{"id":11220,"date":"2015-07-14T19:01:56","date_gmt":"2015-07-14T19:01:56","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11220"},"modified":"2021-11-15T01:44:10","modified_gmt":"2021-11-15T01:44:10","slug":"solutions-logarithmic-properties","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/solutions-logarithmic-properties\/","title":{"raw":"Solutions: Logarithmic Properties","rendered":"Solutions: Logarithmic Properties"},"content":{"raw":"

Solutions to Try Its<\/h2>\r\n1.\u00a0[latex]{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k=3{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k[\/latex]\r\n\r\n2.\u00a0[latex]{\\mathrm{log}}_{3}\\left(x+3\\right)-{\\mathrm{log}}_{3}\\left(x - 1\\right)-{\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex]\r\n\r\n3.\u00a0[latex]2\\mathrm{ln}x[\/latex]\r\n\r\n4.\u00a0[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]\r\n\r\n5.\u00a0[latex]{\\mathrm{log}}_{3}16[\/latex]\r\n\r\n6.\u00a0[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]\r\n\r\n7.\u00a0[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]\r\n\r\n8.\u00a0[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex]\r\n\r\n9. [latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.\r\n\r\n10. [latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]\r\n\r\n11. [latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].\r\n\r\n12. The pH increases by about 0.301.\r\n\r\n13. [latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]\r\n\r\n14. [latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]\r\n

Solutions to Odd-Numbered Exercises<\/h2>\r\n1. Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, [latex]{\\mathrm{log}}_{b}\\left({x}^{\\frac{1}{n}}\\right)=\\frac{1}{n}{\\mathrm{log}}_{b}\\left(x\\right)[\/latex].\r\n\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\n25.\u00a0[latex]{\\mathrm{log}}_{7}\\left(15\\right)=\\frac{\\mathrm{ln}\\left(15\\right)}{\\mathrm{ln}\\left(7\\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\n41.\u00a0Let b<\/em>\u00a0and n<\/em>\u00a0be positive integers greater than 1. Then, by the change-of-base formula, [latex]{\\mathrm{log}}_{b}\\left(n\\right)=\\frac{{\\mathrm{log}}_{n}\\left(n\\right)}{{\\mathrm{log}}_{n}\\left(b\\right)}=\\frac{1}{{\\mathrm{log}}_{n}\\left(b\\right)}[\/latex].","rendered":"

Solutions to Try Its<\/h2>\n

1.\u00a0[latex]{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k=3{\\mathrm{log}}_{b}2+{\\mathrm{log}}_{b}k[\/latex]<\/p>\n

2.\u00a0[latex]{\\mathrm{log}}_{3}\\left(x+3\\right)-{\\mathrm{log}}_{3}\\left(x - 1\\right)-{\\mathrm{log}}_{3}\\left(x - 2\\right)[\/latex]<\/p>\n

3.\u00a0[latex]2\\mathrm{ln}x[\/latex]<\/p>\n

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

5.\u00a0[latex]{\\mathrm{log}}_{3}16[\/latex]<\/p>\n

6.\u00a0[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]<\/p>\n

7.\u00a0[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]<\/p>\n

8.\u00a0[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex]<\/p>\n

9. [latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.<\/p>\n

10. [latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]<\/p>\n

11. [latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].<\/p>\n

12. The pH increases by about 0.301.<\/p>\n

13. [latex]\\frac{\\mathrm{ln}8}{\\mathrm{ln}0.5}[\/latex]<\/p>\n

14. [latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]<\/p>\n

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

1. Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, [latex]{\\mathrm{log}}_{b}\\left({x}^{\\frac{1}{n}}\\right)=\\frac{1}{n}{\\mathrm{log}}_{b}\\left(x\\right)[\/latex].<\/p>\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

25.\u00a0[latex]{\\mathrm{log}}_{7}\\left(15\\right)=\\frac{\\mathrm{ln}\\left(15\\right)}{\\mathrm{ln}\\left(7\\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

41.\u00a0Let b<\/em>\u00a0and n<\/em>\u00a0be positive integers greater than 1. Then, by the change-of-base formula, [latex]{\\mathrm{log}}_{b}\\left(n\\right)=\\frac{{\\mathrm{log}}_{n}\\left(n\\right)}{{\\mathrm{log}}_{n}\\left(b\\right)}=\\frac{1}{{\\mathrm{log}}_{n}\\left(b\\right)}[\/latex].<\/p>\n\n\t\t\t

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