{"id":402,"date":"2025-02-13T19:44:37","date_gmt":"2025-02-13T19:44:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/techniques-for-integration-background-youll-need-2\/"},"modified":"2025-02-13T19:44:37","modified_gmt":"2025-02-13T19:44:37","slug":"techniques-for-integration-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/techniques-for-integration-background-youll-need-2\/","title":{"raw":"Techniques for Integration: Background You'll Need 2","rendered":"Techniques for Integration: Background You&#8217;ll Need 2"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li><span data-sheets-root=\"1\" data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Apply properties of exponential and logarithmic functions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4993,&quot;3&quot;:{&quot;1&quot;:0},&quot;10&quot;:2,&quot;11&quot;:4,&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Apply properties of exponential and logarithmic functions<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Exponential Functions<\/h2>\n<p id=\"fs-id1170572246259\">Exponential functions arise in many applications. One common example is <span class=\"no-emphasis\">population growth<\/span>.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572449480\">If a population starts with [latex]P_0[\/latex] individuals and then grows at an annual rate of [latex]2\\%[\/latex], its population after [latex]1[\/latex] year is<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(1)=P_0+0.02P_0=P_0(1+0.02)=P_0(1.02)[\/latex]<\/div>\n<p id=\"fs-id1170572092410\">Its population after [latex]2[\/latex] years is<\/p>\n<div id=\"fs-id1170572177937\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(2)=P(1)+0.02P(1)=P(1)(1.02)=P_0(1.02)^2[\/latex]<\/div>\n<p id=\"fs-id1170572130048\">In general, its population after [latex]t[\/latex] years is<\/p>\n<div id=\"fs-id1170572280288\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(t)=P_0(1.02)^t[\/latex],<\/div>\n<p>which is an exponential function.<\/p>\n<\/section>\n<p>More generally, any function of the form [latex]f(x)=b^x[\/latex], where [latex]b&gt;0, \\, b \\ne 1[\/latex], is an <strong>exponential function<\/strong> with base [latex]b[\/latex] and exponent [latex]x[\/latex]. Exponential functions have constant bases and variable exponents.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>exponential function<\/h3>\n\nFor any real number [latex]x[\/latex], an exponential function is a function with the form\n\n<p style=\"text-align: center;\">[latex]f(x)=ab^x[\/latex]<\/p>\n\nwhere,\n\n<ul>\n\t<li>[latex]a[\/latex] is a non-zero real number called the initial value and<\/li>\n\t<li>[latex]b[\/latex] is any positive real number ([latex]b&gt;0[\/latex]) such that [latex]b\u22601[\/latex].<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<h3>Evaluating Exponential Functions<\/h3>\n<p>To evaluate an exponential function with the form [latex]f(x)=b^x[\/latex], we simply substitute [latex]x[\/latex] with the given value, and calculate the resulting power.<\/p>\n<section class=\"textbox example\">\n<p>Let [latex]f(x)=2^x[\/latex]. What is [latex]f(3)[\/latex]?<\/p>\n<center>[latex]\\begin{array}{rcl} f(x) &amp; = &amp; 2^x \\\\ f(3) &amp; = &amp; 2^3 &amp; \\quad \\text{Substitute } x = 3. \\\\ &amp; = &amp; 8 &amp; \\quad \\text{Evaluate the power.} \\end{array} [\/latex]<\/center>\n<p>&nbsp;<\/p>\n<\/section>\n<p>To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations.<\/p>\n<section class=\"textbox example\">\n<p>Let [latex]f(x)=30(2)^x[\/latex]. What is [latex]f(3)[\/latex]?<\/p>\n<center>[latex]\\begin{array}{rcll} f(x) &amp; = &amp; 30(2)^x &amp; \\\\ f(3) &amp; = &amp; 30(2)^3 &amp; \\quad \\text{Substitute } x = 3. \\\\ &amp; = &amp; 30(8) &amp; \\quad \\text{Simplify the power first.} \\\\ &amp; = &amp; 240 &amp; \\quad \\text{Multiply.} \\end{array} [\/latex]<\/center>Note that if the order of operations were not followed, the result would be incorrect:<center>[latex]f(3)=30(2)^3\u226060^3=216,000[\/latex]<\/center><\/section>\n<section class=\"textbox questionHelp\">\n<p><b>How To: Evaluating Exponential Functions<\/b><\/p>\n<ol>\n\t<li>Given an exponential function, identify [latex]a[\/latex], [latex]b[\/latex], and the value of [latex]x[\/latex] you're being asked to substitute into the function.<\/li>\n\t<li>Replace the variable [latex]x[\/latex] in the function with the given number.<\/li>\n\t<li>Compute the value of [latex]b^x[\/latex]. This means raising the base [latex]b[\/latex] to the power of [latex]x[\/latex].<\/li>\n\t<li>If there is a coefficient [latex]a[\/latex] in front of the base, multiply the result of [latex]b^x[\/latex] by [latex]a[\/latex]. If [latex]a[\/latex] is [latex]1[\/latex], this step does not change the value.<\/li>\n\t<li>Simplify the expression if necessary. This could involve performing any additional multiplication or addition\/subtraction if the function has more terms.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]f(x)=5(3)^x+1[\/latex]. Evaluate [latex]f(2)[\/latex] without using a calculator.<\/p>\n<p><br>\n[reveal-answer q=\"586760\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"586760\"]Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\n<center>[latex]\\begin{array}{rcll} f(x) &amp; = &amp; 5(3)^{x+1} &amp; \\\\ f(2) &amp; = &amp; 5(3)^{2+1} &amp; \\quad \\text{Substitute } x = 2. \\\\ &amp; = &amp; 5(3)^3 &amp; \\quad \\text{Add the exponents.} \\\\ &amp; = &amp; 5(27) &amp; \\quad \\text{Simplify the power.} \\\\ &amp; = &amp; 135 &amp; \\quad \\text{Multiply.} \\end{array} [\/latex]<\/center>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]284250[\/ohm_question]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572169653\">Suppose a particular population of bacteria is known to double in size every [latex]4[\/latex] hours. If a culture starts with [latex]1000[\/latex] bacteria, the number of bacteria after [latex]4[\/latex] hours is [latex]n(4)=1000\u00b72[\/latex]. The number of bacteria after [latex]8[\/latex] hours is [latex]n(8)=n(4)\u00b72=1000\u00b72^2[\/latex].<\/p>\n<p>In general, the number of bacteria after [latex]4m[\/latex] hours is [latex]n(4m)=1000\u00b72^m[\/latex]. Letting [latex]t=4m[\/latex], we see that the number of bacteria after [latex]t[\/latex] hours is [latex]n(t)=1000\u00b72^{t\/4}[\/latex].<\/p>\n<p>Find the number of bacteria after [latex]6[\/latex] hours, [latex]10[\/latex] hours, and [latex]24[\/latex] hours.<\/p>\n<p>[reveal-answer q=\"fs-id1170572550969\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572550969\"]<\/p>\n<p id=\"fs-id1170572550969\">The number of bacteria after [latex]6[\/latex] hours is given by [latex]n(6)=1000\u00b72^{6\/4} \\approx 2828[\/latex] bacteria.<\/p>\n<p>The number of bacteria after [latex]10[\/latex] hours is given by [latex]n(10)=1000\u00b72^{10\/4} \\approx 5657[\/latex] bacteria.<\/p>\n<p>The number of bacteria after [latex]24[\/latex] hours is given by [latex]n(24)=1000\u00b72^{24\/4}=1000\u00b72^6=64,000[\/latex] bacteria.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h3>Laws of Exponents<\/h3>\n<p id=\"fs-id1170572481226\">The Laws of Exponents are fundamental rules that govern the operations involving powers. These rules are essential for simplifying expressions and are foundational for higher-level math.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>laws of exponents<\/h3>\n<ol id=\"fs-id1170572481268\">\n\t<li>The <strong>Product of Powers<\/strong> rule states that when you multiply two exponents with the same base, you can add the exponents.<center>[latex]b^x\u00b7b^y=b^{x+y}[\/latex]<\/center><\/li>\n\t<li>The <strong>Quotient of Powers<\/strong> rule tells us that when dividing exponents with the same base, we subtract the exponents.<center>[latex]\\large\\frac{b^x}{b^y} \\normalsize = b^{x-y}[\/latex]<\/center><\/li>\n\t<li>The <strong>Power of a Power<\/strong> rule shows that when taking an exponent to another exponent, we multiply the exponents.<center>[latex](b^x)^y=b^{xy}[\/latex]<\/center><\/li>\n\t<li>The <strong>Power of a Product<\/strong> rule lets us know that when raising a product to an exponent, each factor in the product is raised to the exponent.<center>[latex](ab)^x=a^x b^x[\/latex]<\/center><\/li>\n\t<li>The <strong>Power of a Quotient<\/strong> rule indicates that when a quotient is raised to an exponent, both the numerator and the denominator are raised to the exponent.<center>[latex]\\dfrac{a^x}{b^x} =\\left(\\dfrac{a}{b}\\right)^x[\/latex]<\/center><\/li>\n<\/ol>\n<p><em>Note: This is true for any constants [latex]a&gt;0, \\, b&gt;0[\/latex], and for all [latex]x[\/latex] and [latex]y[\/latex]<\/em><\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572440102\">Use the laws of exponents to simplify each of the following expressions.<\/p>\n<ol id=\"fs-id1170572440106\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\large \\frac{(2x^{2\/3})^3}{(4x^{-1\/3})^2}[\/latex]<\/li>\n\t<li>[latex]\\large \\frac{(x^3 y^{-1})^2}{(xy^2)^{-2}}[\/latex]<\/li>\n<\/ol>\n<p>[reveal-answer q=\"fs-id1170572453127\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572453127\"]<\/p>\n<ol id=\"fs-id1170572453127\" style=\"list-style-type: lower-alpha;\">\n\t<li>We can simplify as follows:<br>\n<div id=\"fs-id1170570966957\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\large \\frac{(2x^{2\/3})^3}{(4x^{-1\/3})^2} \\normalsize = \\large \\frac{2^3(x^{2\/3})^3}{4^2(x^{-1\/3})^2} \\normalsize = \\large \\frac{8x^2}{16x^{-2\/3}} \\normalsize = \\large \\frac{x^2x^{2\/3}}{2} \\normalsize = \\large \\frac{x^{8\/3}}{2}[\/latex]<\/div>\n<\/li>\n\t<li>We can simplify as follows:<br>\n<div id=\"fs-id1170573582280\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\large \\frac{(x^3y^{-1})^2}{(xy^2)^{-2}} \\normalsize = \\large \\frac{(x^3)^2(y^{-1})^2}{x^{-2}(y^2)^{-2}} \\normalsize = \\large \\frac{x^6y^{-2}}{x^{-2}y^{-4}} \\normalsize = x^6x^2y^{-2}y^4 = x^8y^2[\/latex]<\/div>\n<\/li>\n<\/ol>\n\nWatch the following video to see the worked solution to this example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=212&amp;end=380&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\n\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions212to380_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox proTip\">\n<p>When you encounter a negative exponent on a term in the denominator of a fraction, you can transform it into a positive exponent by moving the term to the numerator.<\/p>\n<center>[latex]\\frac{1}{a^-n}=a^{n}[\/latex]<\/center>Using this rule can significantly simplify expressions involving exponents.<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]123515[\/ohm_question]<\/p>\n<\/section>\n<h2>Logarithmic Functions<\/h2>\n<p id=\"fs-id1170572455248\">Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.&nbsp;<\/p>\n<section class=\"textbox recall\">\n<p><strong>Inverse Functions<\/strong><\/p>\n<p>For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex].&nbsp;<\/p>\n<p>The notation [latex]{f}^{-1}[\/latex] is read \"[latex]f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]\".<\/p>\n<\/section>\n<p>Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.<\/p>\n<p id=\"fs-id1170572455254\">The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the <strong>logarithmic function<\/strong><em> with base<\/em> [latex]b[\/latex].<\/p>\n<p>For any [latex]b&gt;0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies<\/p>\n<p style=\"text-align: center;\">[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>logarithmic functions<\/h3>\n<p>A logarithmic function is the inverse of an exponential function and is written as [latex]log_{b}(x)[\/latex]. For a given base [latex]b[\/latex], it tells us the power to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<div id=\"fs-id1170572545103\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill &amp; &amp; &amp; \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b&gt;0.\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1170572169184\">The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the<strong> natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex].<\/p>\n<section class=\"textbox example\"><center>[latex]\\begin{array}{l}\\ln (e)=\\log_e (e)=1 \\\\ \\ln(e^3)=\\log_e (e^3)=3 \\\\ \\ln(1)=\\log_e (1)=0\\end{array}[\/latex]<\/center><\/section>\n<section class=\"textbox recall\">\n<p>Euler's number, denoted as [latex]e[\/latex], is a fundamental mathematical constant approximately equal to [latex]2.71828[\/latex]. It is the base of the natural logarithm and the natural exponential function, known for its unique properties in calculus, especially in relation to growth processes and compound interest calculations.<\/p>\n<\/section>\n<p id=\"fs-id1170572482697\">Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Properties of Logarithms<\/h3>\n<p id=\"fs-id1170572482707\">If [latex]a,b,c&gt;0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill &amp; &amp; &amp; \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill &amp; &amp; &amp; \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill &amp; &amp; &amp; \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572174684\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572174692\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\n\t<li>[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\n\t<li>[latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\n<\/ol>\n<p>[reveal-answer q=\"fs-id1170572174799\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572174799\"]<\/p>\n<ol id=\"fs-id1170572174799\" style=\"list-style-type: lower-alpha;\">\n\t<li>By the definition of the natural logarithm function,<br>\n<div id=\"fs-id1170573425282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\n<p>Therefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/p>\n<\/li>\n\t<li>Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as<br>\n<div id=\"fs-id1170573416245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\n<p>Therefore, the equation can be rewritten as<\/p>\n<div id=\"fs-id1170571053549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\n<p>The solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/p>\n<\/li>\n\t<li>Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].<br>\nUsing the quotient property, this becomes\n\n<div id=\"fs-id1170573426389\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\n<p>Therefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex]. We should then check for any extraneous solutions.<\/p>\n<\/li>\n<\/ol>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572551980\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572551988\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]5^x=2[\/latex]<\/li>\n\t<li>[latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\n<\/ol>\n<p>[reveal-answer q=\"fs-id1170572550555\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572550555\"]<\/p>\n<ol id=\"fs-id1170572550555\" style=\"list-style-type: lower-alpha;\">\n\t<li>Applying the natural logarithm function to both sides of the equation, we have<br>\n<div id=\"fs-id1170571071223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\n<p>Using the power property of logarithms,<\/p>\n<div id=\"fs-id1170571277779\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln 5=\\ln 2[\/latex]<\/div>\n<p>Therefore, [latex]x=\\frac{\\ln 2 }{\\ln 5}[\/latex].<\/p>\n<\/li>\n\t<li>Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation<br>\n<div id=\"fs-id1170571301573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}+6=5e^x[\/latex]<\/div>\n<p>Rewriting this equation as<\/p>\n<div id=\"fs-id1170573367583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\n<p>we can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:<\/p>\n<div id=\"fs-id1170570976384\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\n<p>Now we can solve the quadratic equation. Factoring this equation, we obtain<\/p>\n<div id=\"fs-id1170573400246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\n<p>Therefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions<\/p>\n<div style=\"text-align: center;\">[latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div>\n<\/li>\n<\/ol>\n\nWatch the following video to see the worked solution to this example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=640&amp;end=823&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\n\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions640to823_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]217547[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-root=\"1\" data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Apply properties of exponential and logarithmic functions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4993,&quot;3&quot;:{&quot;1&quot;:0},&quot;10&quot;:2,&quot;11&quot;:4,&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Apply properties of exponential and logarithmic functions<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Exponential Functions<\/h2>\n<p id=\"fs-id1170572246259\">Exponential functions arise in many applications. One common example is <span class=\"no-emphasis\">population growth<\/span>.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572449480\">If a population starts with [latex]P_0[\/latex] individuals and then grows at an annual rate of [latex]2\\%[\/latex], its population after [latex]1[\/latex] year is<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(1)=P_0+0.02P_0=P_0(1+0.02)=P_0(1.02)[\/latex]<\/div>\n<p id=\"fs-id1170572092410\">Its population after [latex]2[\/latex] years is<\/p>\n<div id=\"fs-id1170572177937\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(2)=P(1)+0.02P(1)=P(1)(1.02)=P_0(1.02)^2[\/latex]<\/div>\n<p id=\"fs-id1170572130048\">In general, its population after [latex]t[\/latex] years is<\/p>\n<div id=\"fs-id1170572280288\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(t)=P_0(1.02)^t[\/latex],<\/div>\n<p>which is an exponential function.<\/p>\n<\/section>\n<p>More generally, any function of the form [latex]f(x)=b^x[\/latex], where [latex]b>0, \\, b \\ne 1[\/latex], is an <strong>exponential function<\/strong> with base [latex]b[\/latex] and exponent [latex]x[\/latex]. Exponential functions have constant bases and variable exponents.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>exponential function<\/h3>\n<p>For any real number [latex]x[\/latex], an exponential function is a function with the form<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=ab^x[\/latex]<\/p>\n<p>where,<\/p>\n<ul>\n<li>[latex]a[\/latex] is a non-zero real number called the initial value and<\/li>\n<li>[latex]b[\/latex] is any positive real number ([latex]b>0[\/latex]) such that [latex]b\u22601[\/latex].<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<h3>Evaluating Exponential Functions<\/h3>\n<p>To evaluate an exponential function with the form [latex]f(x)=b^x[\/latex], we simply substitute [latex]x[\/latex] with the given value, and calculate the resulting power.<\/p>\n<section class=\"textbox example\">\n<p>Let [latex]f(x)=2^x[\/latex]. What is [latex]f(3)[\/latex]?<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcl} f(x) & = & 2^x \\\\ f(3) & = & 2^3 & \\quad \\text{Substitute } x = 3. \\\\ & = & 8 & \\quad \\text{Evaluate the power.} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/section>\n<p>To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations.<\/p>\n<section class=\"textbox example\">\n<p>Let [latex]f(x)=30(2)^x[\/latex]. What is [latex]f(3)[\/latex]?<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcll} f(x) & = & 30(2)^x & \\\\ f(3) & = & 30(2)^3 & \\quad \\text{Substitute } x = 3. \\\\ & = & 30(8) & \\quad \\text{Simplify the power first.} \\\\ & = & 240 & \\quad \\text{Multiply.} \\end{array}[\/latex]<\/div>\n<p>Note that if the order of operations were not followed, the result would be incorrect:<\/p>\n<div style=\"text-align: center;\">[latex]f(3)=30(2)^3\u226060^3=216,000[\/latex]<\/div>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><b>How To: Evaluating Exponential Functions<\/b><\/p>\n<ol>\n<li>Given an exponential function, identify [latex]a[\/latex], [latex]b[\/latex], and the value of [latex]x[\/latex] you&#8217;re being asked to substitute into the function.<\/li>\n<li>Replace the variable [latex]x[\/latex] in the function with the given number.<\/li>\n<li>Compute the value of [latex]b^x[\/latex]. This means raising the base [latex]b[\/latex] to the power of [latex]x[\/latex].<\/li>\n<li>If there is a coefficient [latex]a[\/latex] in front of the base, multiply the result of [latex]b^x[\/latex] by [latex]a[\/latex]. If [latex]a[\/latex] is [latex]1[\/latex], this step does not change the value.<\/li>\n<li>Simplify the expression if necessary. This could involve performing any additional multiplication or addition\/subtraction if the function has more terms.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]f(x)=5(3)^x+1[\/latex]. Evaluate [latex]f(2)[\/latex] without using a calculator.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q586760\">Show Answer<\/button><\/p>\n<div id=\"q586760\" class=\"hidden-answer\" style=\"display: none\">Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcll} f(x) & = & 5(3)^{x+1} & \\\\ f(2) & = & 5(3)^{2+1} & \\quad \\text{Substitute } x = 2. \\\\ & = & 5(3)^3 & \\quad \\text{Add the exponents.} \\\\ & = & 5(27) & \\quad \\text{Simplify the power.} \\\\ & = & 135 & \\quad \\text{Multiply.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm284250\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=284250&theme=lumen&iframe_resize_id=ohm284250&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572169653\">Suppose a particular population of bacteria is known to double in size every [latex]4[\/latex] hours. If a culture starts with [latex]1000[\/latex] bacteria, the number of bacteria after [latex]4[\/latex] hours is [latex]n(4)=1000\u00b72[\/latex]. The number of bacteria after [latex]8[\/latex] hours is [latex]n(8)=n(4)\u00b72=1000\u00b72^2[\/latex].<\/p>\n<p>In general, the number of bacteria after [latex]4m[\/latex] hours is [latex]n(4m)=1000\u00b72^m[\/latex]. Letting [latex]t=4m[\/latex], we see that the number of bacteria after [latex]t[\/latex] hours is [latex]n(t)=1000\u00b72^{t\/4}[\/latex].<\/p>\n<p>Find the number of bacteria after [latex]6[\/latex] hours, [latex]10[\/latex] hours, and [latex]24[\/latex] hours.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572550969\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572550969\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572550969\">The number of bacteria after [latex]6[\/latex] hours is given by [latex]n(6)=1000\u00b72^{6\/4} \\approx 2828[\/latex] bacteria.<\/p>\n<p>The number of bacteria after [latex]10[\/latex] hours is given by [latex]n(10)=1000\u00b72^{10\/4} \\approx 5657[\/latex] bacteria.<\/p>\n<p>The number of bacteria after [latex]24[\/latex] hours is given by [latex]n(24)=1000\u00b72^{24\/4}=1000\u00b72^6=64,000[\/latex] bacteria.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Laws of Exponents<\/h3>\n<p id=\"fs-id1170572481226\">The Laws of Exponents are fundamental rules that govern the operations involving powers. These rules are essential for simplifying expressions and are foundational for higher-level math.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>laws of exponents<\/h3>\n<ol id=\"fs-id1170572481268\">\n<li>The <strong>Product of Powers<\/strong> rule states that when you multiply two exponents with the same base, you can add the exponents.\n<div style=\"text-align: center;\">[latex]b^x\u00b7b^y=b^{x+y}[\/latex]<\/div>\n<\/li>\n<li>The <strong>Quotient of Powers<\/strong> rule tells us that when dividing exponents with the same base, we subtract the exponents.\n<div style=\"text-align: center;\">[latex]\\large\\frac{b^x}{b^y} \\normalsize = b^{x-y}[\/latex]<\/div>\n<\/li>\n<li>The <strong>Power of a Power<\/strong> rule shows that when taking an exponent to another exponent, we multiply the exponents.\n<div style=\"text-align: center;\">[latex](b^x)^y=b^{xy}[\/latex]<\/div>\n<\/li>\n<li>The <strong>Power of a Product<\/strong> rule lets us know that when raising a product to an exponent, each factor in the product is raised to the exponent.\n<div style=\"text-align: center;\">[latex](ab)^x=a^x b^x[\/latex]<\/div>\n<\/li>\n<li>The <strong>Power of a Quotient<\/strong> rule indicates that when a quotient is raised to an exponent, both the numerator and the denominator are raised to the exponent.\n<div style=\"text-align: center;\">[latex]\\dfrac{a^x}{b^x} =\\left(\\dfrac{a}{b}\\right)^x[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p><em>Note: This is true for any constants [latex]a>0, \\, b>0[\/latex], and for all [latex]x[\/latex] and [latex]y[\/latex]<\/em><\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572440102\">Use the laws of exponents to simplify each of the following expressions.<\/p>\n<ol id=\"fs-id1170572440106\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\large \\frac{(2x^{2\/3})^3}{(4x^{-1\/3})^2}[\/latex]<\/li>\n<li>[latex]\\large \\frac{(x^3 y^{-1})^2}{(xy^2)^{-2}}[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572453127\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572453127\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572453127\" style=\"list-style-type: lower-alpha;\">\n<li>We can simplify as follows:\n<div id=\"fs-id1170570966957\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\large \\frac{(2x^{2\/3})^3}{(4x^{-1\/3})^2} \\normalsize = \\large \\frac{2^3(x^{2\/3})^3}{4^2(x^{-1\/3})^2} \\normalsize = \\large \\frac{8x^2}{16x^{-2\/3}} \\normalsize = \\large \\frac{x^2x^{2\/3}}{2} \\normalsize = \\large \\frac{x^{8\/3}}{2}[\/latex]<\/div>\n<\/li>\n<li>We can simplify as follows:\n<div id=\"fs-id1170573582280\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\large \\frac{(x^3y^{-1})^2}{(xy^2)^{-2}} \\normalsize = \\large \\frac{(x^3)^2(y^{-1})^2}{x^{-2}(y^2)^{-2}} \\normalsize = \\large \\frac{x^6y^{-2}}{x^{-2}y^{-4}} \\normalsize = x^6x^2y^{-2}y^4 = x^8y^2[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=212&amp;end=380&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions212to380_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\">\n<p>When you encounter a negative exponent on a term in the denominator of a fraction, you can transform it into a positive exponent by moving the term to the numerator.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{1}{a^-n}=a^{n}[\/latex]<\/div>\n<p>Using this rule can significantly simplify expressions involving exponents.<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm123515\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=123515&theme=lumen&iframe_resize_id=ohm123515&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h2>Logarithmic Functions<\/h2>\n<p id=\"fs-id1170572455248\">Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.&nbsp;<\/p>\n<section class=\"textbox recall\">\n<p><strong>Inverse Functions<\/strong><\/p>\n<p>For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex].&nbsp;<\/p>\n<p>The notation [latex]{f}^{-1}[\/latex] is read &#8220;[latex]f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]&#8220;.<\/p>\n<\/section>\n<p>Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.<\/p>\n<p id=\"fs-id1170572455254\">The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the <strong>logarithmic function<\/strong><em> with base<\/em> [latex]b[\/latex].<\/p>\n<p>For any [latex]b>0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies<\/p>\n<p style=\"text-align: center;\">[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>logarithmic functions<\/h3>\n<p>A logarithmic function is the inverse of an exponential function and is written as [latex]log_{b}(x)[\/latex]. For a given base [latex]b[\/latex], it tells us the power to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<div id=\"fs-id1170572545103\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b>0.\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1170572169184\">The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the<strong> natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex].<\/p>\n<section class=\"textbox example\">\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\ln (e)=\\log_e (e)=1 \\\\ \\ln(e^3)=\\log_e (e^3)=3 \\\\ \\ln(1)=\\log_e (1)=0\\end{array}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox recall\">\n<p>Euler&#8217;s number, denoted as [latex]e[\/latex], is a fundamental mathematical constant approximately equal to [latex]2.71828[\/latex]. It is the base of the natural logarithm and the natural exponential function, known for its unique properties in calculus, especially in relation to growth processes and compound interest calculations.<\/p>\n<\/section>\n<p id=\"fs-id1170572482697\">Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Properties of Logarithms<\/h3>\n<p id=\"fs-id1170572482707\">If [latex]a,b,c>0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill & & & \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill & & & \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill & & & \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572174684\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572174692\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\n<li>[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\n<li>[latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572174799\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572174799\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572174799\" style=\"list-style-type: lower-alpha;\">\n<li>By the definition of the natural logarithm function,\n<div id=\"fs-id1170573425282\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\n<p>Therefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/p>\n<\/li>\n<li>Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as\n<div id=\"fs-id1170573416245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\n<p>Therefore, the equation can be rewritten as<\/p>\n<div id=\"fs-id1170571053549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\n<p>The solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/p>\n<\/li>\n<li>Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].<br \/>\nUsing the quotient property, this becomes<\/p>\n<div id=\"fs-id1170573426389\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\n<p>Therefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex]. We should then check for any extraneous solutions.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572551980\">Solve each of the following equations for [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572551988\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]5^x=2[\/latex]<\/li>\n<li>[latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572550555\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572550555\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572550555\" style=\"list-style-type: lower-alpha;\">\n<li>Applying the natural logarithm function to both sides of the equation, we have\n<div id=\"fs-id1170571071223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\n<p>Using the power property of logarithms,<\/p>\n<div id=\"fs-id1170571277779\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x \\ln 5=\\ln 2[\/latex]<\/div>\n<p>Therefore, [latex]x=\\frac{\\ln 2 }{\\ln 5}[\/latex].<\/p>\n<\/li>\n<li>Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation\n<div id=\"fs-id1170571301573\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}+6=5e^x[\/latex]<\/div>\n<p>Rewriting this equation as<\/p>\n<div id=\"fs-id1170573367583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\n<p>we can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:<\/p>\n<div id=\"fs-id1170570976384\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\n<p>Now we can solve the quadratic equation. Factoring this equation, we obtain<\/p>\n<div id=\"fs-id1170573400246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\n<p>Therefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions<\/p>\n<div style=\"text-align: center;\">[latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=640&amp;end=823&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExponentialAndLogarithmicFunctions640to823_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm217547\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=217547&theme=lumen&iframe_resize_id=ohm217547&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":399,"module-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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/402"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/402\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/399"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/402\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=402"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=402"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=402"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=402"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}