decay of Sodium-22 Replace y with 0.1a. Divide each side by a. In 0.1= ln e–0.2666t Property of Equality for Logarithms ln 0.1= –0.2666t Inverse Property for Exponents and Logarithms Divide each side by –0.2666. 8.64 ≈ t Use a calculator. Carbon Dating Answer: It reached the surface of Earth about 8.6 years ago. Check Use the formula to find the amount of the sample remaining after 8.6 years. Use an original amount of 1. y = ae-0.2666t Original equation = 1e-0.2666(8.6) a = 1 and t = 8.6 ≈ 0.101 or 10% Use a calculator. HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. A doctor wants to know when the amount of radioactive iodine in a patient’s body is 20% of the original amount. When will this occur? A. about 0.05 hour later B. about 0.39 hour later C. about 2.58 hours later D. about 18.58 hours later Continuous Exponential Growth A. POPULATION In 2007, the population of China was 1.32 billion. In 2000, it was 1.26 billion. Determine the value of k, China’s relative rate of growth. Formula for continuous exponential growth y = 1.32, a = 1.26, and t = 2007 – 2000 or 7 Divide each side by 1.26. Continuous Exponential Growth Property of Equality for Logarithmic Equations ln e x = x Divide each side by 7. Use a calculator. Answer: China’s relative rate of growth is about 0.0066, or about 0.66%. Continuous Exponential Growth B. POPULATION In 2007, the population of China was 1.32 billion. In 2000, it was 1.26 billion. When will China’s population reach 1.5 billion? Formula for continuous exponential growth y = 1.5, a = 1.26, and k = 0.0066 ≈ Divide each side by 1.26. ≈ Property of Equality for Logarithmic Functions Continuous Exponential Growth In 1.1905 ≈ 0.0066t ln e x = x Divide each side by 0.0066. Use a calculator. Answer: China’s population will reach 1.5 billion in 2026. Continuous Exponential Growth C. POPULATION In 2007, the population of China was 1.32 billion. India’s population in 2007 was 1.13 billion and can be modeled by y = 1.13e0.015t. Determine when India’s population will surpass China’s. (Note: t represents years after 2007.) Formula for exponential growth Property of Inequality for Logarithms Continuous Exponential Growth In 1.32 + In e0.0066t < In 1.13 + In e0.015t In 1.32 + 0.0066t < In 1.13 + 0.015t In 1.32 – In 1.13 < 0.0084t Product Property of Logarithms ln e x = x Subtract (0.0066t + ln 1.13) from each side. Divide each side by 0.0084. Use a calculator. Answer: India’s population will surpass China’s in 18.5 years, or midway through 2025. Logistic Growth A. A city’s population in millions is modeled by , where t is the number of years since 2000. Graph the function. Answer: Logistic Growth B. A city’s population in millions is modeled by , where t is the number of years since 2000. What is the horizontal asymptote? Answer: The horizontal asymptote is at f(t) = 1.432. Logistic Growth C. A city’s population in millions is modeled by , where t is the number of years since 2000. What will be the maximum population? Answer: The population will reach a maximum of a little less than 1,432,000 people. Logistic Growth D. A city’s population in millions is modeled by , where t is the number of years since 2000. According to the function, when will the city’s population reach 1 million? Answer: The graph indicates the population will reach 1 million people at t ≈ 3. Replacing f(t) with 1 and solving for t in the equation yields t = 2.78 years. So, the population of the city will reach 1 million people by 2003. A. A city’s population in millions is modeled by where t is the number of years since 2000. Graph the function. A. B. C. D. B. A city’s population in millions is modeled by where t is the number of years since 2000. What is the horizontal asymptote? A. f(t) = 2.971 B. f(t) = 1.13 C. f(t) = –0.28 D. f(t) = 1.563 C. A city’s population in millions is modeled by where t is the number of years since 2000. What will be the maximum population? A. 1 million people B. 1.563 million people C. 1.13 million people D. 0.28 million people D. A city’s population in millions is modeled by where t is the number of years since 2000. According to the function, when will the city’s population reach 1.5 million? A. by the year 2008 B. by the year 2010 C. by the year 2012 D. by the year 2014 Homework: P. 513 # 1 - 11