{"id":312,"date":"2026-02-02T16:57:09","date_gmt":"2026-02-02T16:57:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&p=312"},"modified":"2026-02-02T18:23:11","modified_gmt":"2026-02-02T18:23:11","slug":"exponential-and-logarithmic-equations-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/exponential-and-logarithmic-equations-get-stronger-answer-key\/","title":{"raw":"Exponential and Logarithmic Equations Get Stronger Answer Key","rendered":"Exponential and Logarithmic Equations Get Stronger Answer Key"},"content":{"raw":"

Exponential and Logarithmic Equations: Get Stronger Key<\/h1>\r\n

Logarithmic Properties<\/h2>\r\n3.\u00a0[latex]{\\mathrm{log}}_{b}\\left(2\\right)+{\\mathrm{log}}_{b}\\left(7\\right)+{\\mathrm{log}}_{b}\\left(x\\right)+{\\mathrm{log}}_{b}\\left(y\\right)[\/latex]\r\n\r\n5.\u00a0[latex]{\\mathrm{log}}_{b}\\left(13\\right)-{\\mathrm{log}}_{b}\\left(17\\right)[\/latex]\r\n\r\n7.\u00a0[latex]-k\\mathrm{ln}\\left(4\\right)[\/latex]\r\n\r\n9.\u00a0[latex]\\mathrm{ln}\\left(7xy\\right)[\/latex]\r\n\r\n11.\u00a0[latex]{\\mathrm{log}}_{b}\\left(4\\right)[\/latex]\r\n\r\n13.\u00a0[latex]{\\text{log}}_{b}\\left(7\\right)[\/latex]\r\n\r\n15.\u00a0[latex]15\\mathrm{log}\\left(x\\right)+13\\mathrm{log}\\left(y\\right)-19\\mathrm{log}\\left(z\\right)[\/latex]\r\n\r\n17.\u00a0[latex]\\frac{3}{2}\\mathrm{log}\\left(x\\right)-2\\mathrm{log}\\left(y\\right)[\/latex]\r\n\r\n19.\u00a0[latex]\\frac{8}{3}\\mathrm{log}\\left(x\\right)+\\frac{14}{3}\\mathrm{log}\\left(y\\right)[\/latex]\r\n\r\n21.\u00a0[latex]\\mathrm{ln}\\left(2{x}^{7}\\right)[\/latex]\r\n\r\n23.\u00a0[latex]\\mathrm{log}\\left(\\frac{x{z}^{3}}{\\sqrt{y}}\\right)[\/latex]\r\n\r\n27.\u00a0[latex]{\\mathrm{log}}_{11}\\left(5\\right)=\\frac{{\\mathrm{log}}_{5}\\left(5\\right)}{{\\mathrm{log}}_{5}\\left(11\\right)}=\\frac{1}{b}[\/latex]\r\n\r\n29.\u00a0[latex]{\\mathrm{log}}_{11}\\left(\\frac{6}{11}\\right)=\\frac{{\\mathrm{log}}_{5}\\left(\\frac{6}{11}\\right)}{{\\mathrm{log}}_{5}\\left(11\\right)}=\\frac{{\\mathrm{log}}_{5}\\left(6\\right)-{\\mathrm{log}}_{5}\\left(11\\right)}{{\\mathrm{log}}_{5}\\left(11\\right)}=\\frac{a-b}{b}=\\frac{a}{b}-1[\/latex]\r\n\r\n31.\u00a03\r\n\r\n33. 2.81359\r\n\r\n35. 0.93913\r\n\r\n37. \u20132.23266\r\n\r\n39.\u00a0x\u00a0<\/em>= 4; By the quotient rule: [latex]{\\mathrm{log}}_{6}\\left(x+2\\right)-{\\mathrm{log}}_{6}\\left(x - 3\\right)={\\mathrm{log}}_{6}\\left(\\frac{x+2}{x - 3}\\right)=1[\/latex].\r\n

Rewriting as an exponential equation and solving for x<\/em>:<\/p>\r\n

[latex]\\begin{cases}{6}^{1}\\hfill & =\\frac{x+2}{x - 3}\\hfill \\\\ 0\\hfill & =\\frac{x+2}{x - 3}-6\\hfill \\\\ 0\\hfill & =\\frac{x+2}{x - 3}-\\frac{6\\left(x - 3\\right)}{\\left(x - 3\\right)}\\hfill \\\\ 0\\hfill & =\\frac{x+2 - 6x+18}{x - 3}\\hfill \\\\ 0\\hfill & =\\frac{x - 4}{x - 3}\\hfill \\\\ \\text{ }x\\hfill & =4\\hfill \\end{cases}[\/latex]<\/p>\r\n

Checking, we find that [latex]{\\mathrm{log}}_{6}\\left(4+2\\right)-{\\mathrm{log}}_{6}\\left(4 - 3\\right)={\\mathrm{log}}_{6}\\left(6\\right)-{\\mathrm{log}}_{6}\\left(1\\right)[\/latex] is defined, so x\u00a0<\/em>= 4.<\/p>\r\n\r\n

Exponential and Logarithmic Equations<\/h2>\r\n2. An extraneous solution occurs when solving the equation results in a value that does not satisfy the original equation. It can be recognized by substituting the solution back into the original equation and seeing that it makes the equation false or violates a domain restriction (for example, giving a negative input to a logarithm or an even root, or causing division by zero).\r\n\r\n3.\u00a0The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.\r\n\r\n5.\u00a0[latex]x=-\\frac{1}{3}[\/latex]\r\n\r\n7. n\u00a0<\/em>= \u20131\r\n\r\n9.\u00a0[latex]b=\\frac{6}{5}[\/latex]\r\n\r\n11. x\u00a0<\/em>= 10\r\n\r\n13.\u00a0No solution\r\n\r\n15.\u00a0[latex]p=\\mathrm{log}\\left(\\frac{17}{8}\\right)-7[\/latex]\r\n\r\n21.\u00a0[latex]x=\\mathrm{ln}12[\/latex]\r\n\r\n27.\u00a0[latex]x=\\mathrm{ln}\\left(3\\right)[\/latex]\r\n\r\n31. n\u00a0<\/em>= 49\r\n\r\n33.\u00a0[latex]k=\\frac{1}{36}[\/latex]\r\n\r\n35.\u00a0[latex]x=\\frac{9-e}{8}[\/latex]\r\n\r\n37. n\u00a0<\/em>= 1\r\n\r\n39.\u00a0No solution\r\n\r\n41.\u00a0No solution\r\n\r\n43.\u00a0[latex]x=\\pm \\frac{10}{3}[\/latex]\r\n\r\n51. x\u00a0<\/em>= 9\r\n\"Graph\r\n\r\n53.\u00a0[latex]x=\\frac{{e}^{2}}{3}\\approx 2.5[\/latex]\r\n\"Graph\r\n\r\n55. x\u00a0<\/em>= \u20135\r\n\"Graph\r\n\r\n57.\u00a0[latex]x=\\frac{e+10}{4}\\approx 3.2[\/latex]\r\n\"Graph\r\n\r\n59.\u00a0No solution\r\n\"Graph\r\n\r\n61.\u00a0[latex]x=\\frac{11}{5}\\approx 2.2[\/latex]\r\n\"Graph\r\n\r\n65.\u00a0about $27,710.24\r\n\r\n\"Graph\r\n\r\n67.\u00a0about 5 years\r\n\r\n69.\u00a0[latex]\\frac{\\mathrm{ln}\\left(17\\right)}{5}\\approx 0.567[\/latex]\r\n\r\n71.\u00a0[latex]x=\\frac{\\mathrm{log}\\left(38\\right)+5\\mathrm{log}\\left(3\\right)\\text{ }}{4\\mathrm{log}\\left(3\\right)}\\approx 2.078[\/latex]\r\n\r\n73.\u00a0[latex]x\\approx 2.2401[\/latex]\r\n\r\n75.\u00a0[latex]x\\approx -44655.7143[\/latex]\r\n\r\n77.\u00a0about 5.83\r\n\r\n81.\u00a0[latex]t=\\mathrm{ln}\\left({\\left(\\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\\right)}^{-\\frac{1}{k}}\\right)[\/latex]\r\n

Exponential and Logarithmic Models<\/h2>\r\n1. Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.\r\n\r\n7.\u00a0[latex]f\\left(0\\right)\\approx 16.7[\/latex]; The amount initially present is about 16.7 units.\r\n\r\n9.\u00a0150\r\n\r\n11. exponential; [latex]f\\left(x\\right)={1.2}^{x}[\/latex]\r\n\r\n13. logarithmic\r\n\"Graph\r\n\r\n15. logarithmic\r\n\"Graph\r\n\r\n17.\r\n\"Graph\r\n\r\n21. about 7.3 years\r\n\r\n23. 4 half-lives; 8.18 minutes\r\n\r\n25. [latex]\\begin{cases}\\text{ }M=\\frac{2}{3}\\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)\\hfill \\\\ \\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)=\\frac{3}{2}M\\hfill \\\\ \\text{ }\\frac{S}{{S}_{0}}={10}^{\\frac{3M}{2}}\\hfill \\\\ \\text{ }S={S}_{0}{10}^{\\frac{3M}{2}}\\hfill \\end{cases}\/\/[\/latex]\r\n\r\n28. 2 hours.\r\n\r\n29.\u00a0[latex]A=125{e}^{\\left(-0.3567t\\right)};A\\approx 43[\/latex] mg\r\n\r\n31.\u00a0about 60 days\r\n\r\n32. [latex]A(t)=0.5(0.9885)^t[\/latex]. After 60 days, [latex]A(60)\\approx 0.3[\/latex] grams.\r\n\r\n33.\u00a0[latex]f\\left(t\\right)=250{e}^{\\left(-0.00914t\\right)}[\/latex]; half-life: about [latex]\\text{76}[\/latex] minutes\r\n\r\n35.\u00a0[latex]r\\approx -0.0667[\/latex], So the hourly decay rate is about 6.67%\r\n\r\n37.\u00a0[latex]f\\left(t\\right)=1350{e}^{\\left(0.03466t\\right)}[\/latex]; after 3 hours: [latex]P\\left(180\\right)\\approx 691,200[\/latex]\r\n\r\n40.\u00a0 [latex]T(t)=69+31e^{\\left(\\frac{1}{15}\\ln\\left(\\frac{26}{31}\\right)\\right)t}[\/latex].\r\n\r\n41.\u00a0about 88 minutes\r\n

Fitting Exponential Models to Data<\/h2>\r\n4. A scatterplot that is best described by a logarithmic model would show a rapid increase or decrease at first, followed by a gradual leveling off as the input values increase. The data points would form a curve that rises (or falls) quickly and then flattens.\r\n\r\n6. E\r\n\r\n7. C\r\n\r\n8. A\r\n\r\n9. B\r\n\r\n10. D\r\n\r\n11.\u00a0[latex]P\\left(0\\right)=22[\/latex] ; 175\r\n\r\n13.\u00a0[latex]p\\approx 2.67[\/latex]\r\n\r\n16. \"The\r\n\r\n17. 4 koi\r\n\r\n18. 6.8 months\r\n\r\n20. \"The\r\n\r\n21. 10 wolves\r\n\r\n23.\u00a0about 5.4 years.\r\n\r\n25.\r\n\"Graph\r\n\r\n26. [latex]f(x)\\approx 795(1.415)^x[\/latex]<\/span>\r\n\r\n31.\u00a0[latex]f\\left(x\\right)=731.92{\\left(0.738\\right)}^{x}[\/latex]\r\n\r\n35.\r\n\"Graph\r\n\r\n36. [latex]y\\approx 5.06+1.93\\ln(x)[\/latex]\r\n\r\n37.\u00a0[latex]f\\left(10\\right)\\approx 9.5[\/latex]\r\n\r\n40. \"Graph\r\n\r\n41.\u00a0[latex]f\\left(x\\right)=7.544 - 2.268\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\n45.\r\n\"Graph\r\n\r\n46. [latex]y\\approx 26.91-24.20e^{-0.264x}[\/latex]\r\n\r\n51.\u00a0[latex]f\\left(x\\right)=\\frac{136.068}{1+10.324{e}^{-0.480x}}[\/latex]\r\n\r\n53.\u00a0about 136","rendered":"

Exponential and Logarithmic Equations: Get Stronger Key<\/h1>\n

Logarithmic Properties<\/h2>\n

3.\u00a0[latex]{\\mathrm{log}}_{b}\\left(2\\right)+{\\mathrm{log}}_{b}\\left(7\\right)+{\\mathrm{log}}_{b}\\left(x\\right)+{\\mathrm{log}}_{b}\\left(y\\right)[\/latex]<\/p>\n

5.\u00a0[latex]{\\mathrm{log}}_{b}\\left(13\\right)-{\\mathrm{log}}_{b}\\left(17\\right)[\/latex]<\/p>\n

7.\u00a0[latex]-k\\mathrm{ln}\\left(4\\right)[\/latex]<\/p>\n

9.\u00a0[latex]\\mathrm{ln}\\left(7xy\\right)[\/latex]<\/p>\n

11.\u00a0[latex]{\\mathrm{log}}_{b}\\left(4\\right)[\/latex]<\/p>\n

13.\u00a0[latex]{\\text{log}}_{b}\\left(7\\right)[\/latex]<\/p>\n

15.\u00a0[latex]15\\mathrm{log}\\left(x\\right)+13\\mathrm{log}\\left(y\\right)-19\\mathrm{log}\\left(z\\right)[\/latex]<\/p>\n

17.\u00a0[latex]\\frac{3}{2}\\mathrm{log}\\left(x\\right)-2\\mathrm{log}\\left(y\\right)[\/latex]<\/p>\n

19.\u00a0[latex]\\frac{8}{3}\\mathrm{log}\\left(x\\right)+\\frac{14}{3}\\mathrm{log}\\left(y\\right)[\/latex]<\/p>\n

21.\u00a0[latex]\\mathrm{ln}\\left(2{x}^{7}\\right)[\/latex]<\/p>\n

23.\u00a0[latex]\\mathrm{log}\\left(\\frac{x{z}^{3}}{\\sqrt{y}}\\right)[\/latex]<\/p>\n

27.\u00a0[latex]{\\mathrm{log}}_{11}\\left(5\\right)=\\frac{{\\mathrm{log}}_{5}\\left(5\\right)}{{\\mathrm{log}}_{5}\\left(11\\right)}=\\frac{1}{b}[\/latex]<\/p>\n

29.\u00a0[latex]{\\mathrm{log}}_{11}\\left(\\frac{6}{11}\\right)=\\frac{{\\mathrm{log}}_{5}\\left(\\frac{6}{11}\\right)}{{\\mathrm{log}}_{5}\\left(11\\right)}=\\frac{{\\mathrm{log}}_{5}\\left(6\\right)-{\\mathrm{log}}_{5}\\left(11\\right)}{{\\mathrm{log}}_{5}\\left(11\\right)}=\\frac{a-b}{b}=\\frac{a}{b}-1[\/latex]<\/p>\n

31.\u00a03<\/p>\n

33. 2.81359<\/p>\n

35. 0.93913<\/p>\n

37. \u20132.23266<\/p>\n

39.\u00a0x\u00a0<\/em>= 4; By the quotient rule: [latex]{\\mathrm{log}}_{6}\\left(x+2\\right)-{\\mathrm{log}}_{6}\\left(x - 3\\right)={\\mathrm{log}}_{6}\\left(\\frac{x+2}{x - 3}\\right)=1[\/latex].<\/p>\n

Rewriting as an exponential equation and solving for x<\/em>:<\/p>\n

[latex]\\begin{cases}{6}^{1}\\hfill & =\\frac{x+2}{x - 3}\\hfill \\\\ 0\\hfill & =\\frac{x+2}{x - 3}-6\\hfill \\\\ 0\\hfill & =\\frac{x+2}{x - 3}-\\frac{6\\left(x - 3\\right)}{\\left(x - 3\\right)}\\hfill \\\\ 0\\hfill & =\\frac{x+2 - 6x+18}{x - 3}\\hfill \\\\ 0\\hfill & =\\frac{x - 4}{x - 3}\\hfill \\\\ \\text{ }x\\hfill & =4\\hfill \\end{cases}[\/latex]<\/p>\n

Checking, we find that [latex]{\\mathrm{log}}_{6}\\left(4+2\\right)-{\\mathrm{log}}_{6}\\left(4 - 3\\right)={\\mathrm{log}}_{6}\\left(6\\right)-{\\mathrm{log}}_{6}\\left(1\\right)[\/latex] is defined, so x\u00a0<\/em>= 4.<\/p>\n

Exponential and Logarithmic Equations<\/h2>\n

2. An extraneous solution occurs when solving the equation results in a value that does not satisfy the original equation. It can be recognized by substituting the solution back into the original equation and seeing that it makes the equation false or violates a domain restriction (for example, giving a negative input to a logarithm or an even root, or causing division by zero).<\/p>\n

3.\u00a0The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.<\/p>\n

5.\u00a0[latex]x=-\\frac{1}{3}[\/latex]<\/p>\n

7. n\u00a0<\/em>= \u20131<\/p>\n

9.\u00a0[latex]b=\\frac{6}{5}[\/latex]<\/p>\n

11. x\u00a0<\/em>= 10<\/p>\n

13.\u00a0No solution<\/p>\n

15.\u00a0[latex]p=\\mathrm{log}\\left(\\frac{17}{8}\\right)-7[\/latex]<\/p>\n

21.\u00a0[latex]x=\\mathrm{ln}12[\/latex]<\/p>\n

27.\u00a0[latex]x=\\mathrm{ln}\\left(3\\right)[\/latex]<\/p>\n

31. n\u00a0<\/em>= 49<\/p>\n

33.\u00a0[latex]k=\\frac{1}{36}[\/latex]<\/p>\n

35.\u00a0[latex]x=\\frac{9-e}{8}[\/latex]<\/p>\n

37. n\u00a0<\/em>= 1<\/p>\n

39.\u00a0No solution<\/p>\n

41.\u00a0No solution<\/p>\n

43.\u00a0[latex]x=\\pm \\frac{10}{3}[\/latex]<\/p>\n

51. x\u00a0<\/em>= 9
\n\"Graph<\/p>\n

53.\u00a0[latex]x=\\frac{{e}^{2}}{3}\\approx 2.5[\/latex]
\n\"Graph<\/p>\n

55. x\u00a0<\/em>= \u20135
\n\"Graph<\/p>\n

57.\u00a0[latex]x=\\frac{e+10}{4}\\approx 3.2[\/latex]
\n\"Graph<\/p>\n

59.\u00a0No solution
\n\"Graph<\/p>\n

61.\u00a0[latex]x=\\frac{11}{5}\\approx 2.2[\/latex]
\n\"Graph<\/p>\n

65.\u00a0about $27,710.24<\/p>\n

\"Graph<\/p>\n

67.\u00a0about 5 years<\/p>\n

69.\u00a0[latex]\\frac{\\mathrm{ln}\\left(17\\right)}{5}\\approx 0.567[\/latex]<\/p>\n

71.\u00a0[latex]x=\\frac{\\mathrm{log}\\left(38\\right)+5\\mathrm{log}\\left(3\\right)\\text{ }}{4\\mathrm{log}\\left(3\\right)}\\approx 2.078[\/latex]<\/p>\n

73.\u00a0[latex]x\\approx 2.2401[\/latex]<\/p>\n

75.\u00a0[latex]x\\approx -44655.7143[\/latex]<\/p>\n

77.\u00a0about 5.83<\/p>\n

81.\u00a0[latex]t=\\mathrm{ln}\\left({\\left(\\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\\right)}^{-\\frac{1}{k}}\\right)[\/latex]<\/p>\n

Exponential and Logarithmic Models<\/h2>\n

1. Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.<\/p>\n

7.\u00a0[latex]f\\left(0\\right)\\approx 16.7[\/latex]; The amount initially present is about 16.7 units.<\/p>\n

9.\u00a0150<\/p>\n

11. exponential; [latex]f\\left(x\\right)={1.2}^{x}[\/latex]<\/p>\n

13. logarithmic
\n\"Graph<\/p>\n

15. logarithmic
\n\"Graph<\/p>\n

17.
\n\"Graph<\/p>\n

21. about 7.3 years<\/p>\n

23. 4 half-lives; 8.18 minutes<\/p>\n

25. [latex]\\begin{cases}\\text{ }M=\\frac{2}{3}\\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)\\hfill \\\\ \\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)=\\frac{3}{2}M\\hfill \\\\ \\text{ }\\frac{S}{{S}_{0}}={10}^{\\frac{3M}{2}}\\hfill \\\\ \\text{ }S={S}_{0}{10}^{\\frac{3M}{2}}\\hfill \\end{cases}\/\/[\/latex]<\/p>\n

28. 2 hours.<\/p>\n

29.\u00a0[latex]A=125{e}^{\\left(-0.3567t\\right)};A\\approx 43[\/latex] mg<\/p>\n

31.\u00a0about 60 days<\/p>\n

32. [latex]A(t)=0.5(0.9885)^t[\/latex]. After 60 days, [latex]A(60)\\approx 0.3[\/latex] grams.<\/p>\n

33.\u00a0[latex]f\\left(t\\right)=250{e}^{\\left(-0.00914t\\right)}[\/latex]; half-life: about [latex]\\text{76}[\/latex] minutes<\/p>\n

35.\u00a0[latex]r\\approx -0.0667[\/latex], So the hourly decay rate is about 6.67%<\/p>\n

37.\u00a0[latex]f\\left(t\\right)=1350{e}^{\\left(0.03466t\\right)}[\/latex]; after 3 hours: [latex]P\\left(180\\right)\\approx 691,200[\/latex]<\/p>\n

40.\u00a0 [latex]T(t)=69+31e^{\\left(\\frac{1}{15}\\ln\\left(\\frac{26}{31}\\right)\\right)t}[\/latex].<\/p>\n

41.\u00a0about 88 minutes<\/p>\n

Fitting Exponential Models to Data<\/h2>\n

4. A scatterplot that is best described by a logarithmic model would show a rapid increase or decrease at first, followed by a gradual leveling off as the input values increase. The data points would form a curve that rises (or falls) quickly and then flattens.<\/p>\n

6. E<\/p>\n

7. C<\/p>\n

8. A<\/p>\n

9. B<\/p>\n

10. D<\/p>\n

11.\u00a0[latex]P\\left(0\\right)=22[\/latex] ; 175<\/p>\n

13.\u00a0[latex]p\\approx 2.67[\/latex]<\/p>\n

16. \"The<\/p>\n

17. 4 koi<\/p>\n

18. 6.8 months<\/p>\n

20. \"The<\/p>\n

21. 10 wolves<\/p>\n

23.\u00a0about 5.4 years.<\/p>\n

25.
\n\"Graph<\/p>\n

26. [latex]f(x)\\approx 795(1.415)^x[\/latex]<\/span><\/p>\n

31.\u00a0[latex]f\\left(x\\right)=731.92{\\left(0.738\\right)}^{x}[\/latex]<\/p>\n

35.
\n\"Graph<\/p>\n

36. [latex]y\\approx 5.06+1.93\\ln(x)[\/latex]<\/p>\n

37.\u00a0[latex]f\\left(10\\right)\\approx 9.5[\/latex]<\/p>\n

40. \"Graph<\/p>\n

41.\u00a0[latex]f\\left(x\\right)=7.544 - 2.268\\mathrm{ln}\\left(x\\right)[\/latex]<\/p>\n

45.
\n\"Graph<\/p>\n

46. [latex]y\\approx 26.91-24.20e^{-0.264x}[\/latex]<\/p>\n

51.\u00a0[latex]f\\left(x\\right)=\\frac{136.068}{1+10.324{e}^{-0.480x}}[\/latex]<\/p>\n

53.\u00a0about 136<\/p>\n","protected":false},"author":13,"menu_order":8,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":224,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/312"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/312\/revisions"}],"predecessor-version":[{"id":317,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/312\/revisions\/317"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/224"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/312\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=312"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=312"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=312"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=312"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}