{"id":11240,"date":"2015-07-14T19:08:51","date_gmt":"2015-07-14T19:08:51","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&#038;p=11240"},"modified":"2021-11-15T02:39:28","modified_gmt":"2021-11-15T02:39:28","slug":"solutions-exponential-and-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/solutions-exponential-and-logarithmic-equations\/","title":{"raw":"Solutions: Exponential and Logarithmic Equations","rendered":"Solutions: Exponential and Logarithmic Equations"},"content":{"raw":"<h2>Solutions to Try Its<\/h2>\r\n1. <em>x\u00a0<\/em>= \u20132\r\n\r\n2. <em>x\u00a0<\/em>= \u20131\r\n\r\n3. [latex]x=\\frac{1}{2}[\/latex]\r\n\r\n4. The equation has no solution.\r\n\r\n5.\u00a0[latex]x=\\frac{\\mathrm{ln}3}{\\mathrm{ln}}\\left(23\\right)[\/latex]\r\n\r\n6.\u00a0[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]\r\n\r\n7. [latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]\r\n\r\n8.\u00a0[latex]x=\\mathrm{ln}2[\/latex]\r\n\r\n9. [latex]x={e}^{4}[\/latex]\r\n\r\n10. [latex]x={e}^{5}-1[\/latex]\r\n\r\n11. [latex]x\\approx 9.97[\/latex]\r\n\r\n12. <em>x\u00a0<\/em>= 1 or <em>x\u00a0<\/em>= \u20131\r\n\r\n13. [latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].\r\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\r\n1.\u00a0Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.\r\n\r\n3.\u00a0The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.\r\n\r\n5.\u00a0[latex]x=-\\frac{1}{3}[\/latex]\r\n\r\n7. <em>n\u00a0<\/em>= \u20131\r\n\r\n9.\u00a0[latex]b=\\frac{6}{5}[\/latex]\r\n\r\n11. <em>x\u00a0<\/em>= 10\r\n\r\n13.\u00a0No solution\r\n\r\n15.\u00a0[latex]p=\\mathrm{log}\\left(\\frac{17}{8}\\right)-7[\/latex]\r\n\r\n17.\u00a0[latex]k=-\\frac{\\mathrm{ln}\\left(38\\right)}{3}[\/latex]\r\n\r\n19.\u00a0[latex]x=\\frac{\\mathrm{ln}\\left(\\frac{38}{3}\\right)-8}{9}[\/latex]\r\n\r\n21.\u00a0[latex]x=\\mathrm{ln}12[\/latex]\r\n\r\n23.\u00a0[latex]x=\\frac{\\mathrm{ln}\\left(\\frac{3}{5}\\right)-3}{8}[\/latex]\r\n\r\n25.\u00a0no solution\r\n\r\n27.\u00a0[latex]x=\\mathrm{ln}\\left(3\\right)[\/latex]\r\n\r\n29.\u00a0[latex]{10}^{-2}=\\frac{1}{100}[\/latex]\r\n\r\n31. <em>n\u00a0<\/em>= 49\r\n\r\n33.\u00a0[latex]k=\\frac{1}{36}[\/latex]\r\n\r\n35.\u00a0[latex]x=\\frac{9-e}{8}[\/latex]\r\n\r\n37. <em>n\u00a0<\/em>= 1\r\n\r\n39.\u00a0No solution\r\n\r\n41.\u00a0No solution\r\n\r\n43.\u00a0[latex]x=\\pm \\frac{10}{3}[\/latex]\r\n\r\n45. <em>x\u00a0<\/em>= 10\r\n\r\n47. <em>x\u00a0<\/em>= 0\r\n\r\n49.\u00a0[latex]x=\\frac{3}{4}[\/latex]\r\n\r\n51. <em>x\u00a0<\/em>= 9\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005327\/CNX_PreCalc_Figure_04_06_201.jpg\" alt=\"Graph of log_9(x)-5=y and y=-4.\" \/>\r\n\r\n53.\u00a0[latex]x=\\frac{{e}^{2}}{3}\\approx 2.5[\/latex]\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_203.jpg\" alt=\"Graph of ln(3x)=y and y=2.\" \/>\r\n\r\n55. <em>x\u00a0<\/em>= \u20135\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_205.jpg\" alt=\"Graph of log(4)+log(-5x)=y and y=2.\" \/>\r\n\r\n57.\u00a0[latex]x=\\frac{e+10}{4}\\approx 3.2[\/latex]\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_207.jpg\" alt=\"Graph of ln(4x-10)-6=y and y=-5.\" \/>\r\n\r\n59.\u00a0No solution\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_209.jpg\" alt=\"Graph of log_11(-2x^2-7x)=y and y=log_11(x-2).\" \/>\r\n\r\n61.\u00a0[latex]x=\\frac{11}{5}\\approx 2.2[\/latex]\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005329\/CNX_PreCalc_Figure_04_06_211.jpg\" alt=\"Graph of log_9(3-x)=y and y=log_9(4x-8).\" \/>\r\n\r\n63. [latex]x=\\frac{101}{11}\\approx 9.2[\/latex]\r\n\r\n65.\u00a0about $27,710.24\r\n\r\n67.\u00a0about 5 years\r\n\r\n69.\u00a0[latex]\\frac{\\mathrm{ln}\\left(17\\right)}{5}\\approx 0.567[\/latex]\r\n\r\n71.\u00a0[latex]x=\\frac{\\mathrm{log}\\left(38\\right)+5\\mathrm{log}\\left(3\\right)\\text{ }}{4\\mathrm{log}\\left(3\\right)}\\approx 2.078[\/latex]\r\n\r\n73.\u00a0[latex]x\\approx 2.2401[\/latex]\r\n\r\n75.\u00a0[latex]x\\approx -44655.7143[\/latex]\r\n\r\n77.\u00a0about 5.83\r\n\r\n79.\u00a0[latex]t=\\mathrm{ln}\\left({\\left(\\frac{y}{A}\\right)}^{\\frac{1}{k}}\\right)[\/latex]\r\n\r\n81.\u00a0[latex]t=\\mathrm{ln}\\left({\\left(\\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\\right)}^{-\\frac{1}{k}}\\right)[\/latex]","rendered":"<h2>Solutions to Try Its<\/h2>\n<p>1. <em>x\u00a0<\/em>= \u20132<\/p>\n<p>2. <em>x\u00a0<\/em>= \u20131<\/p>\n<p>3. [latex]x=\\frac{1}{2}[\/latex]<\/p>\n<p>4. The equation has no solution.<\/p>\n<p>5.\u00a0[latex]x=\\frac{\\mathrm{ln}3}{\\mathrm{ln}}\\left(23\\right)[\/latex]<\/p>\n<p>6.\u00a0[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]<\/p>\n<p>7. [latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]<\/p>\n<p>8.\u00a0[latex]x=\\mathrm{ln}2[\/latex]<\/p>\n<p>9. [latex]x={e}^{4}[\/latex]<\/p>\n<p>10. [latex]x={e}^{5}-1[\/latex]<\/p>\n<p>11. [latex]x\\approx 9.97[\/latex]<\/p>\n<p>12. <em>x\u00a0<\/em>= 1 or <em>x\u00a0<\/em>= \u20131<\/p>\n<p>13. [latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].<\/p>\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\n<p>1.\u00a0Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.<\/p>\n<p>3.\u00a0The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.<\/p>\n<p>5.\u00a0[latex]x=-\\frac{1}{3}[\/latex]<\/p>\n<p>7. <em>n\u00a0<\/em>= \u20131<\/p>\n<p>9.\u00a0[latex]b=\\frac{6}{5}[\/latex]<\/p>\n<p>11. <em>x\u00a0<\/em>= 10<\/p>\n<p>13.\u00a0No solution<\/p>\n<p>15.\u00a0[latex]p=\\mathrm{log}\\left(\\frac{17}{8}\\right)-7[\/latex]<\/p>\n<p>17.\u00a0[latex]k=-\\frac{\\mathrm{ln}\\left(38\\right)}{3}[\/latex]<\/p>\n<p>19.\u00a0[latex]x=\\frac{\\mathrm{ln}\\left(\\frac{38}{3}\\right)-8}{9}[\/latex]<\/p>\n<p>21.\u00a0[latex]x=\\mathrm{ln}12[\/latex]<\/p>\n<p>23.\u00a0[latex]x=\\frac{\\mathrm{ln}\\left(\\frac{3}{5}\\right)-3}{8}[\/latex]<\/p>\n<p>25.\u00a0no solution<\/p>\n<p>27.\u00a0[latex]x=\\mathrm{ln}\\left(3\\right)[\/latex]<\/p>\n<p>29.\u00a0[latex]{10}^{-2}=\\frac{1}{100}[\/latex]<\/p>\n<p>31. <em>n\u00a0<\/em>= 49<\/p>\n<p>33.\u00a0[latex]k=\\frac{1}{36}[\/latex]<\/p>\n<p>35.\u00a0[latex]x=\\frac{9-e}{8}[\/latex]<\/p>\n<p>37. <em>n\u00a0<\/em>= 1<\/p>\n<p>39.\u00a0No solution<\/p>\n<p>41.\u00a0No solution<\/p>\n<p>43.\u00a0[latex]x=\\pm \\frac{10}{3}[\/latex]<\/p>\n<p>45. <em>x\u00a0<\/em>= 10<\/p>\n<p>47. <em>x\u00a0<\/em>= 0<\/p>\n<p>49.\u00a0[latex]x=\\frac{3}{4}[\/latex]<\/p>\n<p>51. <em>x\u00a0<\/em>= 9<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005327\/CNX_PreCalc_Figure_04_06_201.jpg\" alt=\"Graph of log_9(x)-5=y and y=-4.\" \/><\/p>\n<p>53.\u00a0[latex]x=\\frac{{e}^{2}}{3}\\approx 2.5[\/latex]<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_203.jpg\" alt=\"Graph of ln(3x)=y and y=2.\" \/><\/p>\n<p>55. <em>x\u00a0<\/em>= \u20135<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_205.jpg\" alt=\"Graph of log(4)+log(-5x)=y and y=2.\" \/><\/p>\n<p>57.\u00a0[latex]x=\\frac{e+10}{4}\\approx 3.2[\/latex]<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_207.jpg\" alt=\"Graph of ln(4x-10)-6=y and y=-5.\" \/><\/p>\n<p>59.\u00a0No solution<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005328\/CNX_PreCalc_Figure_04_06_209.jpg\" alt=\"Graph of log_11(-2x^2-7x)=y and y=log_11(x-2).\" \/><\/p>\n<p>61.\u00a0[latex]x=\\frac{11}{5}\\approx 2.2[\/latex]<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005329\/CNX_PreCalc_Figure_04_06_211.jpg\" alt=\"Graph of log_9(3-x)=y and y=log_9(4x-8).\" \/><\/p>\n<p>63. [latex]x=\\frac{101}{11}\\approx 9.2[\/latex]<\/p>\n<p>65.\u00a0about $27,710.24<\/p>\n<p>67.\u00a0about 5 years<\/p>\n<p>69.\u00a0[latex]\\frac{\\mathrm{ln}\\left(17\\right)}{5}\\approx 0.567[\/latex]<\/p>\n<p>71.\u00a0[latex]x=\\frac{\\mathrm{log}\\left(38\\right)+5\\mathrm{log}\\left(3\\right)\\text{ }}{4\\mathrm{log}\\left(3\\right)}\\approx 2.078[\/latex]<\/p>\n<p>73.\u00a0[latex]x\\approx 2.2401[\/latex]<\/p>\n<p>75.\u00a0[latex]x\\approx -44655.7143[\/latex]<\/p>\n<p>77.\u00a0about 5.83<\/p>\n<p>79.\u00a0[latex]t=\\mathrm{ln}\\left({\\left(\\frac{y}{A}\\right)}^{\\frac{1}{k}}\\right)[\/latex]<\/p>\n<p>81.\u00a0[latex]t=\\mathrm{ln}\\left({\\left(\\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\\right)}^{-\\frac{1}{k}}\\right)[\/latex]<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-11240\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":20,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-11240","chapter","type-chapter","status-publish","hentry"],"part":11222,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11240","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11240\/revisions"}],"predecessor-version":[{"id":16417,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11240\/revisions\/16417"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/parts\/11222"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11240\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/media?parent=11240"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapter-type?post=11240"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/contributor?post=11240"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/license?post=11240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}