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5.2

5.2. LOGARITHMS AND EXPONENTIAL MODELS. Using Logarithms to Undo Exponents. Example 2 The US population, P, in millions, is currently growing according to the formula P = 299 e 0.009 t , where t is in years since 2006. When is the population predicted to reach 350 million ? Solution

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5.2

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  1. 5.2 LOGARITHMS AND EXPONENTIAL MODELS

  2. 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. The US population is predicted to reach 350 million during the year 2024.

  3. Doubling Time Example 4 (a) Find the time needed for the turtle population described by the function P = 175(1.145)tto 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

  4. 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

  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 btand Q = a ekt Any exponential function can be written in either of the two forms: Q = a btor Q = a ekt. If b = ek, so k = lnb, the two formulas represent the same function.

  7. Converting Between Q = a btand 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. Solution Setting P =N, we get 2(1.02)t = 4+0.5t, which, after dividing by 2 and taking the log of both sides can simplify to tlog 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 Finding the intersection of linear and exponential graphs Population (millions) Shortages start → N = 4+0.5t P = 2(1.02)t t, years

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