{"id":1565,"date":"2015-11-12T18:35:28","date_gmt":"2015-11-12T18:35:28","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1565"},"modified":"2015-11-12T18:35:28","modified_gmt":"2015-11-12T18:35:28","slug":"solutions-35","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/chapter\/solutions-35\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions to Try Its<\/h2>\n1. a.\u00a0[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6\\\\[\/latex] is equivalent to [latex]{10}^{6}=1,000,000\\\\[\/latex]\nb. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2\\\\[\/latex] is equivalent to [latex]{5}^{2}=25\\\\[\/latex]\n\n2. a.\u00a0[latex]{3}^{2}=9\\\\[\/latex] is equivalent to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2\\\\[\/latex]\nb. [latex]{5}^{3}=125\\\\[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3\\\\[\/latex]\nc. [latex]{2}^{-1}=\\frac{1}{2}\\\\[\/latex] is equivalent to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1\\\\[\/latex]\n\n3.\u00a0[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}\\\\[\/latex] (recalling that [latex]\\sqrt{121}={\\left(121\\right)}^{\\frac{1}{2}}=11\\\\[\/latex] )\n\n4.\u00a0[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5\\\\[\/latex]\n\n5.\u00a0It is not possible to take the logarithm of a negative number in the set of real numbers.\n\n6.\u00a0It is not possible to take the logarithm of a negative number in the set of real numbers.\n

Solutions to Odd-Numbered Exercises<\/h2>\n1.\u00a0A logarithm is an exponent. Specifically, it is the exponent to which a base b<\/em>\u00a0is raised to produce a given value. In the expressions given, the base b<\/em>\u00a0has the same value. The exponent, y<\/em>, in the expression [latex]{b}^{y}\\\\[\/latex] can also be written as the logarithm, [latex]{\\mathrm{log}}_{b}x\\\\[\/latex], and the value of x<\/em>\u00a0is the result of raising b<\/em>\u00a0to the power of y<\/em>.\n\n3.\u00a0Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation [latex]{b}^{y}=x[\/latex]\\\\, and then properties of exponents can be applied to solve for x<\/em>.\n\n5.\u00a0The natural logarithm is a special case of the logarithm with base b<\/em>\u00a0in that the natural log always has base e<\/em>. Rather than notating the natural logarithm as [latex]{\\mathrm{log}}_{e}\\left(x\\right)\\\\[\/latex], the notation used is [latex]\\mathrm{ln}\\left(x\\right)\\\\[\/latex].\n\n7.\u00a0[latex]{a}^{c}=b\\\\[\/latex]\n\n9. [latex]{x}^{y}=64\\\\[\/latex]\n\n11.\u00a0[latex]{15}^{b}=a\\\\[\/latex]\n\n13.\u00a0[latex]{13}^{a}=142\\\\[\/latex]\n\n15.\u00a0[latex]{e}^{n}=w\\\\[\/latex]\n\n17.\u00a0[latex]{\\text{log}}_{c}\\left(k\\right)=d\\\\[\/latex]\n\n19.\u00a0[latex]{\\mathrm{log}}_{19}y=x\\\\[\/latex]\n\n21.\u00a0[latex]{\\mathrm{log}}_{n}\\left(103\\right)=4\\\\[\/latex]\n\n23.\u00a0[latex]{\\mathrm{log}}_{y}\\left(\\frac{39}{100}\\right)=x\\\\[\/latex]\n\n25.\u00a0[latex]\\text{ln}\\left(h\\right)=k\\\\[\/latex]\n\n27.\u00a0[latex]x={2}^{-3}=\\frac{1}{8}\\\\[\/latex]\n\n29.\u00a0[latex]x={3}^{3}=27\\\\[\/latex]\n\n31.\u00a0[latex]x={9}^{\\frac{1}{2}}=3\\\\[\/latex]\n\n33.\u00a0[latex]x={6}^{-3}=\\frac{1}{216}\\\\[\/latex]\n\n35.\u00a0[latex]x={e}^{2}\\\\[\/latex]\n\n37. 32\n\n39. 1.06\n\n41. 14.125\n\n43.\u00a0[latex]\\frac{1}{2}\\\\[\/latex]\n\n45.\u00a04\n\n47.\u00a0\u20133\n\n49.\u00a0\u201312\n\n51.\u00a00\n\n53.\u00a010\n\n55. 2.708\n\n57. 0.151\n\n59.\u00a0No, the function has no defined value for x\u00a0<\/em>= 0. To verify, suppose x\u00a0<\/em>= 0 is in the domain of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)\\\\[\/latex]. Then there is some number n<\/em>\u00a0such that [latex]n=\\mathrm{log}\\left(0\\right)\\\\[\/latex]. Rewriting as an exponential equation gives: [latex]{10}^{n}=0\\\\[\/latex], which is impossible since no such real number n<\/em>\u00a0exists. Therefore, x\u00a0<\/em>= 0 is not the domain of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)\\\\[\/latex].\n\n61.\u00a0Yes. Suppose there exists a real number x<\/em>\u00a0such that [latex]\\mathrm{ln}x=2\\\\[\/latex]. Rewriting as an exponential equation gives [latex]x={e}^{2}\\\\[\/latex], which is a real number. To verify, let [latex]x={e}^{2}\\\\[\/latex]. Then, by definition, [latex]\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({e}^{2}\\right)=2\\\\[\/latex].\n\n63.\u00a0No; [latex]\\mathrm{ln}\\left(1\\right)=0\\\\[\/latex], so [latex]\\frac{\\mathrm{ln}\\left({e}^{1.725}\\right)}{\\mathrm{ln}\\left(1\\right)}\\\\[\/latex] is undefined.\n\n65.\u00a02","rendered":"

Solutions to Try Its<\/h2>\n

1. a.\u00a0[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6\\\\[\/latex] is equivalent to [latex]{10}^{6}=1,000,000\\\\[\/latex]
\nb. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2\\\\[\/latex] is equivalent to [latex]{5}^{2}=25\\\\[\/latex]<\/p>\n

2. a.\u00a0[latex]{3}^{2}=9\\\\[\/latex] is equivalent to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2\\\\[\/latex]
\nb. [latex]{5}^{3}=125\\\\[\/latex] is equivalent to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3\\\\[\/latex]
\nc. [latex]{2}^{-1}=\\frac{1}{2}\\\\[\/latex] is equivalent to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1\\\\[\/latex]<\/p>\n

3.\u00a0[latex]{\\mathrm{log}}_{121}\\left(11\\right)=\\frac{1}{2}\\\\[\/latex] (recalling that [latex]\\sqrt{121}={\\left(121\\right)}^{\\frac{1}{2}}=11\\\\[\/latex] )<\/p>\n

4.\u00a0[latex]{\\mathrm{log}}_{2}\\left(\\frac{1}{32}\\right)=-5\\\\[\/latex]<\/p>\n

5.\u00a0It is not possible to take the logarithm of a negative number in the set of real numbers.<\/p>\n

6.\u00a0It is not possible to take the logarithm of a negative number in the set of real numbers.<\/p>\n

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

1.\u00a0A logarithm is an exponent. Specifically, it is the exponent to which a base b<\/em>\u00a0is raised to produce a given value. In the expressions given, the base b<\/em>\u00a0has the same value. The exponent, y<\/em>, in the expression [latex]{b}^{y}\\\\[\/latex] can also be written as the logarithm, [latex]{\\mathrm{log}}_{b}x\\\\[\/latex], and the value of x<\/em>\u00a0is the result of raising b<\/em>\u00a0to the power of y<\/em>.<\/p>\n

3.\u00a0Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation [latex]{b}^{y}=x[\/latex]\\\\, and then properties of exponents can be applied to solve for x<\/em>.<\/p>\n

5.\u00a0The natural logarithm is a special case of the logarithm with base b<\/em>\u00a0in that the natural log always has base e<\/em>. Rather than notating the natural logarithm as [latex]{\\mathrm{log}}_{e}\\left(x\\right)\\\\[\/latex], the notation used is [latex]\\mathrm{ln}\\left(x\\right)\\\\[\/latex].<\/p>\n

7.\u00a0[latex]{a}^{c}=b\\\\[\/latex]<\/p>\n

9. [latex]{x}^{y}=64\\\\[\/latex]<\/p>\n

11.\u00a0[latex]{15}^{b}=a\\\\[\/latex]<\/p>\n

13.\u00a0[latex]{13}^{a}=142\\\\[\/latex]<\/p>\n

15.\u00a0[latex]{e}^{n}=w\\\\[\/latex]<\/p>\n

17.\u00a0[latex]{\\text{log}}_{c}\\left(k\\right)=d\\\\[\/latex]<\/p>\n

19.\u00a0[latex]{\\mathrm{log}}_{19}y=x\\\\[\/latex]<\/p>\n

21.\u00a0[latex]{\\mathrm{log}}_{n}\\left(103\\right)=4\\\\[\/latex]<\/p>\n

23.\u00a0[latex]{\\mathrm{log}}_{y}\\left(\\frac{39}{100}\\right)=x\\\\[\/latex]<\/p>\n

25.\u00a0[latex]\\text{ln}\\left(h\\right)=k\\\\[\/latex]<\/p>\n

27.\u00a0[latex]x={2}^{-3}=\\frac{1}{8}\\\\[\/latex]<\/p>\n

29.\u00a0[latex]x={3}^{3}=27\\\\[\/latex]<\/p>\n

31.\u00a0[latex]x={9}^{\\frac{1}{2}}=3\\\\[\/latex]<\/p>\n

33.\u00a0[latex]x={6}^{-3}=\\frac{1}{216}\\\\[\/latex]<\/p>\n

35.\u00a0[latex]x={e}^{2}\\\\[\/latex]<\/p>\n

37. 32<\/p>\n

39. 1.06<\/p>\n

41. 14.125<\/p>\n

43.\u00a0[latex]\\frac{1}{2}\\\\[\/latex]<\/p>\n

45.\u00a04<\/p>\n

47.\u00a0\u20133<\/p>\n

49.\u00a0\u201312<\/p>\n

51.\u00a00<\/p>\n

53.\u00a010<\/p>\n

55. 2.708<\/p>\n

57. 0.151<\/p>\n

59.\u00a0No, the function has no defined value for x\u00a0<\/em>= 0. To verify, suppose x\u00a0<\/em>= 0 is in the domain of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)\\\\[\/latex]. Then there is some number n<\/em>\u00a0such that [latex]n=\\mathrm{log}\\left(0\\right)\\\\[\/latex]. Rewriting as an exponential equation gives: [latex]{10}^{n}=0\\\\[\/latex], which is impossible since no such real number n<\/em>\u00a0exists. Therefore, x\u00a0<\/em>= 0 is not the domain of the function [latex]f\\left(x\\right)=\\mathrm{log}\\left(x\\right)\\\\[\/latex].<\/p>\n

61.\u00a0Yes. Suppose there exists a real number x<\/em>\u00a0such that [latex]\\mathrm{ln}x=2\\\\[\/latex]. Rewriting as an exponential equation gives [latex]x={e}^{2}\\\\[\/latex], which is a real number. To verify, let [latex]x={e}^{2}\\\\[\/latex]. Then, by definition, [latex]\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({e}^{2}\\right)=2\\\\[\/latex].<\/p>\n

63.\u00a0No; [latex]\\mathrm{ln}\\left(1\\right)=0\\\\[\/latex], so [latex]\\frac{\\mathrm{ln}\\left({e}^{1.725}\\right)}{\\mathrm{ln}\\left(1\\right)}\\\\[\/latex] is undefined.<\/p>\n

65.\u00a02<\/p>\n\n\t\t\t

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