{"id":11199,"date":"2015-07-14T18:47:14","date_gmt":"2015-07-14T18:47:14","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11199"},"modified":"2018-06-28T14:52:44","modified_gmt":"2018-06-28T14:52:44","slug":"solutions-20","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/solutions-20\/","title":{"raw":"Solutions 5.4: Graphs of Logarithmic Functions","rendered":"Solutions 5.4: Graphs of Logarithmic Functions"},"content":{"raw":"

Solutions to Try Its<\/h2>\r\n1.\u00a0[latex]\\left(2,\\infty \\right)[\/latex]\r\n\r\n2.\u00a0[latex]\\left(5,\\infty \\right)[\/latex]\r\n\r\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.\r\n\"Graph<\/span>\r\n\r\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.\r\n\"Graph<\/span>\r\n\r\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.\r\n\"Graph<\/span>\r\n\r\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.\r\n\"Graph<\/span>\r\n\r\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.\r\n
\r\n\r\n \"Graph<\/span>\r\n\r\n<\/div>\r\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.\r\n\"Graph<\/span>\r\n

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

Solutions to Odd-Numbered Exercises<\/h2>\r\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.\r\n\r\n3.\u00a0Shifting the function right or left and reflecting the function about the y-axis will affect its domain.\r\n\r\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.\r\n\r\n7.\u00a0Domain: [latex]\\left(-\\infty ,\\frac{1}{2}\\right)[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]\r\n\r\n9.\u00a0Domain: [latex]\\left(-\\frac{17}{4},\\infty \\right)[\/latex]; Range: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]\r\n\r\n11.\u00a0Domain: [latex]\\left(5,\\infty \\right)[\/latex]; Vertical asymptote: x\u00a0<\/em>= 5\r\n\r\n13.\u00a0Domain: [latex]\\left(-\\frac{1}{3},\\infty \\right)[\/latex]; Vertical asymptote: [latex]x=-\\frac{1}{3}[\/latex]\r\n\r\n15.\u00a0Domain: [latex]\\left(-3,\\infty \\right)[\/latex]; Vertical asymptote: x\u00a0<\/em>= \u20133\r\n\r\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]\r\n\r\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]\r\n\r\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\r\n\r\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\r\n\r\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\r\n\r\n27. B\r\n\r\n29. C\r\n\r\n31. B\r\n\r\n33. C\r\n\r\n35.\r\n\"Graph\r\n\r\n37.\r\n\"Graph\r\n\r\n39. C\r\n\r\n41.\r\n\"Graph\r\n\r\n43.\r\n\"Graph\r\n\r\n45.\r\n\"Graph\r\n\r\n47.\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(-\\left(x - 1\\right)\\right)[\/latex]\r\n\r\n49.\u00a0[latex]f\\left(x\\right)=3{\\mathrm{log}}_{4}\\left(x+2\\right)[\/latex]\r\n\r\n51.\u00a0x\u00a0<\/em>= 2\r\n\r\n53.\u00a0[latex]x\\approx \\text{2}\\text{.303}[\/latex]\r\n\r\n55.\u00a0[latex]x\\approx -0.472[\/latex]\r\n\r\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].\r\n\r\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].\r\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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