{"id":11292,"date":"2015-07-14T19:33:26","date_gmt":"2015-07-14T19:33:26","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11292"},"modified":"2018-06-28T14:50:47","modified_gmt":"2018-06-28T14:50:47","slug":"solutions-25","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/solutions-25\/","title":{"raw":"Solutions 5.1: Exponential Functions","rendered":"Solutions 5.1: Exponential Functions"},"content":{"raw":"

Solutions to Try Its<\/h2>\r\n1. 5.5556\r\n\r\n2.\u00a0About 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.\r\n\r\n3.\u00a0[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]\r\n\r\n4.\u00a0[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]\r\n\r\n5.\u00a0[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}[\/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\\left(1.4142\\right)}^{x}[\/latex].\r\n\r\n6.\u00a0[latex]y\\approx 12\\cdot {1.85}^{x}[\/latex]\r\n\r\n7.\u00a0about $3,644,675.88\r\n\r\n8.\u00a0$13,693\r\n\r\n9.\u00a0[latex]{e}^{-0.5}\\approx 0.60653[\/latex]\r\n\r\n10.\u00a0$3,659,823.44\r\n\r\n11.\u00a03.77E-26 (This is calculator notation for the number written as [latex]3.77\\times {10}^{-26}[\/latex] in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)\r\n

Solutions to Odd-Numbered Exercises<\/h2>\r\n1.\u00a0Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.\r\n\r\n3.\u00a0When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal<\/em>.\r\n\r\n5.\u00a0exponential; the population decreases by a proportional rate.\r\n\r\n7.\u00a0not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function.\r\n\r\n9.\u00a0The forest represented by the function [latex]B\\left(t\\right)=82{\\left(1.029\\right)}^{t}[\/latex].\r\n\r\n11.\u00a0After t\u00a0<\/em>= 20 years, forest A will have 43 more trees than forest B.\r\n\r\n13.\u00a0Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.\r\n\r\n15.\u00a0exponential growth; The growth factor, 1.06, is greater than 1.\r\n\r\n17.\u00a0exponential decay; The decay factor, 0.97, is between 0 and 1.\r\n\r\n19.\u00a0[latex]f\\left(x\\right)=2000{\\left(0.1\\right)}^{x}[\/latex]\r\n\r\n21.\u00a0[latex]f\\left(x\\right)={\\left(\\frac{1}{6}\\right)}^{-\\frac{3}{5}}{\\left(\\frac{1}{6}\\right)}^{\\frac{x}{5}}\\approx 2.93{\\left(0.699\\right)}^{x}[\/latex]\r\n\r\n23. Linear\r\n\r\n25. Neither\r\n\r\n27. Linear\r\n\r\n29. $10,250\r\n\r\n31. $13,268.58\r\n\r\n33.\u00a0[latex]P=A\\left(t\\right)\\cdot {\\left(1+\\frac{r}{n}\\right)}^{-nt}[\/latex]\r\n\r\n35. $4,572.56\r\n\r\n37. 4%\r\n\r\n39.\u00a0continuous growth; the growth rate is greater than 0.\r\n\r\n41.\u00a0continuous decay; the growth rate is less than 0.\r\n\r\n43. $669.42\r\n\r\n45.\u00a0[latex]f\\left(-1\\right)=-4[\/latex]\r\n\r\n47.\u00a0[latex]f\\left(-1\\right)\\approx -0.2707[\/latex]\r\n\r\n49.\u00a0[latex]f\\left(3\\right)\\approx 483.8146[\/latex]\r\n\r\n51.\u00a0[latex]y=3\\cdot {5}^{x}[\/latex]\r\n\r\n53.\u00a0[latex]y\\approx 18\\cdot {1.025}^{x}[\/latex]\r\n\r\n55.\u00a0[latex]y\\approx 0.2\\cdot {1.95}^{x}[\/latex]\r\n\r\n57.\u00a0[latex]\\text{APY}=\\frac{A\\left(t\\right)-a}{a}=\\frac{a{\\left(1+\\frac{r}{365}\\right)}^{365\\left(1\\right)}-a}{a}=\\frac{a\\left[{\\left(1+\\frac{r}{365}\\right)}^{365}-1\\right]}{a}={\\left(1+\\frac{r}{365}\\right)}^{365}-1[\/latex]; [latex]I\\left(n\\right)={\\left(1+\\frac{r}{n}\\right)}^{n}-1[\/latex]\r\n\r\n59.\u00a0Let f<\/em>\u00a0be the exponential decay function [latex]f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}[\/latex] such that [latex]b>1[\/latex]. Then for some number [latex]n>0[\/latex], [latex]f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}=a{\\left({b}^{-1}\\right)}^{x}=a{\\left({\\left({e}^{n}\\right)}^{-1}\\right)}^{x}=a{\\left({e}^{-n}\\right)}^{x}=a{\\left(e\\right)}^{-nx}[\/latex].\r\n\r\n61. 47,622 fox\r\n\r\n63. 1.39%; $155,368.09\r\n\r\n65. $35,838.76\r\n\r\n67. $82,247.78; $449.75","rendered":"

Solutions to Try Its<\/h2>\n

1. 5.5556<\/p>\n

2.\u00a0About 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.<\/p>\n

3.\u00a0[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]<\/p>\n

4.\u00a0[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]<\/p>\n

5.\u00a0[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}[\/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\\left(1.4142\\right)}^{x}[\/latex].<\/p>\n

6.\u00a0[latex]y\\approx 12\\cdot {1.85}^{x}[\/latex]<\/p>\n

7.\u00a0about $3,644,675.88<\/p>\n

8.\u00a0$13,693<\/p>\n

9.\u00a0[latex]{e}^{-0.5}\\approx 0.60653[\/latex]<\/p>\n

10.\u00a0$3,659,823.44<\/p>\n

11.\u00a03.77E-26 (This is calculator notation for the number written as [latex]3.77\\times {10}^{-26}[\/latex] in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)<\/p>\n

Solutions to Odd-Numbered Exercises<\/h2>\n

1.\u00a0Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.<\/p>\n

3.\u00a0When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal<\/em>.<\/p>\n

5.\u00a0exponential; the population decreases by a proportional rate.<\/p>\n

7.\u00a0not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function.<\/p>\n

9.\u00a0The forest represented by the function [latex]B\\left(t\\right)=82{\\left(1.029\\right)}^{t}[\/latex].<\/p>\n

11.\u00a0After t\u00a0<\/em>= 20 years, forest A will have 43 more trees than forest B.<\/p>\n

13.\u00a0Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.<\/p>\n

15.\u00a0exponential growth; The growth factor, 1.06, is greater than 1.<\/p>\n

17.\u00a0exponential decay; The decay factor, 0.97, is between 0 and 1.<\/p>\n

19.\u00a0[latex]f\\left(x\\right)=2000{\\left(0.1\\right)}^{x}[\/latex]<\/p>\n

21.\u00a0[latex]f\\left(x\\right)={\\left(\\frac{1}{6}\\right)}^{-\\frac{3}{5}}{\\left(\\frac{1}{6}\\right)}^{\\frac{x}{5}}\\approx 2.93{\\left(0.699\\right)}^{x}[\/latex]<\/p>\n

23. Linear<\/p>\n

25. Neither<\/p>\n

27. Linear<\/p>\n

29. $10,250<\/p>\n

31. $13,268.58<\/p>\n

33.\u00a0[latex]P=A\\left(t\\right)\\cdot {\\left(1+\\frac{r}{n}\\right)}^{-nt}[\/latex]<\/p>\n

35. $4,572.56<\/p>\n

37. 4%<\/p>\n

39.\u00a0continuous growth; the growth rate is greater than 0.<\/p>\n

41.\u00a0continuous decay; the growth rate is less than 0.<\/p>\n

43. $669.42<\/p>\n

45.\u00a0[latex]f\\left(-1\\right)=-4[\/latex]<\/p>\n

47.\u00a0[latex]f\\left(-1\\right)\\approx -0.2707[\/latex]<\/p>\n

49.\u00a0[latex]f\\left(3\\right)\\approx 483.8146[\/latex]<\/p>\n

51.\u00a0[latex]y=3\\cdot {5}^{x}[\/latex]<\/p>\n

53.\u00a0[latex]y\\approx 18\\cdot {1.025}^{x}[\/latex]<\/p>\n

55.\u00a0[latex]y\\approx 0.2\\cdot {1.95}^{x}[\/latex]<\/p>\n

57.\u00a0[latex]\\text{APY}=\\frac{A\\left(t\\right)-a}{a}=\\frac{a{\\left(1+\\frac{r}{365}\\right)}^{365\\left(1\\right)}-a}{a}=\\frac{a\\left[{\\left(1+\\frac{r}{365}\\right)}^{365}-1\\right]}{a}={\\left(1+\\frac{r}{365}\\right)}^{365}-1[\/latex]; [latex]I\\left(n\\right)={\\left(1+\\frac{r}{n}\\right)}^{n}-1[\/latex]<\/p>\n

59.\u00a0Let f<\/em>\u00a0be the exponential decay function [latex]f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}[\/latex] such that [latex]b>1[\/latex]. Then for some number [latex]n>0[\/latex], [latex]f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}=a{\\left({b}^{-1}\\right)}^{x}=a{\\left({\\left({e}^{n}\\right)}^{-1}\\right)}^{x}=a{\\left({e}^{-n}\\right)}^{x}=a{\\left(e\\right)}^{-nx}[\/latex].<\/p>\n

61. 47,622 fox<\/p>\n

63. 1.39%; $155,368.09<\/p>\n

65. $35,838.76<\/p>\n

67. $82,247.78; $449.75<\/p>\n\n\t\t\t

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