Week #1 The Exponential and Logarithm Functions Section 1.4

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1 Week #1 The Exponential and Logarithm Functions Section 1.4 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS For Exercises 7-17, solve for x using logs = 50(1.04) x Take logs: log(20) = log(50(1.04) x ) log(20) = log(50) + log((1.04) x ) log(20) log(50) = x log(1.04) x = Alternatively, simplify before you take logs: log(20) log(50) log(1.04) x = e x+1 20 = 50(1.04) x 2 5 = (1.04)x then take logs log(2/5) = x log(1.04) x = log(2/5) log(1.04) Since the problem has an e in it, take natural log of both sides: xln(2) = (x + 1)ln(e) but ln(e) = 1 xln(2) = x + 1 Now gather x terms: ln(2)x x = 1 x is a common factor: (ln(2) 1)x = 1 1 x = ln(2)

2 13. 2e 3x = 4e 5x There are several routes to solve this. We will simplify first, then take logarithms. e 3x = 2e 5x Multiply both sides by e 5x : e 3x 5x = 2 e 2x = 2 Take ln of both sides: 2x = ln(2) x = ln(2)/ x+3 = 5e 7 x This is complicated enough just to go straight to taking logs. Gather x terms on the left: ln(10 x+3 ) = ln(5e 7 x ) (x + 3)ln(10) = ln(5) + 7 x ln(10)x + 3ln(10) = ln(5) + 7 x ln(10)x + x = ln(5) ln(10) (ln(10) + 1)x = ln(5) + 7 3ln(10) x = ln(5) + 7 3ln(10) ln(10) QUIZ PREPARATION PROBLEMS 41. A cup of coffee contains 100 mg of caffeine, which leaves the body at a continuous rate of 17% per hour. (a) Write a formula for the amount, A mg, of caffeine in the body t hours after drinking a cup of coffee. (b) Graph the function from part (a) Use the graph to estimate the half-life of caffeine. (c) Use logarithms to find the half-life of caffeine. (a) Initial amount = 100 (mg). That amount is leaving (decaying) at a rate of 17% per hour. Just as increasing continuously at a rate of 17% would have the form e.17t, leaving at 17% has the same structure, but the sign is negative: A(t) = 100e (.17t) NOTE: this question is ambiguous, in that the phrase rate 17% per hour can also be interpreted as 83% is left when t = 1. This would produce the function A 2 (t) = (1 0.71) t = 0.83 t. The two versions are similar since A(t) = e (.17)t (0.843) t ). Obviously, these two similar formulas produce similar, but not identical, answers to (b) and (c). For convention in this course, the phrase continuous rate of r will always indiate the use of e rt. 2

3 (b) Use a couple of points to draw the graph: A(0) = 100 A(1) = 100e A(2) = 100e A (mg) t (hours) (c) If you know the half-life formula, feel free to use it. A more general way to calculate the half-life is by setting A(t 2 ) = 1 2 A(0) and solving for t 2: 100e.17 t 2 = e0 e.17 t 2 = 1 2 Take ln of both sides:.17 t 2 = ln(1/2) t 2 = ln(1/2) hours 42. In 1980, there were about 170 million vehicles (cars and trucks) and about 227 million people in the United States. The number of vehicles has been growing at 4% a year, while the population has been growing at 1% a year. When was there, on average, one vehicle per person? NOTE: this question has an interpretation difficulty similar to #41, in that 4% per year could mean a growth function of either e 0.04t or (1.04) t. Here, because the phrase continuous rate was absent, we took the more common-place interpretation: after one year, the number of cars has gone up by exactly 4%, given by the formula (1.04) t. We would also accept a full solution using e 0.04t ( ) t as correct in this case, even though it would produce slightly different numerical values. Let t = number of years since Then the number of vehicles, V, in millions, at time t is given by V = 170(1.04) t 3

4 and the number of people, P, in miollions, at time t is given by P = 227(1.01) t. There is an average of one vehicle per person when V = P. Thus we must solve 170(1.04) t = 227(1.01) t As with all log problems like this, you may proceed in several different ways. One way is shown below. Taking log of both sides: 170(1.04) t = 227(1.01) t 1.04 t = 227 ( 1.01 ) 170 ( ) t log = log t = log ( ) log ( ) To check, ensure that both populations (cars and people) are at the same level near that year. V (9.879) = 170(1.04) ( 9.879) P(9.879) = 227(1.01) ( 9.879) This confirms that our estimate that at t = 9.879, or near the end of 1989, the car and human populations would equal each other in this model. 43. The air in a factory is being filtered so that the quantity of a pollutant, P (in mg/liter), is decreasing according to the function P = P 0 e kt, where t is time in hours. If 10% of the pollution is removed in the first five hours: (a) What percentage of the pollution is left after 10 hours? (b) How long is it before the pollution is reduced by 50%? (c) Plot a graph of pollution against time. Show the results of your calculations on the graph. (d) Explain why the quantity of pollutant might decrease in this way. (a) We know the decay follows the equation P = P 0 e kt 4

5 and that 10% of the pollution is removed after 5 hours (meaning that 90% is left). Therefore, 0.90P 0 = P 0 e k 5 k = 1 5 ln(0.90) Thus after 10 hours, P = P 0 e ln(0.9) = P 0 e 2ln(0.9) = P 0 (0.9) 2 = 0.81P 0 or 81% of the original amount of pollutant. Note that we can simplify our formula for P to make it easier to use: P = P 0 e 1 5 ln(0.9)t = P 0 ( e ln(0.9)) t/5 = P 0 (0.9) t/5 Using this formula, we can see that waiting 10 hours would leave the following amount of contaminant: P(10) = P 0 (0.9) 10/5 = P 0 (0.9) 2 = P 0 (0.81), or 81% of the original amount. (b) Using our easier formula, we want to solve for t when 0.5P 0 = P 0 (0.9) t/5 Cancelling P o, taking logs: Solving for t: log(0.5) = t 5 log(0.9) t = 5 log(0.5) 32.9 hours log(0.9) (c) Graph below. 5

6 P 0.5 Po.81 Po t (hours) (d) When highly polluted air is filtered, there is more pollutant per liter of air to remove. As the air gets cleaner and cleaner, the filter will collect pollutant at a slower rate. 6