{"id":1994,"date":"2021-08-19T16:07:01","date_gmt":"2021-08-19T16:07:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-first-order-linear-equations\/"},"modified":"2022-04-19T20:53:36","modified_gmt":"2022-04-19T20:53:36","slug":"skills-review-for-first-order-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-first-order-linear-equations\/","title":{"raw":"Skills Review for First-order Linear Equations","rendered":"Skills Review for First-order Linear Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve rational equations by clearing denominators<\/li>\r\n \t<li>Apply the inverse property to simplify logarithms<\/li>\r\n \t<li>Define and use the power rule for logarithms to rewrite expressions<\/li>\r\n \t<li>Write function equations using given conditions<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the First-order Linear Equations section, we will explore how to solve yet another special type of differential equation. Here we will review how to clear fractions from and manipulate rational equations.\r\n<h2>Manipulate Rational Equations<\/h2>\r\nEquations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex] \\frac{2y+1}{4}=\\frac{x}{3}[\/latex] is a rational equation (of two variables).\r\n\r\nOne of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.\r\n\r\nSolve for [latex]y[\/latex] in the equation [latex]\\frac{1}{2}y-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.\r\n\r\nMultiply both sides of the equation by\u00a0[latex]4[\/latex], the common denominator of the fractional coefficients.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{1}{2}y-3=2-\\frac{3}{4}x\\\\ 4\\left(\\frac{1}{2}y-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\\\\\text{}\\,\\,\\,\\,4\\left(\\frac{1}{2}y\\right)-4\\left(3\\right)=4\\left(2\\right)+4\\left(-\\frac{3}{4}x\\right)\\\\2y-12=8-3x\\\\\\underline{+12}\\,\\,\\,\\,\\,\\,\\underline{+12}\\\\ 2y=20-3x\\\\ \\frac{2y}{2}=\\frac{20-3x}{2} \\\\ y=10-\\frac{3}{2}x\\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Manipulating a Rational Equation<\/h3>\r\nSolve the equation [latex] \\frac{x+5}{8}=\\frac{7}{y}[\/latex] for [latex]y[\/latex].\r\n[reveal-answer q=\"425621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425621\"]\r\n\r\nFind the least common denominator of\u00a0[latex]8[\/latex] and\u00a0[latex]y[\/latex]. [latex]8y[\/latex] will be the LCD.\r\n\r\nMultiply both sides of the equation by the common denominator,\u00a0[latex]8y[\/latex], to keep the equation balanced and to eliminate the denominators.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8y\\cdot \\frac{x+5}{8}=\\frac{7}{y}\\cdot 8y\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y(x+5)}{8}=\\frac{7(8y)}{y}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y}{8}\\cdot (x+5)=\\frac{7(8y)}{y}\\\\\\\\\\frac{8}{8}\\cdot y(x+5)=7\\cdot 8\\cdot \\frac{y}{y}\\\\\\\\1\\cdot y(x+5)=56\\cdot 1\\,\\,\\,\\\\\\\\ y(x+5)=56\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ \\frac{y(x+5)}{x+5}=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ y=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example:\u00a0Manipulating a Rational Equation<\/h3>\r\nSolve the equation [latex] x=\\frac{4}{3y+1}[\/latex] for [latex]y[\/latex].\r\n\r\n[reveal-answer q=\"331190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"331190\"]\r\n\r\nClear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3y+1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{3y+1}\\right)\\left(3y+1\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Simplify common factors.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{\\cancel{3y+1}}\\right)\\left(\\cancel{3y+1}\\right)\\\\\\left(3y+1\\right)\\left(x\\right)=4\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We can now use the addition and multiplication properties of equality to solve it.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=4 \\\\ \\\\ \\frac{(3y+1)(x)}{x}=\\frac{4}{x} \\\\ \\\\ 3y+1=\\frac{4}{x} \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\,\\,\\,\\underline{-1} \\\\ 3y=\\frac{4}{x}-1\\\\ \\\\ \\frac{3y}{3}=\\frac{4}{3x}-\\frac{1}{3}\\\\ \\\\ y=\\frac{4}{3}x-\\frac{1}{3}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Apply the Inverse Property to Simplify Logarithms<\/h2>\r\nRecall that the logarithmic and exponential functions \"undo\" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}1=0\\\\{\\mathrm{log}}_{b}b=1\\end{array}[\/latex]<\/p>\r\nFor example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex] and [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].\r\n\r\nNext, we have the inverse property.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x&gt;0\\hfill \\end{array}[\/latex]<\/p>\r\nFor example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex] and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].\r\n\r\nTo evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex] and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Simplify Logarithms<\/h3>\r\nSimplify each of the following.\r\n\r\na) [latex]e^{\\ln(7x)}[\/latex]\r\n\r\nb)\u00a0[latex]\\log_9(9^4)[\/latex]\r\n\r\nc)\u00a0[latex]5^{\\log_5(3y)}[\/latex]\r\n\r\n[reveal-answer q=\"9797650\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"9797650\"]\r\n\r\na) Using the second inverse property,\u00a0[latex]e^{\\ln(7x)}=7x[\/latex]\r\n\r\nb) Using the first inverse property,\u00a0[latex]\\log_9(9^4)=4[\/latex]\r\n\r\nc) Using the second inverse property,\u00a0[latex]5^{\\log_5(3y)}=3y[\/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\nSimplify each of the following.\r\n\r\na)[latex]b^{\\log_b(15x)}[\/latex]\r\n\r\nb)\u00a0[latex]\\log_2(2^a)[\/latex]\r\n\r\nc) [latex]e^{\\ln(4y)}[\/latex]\r\n\r\n[reveal-answer q=\"3839720\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"3839720\"]\r\n\r\na) [latex]15x[\/latex]\r\n\r\nb) [latex]a[\/latex]\r\n\r\nc) [latex]4y[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use the Power Property for Logarithms<\/h2>\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\n<div class=\"textbox exercises\">\r\n<h3>Example: Expanding a Logarithm with Powers<\/h3>\r\nRewrite [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].\r\n\r\n[reveal-answer q=\"979765\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979765\"]\r\n\r\nThe argument is already written as a power, so we identify the exponent, 5, and the base, <em>x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [latex]\\mathrm{ln}{x}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"383972\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"383972\"]\r\n\r\n[latex]2\\mathrm{ln}x[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\r\nRewrite [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.\r\n\r\n[reveal-answer q=\"984289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"984289\"]\r\n\r\nExpressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].\r\n\r\nNext we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"947582\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"947582\"]\r\n\r\n[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Write Function Equations Using Given Conditions<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\">Module 4, Skills Review for Basics of Differential Equations<\/a>)<\/em><\/strong>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve rational equations by clearing denominators<\/li>\n<li>Apply the inverse property to simplify logarithms<\/li>\n<li>Define and use the power rule for logarithms to rewrite expressions<\/li>\n<li>Write function equations using given conditions<\/li>\n<\/ul>\n<\/div>\n<p>In the First-order Linear Equations section, we will explore how to solve yet another special type of differential equation. Here we will review how to clear fractions from and manipulate rational equations.<\/p>\n<h2>Manipulate Rational Equations<\/h2>\n<p>Equations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex]\\frac{2y+1}{4}=\\frac{x}{3}[\/latex] is a rational equation (of two variables).<\/p>\n<p>One of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.<\/p>\n<p>Solve for [latex]y[\/latex] in the equation [latex]\\frac{1}{2}y-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.<\/p>\n<p>Multiply both sides of the equation by\u00a0[latex]4[\/latex], the common denominator of the fractional coefficients.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{1}{2}y-3=2-\\frac{3}{4}x\\\\ 4\\left(\\frac{1}{2}y-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\\\\\text{}\\,\\,\\,\\,4\\left(\\frac{1}{2}y\\right)-4\\left(3\\right)=4\\left(2\\right)+4\\left(-\\frac{3}{4}x\\right)\\\\2y-12=8-3x\\\\\\underline{+12}\\,\\,\\,\\,\\,\\,\\underline{+12}\\\\ 2y=20-3x\\\\ \\frac{2y}{2}=\\frac{20-3x}{2} \\\\ y=10-\\frac{3}{2}x\\end{array}[\/latex]<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Manipulating a Rational Equation<\/h3>\n<p>Solve the equation [latex]\\frac{x+5}{8}=\\frac{7}{y}[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q425621\">Show Solution<\/span><\/p>\n<div id=\"q425621\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the least common denominator of\u00a0[latex]8[\/latex] and\u00a0[latex]y[\/latex]. [latex]8y[\/latex] will be the LCD.<\/p>\n<p>Multiply both sides of the equation by the common denominator,\u00a0[latex]8y[\/latex], to keep the equation balanced and to eliminate the denominators.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8y\\cdot \\frac{x+5}{8}=\\frac{7}{y}\\cdot 8y\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y(x+5)}{8}=\\frac{7(8y)}{y}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y}{8}\\cdot (x+5)=\\frac{7(8y)}{y}\\\\\\\\\\frac{8}{8}\\cdot y(x+5)=7\\cdot 8\\cdot \\frac{y}{y}\\\\\\\\1\\cdot y(x+5)=56\\cdot 1\\,\\,\\,\\\\\\\\ y(x+5)=56\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ \\frac{y(x+5)}{x+5}=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ y=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example:\u00a0Manipulating a Rational Equation<\/h3>\n<p>Solve the equation [latex]x=\\frac{4}{3y+1}[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q331190\">Show Solution<\/span><\/p>\n<div id=\"q331190\" class=\"hidden-answer\" style=\"display: none\">\n<p>Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3y+1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{3y+1}\\right)\\left(3y+1\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Simplify common factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{\\cancel{3y+1}}\\right)\\left(\\cancel{3y+1}\\right)\\\\\\left(3y+1\\right)\\left(x\\right)=4\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We can now use the addition and multiplication properties of equality to solve it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=4 \\\\ \\\\ \\frac{(3y+1)(x)}{x}=\\frac{4}{x} \\\\ \\\\ 3y+1=\\frac{4}{x} \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\,\\,\\,\\underline{-1} \\\\ 3y=\\frac{4}{x}-1\\\\ \\\\ \\frac{3y}{3}=\\frac{4}{3x}-\\frac{1}{3}\\\\ \\\\ y=\\frac{4}{3}x-\\frac{1}{3}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Apply the Inverse Property to Simplify Logarithms<\/h2>\n<p>Recall that the logarithmic and exponential functions &#8220;undo&#8221; each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}1=0\\\\{\\mathrm{log}}_{b}b=1\\end{array}[\/latex]<\/p>\n<p>For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex] and [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\n<p>Next, we have the inverse property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x>0\\hfill \\end{array}[\/latex]<\/p>\n<p>For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex] and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\n<p>To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex] and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Simplify Logarithms<\/h3>\n<p>Simplify each of the following.<\/p>\n<p>a) [latex]e^{\\ln(7x)}[\/latex]<\/p>\n<p>b)\u00a0[latex]\\log_9(9^4)[\/latex]<\/p>\n<p>c)\u00a0[latex]5^{\\log_5(3y)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9797650\">Show Solution<\/span><\/p>\n<div id=\"q9797650\" class=\"hidden-answer\" style=\"display: none\">\n<p>a) Using the second inverse property,\u00a0[latex]e^{\\ln(7x)}=7x[\/latex]<\/p>\n<p>b) Using the first inverse property,\u00a0[latex]\\log_9(9^4)=4[\/latex]<\/p>\n<p>c) Using the second inverse property,\u00a0[latex]5^{\\log_5(3y)}=3y[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify each of the following.<\/p>\n<p>a)[latex]b^{\\log_b(15x)}[\/latex]<\/p>\n<p>b)\u00a0[latex]\\log_2(2^a)[\/latex]<\/p>\n<p>c) [latex]e^{\\ln(4y)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3839720\">Show Solution<\/span><\/p>\n<div id=\"q3839720\" class=\"hidden-answer\" style=\"display: none\">\n<p>a) [latex]15x[\/latex]<\/p>\n<p>b) [latex]a[\/latex]<\/p>\n<p>c) [latex]4y[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use the Power Property for Logarithms<\/h2>\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<div class=\"textbox exercises\">\n<h3>Example: Expanding a Logarithm with Powers<\/h3>\n<p>Rewrite [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979765\">Show Solution<\/span><\/p>\n<div id=\"q979765\" class=\"hidden-answer\" style=\"display: none\">\n<p>The argument is already written as a power, so we identify the exponent, 5, and the base, <em>x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Rewrite [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q383972\">Show Solution<\/span><\/p>\n<div id=\"q383972\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{ln}x[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\n<p>Rewrite [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q984289\">Show Solution<\/span><\/p>\n<div id=\"q984289\" class=\"hidden-answer\" style=\"display: none\">\n<p>Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].<\/p>\n<p>Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Rewrite [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q947582\">Show Solution<\/span><\/p>\n<div id=\"q947582\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-2\\mathrm{ln}\\left(x\\right)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Write Function Equations Using Given Conditions<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\">Module 4, Skills Review for Basics of Differential Equations<\/a>)<\/em><\/strong><\/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-1994\">\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 Corequisite. <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":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen 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