{"id":3584,"date":"2017-03-08T00:39:25","date_gmt":"2017-03-08T00:39:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=3584"},"modified":"2017-04-21T18:23:27","modified_gmt":"2017-04-21T18:23:27","slug":"expand-and-condense-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/expand-and-condense-logarithms\/","title":{"raw":"Expand and Condense Logarithms","rendered":"Expand and Condense Logarithms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Expand a logarithm\u00a0using a combination of logarithm rules<\/li>\r\n \t<li>Condense a logarithmic expression into one logarithm<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\nTaken together, the product rule, quotient rule, and power rule are often called \"laws of logs.\" Sometimes we apply more than one rule in order to simplify an expression. For example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/p>\r\nWe can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right)\\hfill &amp; ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right)\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right)\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C\\hfill \\end{array}[\/latex]<\/p>\r\nWe can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.\r\n\r\nWith practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a combination of the rules for logarithms to expand a logarithm<\/h3>\r\nRewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.\r\n\r\n[reveal-answer q=\"526416\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"526416\"]\r\n\r\nFirst, because we have a quotient of two expressions, we can use the quotient rule:\r\n\r\n[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]\r\n\r\nThen seeing the product in the first term, we use the product rule:\r\n\r\n[latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]\r\n\r\nFinally, we use the power rule on the first term:\r\n\r\n[latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"722800\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"722800\"]\r\n\r\n[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=35034&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nIn the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\r\nExpand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"914877\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"914877\"]\r\n\r\n[latex]\\begin{array}{l}\\mathrm{log}\\left(\\sqrt{x}\\right)\\hfill &amp; =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)}\\hfill \\\\ \\hfill &amp; =\\frac{1}{2}\\mathrm{log}x\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"2296\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"2296\"][latex]\\frac{2}{3}\\mathrm{ln}x[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129752&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h4>Can we expand [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/h4>\r\n<em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em>\r\n\r\n<\/div>\r\nNow we will provide\u00a0some examples that will require careful attention.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Expanding Complex Logarithmic Expressions<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"324399\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"324399\"]\r\n\r\nWe can expand by applying the Product and Quotient Rules.\r\n\r\n[latex]\\begin{array}{l}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)\\hfill &amp; ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill &amp; \\text{Apply the Quotient Rule}.\\hfill \\\\ \\hfill &amp; ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill &amp; {\\text{Simplify by writing 64 as 2}}^{6}.\\hfill \\\\ \\hfill &amp; =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill &amp; \\text{Apply the Power Rule}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It 8<\/h3>\r\nExpand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"767425\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"767425\"][latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129764&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Condense logarithmic expressions<\/h2>\r\nWe can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\r\n<ol>\r\n \t<li>Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\r\n \t<li>Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\r\n \t<li>Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Power Rule in Reverse<\/h3>\r\nRewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.\r\n\r\n[reveal-answer q=\"959478\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"959478\"]\r\n\r\nBecause the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base, and rewrite the product as a logarithm of a power:\r\n\r\n[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.\r\n\r\n[reveal-answer q=\"720709\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"720709\"][latex]{\\mathrm{log}}_{3}16[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next examples we will use a combination of logarithm rules to condense logarithms.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\r\nWrite [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.\r\n\r\n[reveal-answer q=\"484876\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"484876\"]\r\n\r\nUsing the product and quotient rules\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]\r\n\r\nThis reduces our original expression to\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]\r\n\r\nThen, using the quotient rule\r\n\r\n[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nCondense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].\r\n\r\n[reveal-answer q=\"52020\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"52020\"][latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129766&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Condensing Complex Logarithmic Expressions<\/h3>\r\nCondense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"841660\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"841660\"]\r\n\r\nWe apply the power rule first:\r\n\r\n[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]\r\n\r\nNext we apply the product rule to the sum:\r\n\r\n[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]\r\n\r\nFinally, we apply the quotient rule to the difference:\r\n\r\n[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Rewriting as a Single Logarithm<\/h3>\r\nRewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.\r\n\r\n[reveal-answer q=\"598036\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"598036\"]\r\n\r\nWe apply the power rule first:\r\n\r\n[latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]\r\n\r\nNext we apply the product rule to the sum:\r\n\r\n[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]\r\n\r\nFinally, we apply the quotient rule to the difference:\r\n\r\n[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.\r\n\r\n[reveal-answer q=\"812624\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"812624\"][latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex][\/hidden-answer]\r\n\r\nCondense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"622494\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"622494\"][latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].[\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129768&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Applications of Laws of Logarithms<\/h2>\r\nIn chemistry, pH is a measure of how acidic or basic a liquid is. \u00a0It is essentially a measure of the concentration of hydrogen ions in a solution. \u00a0The scale for measuring pH is standardized across the world, the scientific community having agreed upon it's values and methods for acquiring them.\r\n\r\nMeasurements of pH can help scientists, farmers, doctors, and engineers solve problems, and identify sources of problems.\r\n<p style=\"text-align: center;\">pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, [latex]a_{H}+[\/latex], in a solution.\r\n[latex]{pH} =-\\log _{10}(a_{{\\text{H}}^{+}})=\\log _{10}\\left({\\frac {1}{a_{{\\text{H}}^{+}}}}\\right)[\/latex]\r\nFor example, a solution with a hydrogen ion activity of [latex]2\u00d710^{-5}=\\frac{1}{(2\u00d710^5)}[\/latex] (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of [latex]\\log_{10}(2\u00d710^5)=5.3[\/latex]<\/p>\r\n<p style=\"text-align: left;\">In the next examples, we will solve some problems involving pH.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying of the Laws of Logs<\/h3>\r\nRecall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?\r\n\r\n[reveal-answer q=\"92345\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"92345\"]\r\n\r\nSuppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]\r\n\r\nUsing the product rule of logs\r\n\r\n[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]\r\n\r\nSince [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is\r\n\r\n[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]\r\n\r\nWhen the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.\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\nHow does the pH change when the concentration of positive hydrogen ions is decreased by half?\r\n\r\n[reveal-answer q=\"569571\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"569571\"]The pH increases by about 0.301.[\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129783&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Expand a logarithm\u00a0using a combination of logarithm rules<\/li>\n<li>Condense a logarithmic expression into one logarithm<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\nTaken together, the product rule, quotient rule, and power rule are often called &#8220;laws of logs.&#8221; Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/p>\n<p>We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C\\hfill \\end{array}[\/latex]<\/p>\n<p>We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\n<p>With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using a combination of the rules for logarithms to expand a logarithm<\/h3>\n<p>Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q526416\">Solution<\/span><\/p>\n<div id=\"q526416\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\n<p>[latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<p>Then seeing the product in the first term, we use the product rule:<\/p>\n<p>[latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<p>Finally, we use the power rule on the first term:<\/p>\n<p>[latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Expand [latex]\\mathrm{log}\\left(\\frac{{x}^{2}{y}^{3}}{{z}^{4}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q722800\">Solution<\/span><\/p>\n<div id=\"q722800\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]2\\mathrm{log}x+3\\mathrm{log}y - 4\\mathrm{log}z[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=35034&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>In the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression<\/h3>\n<p>Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q914877\">Solution<\/span><\/p>\n<div id=\"q914877\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\mathrm{log}\\left(\\sqrt{x}\\right)\\hfill & =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)}\\hfill \\\\ \\hfill & =\\frac{1}{2}\\mathrm{log}x\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Expand [latex]\\mathrm{ln}\\left(\\sqrt[3]{{x}^{2}}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2296\">Solution<\/span><\/p>\n<div id=\"q2296\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{2}{3}\\mathrm{ln}x[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129752&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Can we expand [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]?<\/h4>\n<p><em>No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.<\/em><\/p>\n<\/div>\n<p>Now we will provide\u00a0some examples that will require careful attention.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Expanding Complex Logarithmic Expressions<\/h3>\n<p>Expand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324399\">Solution<\/span><\/p>\n<div id=\"q324399\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can expand by applying the Product and Quotient Rules.<\/p>\n<p>[latex]\\begin{array}{l}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)\\hfill & ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & \\text{Apply the Quotient Rule}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & {\\text{Simplify by writing 64 as 2}}^{6}.\\hfill \\\\ \\hfill & =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & \\text{Apply the Power Rule}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It 8<\/h3>\n<p>Expand [latex]\\mathrm{ln}\\left(\\frac{\\sqrt{\\left(x - 1\\right){\\left(2x+1\\right)}^{2}}}{\\left({x}^{2}-9\\right)}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q767425\">Solution<\/span><\/p>\n<div id=\"q767425\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{1}{2}\\mathrm{ln}\\left(x - 1\\right)+\\mathrm{ln}\\left(2x+1\\right)-\\mathrm{ln}\\left(x+3\\right)-\\mathrm{ln}\\left(x - 3\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129764&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Condense logarithmic expressions<\/h2>\n<p>We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\n<ol>\n<li>Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\n<li>Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\n<li>Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Power Rule in Reverse<\/h3>\n<p>Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q959478\">Solution<\/span><\/p>\n<div id=\"q959478\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor, 4, as the exponent and the argument, <em>x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\n<p>[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q720709\">Solution<\/span><\/p>\n<div id=\"q720709\" class=\"hidden-answer\" style=\"display: none\">[latex]{\\mathrm{log}}_{3}16[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>In our next examples we will use a combination of logarithm rules to condense logarithms.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\n<p>Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484876\">Solution<\/span><\/p>\n<div id=\"q484876\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the product and quotient rules<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/p>\n<p>This reduces our original expression to<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/p>\n<p>Then, using the quotient rule<\/p>\n<p>[latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q52020\">Solution<\/span><\/p>\n<div id=\"q52020\" class=\"hidden-answer\" style=\"display: none\">[latex]\\mathrm{log}\\left(\\frac{3\\cdot 5}{4\\cdot 6}\\right)[\/latex]; can also be written [latex]\\mathrm{log}\\left(\\frac{5}{8}\\right)[\/latex] by reducing the fraction to lowest terms.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129766&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Condensing Complex Logarithmic Expressions<\/h3>\n<p>Condense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q841660\">Solution<\/span><\/p>\n<div id=\"q841660\" class=\"hidden-answer\" style=\"display: none\">\n<p>We apply the power rule first:<\/p>\n<p>[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\n<p>Next we apply the product rule to the sum:<\/p>\n<p>[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/p>\n<p>Finally, we apply the quotient rule to the difference:<\/p>\n<p>[latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Rewriting as a Single Logarithm<\/h3>\n<p>Rewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q598036\">Solution<\/span><\/p>\n<div id=\"q598036\" class=\"hidden-answer\" style=\"display: none\">\n<p>We apply the power rule first:<\/p>\n<p>[latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p>Next we apply the product rule to the sum:<\/p>\n<p>[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/p>\n<p>Finally, we apply the quotient rule to the difference:<\/p>\n<p>[latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\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{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812624\">Solution<\/span><\/p>\n<div id=\"q812624\" class=\"hidden-answer\" style=\"display: none\">[latex]\\mathrm{log}\\left(\\frac{5{\\left(x - 1\\right)}^{3}\\sqrt{x}}{\\left(7x - 1\\right)}\\right)[\/latex]<\/div>\n<\/div>\n<p>Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q622494\">Solution<\/span><\/p>\n<div id=\"q622494\" class=\"hidden-answer\" style=\"display: none\">[latex]\\mathrm{log}\\frac{{x}^{12}{\\left(x+5\\right)}^{4}}{{\\left(2x+3\\right)}^{4}}[\/latex]; this answer could also be written [latex]\\mathrm{log}{\\left(\\frac{{x}^{3}\\left(x+5\\right)}{\\left(2x+3\\right)}\\right)}^{4}[\/latex].<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129768&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Applications of Laws of Logarithms<\/h2>\n<p>In chemistry, pH is a measure of how acidic or basic a liquid is. \u00a0It is essentially a measure of the concentration of hydrogen ions in a solution. \u00a0The scale for measuring pH is standardized across the world, the scientific community having agreed upon it&#8217;s values and methods for acquiring them.<\/p>\n<p>Measurements of pH can help scientists, farmers, doctors, and engineers solve problems, and identify sources of problems.<\/p>\n<p style=\"text-align: center;\">pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, [latex]a_{H}+[\/latex], in a solution.<br \/>\n[latex]{pH} =-\\log _{10}(a_{{\\text{H}}^{+}})=\\log _{10}\\left({\\frac {1}{a_{{\\text{H}}^{+}}}}\\right)[\/latex]<br \/>\nFor example, a solution with a hydrogen ion activity of [latex]2\u00d710^{-5}=\\frac{1}{(2\u00d710^5)}[\/latex] (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of [latex]\\log_{10}(2\u00d710^5)=5.3[\/latex]<\/p>\n<p style=\"text-align: left;\">In the next examples, we will solve some problems involving pH.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Applying of the Laws of Logs<\/h3>\n<p>Recall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q92345\">Solution<\/span><\/p>\n<div id=\"q92345\" class=\"hidden-answer\" style=\"display: none\">\n<p>Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/p>\n<p>Using the product rule of logs<\/p>\n<p>[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/p>\n<p>Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\n<p>[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/p>\n<p>When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>How does the pH change when the concentration of positive hydrogen ions is decreased by half?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q569571\">Solution<\/span><\/p>\n<div id=\"q569571\" class=\"hidden-answer\" style=\"display: none\">The pH increases by about 0.301.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129783&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3584\">\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>REvision and Adaptation. <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>pH. <strong>Authored by<\/strong>: Wikipedia. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/en.wikipedia.org\/wiki\/PH\">https:\/\/en.wikipedia.org\/wiki\/PH<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 35034. <strong>Authored by<\/strong>: Smart,Jim. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 129752, 129768, 129766, 129783. <strong>Authored by<\/strong>: Day,Alyson. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/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":21,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"pH\",\"author\":\"Wikipedia\",\"organization\":\"\",\"url\":\"https:\/\/en.wikipedia.org\/wiki\/PH\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"REvision and Adaptation\",\"author\":\"\",\"organization\":\"lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 35034\",\"author\":\"Smart,Jim\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 129752, 129768, 129766, 129783\",\"author\":\"Day,Alyson\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"a6718ccc-8694-4a98-8d1d-9b5a753b5ae6","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3584","chapter","type-chapter","status-publish","hentry"],"part":2074,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3584","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":14,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3584\/revisions"}],"predecessor-version":[{"id":4282,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3584\/revisions\/4282"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/2074"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3584\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=3584"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=3584"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=3584"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=3584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}