5 2 LOGARITHMS AND EXPONENTIAL MODELS 1 Using

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 = 299 e 0. 009 t, 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: 299 e 0. 009 t = 350. 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. 145 t = 2 log 1. 145 t = 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

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Half-Life Example 8 The quantity, Q, of a substance decays according to the formula Q = Q 0 e−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 Q 0. Thus, solving for k, we get Q 0 e−k· 11 = ½ Q 0 e− 11 k = ½ − 11 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 = 175 e 0. 1354 t. Example 10 Convert the formula Q = 7 e 0. 3 t to the form Q = a bt. Solution Using the properties of exponents, Q = 7 e 0. 3 t = 7(e 0. 3)t. Using a calculator, we find e 0. 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. 5 t. 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. 5 t, Population which, after dividing by 2 and taking (millions) the log of both sides can simplify to t log 1. 02 = log(2 + 0. 25 t). 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. 5 t P = 2(1. 02)t t, years 9

The growth factor b and the doubling time T are related by: 10

The growth factor k and the doubling time T are related by: 11