{"id":1608,"date":"2015-11-12T18:35:27","date_gmt":"2015-11-12T18:35:27","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1608"},"modified":"2015-11-12T18:35:27","modified_gmt":"2015-11-12T18:35:27","slug":"solutions-33","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/chapter\/solutions-33\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions to Try Its<\/h2>\n1.\u00a0[latex]\\left(2,\\infty \\right)\\\\[\/latex]\n\n2.\u00a0[latex]\\left(5,\\infty \\right)\\\\[\/latex]\n\n3.\u00a0The domain is [latex]\\left(0,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.\n\"Graph<\/span>\n\n4.\u00a0The domain is [latex]\\left(-4,\\infty \\right)\\\\[\/latex], the range [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the asymptote x\u00a0<\/em>= \u20134.\n\"Graph<\/span>\n\n5.\u00a0The domain is [latex]\\left(0,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.\n\"Graph<\/span>\n\n6.\u00a0The domain is [latex]\\left(0,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.\n\"Graph<\/span>\n\n7.\u00a0The domain is [latex]\\left(2,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 2.\n
\n\n \"Graph<\/span>\n\n<\/div>\n8.\u00a0The domain is [latex]\\left(-\\infty ,0\\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.\n\"Graph<\/span>\n

9.\u00a0[latex]x\\approx 3.049\\\\[\/latex]<\/p>\n10.\u00a0x\u00a0<\/em>= 1\n\n11.\u00a0[latex]f\\left(x\\right)=2\\mathrm{ln}\\left(x+3\\right)-1\\\\[\/latex]\n

Solutions to Odd-Numbered Exercises<\/h2>\n1.\u00a0Since the functions are inverses, their graphs are mirror images about the line y\u00a0<\/em>= x<\/em>. So for every point [latex]\\left(a,b\\right)\\\\[\/latex] on the graph of a logarithmic function, there is a corresponding point [latex]\\left(b,a\\right)\\\\[\/latex] on the graph of its inverse exponential function.\n\n3.\u00a0Shifting the function right or left and reflecting the function about the y-axis will affect its domain.\n\n5.\u00a0No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.\n\n7.\u00a0Domain: [latex]\\left(-\\infty ,\\frac{1}{2}\\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]\n\n9.\u00a0Domain: [latex]\\left(-\\frac{17}{4},\\infty \\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]\n\n11.\u00a0Domain: [latex]\\left(5,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 5\n\n13.\u00a0Domain: [latex]\\left(-\\frac{1}{3},\\infty \\right)\\\\[\/latex]; Vertical asymptote: [latex]x=-\\frac{1}{3}\\\\[\/latex]\n\n15.\u00a0Domain: [latex]\\left(-3,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= \u20133\n\n17.\u00a0Domain: [latex]\\left(\\frac{3}{7},\\infty \\right)\\\\[\/latex];\u00a0Vertical asymptote: [latex]x=\\frac{3}{7}\\\\[\/latex] ; End behavior: as [latex]x\\to {\\left(\\frac{3}{7}\\right)}^{+},f\\left(x\\right)\\to -\\infty \\\\[\/latex] and as [latex]x\\to \\infty ,f\\left(x\\right)\\to \\infty \\\\[\/latex]\n\n19.\u00a0Domain: [latex]\\left(-3,\\infty \\right)\\\\[\/latex] ; Vertical asymptote: x\u00a0<\/em>= \u20133;\u00a0End behavior: as [latex]x\\to -{3}^{+}\\\\[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty \\\\[\/latex] and as [latex]x\\to \\infty \\\\[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty \\\\[\/latex]\n\n21.\u00a0Domain: [latex]\\left(1,\\infty \\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 1; x<\/em>-intercept: [latex]\\left(\\frac{5}{4},0\\right)\\\\[\/latex]; y<\/em>-intercept: DNE\n\n23.\u00a0Domain: [latex]\\left(-\\infty ,0\\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 0; x-intercept: [latex]\\left(-{e}^{2},0\\right)\\\\[\/latex]; y<\/em>-intercept: DNE\n\n25.\u00a0Domain: [latex]\\left(0,\\infty \\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 0; x<\/em>-intercept: [latex]\\left({e}^{3},0\\right)[\/latex]; y<\/em>-intercept: DNE\n\n27. B\n\n29. C\n\n31. B\n\n33. C\n\n35.\n\"Graph\n\n37.\n\"Graph\n\n39. C\n\n41.\n\"Graph\n\n43.\n\"Graph\n\n45.\n\"Graph\n\n47.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(-\\left(x - 1\\right)\\right)\\\\[\/latex]\n\n49.\u00a0[latex]f\\left(x\\right)=3{\\mathrm{log}}_{4}\\left(x+2\\right)\\\\[\/latex]\n\n51.\u00a0x\u00a0<\/em>= 2\n\n53.\u00a0[latex]x\\approx \\text{2}\\text{.303}\\\\[\/latex]\n\n55.\u00a0[latex]x\\approx -0.472\\\\[\/latex]\n\n57.\u00a0The graphs of [latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{2}}\\left(x\\right)\\\\[\/latex] and [latex]g\\left(x\\right)=-{\\mathrm{log}}_{2}\\left(x\\right)\\\\[\/latex] appear to be the same; Conjecture: for any positive base [latex]b\\ne 1\\\\[\/latex], [latex]{\\mathrm{log}}_{b}\\left(x\\right)=-{\\mathrm{log}}_{\\frac{1}{b}}\\left(x\\right)\\\\[\/latex].\n\n59.\u00a0Recall that the argument of a logarithmic function must be positive, so we determine where [latex]\\frac{x+2}{x - 4}>0\\\\[\/latex] . From the graph of the function [latex]f\\left(x\\right)=\\frac{x+2}{x - 4}\\\\[\/latex], note that the graph lies above the x-axis on the interval [latex]\\left(-\\infty ,-2\\right)\\\\[\/latex] and again to the right of the vertical asymptote, that is [latex]\\left(4,\\infty \\right)\\\\[\/latex]. Therefore, the domain is [latex]\\left(-\\infty ,-2\\right)\\cup \\left(4,\\infty \\right)\\\\[\/latex].\n\"\"","rendered":"

Solutions to Try Its<\/h2>\n

1.\u00a0[latex]\\left(2,\\infty \\right)\\\\[\/latex]<\/p>\n

2.\u00a0[latex]\\left(5,\\infty \\right)\\\\[\/latex]<\/p>\n

3.\u00a0The domain is [latex]\\left(0,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.
\n\"Graph<\/span><\/p>\n

4.\u00a0The domain is [latex]\\left(-4,\\infty \\right)\\\\[\/latex], the range [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the asymptote x\u00a0<\/em>= \u20134.
\n\"Graph<\/span><\/p>\n

5.\u00a0The domain is [latex]\\left(0,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.
\n\"Graph<\/span><\/p>\n

6.\u00a0The domain is [latex]\\left(0,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.
\n\"Graph<\/span><\/p>\n

7.\u00a0The domain is [latex]\\left(2,\\infty \\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 2.<\/p>\n

\n

\"Graph<\/span><\/p>\n<\/div>\n

8.\u00a0The domain is [latex]\\left(-\\infty ,0\\right)\\\\[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex], and the vertical asymptote is x\u00a0<\/em>= 0.
\n\"Graph<\/span><\/p>\n

9.\u00a0[latex]x\\approx 3.049\\\\[\/latex]<\/p>\n

10.\u00a0x\u00a0<\/em>= 1<\/p>\n

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

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

1.\u00a0Since the functions are inverses, their graphs are mirror images about the line y\u00a0<\/em>= x<\/em>. So for every point [latex]\\left(a,b\\right)\\\\[\/latex] on the graph of a logarithmic function, there is a corresponding point [latex]\\left(b,a\\right)\\\\[\/latex] on the graph of its inverse exponential function.<\/p>\n

3.\u00a0Shifting the function right or left and reflecting the function about the y-axis will affect its domain.<\/p>\n

5.\u00a0No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.<\/p>\n

7.\u00a0Domain: [latex]\\left(-\\infty ,\\frac{1}{2}\\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]<\/p>\n

9.\u00a0Domain: [latex]\\left(-\\frac{17}{4},\\infty \\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]<\/p>\n

11.\u00a0Domain: [latex]\\left(5,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 5<\/p>\n

13.\u00a0Domain: [latex]\\left(-\\frac{1}{3},\\infty \\right)\\\\[\/latex]; Vertical asymptote: [latex]x=-\\frac{1}{3}\\\\[\/latex]<\/p>\n

15.\u00a0Domain: [latex]\\left(-3,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= \u20133<\/p>\n

17.\u00a0Domain: [latex]\\left(\\frac{3}{7},\\infty \\right)\\\\[\/latex];\u00a0Vertical asymptote: [latex]x=\\frac{3}{7}\\\\[\/latex] ; End behavior: as [latex]x\\to {\\left(\\frac{3}{7}\\right)}^{+},f\\left(x\\right)\\to -\\infty \\\\[\/latex] and as [latex]x\\to \\infty ,f\\left(x\\right)\\to \\infty \\\\[\/latex]<\/p>\n

19.\u00a0Domain: [latex]\\left(-3,\\infty \\right)\\\\[\/latex] ; Vertical asymptote: x\u00a0<\/em>= \u20133;\u00a0End behavior: as [latex]x\\to -{3}^{+}\\\\[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty \\\\[\/latex] and as [latex]x\\to \\infty \\\\[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty \\\\[\/latex]<\/p>\n

21.\u00a0Domain: [latex]\\left(1,\\infty \\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 1; x<\/em>-intercept: [latex]\\left(\\frac{5}{4},0\\right)\\\\[\/latex]; y<\/em>-intercept: DNE<\/p>\n

23.\u00a0Domain: [latex]\\left(-\\infty ,0\\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 0; x-intercept: [latex]\\left(-{e}^{2},0\\right)\\\\[\/latex]; y<\/em>-intercept: DNE<\/p>\n

25.\u00a0Domain: [latex]\\left(0,\\infty \\right)\\\\[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)\\\\[\/latex]; Vertical asymptote: x\u00a0<\/em>= 0; x<\/em>-intercept: [latex]\\left({e}^{3},0\\right)[\/latex]; y<\/em>-intercept: DNE<\/p>\n

27. B<\/p>\n

29. C<\/p>\n

31. B<\/p>\n

33. C<\/p>\n

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

37.
\n\"Graph<\/p>\n

39. C<\/p>\n

41.
\n\"Graph<\/p>\n

43.
\n\"Graph<\/p>\n

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

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

49.\u00a0[latex]f\\left(x\\right)=3{\\mathrm{log}}_{4}\\left(x+2\\right)\\\\[\/latex]<\/p>\n

51.\u00a0x\u00a0<\/em>= 2<\/p>\n

53.\u00a0[latex]x\\approx \\text{2}\\text{.303}\\\\[\/latex]<\/p>\n

55.\u00a0[latex]x\\approx -0.472\\\\[\/latex]<\/p>\n

57.\u00a0The graphs of [latex]f\\left(x\\right)={\\mathrm{log}}_{\\frac{1}{2}}\\left(x\\right)\\\\[\/latex] and [latex]g\\left(x\\right)=-{\\mathrm{log}}_{2}\\left(x\\right)\\\\[\/latex] appear to be the same; Conjecture: for any positive base [latex]b\\ne 1\\\\[\/latex], [latex]{\\mathrm{log}}_{b}\\left(x\\right)=-{\\mathrm{log}}_{\\frac{1}{b}}\\left(x\\right)\\\\[\/latex].<\/p>\n

59.\u00a0Recall that the argument of a logarithmic function must be positive, so we determine where [latex]\\frac{x+2}{x - 4}>0\\\\[\/latex] . From the graph of the function [latex]f\\left(x\\right)=\\frac{x+2}{x - 4}\\\\[\/latex], note that the graph lies above the x-axis on the interval [latex]\\left(-\\infty ,-2\\right)\\\\[\/latex] and again to the right of the vertical asymptote, that is [latex]\\left(4,\\infty \\right)\\\\[\/latex]. Therefore, the domain is [latex]\\left(-\\infty ,-2\\right)\\cup \\left(4,\\infty \\right)\\\\[\/latex].
\n\"\"<\/p>\n\n\t\t\t

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