{"id":11236,"date":"2015-07-14T19:08:24","date_gmt":"2015-07-14T19:08:24","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&#038;p=11236"},"modified":"2015-09-09T21:44:35","modified_gmt":"2015-09-09T21:44:35","slug":"key-terms-glossary-10","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/key-terms-glossary-10\/","title":{"raw":"Key Concepts &amp; Glossary","rendered":"Key Concepts &amp; Glossary"},"content":{"raw":"<section id=\"fs-id1165135255941\" class=\"key-equations\" data-depth=\"1\">\r\n<h1 data-type=\"title\">Key Equations<\/h1>\r\n<table id=\"fs-id2868258\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>One-to-one property for exponential functions<\/td>\r\n<td>For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>\u00a0and any positive real number <em>b<\/em>, where[latex]{b}^{S}={b}^{T}[\/latex] if and only if <em>S<\/em>\u00a0=\u00a0<em>T<\/em>.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Definition of a logarithm<\/td>\r\n<td>For any algebraic expression <em data-effect=\"italics\">S<\/em> and positive real numbers <em>b<\/em>\u00a0and <em>c<\/em>, where [latex]b\\ne 1[\/latex],[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex] if and only if [latex]{b}^{c}=S[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One-to-one property for logarithmic functions<\/td>\r\n<td>For any algebraic expressions <em data-effect=\"italics\">S<\/em> and <em data-effect=\"italics\">T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],\r\n<div data-type=\"newline\"><\/div>\r\n[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex] if and only if <em>S<\/em> =\u00a0<em>T<\/em>.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165137431646\" class=\"key-concepts\" data-depth=\"1\">\r\n<h1 data-type=\"title\">Key Concepts<\/h1>\r\n<ul id=\"fs-id1165137450888\">\r\n\t<li>We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.<\/li>\r\n\t<li>When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown.<\/li>\r\n\t<li>When we are given an exponential equation where the bases are <em data-effect=\"italics\">not<\/em> explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown.<\/li>\r\n\t<li>When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side.<\/li>\r\n\t<li>We can solve exponential equations with base <em>e<\/em>, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.<\/li>\r\n\t<li>After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.<\/li>\r\n\t<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex], where <em>S<\/em>\u00a0is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[\/latex], and solve for the unknown.<\/li>\r\n\t<li>We can also use graphing to solve equations with the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex]. We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and <em>y\u00a0<\/em>= <em>c<\/em> on the same coordinate plane and identify the solution as the <em data-effect=\"italics\">x-<\/em>value of the intersecting point.<\/li>\r\n\t<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex], where <em>S<\/em>\u00a0and <em>T<\/em>\u00a0are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation <em>S\u00a0<\/em>= <em>T<\/em>\u00a0for the unknown.<\/li>\r\n\t<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\r\n<\/ul>\r\n<div data-type=\"glossary\">\r\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\r\n<dl id=\"fs-id1165135149311\" class=\"definition\"><dt><strong>extraneous solution<\/strong><\/dt><dd id=\"fs-id1165134059776\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd><\/dl><\/div>\r\n<\/section>","rendered":"<section id=\"fs-id1165135255941\" class=\"key-equations\" data-depth=\"1\">\n<h1 data-type=\"title\">Key Equations<\/h1>\n<table id=\"fs-id2868258\" summary=\"...\">\n<tbody>\n<tr>\n<td>One-to-one property for exponential functions<\/td>\n<td>For any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>\u00a0and any positive real number <em>b<\/em>, where[latex]{b}^{S}={b}^{T}[\/latex] if and only if <em>S<\/em>\u00a0=\u00a0<em>T<\/em>.<\/td>\n<\/tr>\n<tr>\n<td>Definition of a logarithm<\/td>\n<td>For any algebraic expression <em data-effect=\"italics\">S<\/em> and positive real numbers <em>b<\/em>\u00a0and <em>c<\/em>, where [latex]b\\ne 1[\/latex],[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex] if and only if [latex]{b}^{c}=S[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>One-to-one property for logarithmic functions<\/td>\n<td>For any algebraic expressions <em data-effect=\"italics\">S<\/em> and <em data-effect=\"italics\">T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\n<div data-type=\"newline\"><\/div>\n<p>[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex] if and only if <em>S<\/em> =\u00a0<em>T<\/em>.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165137431646\" class=\"key-concepts\" data-depth=\"1\">\n<h1 data-type=\"title\">Key Concepts<\/h1>\n<ul id=\"fs-id1165137450888\">\n<li>We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.<\/li>\n<li>When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown.<\/li>\n<li>When we are given an exponential equation where the bases are <em data-effect=\"italics\">not<\/em> explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown.<\/li>\n<li>When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side.<\/li>\n<li>We can solve exponential equations with base <em>e<\/em>, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.<\/li>\n<li>After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.<\/li>\n<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex], where <em>S<\/em>\u00a0is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[\/latex], and solve for the unknown.<\/li>\n<li>We can also use graphing to solve equations with the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex]. We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and <em>y\u00a0<\/em>= <em>c<\/em> on the same coordinate plane and identify the solution as the <em data-effect=\"italics\">x-<\/em>value of the intersecting point.<\/li>\n<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex], where <em>S<\/em>\u00a0and <em>T<\/em>\u00a0are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation <em>S\u00a0<\/em>= <em>T<\/em>\u00a0for the unknown.<\/li>\n<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\n<\/ul>\n<div data-type=\"glossary\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl id=\"fs-id1165135149311\" class=\"definition\">\n<dt><strong>extraneous solution<\/strong><\/dt>\n<dd id=\"fs-id1165134059776\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\n<\/dl>\n<\/div>\n<\/section>\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-11236\">\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":7,"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-11236","chapter","type-chapter","status-publish","hentry"],"part":11222,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/chapters\/11236","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/chapters\/11236\/revisions"}],"predecessor-version":[{"id":13009,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/chapters\/11236\/revisions\/13009"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/parts\/11222"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/chapters\/11236\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/wp\/v2\/media?parent=11236"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/pressbooks\/v2\/chapter-type?post=11236"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/wp\/v2\/contributor?post=11236"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/wp-json\/wp\/v2\/license?post=11236"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}