{"id":299,"date":"2026-02-02T15:42:50","date_gmt":"2026-02-02T15:42:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&#038;p=299"},"modified":"2026-02-02T16:52:40","modified_gmt":"2026-02-02T16:52:40","slug":"exponential-and-logarithmic-functions-get-stronger-answer-key-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/exponential-and-logarithmic-functions-get-stronger-answer-key-2\/","title":{"raw":"Exponential and Logarithmic Functions Get Stronger Answer Key","rendered":"Exponential and Logarithmic Functions Get Stronger Answer Key"},"content":{"raw":"<div class=\"bc-section section\">\r\n<div id=\"chapter-gs-exp-log\" class=\"chapter standard\" title=\"Exponential and Logarithmic Functions: Get Stronger Key\">\r\n<div class=\"chapter-title-wrap\">\r\n<h1 class=\"chapter-title\">Exponential and Logarithmic Functions: Get Stronger Key<\/h1>\r\n<\/div>\r\n<div class=\"ugc chapter-ugc\">\r\n<h2>Exponential Functions<\/h2>\r\n2. Yes. In an exponential function written as [latex]f(x)=ab^x[\/latex], the base [latex]b[\/latex] determines growth or decay: if [latex]b&gt;1[\/latex] the function grows; if [latex]0&lt;b&lt;1[\/latex] the function decays.<br>\r\n\r\n5. Exponential function. \u201cDecreases by a factor of [latex]\\frac{1}{8}[\/latex] every 24 hours\u201d means each 24-hour period multiplies the population by a constant ratio [latex]\\frac{1}{8}[\/latex].<br>\r\n\r\n6. Exponential function. Increasing by 3.25% annually means the value is multiplied each year by the constant factor [latex]1.0325[\/latex].<br>\r\n\r\n7. Not exponential. Subtracting $5 each session is a constant difference (linear), not a constant ratio.<br>\r\n\r\n9. Forest [latex]B[\/latex] grows faster because its growth factor is larger: [latex]1.029&gt;1.025[\/latex].<br>\r\n\r\n10. Forest [latex]A[\/latex] had more trees initially: [latex]A(0)=115[\/latex] and [latex]B(0)=82[\/latex], so forest [latex]A[\/latex] had [latex]115-82=33[\/latex] more trees.<br>\r\n\r\n11. After 20 years: [latex]A(20)\\approx 188[\/latex] and [latex]B(20)\\approx 145[\/latex]. Forest [latex]A[\/latex] will have about [latex]188-145=43[\/latex] more trees.<br>\r\n\r\n12. After 100 years: [latex]A(100)\\approx 1359[\/latex] and [latex]B(100)\\approx 1430[\/latex]. Forest [latex]B[\/latex] will have about [latex]1430-1359=71[\/latex] more trees.<br>\r\n\r\n15. Exponential growth because the base is [latex]1.06&gt;1[\/latex].<br>\r\n\r\n17. Exponential decay because the base is [latex]0.97[\/latex] and [latex]0&lt;0.97&lt;1[\/latex].<br>\r\n\r\n19. Let [latex]f(x)=ab^x[\/latex]. From [latex](0,2000)[\/latex], [latex]a=2000[\/latex]. Using [latex](2,20)[\/latex]: [latex]20=2000b^2\\Rightarrow b^2=0.01\\Rightarrow b=0.1[\/latex]. So [latex]f(x)=2000(0.1)^x[\/latex].\r\n<br>\r\n\r\n21. Let [latex]f(x)=ab^x[\/latex]. Using [latex](-2,6)[\/latex] and [latex](3,1)[\/latex], the function is [latex]f(x)=6^{\\frac{3}{5}}\\left(6^{-\\frac{1}{5}}\\right)^x[\/latex] (equivalently [latex]f(x)=6^{\\frac{3-x}{5}}[\/latex]).<br>\r\n\r\n23. Linear. The differences are constant: [latex]40-70=-30[\/latex], [latex]10-40=-30[\/latex], [latex]-20-10=-30[\/latex].<br>\r\n\r\n25. Neither. The differences are not constant and the ratios are not constant, so it is not linear or exponential.<br>\r\n\r\n31. [latex]A(t)=P\\left(1+\\frac{r}{n}\\right)^{nt}[\/latex] with [latex]P=6500[\/latex], [latex]r=0.036[\/latex], [latex]n=2[\/latex], [latex]t=20[\/latex]: [latex]A\\approx 6500(1.018)^{40}\\approx \\$13{,}268.58[\/latex].<br>\r\n\r\n32. Weekly compounding: [latex]n=52[\/latex]. [latex]A\\approx 6500\\left(1+\\frac{0.036}{52}\\right)^{52\\cdot 20}\\approx \\$13{,}350.49[\/latex]. This is about [latex]\\$13{,}350.49-\\$13{,}268.58=\\$81.91[\/latex] more.<br>\r\n\r\n35. If earning interest 5 more years (25 years total):\r\nSemi-annually: [latex]\\$15{,}859.97-\\$13{,}268.58=\\$2{,}591.39[\/latex] more.\r\nWeekly: [latex]\\$15{,}982.44-\\$13{,}350.49=\\$2{,}631.95[\/latex] more.<br>\r\n\r\n42. Continuous compounding: [latex]A=Pe^{rt}=12000e^{0.072\\cdot 30}\\approx \\$104{,}053.65[\/latex].<br>\r\n\r\n43. Monthly compounding: [latex]A\\approx 12000\\left(1+\\frac{0.072}{12}\\right)^{12\\cdot 30}\\approx \\$103{,}384.23[\/latex]. This is about [latex]\\$104{,}053.65-\\$103{,}384.23=\\$669.42[\/latex] less than continuous compounding.<br>\r\n\r\n51. Using [latex]y=ab^x[\/latex]: from [latex](0,3)[\/latex], [latex]a=3[\/latex]. From [latex](3,375)[\/latex], [latex]375=3b^3\\Rightarrow b^3=125\\Rightarrow b=5[\/latex]. So [latex]y=3\\cdot 5^x[\/latex].<br>\r\n\r\n53. Using [latex]y=ab^x[\/latex] through [latex](20,29.495)[\/latex] and [latex](150,730.89)[\/latex], a suitable model is [latex]y\\approx 18(1.025)^x[\/latex].<br>\r\n\r\n55. Using [latex]y=ab^x[\/latex] through [latex](11,310.035)[\/latex] and [latex](25,356.3652)[\/latex], a suitable model is [latex]y\\approx 277.899(1.01)^x[\/latex].<br>\r\n\r\n61. Growth rate 9% per year from 2012 to 2020 (8 years): [latex]23900(1.09)^8\\approx 47{,}622[\/latex] foxes.<br>\r\n\r\n63. From 1985 to 2005 is 20 years: [latex]145000=110000(1+r)^{20}\\Rightarrow r\\approx 0.0139[\/latex] (about 1.39% per year).\r\nValue in 2010 (25 years after 1985): [latex]110000(1+r)^{25}\\approx \\$155{,}368.09[\/latex].<br>\r\n\r\n65. [latex]54000=P\\left(1+\\frac{0.082}{365}\\right)^{365\\cdot 5}\\Rightarrow P\\approx \\$35{,}838.76[\/latex].<br>\r\n\r\n67. From 2000 to 2025 is 25 years, [latex]P=13500[\/latex], [latex]r=0.0725[\/latex].\r\nMonthly: [latex]A\\approx 13500\\left(1+\\frac{0.0725}{12}\\right)^{12\\cdot 25}\\approx \\$82{,}247.78[\/latex].\r\nContinuous: [latex]A\\approx 13500e^{0.0725\\cdot 25}\\approx \\$82{,}697.53[\/latex].\r\nContinuous earns about [latex]\\$82{,}697.53-\\$82{,}247.78=\\$449.75[\/latex] more.<br>\r\n\r\n<h2>Graphs of Exponential Functions<\/h2>\r\n1. The horizontal asymptote tells what [latex]f(x)[\/latex] approaches as [latex]x\\to\\infty[\/latex] (and often also as [latex]x\\to -\\infty[\/latex]). It describes the \u201cend behavior\u201d level the graph gets closer and closer to without reaching.<br>\r\n3. Reflect about the y-axis: [latex]3^{-x}[\/latex]. Stretch vertically by 4: [latex]g(x)=4\\cdot 3^{-x}[\/latex].\r\ny-intercept: [latex]g(0)=4[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](0,\\infty)[\/latex].\r\n\r\n5. Reflect about the x-axis: [latex]-10^x[\/latex]. Shift up 7: [latex]g(x)=-10^x+7[\/latex].\r\ny-intercept: [latex]g(0)=-1+7=6[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](-\\infty,7)[\/latex].\r\n\r\n7. Starting with [latex]\\left(\\frac{1}{4}\\right)^x[\/latex]: left 2 gives [latex]\\left(\\frac{1}{4}\\right)^{x+2}[\/latex]; stretch by 4 gives [latex]4\\left(\\frac{1}{4}\\right)^{x+2}[\/latex]; reflect over x-axis gives [latex]-4\\left(\\frac{1}{4}\\right)^{x+2}[\/latex]; down 4 gives\r\n[latex]g(x)=-4\\left(\\frac{1}{4}\\right)^{x+2}-4[\/latex].\r\ny-intercept: [latex]g(0)=-4\\left(\\frac{1}{4}\\right)^2-4=-\\frac{17}{4}[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](-\\infty,-4)[\/latex].\r\n\r\n11. <img class=\"alignnone size-full wp-image-302\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM.png\" alt=\"The graph shows three exponential functions on the same coordinate plane. All three pass through the point (0, 3). One function is an increasing exponential that grows slowly for negative x and more rapidly for positive x. A second increasing exponential grows more steeply than the first for positive x. The third function is an exponential decay function that decreases as x increases and approaches y equals 0 to the right. All three functions have a horizontal asymptote at y equals 0.\" width=\"724\" height=\"868\" \/>\r\n\r\n13. B\r\n\r\n14. D\r\n\r\n15. A\r\n\r\n16. F\r\n\r\n17. E\r\n\r\n18. C\r\n\r\n26. [latex]f(x)=2^{-x}[\/latex]. Horizontal asymptote: [latex]y=0[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](0,\\infty)[\/latex].\r\n\r\n27. [latex]h(x)=2^x+3[\/latex]. Horizontal asymptote: [latex]y=3[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](3,\\infty)[\/latex].\r\n\r\n28. [latex]f(x)=2^{x-2}[\/latex]. Horizontal asymptote: [latex]y=0[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](0,\\infty)[\/latex].\r\n\r\n39. [latex]y=-2^x+3[\/latex].\r\n\r\n41. [latex]y=-2\\cdot 3^x+7[\/latex].\r\n\r\n47. [latex]116=\\frac14\\left(\\frac18\\right)^x\\Rightarrow \\left(\\frac18\\right)^x=464\\Rightarrow x\\approx -2.953[\/latex].\r\n\r\n49. [latex]5=3\\left(\\frac12\\right)^{x-1}-2\\Rightarrow 7=3\\left(\\frac12\\right)^{x-1}\\Rightarrow x\\approx -0.222[\/latex].\r\n<h2>Logarithmic Functions<\/h2>\r\n3. Use the definition of a logarithm: if [latex]\\log_b(x)=y[\/latex], then [latex]b^y=x[\/latex]. So solving for [latex]x[\/latex] gives [latex]x=b^y[\/latex].\r\n\r\n7. [latex]\\log_a(b)=c \\iff a^c=b[\/latex].\r\n\r\n9. [latex]\\log_x(64)=y \\iff x^y=64[\/latex].\r\n\r\n11. [latex]\\log_{15}(a)=b \\iff 15^b=a[\/latex].\r\n\r\n13. [latex]\\log_{13}(142)=a \\iff 13^a=142[\/latex].\r\n\r\n15. [latex]\\ln(w)=n \\iff e^n=w[\/latex].\r\n\r\n17. [latex]c^d=k \\iff \\log_c(k)=d[\/latex].\r\n\r\n19. [latex]19^x=y \\iff \\log_{19}(y)=x[\/latex].\r\n\r\n21. [latex]n^4=103 \\iff \\log_n(103)=4[\/latex].\r\n\r\n23. [latex]y^x=\\frac{39}{100} \\iff \\log_y\\left(\\frac{39}{100}\\right)=x[\/latex].\r\n\r\n25. [latex]e^k=h \\iff \\ln(h)=k[\/latex].\r\n\r\n27. [latex]\\log_2(x)=-3 \\Rightarrow x=2^{-3}=\\frac18[\/latex].\r\n\r\n29. [latex]\\log_3(x)=3 \\Rightarrow x=3^3=27[\/latex].\r\n\r\n31. [latex]\\log_9(x)=\\frac12 \\Rightarrow x=9^{1\/2}=3[\/latex].\r\n\r\n33. [latex]\\log_6(x)=-3 \\Rightarrow x=6^{-3}=\\frac{1}{216}[\/latex].\r\n\r\n35. [latex]\\ln(x)=2 \\Rightarrow x=e^2[\/latex].\r\n\r\n37. [latex]10^{\\log(32)}=32[\/latex].\r\n\r\n39. [latex]e^{\\ln(1.06)}=1.06[\/latex].\r\n\r\n41. [latex]e^{\\ln(10.125)}+4=10.125+4=14.125[\/latex].\r\n\r\n43. [latex]\\log_6(\\sqrt{6})=\\log_6(6^{1\/2})=\\frac12[\/latex].\r\n\r\n45. [latex]6\\log_8(4)=6\\cdot\\frac{2}{3}=4[\/latex].\r\n\r\n47. [latex]\\log(0.001)=\\log(10^{-3})=-3[\/latex].\r\n\r\n49. [latex]2\\log(100^{-3})=2\\log(10^{-6})=2(-6)=-12[\/latex].\r\n\r\n51. [latex]\\ln(1)=0[\/latex].\r\n\r\n53. [latex]25\\ln\\left(e^{2\/5}\\right)=25\\cdot\\frac{2}{5}=10[\/latex].\r\n\r\n59. No. [latex]x=0[\/latex] is not in the domain of [latex]f(x)=\\log(x)[\/latex] because logarithms require [latex]x&gt;0[\/latex]. So [latex]f(0)[\/latex] is undefined (DNE).\r\n\r\n66. [latex]\\log\\left(\\frac{I_1}{I_2}\\right)=M_1-M_2=9.0-6.1=2.9[\/latex], so [latex]\\frac{I_1}{I_2}=10^{2.9}\\approx 794[\/latex]. The 2011 quake was about 794 times as intense.\r\n<h2>Graphs of Logarithmic Functions<\/h2>\r\n1. If two functions are inverses, their graphs are reflections across the line [latex]y=x[\/latex]. So a point [latex](a,b)[\/latex] on one graph corresponds to [latex](b,a)[\/latex] on the inverse graph.\r\n\r\n2. None. The range of a logarithmic function is all real numbers, and vertical translations do not change that (it remains [latex](-\\infty,\\infty)[\/latex]).\r\n\r\n3. Horizontal translations affect the domain because they move the vertical asymptote left or right (changing where the input becomes valid).\r\n\r\n21. [latex]h(x)=\\log_4(x-1)+1[\/latex]. Domain: [latex](1,\\infty)[\/latex]. Range: [latex](-\\infty,\\infty)[\/latex].\r\nx-intercept: [latex]\\log_4(x-1)+1=0\\Rightarrow x-1=\\frac14\\Rightarrow x=\\frac54[\/latex]. y-intercept: DNE.\r\n\r\n23. [latex]g(x)=\\ln(-x)-2[\/latex]. Domain: [latex](-\\infty,0)[\/latex]. Range: [latex](-\\infty,\\infty)[\/latex].\r\nx-intercept: [latex]\\ln(-x)-2=0\\Rightarrow -x=e^2\\Rightarrow x=-e^2[\/latex]. y-intercept: DNE.\r\n\r\n25. [latex]h(x)=3\\ln(x)-9[\/latex]. Domain: [latex](0,\\infty)[\/latex]. Range: [latex](-\\infty,\\infty)[\/latex].\r\nx-intercept: [latex]3\\ln(x)-9=0\\Rightarrow \\ln(x)=3\\Rightarrow x=e^3[\/latex]. y-intercept: DNE.\r\n\r\n31. C\r\n\r\n32. A\r\n\r\n33. B\r\n\r\n35. <img class=\"alignnone wp-image-303\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM.png\" alt=\"The graph shows two logarithmic functions on the same coordinate plane. One function is log base 10 of x, which increases as x increases. The other function is log base one-half of x, which decreases as x increases. Both functions have a vertical asymptote at x equals 0 and pass through the point (1, 0). For values of x between 0 and 1, log base 10 of x has negative y-values while log base one-half of x has positive y-values. For values of x greater than 1, log base 10 of x has positive y-values while log base one-half of x has negative y-values. Both curves approach the y-axis but never touch it.\" width=\"251\" height=\"253\" \/>\r\n\r\n37. <img class=\"alignnone wp-image-304\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM.png\" alt=\"The graph shows an exponential function and a logarithmic function on the same coordinate plane. The exponential function increases rapidly as x increases and passes through the point (0, 1). It has a horizontal asymptote at y equals 0. The logarithmic function increases slowly as x increases and has a vertical asymptote at x equals 0. It passes through the point (1, 0). The point (1, 0) is highlighted on the logarithmic function, showing where the logarithmic curve crosses the x-axis.\" width=\"250\" height=\"257\" \/>\r\n\r\n38. B\r\n\r\n39. C\r\n\r\n40. A\r\n\r\n41. <img class=\"alignnone wp-image-305\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM.png\" alt=\"The graph is a logarithmic function that increases slowly as x increases. It has a vertical asymptote at x equals 0. The graph passes through the points: (1, 0) and (2, about 0.7) As x approaches 0 from the right, the y-values decrease without bound.\" width=\"252\" height=\"279\" \/>\r\n\r\n43. <img class=\"alignnone wp-image-307\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM.png\" alt=\"The graph is a logarithmic function that decreases as x increases. It has a vertical asymptote at x equals 0. The graph passes through the point: (-1, 0) As x approaches 0 from the right, the y-values decrease without bound.\" width=\"244\" height=\"242\" \/>\r\n\r\n45. <img class=\"alignnone wp-image-306\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM.png\" alt=\"The graph is a logarithmic function that decreases as x increases. It has a vertical asymptote at x equals 2. As x approaches 2 from the left, the y-values decrease without bound.\" width=\"250\" height=\"249\" \/>\r\n\r\n51. Let [latex]u=x-1[\/latex]. Then [latex]\\log(u)=\\ln(u)[\/latex] implies [latex]u=1[\/latex], so [latex]x=2[\/latex].\r\n\r\n53. [latex]\\ln(x-2)=-\\ln(x+1)\\Rightarrow \\ln\\big((x-2)(x+1)\\big)=0\\Rightarrow (x-2)(x+1)=1[\/latex].\r\n[latex]x=\\frac{1+\\sqrt{13}}{2}\\approx 2.303[\/latex].\r\n\r\n55. [latex]\\frac13\\log(1-x)=\\log(x+1)+\\frac13[\/latex] has solution [latex]x\\approx -0.472[\/latex] (nearest thousandth).\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"bc-section section\">\n<div id=\"chapter-gs-exp-log\" class=\"chapter standard\" title=\"Exponential and Logarithmic Functions: Get Stronger Key\">\n<div class=\"chapter-title-wrap\">\n<h1 class=\"chapter-title\">Exponential and Logarithmic Functions: Get Stronger Key<\/h1>\n<\/div>\n<div class=\"ugc chapter-ugc\">\n<h2>Exponential Functions<\/h2>\n<p>2. Yes. In an exponential function written as [latex]f(x)=ab^x[\/latex], the base [latex]b[\/latex] determines growth or decay: if [latex]b>1[\/latex] the function grows; if [latex]0<b<1[\/latex] the function decays.<\/p>\n<p>5. Exponential function. \u201cDecreases by a factor of [latex]\\frac{1}{8}[\/latex] every 24 hours\u201d means each 24-hour period multiplies the population by a constant ratio [latex]\\frac{1}{8}[\/latex].<\/p>\n<p>6. Exponential function. Increasing by 3.25% annually means the value is multiplied each year by the constant factor [latex]1.0325[\/latex].<\/p>\n<p>7. Not exponential. Subtracting $5 each session is a constant difference (linear), not a constant ratio.<\/p>\n<p>9. Forest [latex]B[\/latex] grows faster because its growth factor is larger: [latex]1.029>1.025[\/latex].<\/p>\n<p>10. Forest [latex]A[\/latex] had more trees initially: [latex]A(0)=115[\/latex] and [latex]B(0)=82[\/latex], so forest [latex]A[\/latex] had [latex]115-82=33[\/latex] more trees.<\/p>\n<p>11. After 20 years: [latex]A(20)\\approx 188[\/latex] and [latex]B(20)\\approx 145[\/latex]. Forest [latex]A[\/latex] will have about [latex]188-145=43[\/latex] more trees.<\/p>\n<p>12. After 100 years: [latex]A(100)\\approx 1359[\/latex] and [latex]B(100)\\approx 1430[\/latex]. Forest [latex]B[\/latex] will have about [latex]1430-1359=71[\/latex] more trees.<\/p>\n<p>15. Exponential growth because the base is [latex]1.06>1[\/latex].<\/p>\n<p>17. Exponential decay because the base is [latex]0.97[\/latex] and [latex]0<0.97<1[\/latex].<\/p>\n<p>19. Let [latex]f(x)=ab^x[\/latex]. From [latex](0,2000)[\/latex], [latex]a=2000[\/latex]. Using [latex](2,20)[\/latex]: [latex]20=2000b^2\\Rightarrow b^2=0.01\\Rightarrow b=0.1[\/latex]. So [latex]f(x)=2000(0.1)^x[\/latex].<br \/>\n<\/p>\n<p>21. Let [latex]f(x)=ab^x[\/latex]. Using [latex](-2,6)[\/latex] and [latex](3,1)[\/latex], the function is [latex]f(x)=6^{\\frac{3}{5}}\\left(6^{-\\frac{1}{5}}\\right)^x[\/latex] (equivalently [latex]f(x)=6^{\\frac{3-x}{5}}[\/latex]).<\/p>\n<p>23. Linear. The differences are constant: [latex]40-70=-30[\/latex], [latex]10-40=-30[\/latex], [latex]-20-10=-30[\/latex].<\/p>\n<p>25. Neither. The differences are not constant and the ratios are not constant, so it is not linear or exponential.<\/p>\n<p>31. [latex]A(t)=P\\left(1+\\frac{r}{n}\\right)^{nt}[\/latex] with [latex]P=6500[\/latex], [latex]r=0.036[\/latex], [latex]n=2[\/latex], [latex]t=20[\/latex]: [latex]A\\approx 6500(1.018)^{40}\\approx \\$13{,}268.58[\/latex].<\/p>\n<p>32. Weekly compounding: [latex]n=52[\/latex]. [latex]A\\approx 6500\\left(1+\\frac{0.036}{52}\\right)^{52\\cdot 20}\\approx \\$13{,}350.49[\/latex]. This is about [latex]\\$13{,}350.49-\\$13{,}268.58=\\$81.91[\/latex] more.<\/p>\n<p>35. If earning interest 5 more years (25 years total):<br \/>\nSemi-annually: [latex]\\$15{,}859.97-\\$13{,}268.58=\\$2{,}591.39[\/latex] more.<br \/>\nWeekly: [latex]\\$15{,}982.44-\\$13{,}350.49=\\$2{,}631.95[\/latex] more.<\/p>\n<p>42. Continuous compounding: [latex]A=Pe^{rt}=12000e^{0.072\\cdot 30}\\approx \\$104{,}053.65[\/latex].<\/p>\n<p>43. Monthly compounding: [latex]A\\approx 12000\\left(1+\\frac{0.072}{12}\\right)^{12\\cdot 30}\\approx \\$103{,}384.23[\/latex]. This is about [latex]\\$104{,}053.65-\\$103{,}384.23=\\$669.42[\/latex] less than continuous compounding.<\/p>\n<p>51. Using [latex]y=ab^x[\/latex]: from [latex](0,3)[\/latex], [latex]a=3[\/latex]. From [latex](3,375)[\/latex], [latex]375=3b^3\\Rightarrow b^3=125\\Rightarrow b=5[\/latex]. So [latex]y=3\\cdot 5^x[\/latex].<\/p>\n<p>53. Using [latex]y=ab^x[\/latex] through [latex](20,29.495)[\/latex] and [latex](150,730.89)[\/latex], a suitable model is [latex]y\\approx 18(1.025)^x[\/latex].<\/p>\n<p>55. Using [latex]y=ab^x[\/latex] through [latex](11,310.035)[\/latex] and [latex](25,356.3652)[\/latex], a suitable model is [latex]y\\approx 277.899(1.01)^x[\/latex].<\/p>\n<p>61. Growth rate 9% per year from 2012 to 2020 (8 years): [latex]23900(1.09)^8\\approx 47{,}622[\/latex] foxes.<\/p>\n<p>63. From 1985 to 2005 is 20 years: [latex]145000=110000(1+r)^{20}\\Rightarrow r\\approx 0.0139[\/latex] (about 1.39% per year).<br \/>\nValue in 2010 (25 years after 1985): [latex]110000(1+r)^{25}\\approx \\$155{,}368.09[\/latex].<\/p>\n<p>65. [latex]54000=P\\left(1+\\frac{0.082}{365}\\right)^{365\\cdot 5}\\Rightarrow P\\approx \\$35{,}838.76[\/latex].<\/p>\n<p>67. From 2000 to 2025 is 25 years, [latex]P=13500[\/latex], [latex]r=0.0725[\/latex].<br \/>\nMonthly: [latex]A\\approx 13500\\left(1+\\frac{0.0725}{12}\\right)^{12\\cdot 25}\\approx \\$82{,}247.78[\/latex].<br \/>\nContinuous: [latex]A\\approx 13500e^{0.0725\\cdot 25}\\approx \\$82{,}697.53[\/latex].<br \/>\nContinuous earns about [latex]\\$82{,}697.53-\\$82{,}247.78=\\$449.75[\/latex] more.<\/p>\n<h2>Graphs of Exponential Functions<\/h2>\n<p>1. The horizontal asymptote tells what [latex]f(x)[\/latex] approaches as [latex]x\\to\\infty[\/latex] (and often also as [latex]x\\to -\\infty[\/latex]). It describes the \u201cend behavior\u201d level the graph gets closer and closer to without reaching.<br \/>\n3. Reflect about the y-axis: [latex]3^{-x}[\/latex]. Stretch vertically by 4: [latex]g(x)=4\\cdot 3^{-x}[\/latex].<br \/>\ny-intercept: [latex]g(0)=4[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](0,\\infty)[\/latex].<\/p>\n<p>5. Reflect about the x-axis: [latex]-10^x[\/latex]. Shift up 7: [latex]g(x)=-10^x+7[\/latex].<br \/>\ny-intercept: [latex]g(0)=-1+7=6[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](-\\infty,7)[\/latex].<\/p>\n<p>7. Starting with [latex]\\left(\\frac{1}{4}\\right)^x[\/latex]: left 2 gives [latex]\\left(\\frac{1}{4}\\right)^{x+2}[\/latex]; stretch by 4 gives [latex]4\\left(\\frac{1}{4}\\right)^{x+2}[\/latex]; reflect over x-axis gives [latex]-4\\left(\\frac{1}{4}\\right)^{x+2}[\/latex]; down 4 gives<br \/>\n[latex]g(x)=-4\\left(\\frac{1}{4}\\right)^{x+2}-4[\/latex].<br \/>\ny-intercept: [latex]g(0)=-4\\left(\\frac{1}{4}\\right)^2-4=-\\frac{17}{4}[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](-\\infty,-4)[\/latex].<\/p>\n<p>11. <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-302\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM.png\" alt=\"The graph shows three exponential functions on the same coordinate plane. All three pass through the point (0, 3). One function is an increasing exponential that grows slowly for negative x and more rapidly for positive x. A second increasing exponential grows more steeply than the first for positive x. The third function is an exponential decay function that decreases as x increases and approaches y equals 0 to the right. All three functions have a horizontal asymptote at y equals 0.\" width=\"724\" height=\"868\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM.png 724w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM-250x300.png 250w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM-65x78.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM-225x270.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02160410\/Screenshot-2026-02-02-at-8.47.29%E2%80%AFAM-350x420.png 350w\" sizes=\"(max-width: 724px) 100vw, 724px\" \/><\/p>\n<p>13. B<\/p>\n<p>14. D<\/p>\n<p>15. A<\/p>\n<p>16. F<\/p>\n<p>17. E<\/p>\n<p>18. C<\/p>\n<p>26. [latex]f(x)=2^{-x}[\/latex]. Horizontal asymptote: [latex]y=0[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](0,\\infty)[\/latex].<\/p>\n<p>27. [latex]h(x)=2^x+3[\/latex]. Horizontal asymptote: [latex]y=3[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](3,\\infty)[\/latex].<\/p>\n<p>28. [latex]f(x)=2^{x-2}[\/latex]. Horizontal asymptote: [latex]y=0[\/latex]. Domain: [latex](-\\infty,\\infty)[\/latex]. Range: [latex](0,\\infty)[\/latex].<\/p>\n<p>39. [latex]y=-2^x+3[\/latex].<\/p>\n<p>41. [latex]y=-2\\cdot 3^x+7[\/latex].<\/p>\n<p>47. [latex]116=\\frac14\\left(\\frac18\\right)^x\\Rightarrow \\left(\\frac18\\right)^x=464\\Rightarrow x\\approx -2.953[\/latex].<\/p>\n<p>49. [latex]5=3\\left(\\frac12\\right)^{x-1}-2\\Rightarrow 7=3\\left(\\frac12\\right)^{x-1}\\Rightarrow x\\approx -0.222[\/latex].<\/p>\n<h2>Logarithmic Functions<\/h2>\n<p>3. Use the definition of a logarithm: if [latex]\\log_b(x)=y[\/latex], then [latex]b^y=x[\/latex]. So solving for [latex]x[\/latex] gives [latex]x=b^y[\/latex].<\/p>\n<p>7. [latex]\\log_a(b)=c \\iff a^c=b[\/latex].<\/p>\n<p>9. [latex]\\log_x(64)=y \\iff x^y=64[\/latex].<\/p>\n<p>11. [latex]\\log_{15}(a)=b \\iff 15^b=a[\/latex].<\/p>\n<p>13. [latex]\\log_{13}(142)=a \\iff 13^a=142[\/latex].<\/p>\n<p>15. [latex]\\ln(w)=n \\iff e^n=w[\/latex].<\/p>\n<p>17. [latex]c^d=k \\iff \\log_c(k)=d[\/latex].<\/p>\n<p>19. [latex]19^x=y \\iff \\log_{19}(y)=x[\/latex].<\/p>\n<p>21. [latex]n^4=103 \\iff \\log_n(103)=4[\/latex].<\/p>\n<p>23. [latex]y^x=\\frac{39}{100} \\iff \\log_y\\left(\\frac{39}{100}\\right)=x[\/latex].<\/p>\n<p>25. [latex]e^k=h \\iff \\ln(h)=k[\/latex].<\/p>\n<p>27. [latex]\\log_2(x)=-3 \\Rightarrow x=2^{-3}=\\frac18[\/latex].<\/p>\n<p>29. [latex]\\log_3(x)=3 \\Rightarrow x=3^3=27[\/latex].<\/p>\n<p>31. [latex]\\log_9(x)=\\frac12 \\Rightarrow x=9^{1\/2}=3[\/latex].<\/p>\n<p>33. [latex]\\log_6(x)=-3 \\Rightarrow x=6^{-3}=\\frac{1}{216}[\/latex].<\/p>\n<p>35. [latex]\\ln(x)=2 \\Rightarrow x=e^2[\/latex].<\/p>\n<p>37. [latex]10^{\\log(32)}=32[\/latex].<\/p>\n<p>39. [latex]e^{\\ln(1.06)}=1.06[\/latex].<\/p>\n<p>41. [latex]e^{\\ln(10.125)}+4=10.125+4=14.125[\/latex].<\/p>\n<p>43. [latex]\\log_6(\\sqrt{6})=\\log_6(6^{1\/2})=\\frac12[\/latex].<\/p>\n<p>45. [latex]6\\log_8(4)=6\\cdot\\frac{2}{3}=4[\/latex].<\/p>\n<p>47. [latex]\\log(0.001)=\\log(10^{-3})=-3[\/latex].<\/p>\n<p>49. [latex]2\\log(100^{-3})=2\\log(10^{-6})=2(-6)=-12[\/latex].<\/p>\n<p>51. [latex]\\ln(1)=0[\/latex].<\/p>\n<p>53. [latex]25\\ln\\left(e^{2\/5}\\right)=25\\cdot\\frac{2}{5}=10[\/latex].<\/p>\n<p>59. No. [latex]x=0[\/latex] is not in the domain of [latex]f(x)=\\log(x)[\/latex] because logarithms require [latex]x>0[\/latex]. So [latex]f(0)[\/latex] is undefined (DNE).<\/p>\n<p>66. [latex]\\log\\left(\\frac{I_1}{I_2}\\right)=M_1-M_2=9.0-6.1=2.9[\/latex], so [latex]\\frac{I_1}{I_2}=10^{2.9}\\approx 794[\/latex]. The 2011 quake was about 794 times as intense.<\/p>\n<h2>Graphs of Logarithmic Functions<\/h2>\n<p>1. If two functions are inverses, their graphs are reflections across the line [latex]y=x[\/latex]. So a point [latex](a,b)[\/latex] on one graph corresponds to [latex](b,a)[\/latex] on the inverse graph.<\/p>\n<p>2. None. The range of a logarithmic function is all real numbers, and vertical translations do not change that (it remains [latex](-\\infty,\\infty)[\/latex]).<\/p>\n<p>3. Horizontal translations affect the domain because they move the vertical asymptote left or right (changing where the input becomes valid).<\/p>\n<p>21. [latex]h(x)=\\log_4(x-1)+1[\/latex]. Domain: [latex](1,\\infty)[\/latex]. Range: [latex](-\\infty,\\infty)[\/latex].<br \/>\nx-intercept: [latex]\\log_4(x-1)+1=0\\Rightarrow x-1=\\frac14\\Rightarrow x=\\frac54[\/latex]. y-intercept: DNE.<\/p>\n<p>23. [latex]g(x)=\\ln(-x)-2[\/latex]. Domain: [latex](-\\infty,0)[\/latex]. Range: [latex](-\\infty,\\infty)[\/latex].<br \/>\nx-intercept: [latex]\\ln(-x)-2=0\\Rightarrow -x=e^2\\Rightarrow x=-e^2[\/latex]. y-intercept: DNE.<\/p>\n<p>25. [latex]h(x)=3\\ln(x)-9[\/latex]. Domain: [latex](0,\\infty)[\/latex]. Range: [latex](-\\infty,\\infty)[\/latex].<br \/>\nx-intercept: [latex]3\\ln(x)-9=0\\Rightarrow \\ln(x)=3\\Rightarrow x=e^3[\/latex]. y-intercept: DNE.<\/p>\n<p>31. C<\/p>\n<p>32. A<\/p>\n<p>33. B<\/p>\n<p>35. <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-303\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM.png\" alt=\"The graph shows two logarithmic functions on the same coordinate plane. One function is log base 10 of x, which increases as x increases. The other function is log base one-half of x, which decreases as x increases. Both functions have a vertical asymptote at x equals 0 and pass through the point (1, 0). For values of x between 0 and 1, log base 10 of x has negative y-values while log base one-half of x has positive y-values. For values of x greater than 1, log base 10 of x has positive y-values while log base one-half of x has negative y-values. Both curves approach the y-axis but never touch it.\" width=\"251\" height=\"253\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM.png 726w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM-298x300.png 298w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM-150x150.png 150w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM-65x65.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM-225x226.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164127\/Screenshot-2026-02-02-at-9.07.28%E2%80%AFAM-350x352.png 350w\" sizes=\"(max-width: 251px) 100vw, 251px\" \/><\/p>\n<p>37. <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-304\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM.png\" alt=\"The graph shows an exponential function and a logarithmic function on the same coordinate plane. The exponential function increases rapidly as x increases and passes through the point (0, 1). It has a horizontal asymptote at y equals 0. The logarithmic function increases slowly as x increases and has a vertical asymptote at x equals 0. It passes through the point (1, 0). The point (1, 0) is highlighted on the logarithmic function, showing where the logarithmic curve crosses the x-axis.\" width=\"250\" height=\"257\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM.png 582w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM-292x300.png 292w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM-65x67.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM-225x231.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164230\/Screenshot-2026-02-02-at-9.42.02%E2%80%AFAM-350x360.png 350w\" sizes=\"(max-width: 250px) 100vw, 250px\" \/><\/p>\n<p>38. B<\/p>\n<p>39. C<\/p>\n<p>40. A<\/p>\n<p>41. <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-305\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM.png\" alt=\"The graph is a logarithmic function that increases slowly as x increases. It has a vertical asymptote at x equals 0. The graph passes through the points: (1, 0) and (2, about 0.7) As x approaches 0 from the right, the y-values decrease without bound.\" width=\"252\" height=\"279\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM.png 650w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM-272x300.png 272w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM-65x72.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM-225x249.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164424\/Screenshot-2026-02-02-at-9.43.16%E2%80%AFAM-350x387.png 350w\" sizes=\"(max-width: 252px) 100vw, 252px\" \/><\/p>\n<p>43. <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-307\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM.png\" alt=\"The graph is a logarithmic function that decreases as x increases. It has a vertical asymptote at x equals 0. The graph passes through the point: (-1, 0) As x approaches 0 from the right, the y-values decrease without bound.\" width=\"244\" height=\"242\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM.png 718w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM-300x297.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM-150x150.png 150w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM-65x64.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM-225x223.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164426\/Screenshot-2026-02-02-at-9.43.40%E2%80%AFAM-350x347.png 350w\" sizes=\"(max-width: 244px) 100vw, 244px\" \/><\/p>\n<p>45. <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-306\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM.png\" alt=\"The graph is a logarithmic function that decreases as x increases. It has a vertical asymptote at x equals 2. As x approaches 2 from the left, the y-values decrease without bound.\" width=\"250\" height=\"249\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM.png 728w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM-300x300.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM-150x150.png 150w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM-65x65.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM-225x224.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/02\/02164425\/Screenshot-2026-02-02-at-9.43.55%E2%80%AFAM-350x349.png 350w\" sizes=\"(max-width: 250px) 100vw, 250px\" \/><\/p>\n<p>51. Let [latex]u=x-1[\/latex]. Then [latex]\\log(u)=\\ln(u)[\/latex] implies [latex]u=1[\/latex], so [latex]x=2[\/latex].<\/p>\n<p>53. [latex]\\ln(x-2)=-\\ln(x+1)\\Rightarrow \\ln\\big((x-2)(x+1)\\big)=0\\Rightarrow (x-2)(x+1)=1[\/latex].<br \/>\n[latex]x=\\frac{1+\\sqrt{13}}{2}\\approx 2.303[\/latex].<\/p>\n<p>55. [latex]\\frac13\\log(1-x)=\\log(x+1)+\\frac13[\/latex] has solution [latex]x\\approx -0.472[\/latex] (nearest thousandth).<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":13,"menu_order":7,"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\/299"}],"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":6,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/299\/revisions"}],"predecessor-version":[{"id":311,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/299\/revisions\/311"}],"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\/299\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=299"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=299"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=299"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=299"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}