{"id":1979,"date":"2021-08-19T16:03:44","date_gmt":"2021-08-19T16:03:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-integrals-exponential-functions-and-logarithms\/"},"modified":"2021-11-17T01:53:30","modified_gmt":"2021-11-17T01:53:30","slug":"skills-review-for-integrals-exponential-functions-and-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-integrals-exponential-functions-and-logarithms\/","title":{"raw":"Skills Review for Integrals, Exponential Functions, and Logarithms","rendered":"Skills Review for Integrals, Exponential Functions, and Logarithms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Combine the product, power, and quotient rules to condense logarithmic expressions<\/li>\r\n \t<li>Simplify expressions using the Product Property of Exponents<\/li>\r\n \t<li>Simplifying expressions using the Quotient Property of Exponents<\/li>\r\n \t<li>Simplify expressions using the Power Property of Exponents<\/li>\r\n \t<li>Use logarithms to solve exponential equations whose terms cannot be rewritten with the same base<\/li>\r\n \t<li>Solve exponential equations of the form [latex]y=Ae^{kt}[\/latex] for [latex]t[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Integrals, Exponential Functions, and Logarithms and Exponential Growth and Decay sections, we will look more into differentiating and integrating exponential and logarithmic functions. Here we will review condensing logarithmic expressions, using rules of exponents, and how to solve exponential equations.\r\n<h2>Condense Logarithmic Expressions<\/h2>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Product Rule for Logarithms<\/h3>\r\nThe <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b&gt;0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Quotient Rule for Logarithms<\/h3>\r\nThe <strong>quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Power Rule for Logarithms<\/h3>\r\nThe <strong>power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/p>\r\n\r\n<\/div>\r\nWe can use the rules of logarithms to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm<\/h3>\r\n<ol>\r\n \t<li>Apply the power property first. Identify terms that are products of factors and a logarithm and rewrite each as the logarithm of a power.<\/li>\r\n \t<li>From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Condensing Logarithmic Expressions<\/h3>\r\nUse the power rule for logs to rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] as a single logarithm with a leading coefficient of 1.\r\n\r\n[reveal-answer q=\"959478\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"959478\"]\r\n\r\nBecause the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base and rewrite the product as a logarithm of a power:\r\n\r\n[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the power rule for logs to rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] as a single logarithm with a leading coefficient of 1.\r\n\r\n[reveal-answer q=\"720709\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720709\"]\r\n\r\n[latex]{\\mathrm{log}}_{3}16[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next few examples, we will use a combination of logarithm rules to condense logarithms.\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Condensing Logarithmic Expressions<\/h3>\r\nWrite [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.\r\n\r\n[reveal-answer q=\"484876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484876\"]\r\n\r\nFrom left to right, since we have the addition of two logs, we first use the product rule:\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]\r\n\r\nThis simplifies our original expression to:\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]\r\n\r\nUsing the quotient rule:\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]129766[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use Rules of Exponents<\/h2>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Product Rule of Exponents<\/h3>\r\nFor any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], the product rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Product Rule<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"878162\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"878162\"]\r\nUse the product rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\nAt first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\r\nNotice we get the same result by adding the three exponents in one step.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1961&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Quotient Rule of Exponents<\/h3>\r\nFor any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], such that [latex]m&gt;n[\/latex], the quotient rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Quotient Rule<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{t}^{23}}{{t}^{15}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"717838\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"717838\"]\r\n\r\nUse the quotient rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}={\\left(-2\\right)}^{14 - 9}={\\left(-2\\right)}^{5}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}={\\left(z\\sqrt{2}\\right)}^{5 - 1}={\\left(z\\sqrt{2}\\right)}^{4}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109745&amp;theme=oea&amp;iframe_resize_id=mom60[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Power Rule of Exponents<\/h3>\r\nFor any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{\\left({x}^{2}\\right)}^{7}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"992335\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"992335\"]\r\n\r\nUse the power rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{\\left({x}^{2}\\right)}^{7}={x}^{2\\cdot 7}={x}^{14}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}={\\left(2t\\right)}^{5\\cdot 3}={\\left(2t\\right)}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}={\\left(-3\\right)}^{5\\cdot 11}={\\left(-3\\right)}^{55}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93370&amp;theme=oea&amp;iframe_resize_id=mom80[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93399&amp;theme=oea&amp;iframe_resize_id=mom90[\/embed]\r\n\r\n<\/div>\r\n<h2>Solve Exponential Equations<\/h2>\r\nOne technique that is used to solve exponential equations is to take the logarithm of each side of the equation.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an exponential equation, Solve By Taking the Logarithm of Each Side<\/h3>\r\n<ol>\r\n \t<li>Apply the logarithm to both sides of the equation.\r\n<ul id=\"fs-id1165137824134\">\r\n \t<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\r\n \t<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the rules of logarithms to solve for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Exponential Equations<\/h3>\r\nSolve [latex]{5}^{x+2}={4}^{x}[\/latex].\r\n\r\n[reveal-answer q=\"837781\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"837781\"]\r\n\r\n[latex]\\begin{array}{l}\\text{ }{5}^{x+2}={4}^{x}\\hfill &amp; \\text{There is no easy way to get the powers to have the same base}.\\hfill \\\\ \\text{ }\\mathrm{ln}{5}^{x+2}=\\mathrm{ln}{4}^{x}\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill &amp; \\text{Use the power rule for logs}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5+2\\mathrm{ln}5=x\\mathrm{ln}4\\hfill &amp; \\text{Use the distributive property}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5-x\\mathrm{ln}4=-2\\mathrm{ln}5\\hfill &amp; \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}.\\hfill \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5\\hfill &amp; \\text{On the left hand side, factor out }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill &amp; \\text{Use the properties of logs}.\\hfill \\\\ \\text{ }x=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}\\hfill &amp; \\text{Divide by the coefficient of }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{2}^{x}={3}^{x+1}[\/latex].\r\n\r\n[reveal-answer q=\"311643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"311643\"]\r\n\r\n[latex]x=\\frac{\\mathrm{ln}3}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]98555[\/ohm_question]\r\n\r\n<\/div>\r\nOne common type of exponential equations are those with base <em>e<\/em>. This constant occurs again and again in nature, mathematics, science, engineering, and finance. When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for [latex]t[\/latex]<\/h3>\r\n<ol>\r\n \t<li>Divide both sides of the equation by <em>A<\/em>.<\/li>\r\n \t<li>Apply the natural logarithm to both sides of the equation.<\/li>\r\n \t<li>Divide both sides of the equation by <em>k<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Exponential Equations with Base [latex]e[\/latex]<\/h3>\r\nSolve [latex]100=20{e}^{2t}[\/latex].\r\n\r\n[reveal-answer q=\"7965\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7965\"]\r\n\r\n[latex]\\begin{array}{l}100\\hfill &amp; =20{e}^{2t}\\hfill &amp; \\hfill \\\\ 5\\hfill &amp; ={e}^{2t}\\hfill &amp; \\text{Divide by the coefficient 20}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =\\mathrm{ln}{{e}^{2t}}\\hfill &amp; \\text{Take ln of both sides.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =2t\\hfill &amp; \\text{Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions}\\text{.}\\hfill \\\\ t\\hfill &amp; =\\frac{\\mathrm{ln}5}{2}\\hfill &amp; \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\n&nbsp;\r\n<h4>Analysis of the Solution<\/h4>\r\nUsing laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt{5}[\/latex]. If we want a decimal approximation of the answer, then we use a calculator.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]3{e}^{0.5t}=11[\/latex].\r\n\r\n[reveal-answer q=\"585330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"585330\"]\r\n\r\n[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]98596[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Combine the product, power, and quotient rules to condense logarithmic expressions<\/li>\n<li>Simplify expressions using the Product Property of Exponents<\/li>\n<li>Simplifying expressions using the Quotient Property of Exponents<\/li>\n<li>Simplify expressions using the Power Property of Exponents<\/li>\n<li>Use logarithms to solve exponential equations whose terms cannot be rewritten with the same base<\/li>\n<li>Solve exponential equations of the form [latex]y=Ae^{kt}[\/latex] for [latex]t[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>In the Integrals, Exponential Functions, and Logarithms and Exponential Growth and Decay sections, we will look more into differentiating and integrating exponential and logarithmic functions. Here we will review condensing logarithmic expressions, using rules of exponents, and how to solve exponential equations.<\/p>\n<h2>Condense Logarithmic Expressions<\/h2>\n<div class=\"textbox\">\n<h3>A General Note: The Product Rule for Logarithms<\/h3>\n<p>The <strong>product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b>0[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Quotient Rule for Logarithms<\/h3>\n<p>The <strong>quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Power Rule for Logarithms<\/h3>\n<p>The <strong>power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/p>\n<\/div>\n<p>We can use the rules of logarithms to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm<\/h3>\n<ol>\n<li>Apply the power property first. Identify terms that are products of factors and a logarithm and rewrite each as the logarithm of a power.<\/li>\n<li>From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Condensing Logarithmic Expressions<\/h3>\n<p>Use the power rule for logs to rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] as a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q959478\">Show Solution<\/span><\/p>\n<div id=\"q959478\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base and rewrite the product as a logarithm of a power:<\/p>\n<p>[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the power rule for logs to rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] as a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q720709\">Show Solution<\/span><\/p>\n<div id=\"q720709\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\mathrm{log}}_{3}16[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>In our next few examples, we will use a combination of logarithm rules to condense logarithms.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Condensing Logarithmic Expressions<\/h3>\n<p>Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484876\">Show Solution<\/span><\/p>\n<div id=\"q484876\" class=\"hidden-answer\" style=\"display: none\">\n<p>From left to right, since we have the addition of two logs, we first use the product rule:<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/p>\n<p>This simplifies our original expression to:<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\n<p>Using the quotient rule:<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm129766\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129766&theme=oea&iframe_resize_id=ohm129766&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use Rules of Exponents<\/h2>\n<div class=\"textbox\">\n<h3>A General Note: The Product Rule of Exponents<\/h3>\n<p>For any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], the product rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Product Rule<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\n<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q878162\">Show Solution<\/span><\/p>\n<div id=\"q878162\" class=\"hidden-answer\" style=\"display: none\">\nUse the product rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\n<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<p>At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\n<p>Notice we get the same result by adding the three exponents in one step.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm1961\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1961&#38;theme=oea&#38;iframe_resize_id=ohm1961&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div><\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Quotient Rule of Exponents<\/h3>\n<p>For any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], such that [latex]m>n[\/latex], the quotient rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Quotient Rule<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]\\dfrac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{t}^{23}}{{t}^{15}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q717838\">Show Solution<\/span><\/p>\n<div id=\"q717838\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the quotient rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]\\dfrac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}={\\left(-2\\right)}^{14 - 9}={\\left(-2\\right)}^{5}[\/latex]<\/li>\n<li>[latex]\\dfrac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}={\\left(z\\sqrt{2}\\right)}^{5 - 1}={\\left(z\\sqrt{2}\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm109745\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109745&#38;theme=oea&#38;iframe_resize_id=ohm109745&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Power Rule of Exponents<\/h3>\n<p>For any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{\\left({x}^{2}\\right)}^{7}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q992335\">Show Solution<\/span><\/p>\n<div id=\"q992335\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the power rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{\\left({x}^{2}\\right)}^{7}={x}^{2\\cdot 7}={x}^{14}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}={\\left(2t\\right)}^{5\\cdot 3}={\\left(2t\\right)}^{15}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}={\\left(-3\\right)}^{5\\cdot 11}={\\left(-3\\right)}^{55}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm93370\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93370&#38;theme=oea&#38;iframe_resize_id=ohm93370&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm93399\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93399&#38;theme=oea&#38;iframe_resize_id=ohm93399&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Exponential Equations<\/h2>\n<p>One technique that is used to solve exponential equations is to take the logarithm of each side of the equation.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an exponential equation, Solve By Taking the Logarithm of Each Side<\/h3>\n<ol>\n<li>Apply the logarithm to both sides of the equation.\n<ul id=\"fs-id1165137824134\">\n<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\n<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\n<\/ul>\n<\/li>\n<li>Use the rules of logarithms to solve for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Exponential Equations<\/h3>\n<p>Solve [latex]{5}^{x+2}={4}^{x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q837781\">Show Solution<\/span><\/p>\n<div id=\"q837781\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\text{ }{5}^{x+2}={4}^{x}\\hfill & \\text{There is no easy way to get the powers to have the same base}.\\hfill \\\\ \\text{ }\\mathrm{ln}{5}^{x+2}=\\mathrm{ln}{4}^{x}\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use the power rule for logs}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5+2\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use the distributive property}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5-x\\mathrm{ln}4=-2\\mathrm{ln}5\\hfill & \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}.\\hfill \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5\\hfill & \\text{On the left hand side, factor out }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill & \\text{Use the properties of logs}.\\hfill \\\\ \\text{ }x=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}\\hfill & \\text{Divide by the coefficient of }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]{2}^{x}={3}^{x+1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q311643\">Show Solution<\/span><\/p>\n<div id=\"q311643\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{\\mathrm{ln}3}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm98555\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98555&theme=oea&iframe_resize_id=ohm98555&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>One common type of exponential equations are those with base <em>e<\/em>. This constant occurs again and again in nature, mathematics, science, engineering, and finance. When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for [latex]t[\/latex]<\/h3>\n<ol>\n<li>Divide both sides of the equation by <em>A<\/em>.<\/li>\n<li>Apply the natural logarithm to both sides of the equation.<\/li>\n<li>Divide both sides of the equation by <em>k<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Exponential Equations with Base [latex]e[\/latex]<\/h3>\n<p>Solve [latex]100=20{e}^{2t}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7965\">Show Solution<\/span><\/p>\n<div id=\"q7965\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}100\\hfill & =20{e}^{2t}\\hfill & \\hfill \\\\ 5\\hfill & ={e}^{2t}\\hfill & \\text{Divide by the coefficient 20}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =\\mathrm{ln}{{e}^{2t}}\\hfill & \\text{Take ln of both sides.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =2t\\hfill & \\text{Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions}\\text{.}\\hfill \\\\ t\\hfill & =\\frac{\\mathrm{ln}5}{2}\\hfill & \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Using laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt{5}[\/latex]. If we want a decimal approximation of the answer, then we use a calculator.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]3{e}^{0.5t}=11[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q585330\">Show Solution<\/span><\/p>\n<div id=\"q585330\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm98596\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98596&theme=oea&iframe_resize_id=ohm98596&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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-1979\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision . <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Prerequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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":349141,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Prerequisite\",\"author\":\"\",\"organization\":\"Lumen 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