{"id":2007,"date":"2023-04-18T17:42:48","date_gmt":"2023-04-18T17:42:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2007"},"modified":"2025-08-28T20:53:08","modified_gmt":"2025-08-28T20:53:08","slug":"logarithms-and-logistic-growth-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/logarithms-and-logistic-growth-learn-it-1\/","title":{"raw":"Logarithms and Logistic Growth: Learn It 1","rendered":"Logarithms and Logistic Growth: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Use the properties of logarithms to solve exponential models for time<\/li>\r\n\t<li>Identify the carrying capacity in a logistic growth model<\/li>\r\n\t<li>Use a logistic growth model to predict growth<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Reversing an Exponent<\/h2>\r\n<p>Earlier, we found that since Olympia, WA had a population of [latex]245[\/latex] thousand in 2008 and had been growing at [latex]3\\%[\/latex] per year, the population could be modeled by the equation<\/p>\r\n<p style=\"text-align: center;\">[latex]P_n = (1+0.03)^{n}(245,000)[\/latex], or equivalently, [latex]P_n = 245,000(1.03)^{n}[\/latex].<\/p>\r\n<p>Using this equation, we were able to predict the population in the future.<\/p>\r\n<center>\r\n[caption id=\"attachment_899\" align=\"aligncenter\" width=\"1024\"]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280.jpg\"><img class=\"wp-image-899 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280-1024x426.jpg\" alt=\"A soft-focus landscape photo of a crowd of people. Overlaid on top is a dotted red horizontal line and many vertical blue lines rising above and below the dotted red line, to give the impression of a population growth chart.\" width=\"1024\" height=\"426\" \/><\/a> Figure 1. A soft-focus landscape of a crowd with an overlaid population chart[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>Suppose we wanted to know when the population of Olympia would reach [latex]400[\/latex] thousand, rather than what the population would be after a set amount of time. Since we are looking for the year [latex]n[\/latex] when the population will be [latex]400[\/latex] thousand, we would need to solve the equation<\/p>\r\n<p style=\"text-align: center;\">[latex]400,000 = 245,000(1.03)^{n}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Dividing both sides by\u00a0[latex]245,000[\/latex] gives<\/p>\r\n<p style=\"text-align: center;\">[latex]1.6327 = 1.03^{n}[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>One approach to this problem would be to create a table of values, or to use technology to draw a graph to estimate the solution.<\/p>\r\n<center>\r\n[caption id=\"attachment_368\" align=\"aligncenter\" width=\"350\"]<img class=\"wp-image-368\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11202224\/estimatesolution-300x164.png\" alt=\"Line graph. Vertical measures increments of 50 from 0 to 450. Horizontal measures increments of 1 from 0 to 20. The line starts at 250\/0, and moves in a sloping up trend towards 450\/20.\" width=\"350\" height=\"191\" \/> Figure 2. A graph of a equation [latex]400,000 = 245,000(1.03)^{n}[\/latex][\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>From the graph, we can estimate that the solution will be around\u00a0[latex]16[\/latex] to\u00a0[latex]17[\/latex] years after 2008 (2024 to 2025). This is pretty good, but we\u2019d really like to have an algebraic tool to answer this question. To do that, we need to introduce a new function that will undo exponentials, similar to how a square root undoes a square. For exponentials, the function we need is called a logarithm. It is the inverse of the exponential, meaning it undoes the exponential. While there is a whole family of logarithms with different bases, we will focus on the common log, which is based on the exponential [latex]10^{n}[\/latex].<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>common logarithm<\/h3>\r\n<p>The common logarithm, written [latex]log(x)[\/latex], undoes the exponential [latex]10^{x}[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>This means that\u00a0[latex]log(10^{x}) = x[\/latex], and likewise [latex]10^{log(x)} = x[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>This also means the statement [latex]10^{a} = b[\/latex] is equivalent to the statement [latex]log(b) = a[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>[latex]log(x)[\/latex] is read as \u201clog of [latex]x[\/latex]\u201d, and means \u201cthe logarithm of the value [latex]x[\/latex]\u201d. It is important to note that this is <em>not<\/em> multiplication \u2013 the log doesn\u2019t mean anything by itself, just like \u221a doesn\u2019t mean anything by itself; it has to be applied to a number.<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox recall\"><strong>rules for rewriting expressions containing exponents<\/strong>\r\n<ul>\r\n\t<li>Product Rule [latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/li>\r\n\t<li>Quotient Rule [latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex]<\/li>\r\n\t<li>Power Rule [latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/li>\r\n\t<li>Zero Exponent [latex]{a}^{0}=1[\/latex]<\/li>\r\n\t<li>Negative Exponent [latex]{a}^{-n}=\\dfrac{1}{{a}^{n}} \\text{ and } {a}^{n}=\\dfrac{1}{{a}^{-n}}[\/latex]<\/li>\r\n\t<li>Power of a Product [latex]\\large{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/li>\r\n\t<li>Power of a Quotient [latex]\\large{\\left(\\dfrac{a}{b}\\right)}^{n}=\\dfrac{{a}^{n}}{{b}^{n}}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox proTip\"><strong>exponents and logarithms: a little help<\/strong>\r\n<p>You've seen that\u00a0the common logarithm, written\u00a0[latex]log(x)[\/latex], undoes the exponential\u00a0[latex]10^{x}[\/latex].<\/p>\r\n<p>This works because the common logarithm has a base of\u00a0[latex]10[\/latex], just like the exponential expression[latex]10^{x}[\/latex]. That is, [latex]log(x)[\/latex] undoes [latex]10^{x}[\/latex] because [latex]log(x)[\/latex] is the number we to which we raise [latex]10[\/latex] to obtain the number\u00a0[latex]x[\/latex].<\/p>\r\n<p>In the EXAMPLE below, part (a) asks you to evaluate\u00a0[latex]log(100)[\/latex].<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]log(100) = log(10^{2})[\/latex], which gives us the statement\u00a0[latex]x = 2[\/latex] because\u00a0[latex]2[\/latex] is the number to which we raise\u00a0[latex]10[\/latex] to obtain\u00a0[latex]100[\/latex].<\/p>\r\n<p>Part (d) asks you to evaluate\u00a0[latex]log(1\/100)[\/latex]. First, try rewriting\u00a0[latex]1\/100[\/latex] as the base\u00a0[latex]10[\/latex] raised to a number.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]log(100)[\/latex]<\/li>\r\n\t<li>[latex]log(1000)[\/latex]<\/li>\r\n\t<li>[latex]log(10000)[\/latex]<\/li>\r\n\t<li>[latex]log(\\frac{1}{100})[\/latex]<\/li>\r\n\t<li>[latex]log(1)[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"121444\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"121444\"]<\/p>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]log(100)[\/latex] can be written as [latex]log(10^{2})[\/latex]. Since the log undoes the exponential, [latex]log(10^{2}) = 2[\/latex]<\/li>\r\n\t<li>[latex]log(1000) = log(10^{3}) = 3[\/latex]<\/li>\r\n\t<li>[latex]log(10000) = log(10^{4}) = 4[\/latex]<\/li>\r\n\t<li>Recall that [latex]{{x}^{-n}}=\\frac{1}{{{x}^{n}}}[\/latex].\u00a0\u00a0 [latex]\\log\\left(\\frac{1}{100}\\right)=\\log\\left({{10}^{-2}}\\right)=-2[\/latex]<\/li>\r\n\t<li>Recall that [latex]x^{0} = 1[\/latex].\u00a0\u00a0[latex]log(1) = log(10^{0}) = 0[\/latex]<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>It is helpful to note that from the first three parts of the previous example that the number we\u2019re taking the log of has to get [latex]10[\/latex]<em> times bigger<\/em> for the log to increase in value by [latex]1[\/latex].<\/p>\r\n<p>Of course, most numbers cannot be written as a nice simple power of\u00a0[latex]10[\/latex]. For those numbers, we can evaluate the log using a scientific calculator with a log button.<\/p>\r\n<p>With an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation. This will allow us to solve some simple equations.<\/p>\r\n<section class=\"textbox example\">\r\n<ol>\r\n\t<li>Solve [latex]10^{x}= 1000[\/latex]<\/li>\r\n\t<li>Solve\u00a0[latex]10^{x} = 3[\/latex]<\/li>\r\n\t<li>Solve\u00a0[latex]2(10^{x}) = 8[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"588505\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"588505\"]<\/p>\r\n<ol>\r\n\t<li>Taking the log of both sides gives [latex]log(10^{x}) = log(1000)[\/latex].\r\n\r\n<ul>\r\n\t<li>Since the log undoes the exponential,\u00a0[latex]log(10^{x}) = x[\/latex]. Similarly\u00a0[latex]log(1000) = log(10^{3}) = 3[\/latex].<\/li>\r\n\t<li>The equation simplifies then to\u00a0[latex]x = 3[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li>Taking the log of both sides gives\u00a0[latex]log(10^{x}) = log(3)[\/latex].\r\n\r\n<ul>\r\n\t<li>On the left side,\u00a0[latex]log(10^{x}) = x[\/latex], so\u00a0[latex]x = log(3)[\/latex].<\/li>\r\n\t<li>We can approximate this value with a calculator. [latex]x \u2248 0.477[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li>Here we would first want to isolate the exponential by dividing both sides of the equation by\u00a0[latex]2[\/latex], giving\u00a0[latex]10^{x} = 4[\/latex].\r\n\r\n<ul>\r\n\t<li>Now we can take the log of both sides, giving [latex]log(10^{x}) = log(4)[\/latex], which simplifies to [latex]x = log(4) \u2248 0.602[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>properties of logs: exponent property<\/h3>\r\n<center>[latex]\\log\\left({{A}^{r}}\\right)=r\\log\\left(A\\right)[\/latex]<\/center><\/div>\r\n<\/section>\r\n<p>To show why this is true, we offer a proof.<\/p>\r\n<p>Since the logarithm and exponential undo each other, [latex]{{10}^{\\log{A}}}=A[\/latex].<\/p>\r\n<p style=\"text-align: center;\">So [latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}[\/latex]<\/p>\r\n<p>Utilizing the exponential rule that states [latex]{{\\left({{x}^{a}}\\right)}^{b}}={{x}^{ab}}[\/latex],<\/p>\r\n<p style=\"text-align: center;\">[latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}={{10}^{r\\log{A}}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">So then [latex]\\log\\left({{A}^{r}}\\right)=\\log\\left({{10}^{r\\log{A}}}\\right)[\/latex]<\/p>\r\n<p>Again utilizing the property that the log undoes the exponential on the right side yields the result<\/p>\r\n<p style=\"text-align: center;\">[latex]\\log\\left({{A}^{r}}\\right)=r\\log{A}[\/latex]<\/p>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6923[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use the properties of logarithms to solve exponential models for time<\/li>\n<li>Identify the carrying capacity in a logistic growth model<\/li>\n<li>Use a logistic growth model to predict growth<\/li>\n<\/ul>\n<\/section>\n<h2>Reversing an Exponent<\/h2>\n<p>Earlier, we found that since Olympia, WA had a population of [latex]245[\/latex] thousand in 2008 and had been growing at [latex]3\\%[\/latex] per year, the population could be modeled by the equation<\/p>\n<p style=\"text-align: center;\">[latex]P_n = (1+0.03)^{n}(245,000)[\/latex], or equivalently, [latex]P_n = 245,000(1.03)^{n}[\/latex].<\/p>\n<p>Using this equation, we were able to predict the population in the future.<\/p>\n<div style=\"text-align: center;\">\n<figure id=\"attachment_899\" aria-describedby=\"caption-attachment-899\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-899 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280-1024x426.jpg\" alt=\"A soft-focus landscape photo of a crowd of people. Overlaid on top is a dotted red horizontal line and many vertical blue lines rising above and below the dotted red line, to give the impression of a population growth chart.\" width=\"1024\" height=\"426\" \/><\/a><figcaption id=\"caption-attachment-899\" class=\"wp-caption-text\">Figure 1. A soft-focus landscape of a crowd with an overlaid population chart<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Suppose we wanted to know when the population of Olympia would reach [latex]400[\/latex] thousand, rather than what the population would be after a set amount of time. Since we are looking for the year [latex]n[\/latex] when the population will be [latex]400[\/latex] thousand, we would need to solve the equation<\/p>\n<p style=\"text-align: center;\">[latex]400,000 = 245,000(1.03)^{n}[\/latex]<\/p>\n<p style=\"text-align: left;\">Dividing both sides by\u00a0[latex]245,000[\/latex] gives<\/p>\n<p style=\"text-align: center;\">[latex]1.6327 = 1.03^{n}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>One approach to this problem would be to create a table of values, or to use technology to draw a graph to estimate the solution.<\/p>\n<div style=\"text-align: center;\">\n<figure id=\"attachment_368\" aria-describedby=\"caption-attachment-368\" style=\"width: 350px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-368\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11202224\/estimatesolution-300x164.png\" alt=\"Line graph. Vertical measures increments of 50 from 0 to 450. Horizontal measures increments of 1 from 0 to 20. The line starts at 250\/0, and moves in a sloping up trend towards 450\/20.\" width=\"350\" height=\"191\" \/><figcaption id=\"caption-attachment-368\" class=\"wp-caption-text\">Figure 2. A graph of a equation [latex]400,000 = 245,000(1.03)^{n}[\/latex]<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>From the graph, we can estimate that the solution will be around\u00a0[latex]16[\/latex] to\u00a0[latex]17[\/latex] years after 2008 (2024 to 2025). This is pretty good, but we\u2019d really like to have an algebraic tool to answer this question. To do that, we need to introduce a new function that will undo exponentials, similar to how a square root undoes a square. For exponentials, the function we need is called a logarithm. It is the inverse of the exponential, meaning it undoes the exponential. While there is a whole family of logarithms with different bases, we will focus on the common log, which is based on the exponential [latex]10^{n}[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>common logarithm<\/h3>\n<p>The common logarithm, written [latex]log(x)[\/latex], undoes the exponential [latex]10^{x}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>This means that\u00a0[latex]log(10^{x}) = x[\/latex], and likewise [latex]10^{log(x)} = x[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>This also means the statement [latex]10^{a} = b[\/latex] is equivalent to the statement [latex]log(b) = a[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]log(x)[\/latex] is read as \u201clog of [latex]x[\/latex]\u201d, and means \u201cthe logarithm of the value [latex]x[\/latex]\u201d. It is important to note that this is <em>not<\/em> multiplication \u2013 the log doesn\u2019t mean anything by itself, just like \u221a doesn\u2019t mean anything by itself; it has to be applied to a number.<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox recall\"><strong>rules for rewriting expressions containing exponents<\/strong><\/p>\n<ul>\n<li>Product Rule [latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/li>\n<li>Quotient Rule [latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex]<\/li>\n<li>Power Rule [latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/li>\n<li>Zero Exponent [latex]{a}^{0}=1[\/latex]<\/li>\n<li>Negative Exponent [latex]{a}^{-n}=\\dfrac{1}{{a}^{n}} \\text{ and } {a}^{n}=\\dfrac{1}{{a}^{-n}}[\/latex]<\/li>\n<li>Power of a Product [latex]\\large{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/li>\n<li>Power of a Quotient [latex]\\large{\\left(\\dfrac{a}{b}\\right)}^{n}=\\dfrac{{a}^{n}}{{b}^{n}}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox proTip\"><strong>exponents and logarithms: a little help<\/strong><\/p>\n<p>You&#8217;ve seen that\u00a0the common logarithm, written\u00a0[latex]log(x)[\/latex], undoes the exponential\u00a0[latex]10^{x}[\/latex].<\/p>\n<p>This works because the common logarithm has a base of\u00a0[latex]10[\/latex], just like the exponential expression[latex]10^{x}[\/latex]. That is, [latex]log(x)[\/latex] undoes [latex]10^{x}[\/latex] because [latex]log(x)[\/latex] is the number we to which we raise [latex]10[\/latex] to obtain the number\u00a0[latex]x[\/latex].<\/p>\n<p>In the EXAMPLE below, part (a) asks you to evaluate\u00a0[latex]log(100)[\/latex].<\/p>\n<p style=\"padding-left: 30px;\">[latex]log(100) = log(10^{2})[\/latex], which gives us the statement\u00a0[latex]x = 2[\/latex] because\u00a0[latex]2[\/latex] is the number to which we raise\u00a0[latex]10[\/latex] to obtain\u00a0[latex]100[\/latex].<\/p>\n<p>Part (d) asks you to evaluate\u00a0[latex]log(1\/100)[\/latex]. First, try rewriting\u00a0[latex]1\/100[\/latex] as the base\u00a0[latex]10[\/latex] raised to a number.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]log(100)[\/latex]<\/li>\n<li>[latex]log(1000)[\/latex]<\/li>\n<li>[latex]log(10000)[\/latex]<\/li>\n<li>[latex]log(\\frac{1}{100})[\/latex]<\/li>\n<li>[latex]log(1)[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q121444\">Show Solution<\/button><\/p>\n<div id=\"q121444\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]log(100)[\/latex] can be written as [latex]log(10^{2})[\/latex]. Since the log undoes the exponential, [latex]log(10^{2}) = 2[\/latex]<\/li>\n<li>[latex]log(1000) = log(10^{3}) = 3[\/latex]<\/li>\n<li>[latex]log(10000) = log(10^{4}) = 4[\/latex]<\/li>\n<li>Recall that [latex]{{x}^{-n}}=\\frac{1}{{{x}^{n}}}[\/latex].\u00a0\u00a0 [latex]\\log\\left(\\frac{1}{100}\\right)=\\log\\left({{10}^{-2}}\\right)=-2[\/latex]<\/li>\n<li>Recall that [latex]x^{0} = 1[\/latex].\u00a0\u00a0[latex]log(1) = log(10^{0}) = 0[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<p>It is helpful to note that from the first three parts of the previous example that the number we\u2019re taking the log of has to get [latex]10[\/latex]<em> times bigger<\/em> for the log to increase in value by [latex]1[\/latex].<\/p>\n<p>Of course, most numbers cannot be written as a nice simple power of\u00a0[latex]10[\/latex]. For those numbers, we can evaluate the log using a scientific calculator with a log button.<\/p>\n<p>With an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation. This will allow us to solve some simple equations.<\/p>\n<section class=\"textbox example\">\n<ol>\n<li>Solve [latex]10^{x}= 1000[\/latex]<\/li>\n<li>Solve\u00a0[latex]10^{x} = 3[\/latex]<\/li>\n<li>Solve\u00a0[latex]2(10^{x}) = 8[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q588505\">Show Solution<\/button><\/p>\n<div id=\"q588505\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Taking the log of both sides gives [latex]log(10^{x}) = log(1000)[\/latex].\n<ul>\n<li>Since the log undoes the exponential,\u00a0[latex]log(10^{x}) = x[\/latex]. Similarly\u00a0[latex]log(1000) = log(10^{3}) = 3[\/latex].<\/li>\n<li>The equation simplifies then to\u00a0[latex]x = 3[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Taking the log of both sides gives\u00a0[latex]log(10^{x}) = log(3)[\/latex].\n<ul>\n<li>On the left side,\u00a0[latex]log(10^{x}) = x[\/latex], so\u00a0[latex]x = log(3)[\/latex].<\/li>\n<li>We can approximate this value with a calculator. [latex]x \u2248 0.477[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Here we would first want to isolate the exponential by dividing both sides of the equation by\u00a0[latex]2[\/latex], giving\u00a0[latex]10^{x} = 4[\/latex].\n<ul>\n<li>Now we can take the log of both sides, giving [latex]log(10^{x}) = log(4)[\/latex], which simplifies to [latex]x = log(4) \u2248 0.602[\/latex].<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>properties of logs: exponent property<\/h3>\n<div style=\"text-align: center;\">[latex]\\log\\left({{A}^{r}}\\right)=r\\log\\left(A\\right)[\/latex]<\/div>\n<\/div>\n<\/section>\n<p>To show why this is true, we offer a proof.<\/p>\n<p>Since the logarithm and exponential undo each other, [latex]{{10}^{\\log{A}}}=A[\/latex].<\/p>\n<p style=\"text-align: center;\">So [latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}[\/latex]<\/p>\n<p>Utilizing the exponential rule that states [latex]{{\\left({{x}^{a}}\\right)}^{b}}={{x}^{ab}}[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}={{10}^{r\\log{A}}}[\/latex]<\/p>\n<p style=\"text-align: center;\">So then [latex]\\log\\left({{A}^{r}}\\right)=\\log\\left({{10}^{r\\log{A}}}\\right)[\/latex]<\/p>\n<p>Again utilizing the property that the log undoes the exponential on the right side yields the result<\/p>\n<p style=\"text-align: center;\">[latex]\\log\\left({{A}^{r}}\\right)=r\\log{A}[\/latex]<\/p>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6923\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6923&theme=lumen&iframe_resize_id=ohm6923&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":8093,"module-header":"learn_it","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\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2007"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":22,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2007\/revisions"}],"predecessor-version":[{"id":15884,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2007\/revisions\/15884"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/8093"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2007\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2007"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2007"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2007"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2007"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}