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5.2 LOGARITHMS AND EXPONENTIAL MODELS 1 Using Logarithms to Undo Exponents Example 2 The US population, P, in millions, is currently growing according to the formula P = 299e0.009t, where t is in years since 2006. When is the population predicted to reach 350 million? Solution We want to solve the following equation for t: 299e0.009t = 350. 299 e 0.009 t 350 e 0.009 t 350 / 299 ln e 0.009 t ln(350 / 299 ) 0.009 t ln(350 / 299 ) t ln(350 / 299 ) 17.5 years 0.009 The US population is predicted to reach 350 million during the year 2024. 2 Doubling Time Example 4 (a) Find the time needed for the turtle population described by the function P = 175(1.145)t to double its initial size. Solution (a) The initial size is 175 turtles; doubling this gives 350 turtles. We need to solve the following equation for t: 175(1.145)t = 350 1.145t = 2 log 1.145t = log 2 t · log 1.145 = log 2 t = log 2/log 1.145 ≈ 5.119 years. Note that 175(1.145)5.119 ≈ 350 3 Since the growth rate is 0.000121, the growth rate factor is b = 1 – 0.000121 = 0.999879. Thus after t years the amount left will be 4 5 Half-Life Example 8 The quantity, Q, of a substance decays according to the formula Q = Q0e−kt, where t is in minutes. The half-life of the substance is 11 minutes. What is the value of k? Solution We know that after 11 minutes, Q = ½ Q0. Thus, solving for k, we get Q0e−k·11 = ½ Q0 e−11k = ½ −11k = ln ½ k = ln ½ / (−11) ≈ 0.06301, so k = 0.063 per minute. This substance decays at the continuous rate of 6.301% per minute. 6 Converting Between Q = a bt and Q = a ekt Any exponential function can be written in either of the two forms: Q = a bt or Q = a ekt. If b = ek, so k = ln b, the two formulas represent the same function. 7 Converting Between Q = a bt and Q = a ekt Example 9 Convert the exponential function P = 175(1.145)t to the form P = aekt. Solution The parameter a in both functions represents the initial population. For all t, 175(1.145)t = 175(ek)t, so we must find k such that ek = 1.145. Therefore k is the power of e which gives 1.145. By the definition of ln, we have k = ln 1.145 ≈ 0.1354. Therefore, P = 175e0.1354t. Example 10 Convert the formula Q = 7e0.3t to the form Q = a bt. Solution Using the properties of exponents, Q = 7e0.3t = 7(e0.3)t. Using a calculator, we find e0.3 ≈ 1.3499, so Q = 7(1.3499)t. 8 Exponential Growth Problems That Cannot Be Solved By Logarithms Example 13 With t in years, the population of a country (in millions) is given by P = 2(1.02)t, while the food supply (in millions of people that can be fed) is given by N = 4+0.5t. Determine the year in which the country first experiences food shortages. Finding the intersection of linear and exponential graphs Solution Setting P = N, we get 2(1.02)t = 4+0.5t, Population which, after dividing by 2 and taking (millions) the log of both sides can simplify to t log 1.02 = log(2 + 0.25t). We cannot solve this equation algebraically, but we can estimate its solution numerically or graphically: t ≈ 199.381. So it will be nearly 200 years before shortages are experienced Shortages start → N = 4+0.5t P = 2(1.02)t t, years 199.381 9 Q ab t The growth factor b and the doubling time T are related by: t Q ab 2a = ab 2= b t t ln 2 ln b t t ln b ln 2 D oubling T im e T ln 2 ln b 10 Q ae kt The growth factor k and the doubling time T are kt related by: Q ae 2a = ae 2= e kt kt ln 2 ln e kt kt ln 2 D oubling T im e T ln 2 k 11