{"id":2209,"date":"2022-05-19T17:47:23","date_gmt":"2022-05-19T17:47:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=2209"},"modified":"2026-01-22T19:59:01","modified_gmt":"2026-01-22T19:59:01","slug":"5-3-2-properties-of-exponents-with-zero-and-negative-exponents","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/5-3-2-properties-of-exponents-with-zero-and-negative-exponents\/","title":{"raw":"5.3.2: Properties of Exponents with Zero and Negative Exponents","rendered":"5.3.2: Properties of Exponents with Zero and Negative Exponents"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n\r\n<!-- Le HTML5 shim, for IE6-8 support of HTML5 elements --><!-- [if lt IE 9]>\r\n<script src=\"https:\/\/html5shim.googlecode.com\/svn\/trunk\/html5.js\">\r\n\t<\/script>\r\n<![endif]-->\r\n<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-254\" class=\"standard post-254 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Explain the meaning of a zero exponent<\/li>\r\n \t<li>Explain the meaning of a negative exponent<\/li>\r\n \t<li>Apply the power of product rule for exponents<\/li>\r\n \t<li>Apply the power of quotient rule for exponents<\/li>\r\n \t<li>Simplify exponential expressions<\/li>\r\n \t<li>Use properties of exponents to write equivalent exponential functions in standard form<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Zero Exponents<\/h2>\r\nThe quotient rule for exponents can be used to determine the meaning of [latex]x^0[\/latex]:\r\n\r\nThe\u00a0quotient rule for exponents tells us that by subtracting the exponents:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x^n}{x^n}=x^0[\/latex]<\/p>\r\nBut since the numerator and denominator are identical, we can cancel the terms by division:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x^n}{x^n}=1[\/latex] provided\u00a0[latex]x\\neq0[\/latex] since we can't divide by 0.<\/p>\r\nTherefore,\r\n<p style=\"text-align: center;\">[latex]x^0=1[\/latex],\u00a0[latex]x\\neq0[\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\" style=\"text-align: center;\">\r\n<h3>exponent of zero<\/h3>\r\nFor all real numbers [latex]a\\neq0[\/latex],\r\n<p style=\"text-align: center;\">[latex]a^0=1[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\nFor example,\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2022^0&amp;=1\\\\(pq)^0&amp;=1,\\;p,\\;q\\neq 0 \\\\(2050xy)^{0}&amp;=1,\\;x,\\;y\\neq 0\\end{aligned}[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n<div style=\"text-align: left;\">The sole exception is the expression [latex]{0}^{0}[\/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined.<\/div>\r\n&nbsp;\r\n<div style=\"text-align: left;\"><strong>TIP FOR SIMPLIFYING EXPONENTIAL EXPRESSIONS<\/strong><\/div>\r\n<div style=\"text-align: left;\">When simplifying expressions with exponents, it is sometimes helpful to rely on the rule for multiplying fractions to separate the factors before doing work on them. For example, to simplify the expression\u00a0[latex]\\dfrac{5 a^m z^2}{a^mz}[\/latex]\u00a0using exponent rules, it may be helpful to break the fraction up into a product of fractions, then simplify:<\/div>\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{5 a^m z^2}{a^mz}&amp;=5\\cdot\\dfrac{a^m}{a^m}\\cdot\\dfrac{z^2}{z}&amp;&amp;\\text{Separate into fractions}\\\\&amp;=5 \\cdot a^{m-m}\\cdot z^{2-1}&amp;&amp;\\text{Subtract the exponents}\\\\&amp;=5\\cdot a^0 \\cdot z^1 &amp;&amp;\\text{Simplify }a^0=1\\text{ and }z^1=z\\\\&amp;=5z\\end{aligned}[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nSimplify each expression.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{c}^{3}}{{c}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-3{x}^{5}}{{x}^{5}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\nWe can apply the zero exponent rule and other rules to simplify each expression:\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 100%;\">\r\n<p style=\"text-align: initial; font-size: 1rem;\">Applying the rules of exponents<\/p>\r\n<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 100%;\">\r\n<p style=\"text-align: initial; font-size: 1rem;\">1.<\/p>\r\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align}\\frac{c^{3}}{c^{3}}&amp;=c^{3-3}&amp;&amp;\\text{Apply the quotient rule: subtract exponents} \\\\ &amp; =c^{0}&amp;&amp;\\text{Apply the zero exponent rule} \\\\ &amp; =1\\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 100%;\">\r\n<p style=\"text-align: initial; font-size: 1rem;\">2.<\/p>\r\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align} \\frac{-3{x}^{5}}{{x}^{5}}&amp; = -3\\cdot \\frac{{x}^{5}}{{x}^{5}} \\\\ &amp; = -3\\cdot {x}^{5 - 5}&amp;&amp;\\text{Apply the quotient rule: subtract exponents} \\\\ &amp; = -3\\cdot {x}^{0}&amp;&amp;\\text{Apply the zero exponent rule}\\\\ &amp; = -3\\cdot 1 \\\\ &amp; = -3 \\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 100%;\">\r\n<p style=\"text-align: initial; font-size: 1rem;\">3.<\/p>\r\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align} \\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}&amp; = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{1+3}} &amp;&amp; \\text{Use the product rule in the denominator. The base is }(j^2k). \\\\ &amp; = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}} &amp;&amp; \\text{Use the quotient rule}. \\\\ &amp; = {\\left({j}^{2}k\\right)}^{4 - 4} \\\\ &amp; = {\\left({j}^{2}k\\right)}^{0} &amp;&amp; \\text{Use the zero exponent rule}. \\\\ &amp; = 1 \\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 100%;\">\r\n<p style=\"text-align: initial; font-size: 1rem;\">4.<\/p>\r\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align} \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}&amp; = 5{\\left(r{s}^{2}\\right)}^{2 - 2}&amp;&amp;\\text{Use the quotient rule}. \\\\ &amp; = 5{\\left(r{s}^{2}\\right)}^{0}&amp;&amp;\\text{Use the zero exponent rule}. \\\\ &amp; = 5\\cdot 1 \\\\ &amp; = 5 \\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nSimplify each expression using the zero exponent rule of exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{t}^{7}}{{t}^{7}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(d{e}^{2}\\right)}^{11}}{2{\\left(d{e}^{2}\\right)}^{11}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{w}^{4}\\cdot {w}^{2}}{{w}^{6}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-5{t}^{3}\\cdot {t}^{4}}{{t}^{2}\\cdot {t}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"q703483\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"q703483\"]\r\n<ol>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]-5[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Negative Exponents<\/span>\r\n\r\nThe quotient rule for exponents can also be used to determine what it means to have a negative exponent [latex]x^{-n}[\/latex]. If\u00a0[latex]m&lt;n,\\;m-n&lt;0[\/latex], so\u00a0[latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex] will have a negative exponent.\r\n\r\nConsider, for example,\u00a0[latex]\\dfrac{{x}^{2}}{{x}^{4}}={x}^{2-4}=x^{-2}[\/latex]. We have a negative exponent, but what does that mean?\r\n\r\nAnother way to simplify\u00a0[latex]\\dfrac{{x}^{2}}{{x}^{4}}[\/latex] is to expand the numerator and denominator then cancel common factors:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x^{2}}{x^{4}} &amp;=\\frac{x\\cdot x}{x\\cdot x\\cdot x\\cdot x} \\\\[1mm] &amp;=\\frac{\\cancel{x}\\cdot\\cancel{x}}{\\cancel{x}\\cdot\\cancel{x}\\cdot x\\cdot x} \\\\[1mm] &amp; =\\frac{1}{x^2}\\end{align}[\/latex]<\/p>\r\nConsequently, [latex]x^{-2}=\\dfrac{1}{x^2}[\/latex].\r\n\r\nA factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar\u2014from numerator to denominator. This can be generalized to [latex]a^{-n}=\\dfrac{1}{a^n}[\/latex].\r\n\r\nIf the negative exponent is on the denominator, [latex]\\dfrac{1}{x^{-n}}[\/latex], we can use division of fractions to simplify it:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{1}{x^{-n}}&amp;=1\\div x^{-n}\\\\&amp;=1\\div \\dfrac{1}{x^n}\\\\&amp;=1\\times \\dfrac{x^n}{1}\\\\&amp;=x^n\\end{aligned}[\/latex]<\/p>\r\n\r\n<div style=\"text-align: left;\">A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar\u2014from denominator to\u00a0<span style=\"font-size: 1em;\">numerator<\/span><span style=\"font-size: 1em;\">.<\/span><\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>NEGATIVE EXPONENTS<\/h3>\r\nFor any real numbers [latex]a\\neq0[\/latex] and [latex]n[\/latex],\r\n<p style=\"text-align: center;\">[latex]a^{-n}=\\dfrac{1}{a^n}[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{{a}^{-n}}=a^n[\/latex]<\/p>\r\nA factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nSimplify the expressions. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{x^3}{x^{10}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{z^2\\cdot z}{z^4}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(-5t^3\\right)}^4}{\\left(-5t^3\\right)^8}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\"><span style=\"font-size: 16px; orphans: 1;\">Applying the rules for exponents<\/span>\r\n<span style=\"font-size: 16px; orphans: 1;\">\r\n<\/span><\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\">[latex]1.\\\\ \\begin{aligned}\\dfrac{x^3}{x^{10}}&amp;=x^{3 - 10}&amp;&amp;\\text{Quotient rule}\\\\&amp;=x^{-7}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{1}{x^7}\\end{aligned}[\/latex]<\/span><\/td>\r\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\">[latex]2.\\\\ \\begin{aligned}\\dfrac{z^2\\cdot z}{z^4}&amp;=\\dfrac{z^{2+1}}{z^4}&amp;&amp;\\text{Product rule}\\\\&amp;=\\dfrac{z^3}{z^4}&amp;&amp;\\text{Quotient rule}\\\\&amp;={z}^{3 - 4}\\\\&amp;={z}^{-1}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{1}{z}\\end{aligned}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\"><span style=\"font-size: 16px; orphans: 1;\">[latex]3.\\\\ \\begin{aligned}\\dfrac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}&amp;={\\left(-5{t}^{3}\\right)}^{4 - 8}&amp;&amp;\\text{Product rule: the base is }-5t^3\\\\&amp;={\\left(-5{t}^{3}\\right)}^{-4}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{1}{{\\left(-5{t}^{3}\\right)}^{4}}\\end{aligned}[\/latex]<\/span><\/span><\/td>\r\n<td style=\"width: 50%; vertical-align: top;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nSimplify the expressions. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{\\left(-3t\\right)}^{2}}{{\\left(-3t\\right)}^{8}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{f}^{47}}{{f}^{49}\\cdot f}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{2{k}^{4}}{5{k}^{7}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{\\left(-3x^4\\right)^5}{\\left(-3x^4\\right)^{12}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{5y^{-8}}{y^{-6}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"q85205\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"q85205\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{1}{{\\left(-3t\\right)}^{6}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{f}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{2}{5{k}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{\\left(-3x^4\\right)^7}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{5}{y^2}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{b}^{2}\\cdot {b}^{-8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\"><span style=\"font-size: 16px; orphans: 1;\">Applying the rules for exponents<\/span><\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\">1.<\/span>\r\n\r\n[latex]\\begin{aligned}{b}^{2}\\cdot {b}^{-8}&amp;={b}^{2 + (-8)}&amp;&amp;\\text{Product rule}\\\\&amp;={b}^{-6}\\\\&amp;=\\frac{1}{{b}^{6}}&amp;&amp;\\text{Negative exponent rule}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%; vertical-align: top;\">2.\r\n\r\n[latex]\\begin{aligned}{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}&amp;={\\left(-x\\right)}^{5+(-5)}&amp;&amp;\\text{Product rule}\\\\&amp;={\\left(-x\\right)}^{0}\\\\&amp;=1&amp;&amp;\\text{Zero exponent rule}\\end{aligned}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">3.\r\n\r\n[latex]\\begin{aligned}\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}&amp;=\\dfrac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}\\\\&amp;={\\left(-7z\\right)}^{1 - 5}&amp;&amp;\\text{Quotient rule}\\\\&amp;={\\left(-7z\\right)}^{-4}\\\\&amp;=\\dfrac{1}{{\\left(-7z\\right)}^{4}}&amp;&amp;\\text{Negative exponent rule}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%; vertical-align: top;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nSimplify. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{t}^{-11}\\cdot {t}^{6}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{25}^{12}}{{25}^{13}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"q592096\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"q592096\"]\r\n<ol>\r\n \t<li>[latex]{t}^{-5}=\\dfrac{1}{{t}^{5}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{25}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Power of a Product Rule<\/h2>\r\nTo simplify the power of a product of two exponential expressions, we can use the\u00a0<em><strong>power of a product rule of exponents<\/strong>,\u00a0<\/em>which breaks up the power of a product of factors into the product of the powers of the factors. For example, consider [latex]{\\left(pq\\right)}^{3}[\/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors:\r\n<div style=\"text-align: center;\">[latex]\\begin{align} {\\left(pq\\right)}^{3}&amp; = \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}} \\\\ &amp; = p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q \\\\&amp; = \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}} \\\\ &amp; = {p}^{3}\\cdot {q}^{3} \\end{align}[\/latex]<\/div>\r\nIn other words, [latex]{\\left(pq\\right)}^{3}={p}^{3}\\cdot {q}^{3}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>The Power of a Product Rule<\/h3>\r\nFor any real numbers [latex]a,\\;b[\/latex] and [latex]n[\/latex], the power of a product rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]\\large{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nSimplify each of the following products as much as possible. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(2t\\right)}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\nWe can use the product and quotient rules and the new definitions to simplify each expression. If a number is raised to a power, we can evaluate it.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%;\" colspan=\"2\">Applying the rules for exponents<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">1.\r\n\r\n[latex]\\begin{aligned}{\\left(a{b}^{2}\\right)}^{3}&amp;={\\left(a\\right)}^{3}\\cdot {\\left({b}^{2}\\right)}^{3}&amp;&amp;\\text{Power of a product rule}\\\\&amp;={a}^{1\\cdot 3}\\cdot {b}^{2\\cdot 3}\\\\&amp;={a}^{3}{b}^{6}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%;\">2.\r\n\r\n[latex]\\begin{aligned}(2{t})^{15}&amp;={\\left(2\\right)}^{15}\\cdot {\\left(t\\right)}^{15}&amp;&amp;\\text{Power of a product rule}\\\\&amp;={2}^{15}{t}^{15}&amp;&amp;\\text{Evaluate }2^{15}\\text{ using a calculator}\\\\&amp;=32,768{t}^{15}\\end{aligned}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">3.\r\n\r\n[latex]\\begin{aligned}{\\left(-2{w}^{3}\\right)}^{3}&amp;={\\left(-2\\right)}^{3}\\cdot {\\left({w}^{3}\\right)}^{3}&amp;&amp;\\text{Power of a product rule}\\\\&amp;=-8\\cdot {w}^{3\\cdot 3}&amp;&amp;{(-2)}^3=-8\\\\&amp;=-8{w}^{9}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%;\">4.\r\n\r\n[latex]\\begin{aligned}\\dfrac{1}{{\\left(-7z\\right)}^{4}}&amp;=\\dfrac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}&amp;&amp;\\text{Power of a product rule}\\\\&amp;=\\dfrac{1}{2,401{z}^{4}}&amp;&amp;{(-7)}^4\\text{ is evaluated}\\end{aligned}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">5.\r\n\r\n[latex]\\begin{aligned}{\\left({e}^{-2}{f}^{2}\\right)}^{7}&amp;={\\left({e}^{-2}\\right)}^{7}\\cdot {\\left({f}^{2}\\right)}^{7}&amp;&amp;\\text{Power of a product rule}\\\\&amp;={e}^{-2\\cdot 7}\\cdot {f}^{2\\cdot 7}\\\\&amp;={e}^{-14}{f}^{14}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{{f}^{14}}{{e}^{14}}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nSimplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left({g}^{2}{h}^{3}\\right)}^{5}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(5t\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-3{y}^{5}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{\\left({a}^{6}{b}^{7}\\right)}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({r}^{3}{s}^{-2}\\right)}^{4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"q540817\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"q540817\"]\r\n<ol>\r\n \t<li>[latex]{g}^{10}{h}^{15}[\/latex]<\/li>\r\n \t<li>[latex]125{t}^{3}[\/latex]<\/li>\r\n \t<li>[latex]-27{y}^{15}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{a}^{18}{b}^{21}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{r}^{12}}{{s}^{8}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Power of a Quotient Rule<\/h2>\r\nTo simplify the power of a quotient of two expressions, we can use the\u00a0<em><strong>power of a quotient rule<\/strong>,\u00a0<\/em>which states that the power of a quotient of factors is the quotient of the powers of the factors.The power of a quotient rule is an extension of the power of a product rule since a quotient can be written as a product:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(\\dfrac{a}{b}\\right)^n&amp;=\\left(a\\times\\dfrac{1}{b}\\right)^n\\\\&amp;=a^n\\times \\left(b^{-1}\\right)^n\\\\&amp;=a^n\\times b^{-n}\\\\&amp;=a^n\\times \\dfrac{1}{b^n}\\\\&amp;=\\dfrac{a^n}{b^n}\\end{aligned}[\/latex].<\/p>\r\n\r\n<div style=\"text-align: center;\"><\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>The Power of a Quotient Rule<\/h3>\r\nFor any real numbers [latex]a,\\;b[\/latex] and [latex]n[\/latex], provided [latex]b\\neq0[\/latex], the power of a quotient rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]\\large{\\left(\\dfrac{a}{b}\\right)}^{n}=\\dfrac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nSimplify each of the following quotients as much as possible. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%;\" colspan=\"2\">Applying the rules for exponents<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">1.\r\n\r\n[latex]\\begin{aligned}{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}&amp;=\\dfrac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}&amp;&amp;\\text{Quotient to a power rule}\\\\&amp;=\\dfrac{64}{{z}^{11\\cdot 3}}&amp;&amp;\\text{Evaluate }4^3=64\\text{. Power to a power rule.} \\\\&amp;=\\dfrac{64}{{z}^{33}}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%;\">2.\r\n\r\n[latex]\\begin{aligned}{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}&amp;=\\dfrac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}&amp;&amp;\\text{Power of a quotient rule}\\\\&amp;=\\dfrac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}&amp;&amp;\\text{Power to a power rule}\\\\&amp;=\\dfrac{{p}^{6}}{{q}^{18}}\\end{aligned}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">3.\r\n\r\n[latex]\\begin{aligned}{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}&amp;=\\dfrac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}&amp;&amp;\\text{Power of a quotient rule}\\\\&amp;=\\dfrac{-1}{{t}^{2\\cdot 27}}&amp;&amp;\\text{Power of a power rule}\\\\&amp;=\\dfrac{-1}{{t}^{54}}&amp;&amp;\\text{Put the negative sign in front of the fraction}\\\\&amp;=-\\dfrac{1}{{t}^{54}}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%;\">4.\r\n\r\n[latex]\\begin{aligned}{\\left({j}^{3}{k}^{-2}\\right)}^{4}&amp;={\\left(\\dfrac{{j}^{3}}{{k}^{2}}\\right)}^{4}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}&amp;&amp;\\text{Power of a quotient rule}\\\\&amp;=\\dfrac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}&amp;&amp;\\text{Power to a power rule}\\\\&amp;=\\dfrac{{j}^{12}}{{k}^{8}}\\end{aligned}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">5.\r\n\r\n[latex]\\begin{aligned}{\\left({m}^{-2}{n}^{-2}\\right)}^{3}&amp;={\\left(\\dfrac{1}{{m}^{2}{n}^{2}}\\right)}^{3}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}&amp;&amp;\\text{Power of a quotient rule}\\\\&amp;=\\dfrac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}&amp;&amp;\\text{Evaluate }1^3=1\\\\&amp;=\\dfrac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}&amp;&amp;\\text{Power to a power rule}\\\\&amp;=\\dfrac{1}{{m}^{6}{n}^{6}}\\end{aligned}[\/latex]<\/td>\r\n<td style=\"width: 50%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nSimplify each of the following quotients as much as possible. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left(\\dfrac{{b}^{5}}{c}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{5}{{u}^{8}}\\right)}^{4}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{-1}{{w}^{3}}\\right)}^{35}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({p}^{-4}{q}^{3}\\right)}^{8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({c}^{-5}{d}^{-3}\\right)}^{4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"q815731\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"q815731\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{{b}^{15}}{{c}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{625}{{u}^{32}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-1}{{w}^{105}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{q}^{24}}{{p}^{32}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{c}^{20}{d}^{12}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Simplifying Exponential Expressions<\/h2>\r\nRecall that to simplify an expression means to rewrite it by combining terms or exponents. Evaluating an expression means to get a numerical answer. The rules for exponents can be combined to simplify expressions.\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nSimplify each expression and write the answer with positive exponents only.\r\n<ol>\r\n \t<li>[latex]{\\left(6{m}^{2}{n}^{-1}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-2{a}^{3}{b}^{-1}\\right)\\left(5{a}^{-2}{b}^{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n<table style=\"border-collapse: collapse; width: 99.0613%; height: 850px;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\">\r\n<p style=\"text-align: initial;\">Applying the rules for exponents<\/p>\r\n<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">\r\n<p style=\"text-align: initial;\">1.<\/p>\r\n<p style=\"text-align: initial;\">[latex]\\begin{align} {\\left(6{m}^{2}{n}^{-1}\\right)}^{3}&amp; = {\\left(6\\right)}^{3}{\\left({m}^{2}\\right)}^{3}{\\left({n}^{-1}\\right)}^{3}&amp;&amp; \\text{Power of a product rule} \\\\ &amp; = {6}^{3}{m}^{2\\cdot 3}{n}^{-1\\cdot 3}&amp;&amp; \\text{Power rule} \\\\ &amp; = 216{m}^{6}{n}^{-3}&amp;&amp;\\text{Evaluate: }2^6=216\\text{ and simplify}. \\\\ &amp; = \\frac{216{m}^{6}}{{n}^{3}}&amp;&amp; \\text{Negative exponent rule} \\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<td style=\"width: 50%; vertical-align: top;\">\r\n<p style=\"text-align: initial;\">2.<\/p>\r\n<p style=\"text-align: initial;\">[latex]\\begin{align} {17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}&amp;=17^{5 +(-4)+(-3)}&amp;&amp; \\text{Product rule} \\\\ &amp; = {17}^{-2}&amp;&amp;\\text{Negative exponent rule}\\\\&amp;=\\dfrac{1}{{17}^{2}}&amp;&amp;\\text{Evaluate}\\\\&amp;=\\dfrac{1}{289}\\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">\r\n<p style=\"text-align: initial;\">3.<\/p>\r\n<p style=\"text-align: initial;\">[latex]\\begin{align} {\\left(\\dfrac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}&amp;=\\dfrac{{\\left({u}^{-1}v\\right)}^{2}}{{\\left({v}^{-1}\\right)}^{2}}&amp;&amp;\\text{Power of a quotient rule}\\\\&amp;=\\dfrac{{u}^{-2}{v}^{2}}{{v}^{-2}}&amp;&amp;\\text{Power of a product rule}\\\\&amp;={u}^{-2}{v}^{2-(-2)}&amp;&amp;\\text{Quotient rule}\\\\&amp;={u}^{-2}{v}^{4}&amp;&amp;\\text{Evaluate: }2-(-2)=2+2=4\\\\&amp;=\\dfrac{{v}^{4}}{{u}^{2}}&amp;&amp;\\text{Negative exponent rule}\\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<td style=\"width: 50%; vertical-align: top;\">\r\n<p style=\"text-align: initial;\">4.<\/p>\r\n<p style=\"text-align: initial;\">[latex]\\begin{align}\\left(-2a^3b^{-1}\\right)\\left(5a^{-2}b^2 \\right)&amp;=\\left(-2\\cdot 5\\right)\\left({a}^{3}\\cdot {a}^{-2}\\right)\\left( {b}^{-1}\\cdot {b}^{2}\\right)&amp;&amp; \\text{Commutative\/associative properties} \\\\ &amp; = -10{a}^{3+(-2)}{b}^{-1+2}&amp;&amp; \\text{Product rule} \\\\ &amp; = -10a^1b^1\\\\&amp;=-10ab \\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">\r\n<p style=\"text-align: initial;\">5.<\/p>\r\n<p style=\"text-align: initial;\">[latex]\\begin{align} {\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}&amp; = {\\left({x}^{2}\\sqrt{2}\\right)}^{4+(-4)} &amp;&amp; \\text{Product rule: base is }(x^2\\sqrt{2}) \\\\ &amp; = {\\left({x}^{2}\\sqrt{2}\\right)}^{0}&amp;&amp; \\text{Zero exponent rule}\\\\ &amp; = 1 \\end{align}[\/latex]<\/p>\r\n<\/td>\r\n<td style=\"width: 50%; vertical-align: top;\">\r\n<p style=\"text-align: initial;\">6.<\/p>\r\n<span style=\"text-align: initial;\">[latex]\\begin{align} \\frac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}&amp; = \\frac{{\\left(3\\right)}^{5}\\cdot {\\left({w}^{2}\\right)}^{5}}{{\\left(6\\right)}^{2}\\cdot {\\left({w}^{-2}\\right)}^{2}}&amp;&amp; \\text{Power of a product rule} \\\\ &amp; = \\frac{{3}^{5}{w}^{2\\cdot 5}}{{6}^{2}{w}^{-2\\cdot 2}}&amp;&amp; \\text{Power rule} \\\\ &amp; = \\frac{243{w}^{10}}{36{w}^{-4}} &amp;&amp; \\text{Evaluate: }3^5=243\\text{ and }6^2=36 \\\\ &amp; = \\frac{27{w}^{10-\\left(-4\\right)}}{4}&amp;&amp; \\text{Quotient rule and simplify fraction} \\\\ &amp; = \\frac{27{w}^{14}}{4}\\end{align}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 6<\/h3>\r\nSimplify each expression and write the answer with positive exponents only.\r\n<ol>\r\n \t<li>[latex]{\\left(2u{v}^{-2}\\right)}^{-3}[\/latex]<\/li>\r\n \t<li>[latex]{x}^{8}\\cdot {x}^{-12}\\cdot x[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{{e}^{2}{f}^{-3}}{{f}^{-1}}\\right)}^{2}[\/latex]<\/li>\r\n \t<li>[latex]\\left(9{r}^{-5}{s}^{3}\\right)\\left(3{r}^{6}{s}^{-4}\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{4}{9}t{w}^{-2}\\right)}^{-3}{\\left(\\dfrac{4}{9}t{w}^{-2}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(2{h}^{2}k\\right)}^{4}}{{\\left(7{h}^{-1}{k}^{2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"31244\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"31244\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{{v}^{6}}{8{u}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{x}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{e}^{4}}{{f}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{27r}{s}[\/latex]<\/li>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{16{h}^{10}}{49}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>TRY IT 7<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43896&amp;amp;theme=oea&amp;amp;iframe_resize_id=mom95[\/embed] title= \"Interactive problem\"\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 8<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14060&amp;amp;theme=oea&amp;amp;iframe_resize_id=mom90[\/embed] title= \"Interactive problem\"\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 9<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14056&amp;theme=oea&amp;iframe_resize_id=mom70[\/embed] title= \"Interactive problem\"\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 10<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14057&amp;theme=oea&amp;iframe_resize_id=mom80[\/embed] title= \"Interactive problem\"\r\n\r\n<\/div>\r\n<h2>Simplifying Exponential Functions<\/h2>\r\nEach of the properties of exponents can be used to simplify and write equivalent exponential functions. For example, the function [latex]f(x)=2^{x+3}[\/latex] can be written as the equivalent exponential function [latex]f(x)=8\\left(2^x\\right)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=2^{x+3}\\\\&amp;=2^x\\cdot 2^3\\\\&amp;=2^x\\cdot 8\\\\&amp;=8\\left(2^x\\right)\\end{aligned}[\/latex]<\/p>\r\nBeing able to simplify an exponential function into the standard form [latex]f(x)=ar^{x-h}+k[\/latex] makes the function easier to graph and allows us to determine the transformations that were made from the parent function [latex]f(x)=r^x[\/latex]. In the case of\u00a0[latex]f(x)=2^{x+3}=8\\left(2^x\\right)[\/latex], the 8 tells us that the graph of [latex]f(x)=2^x[\/latex] has been stretched by a factor of 8.\r\n<div class=\"textbox examples\">\r\n<h3>Example 7<\/h3>\r\n<span class=\"TextRun SCXW130104365 BCX9\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"auto\"><span class=\"NormalTextRun SCXW130104365 BCX9\">Use exponential rules to <\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">write equivalent<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\"> exponential functions<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">.<\/span>\u00a0<\/span>\r\n<ol>\r\n \t<li>[latex]f(x)=3^{x+2}[\/latex]<\/li>\r\n \t<li>[latex]g(x)=5^{2x+1}[\/latex]<\/li>\r\n \t<li>[latex]h(x)=4^{3x-2}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n1.\r\n\r\n[latex]\\begin{aligned}f(x)&amp;=3^{x+2}\\\\&amp;=3^x\\cdot 3^2&amp;&amp;\\text{Product rule (in reverse)}\\\\&amp;=3^x\\cdot 9&amp;&amp;\\text{Evaluate }3^2=9\\\\&amp;=9\\left(3^x\\right)&amp;&amp;\\text{Write in standard form}\\end{aligned}[\/latex]\r\n\r\n2.\r\n\r\n[latex]\\begin{aligned}g(x)&amp;=5^{2x+1}\\\\&amp;=5^{2x}\\cdot 5^1&amp;&amp;\\text{Product rule (in reverse)}\\\\&amp;=\\left(5^2\\right)^x\\cdot 5&amp;&amp;\\text{Evaluate }5^1=5\\\\&amp;=5\\left(25^x\\right)&amp;&amp;\\text{Write in standard form}\\end{aligned}[\/latex]\r\n\r\n3.\r\n\r\n[latex]\\begin{aligned}h(x)&amp;=4^{3x-2}\\\\&amp;=4^{3x}\\cdot 4^{-2}&amp;&amp;\\text{Product rule (in reverse)}\\\\&amp;=\\left(4^3\\right)^x\\cdot \\dfrac{1}{4^2}&amp;&amp;\\text{Power to a power rule (in reverse) and negative exponent rule}\\\\&amp;=64^x\\cdot \\dfrac{1}{16}&amp;&amp;\\text{Evaluate: }4^3=64\\text{ and }4^2=16\\\\&amp;=\\dfrac{1}{16}\\left(64^x\\right)&amp;&amp;\\text{Write in standard form}\\end{aligned}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 11<\/h3>\r\n<span class=\"TextRun SCXW130104365 BCX9\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"auto\"><span class=\"NormalTextRun SCXW130104365 BCX9\">Use exponential rules to <\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">write equivalent<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\"> exponential functions<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">.<\/span>\u00a0<\/span>\r\n<ol>\r\n \t<li>[latex]f(x)=2^{x+4}[\/latex]<\/li>\r\n \t<li>[latex]g(x)=3^{x-2}[\/latex]<\/li>\r\n \t<li>[latex]h(x)=5^{2x-1}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm292\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm292\"]\r\n<ol>\r\n \t<li>[latex]f(x)=16\\left(2^x\\right)[\/latex]<\/li>\r\n \t<li>[latex]g(x)=\\dfrac{1}{9}\\left(3^x\\right)[\/latex]<\/li>\r\n \t<li>[latex]h(x)=\\dfrac{1}{5}\\left(25^{x}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<p><!-- Le HTML5 shim, for IE6-8 support of HTML5 elements --><!-- [if lt IE 9]>\n<script src=\"https:\/\/html5shim.googlecode.com\/svn\/trunk\/html5.js\">\n\t<\/script>\n<![endif] --><\/p>\n<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-254\" class=\"standard post-254 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Explain the meaning of a zero exponent<\/li>\n<li>Explain the meaning of a negative exponent<\/li>\n<li>Apply the power of product rule for exponents<\/li>\n<li>Apply the power of quotient rule for exponents<\/li>\n<li>Simplify exponential expressions<\/li>\n<li>Use properties of exponents to write equivalent exponential functions in standard form<\/li>\n<\/ul>\n<\/div>\n<h2>Zero Exponents<\/h2>\n<p>The quotient rule for exponents can be used to determine the meaning of [latex]x^0[\/latex]:<\/p>\n<p>The\u00a0quotient rule for exponents tells us that by subtracting the exponents:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x^n}{x^n}=x^0[\/latex]<\/p>\n<p>But since the numerator and denominator are identical, we can cancel the terms by division:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x^n}{x^n}=1[\/latex] provided\u00a0[latex]x\\neq0[\/latex] since we can&#8217;t divide by 0.<\/p>\n<p>Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]x^0=1[\/latex],\u00a0[latex]x\\neq0[\/latex]<\/p>\n<div class=\"textbox shaded\" style=\"text-align: center;\">\n<h3>exponent of zero<\/h3>\n<p>For all real numbers [latex]a\\neq0[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]a^0=1[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p>For example,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2022^0&=1\\\\(pq)^0&=1,\\;p,\\;q\\neq 0 \\\\(2050xy)^{0}&=1,\\;x,\\;y\\neq 0\\end{aligned}[\/latex]<\/p>\n<div><\/div>\n<div style=\"text-align: left;\">The sole exception is the expression [latex]{0}^{0}[\/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined.<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: left;\"><strong>TIP FOR SIMPLIFYING EXPONENTIAL EXPRESSIONS<\/strong><\/div>\n<div style=\"text-align: left;\">When simplifying expressions with exponents, it is sometimes helpful to rely on the rule for multiplying fractions to separate the factors before doing work on them. For example, to simplify the expression\u00a0[latex]\\dfrac{5 a^m z^2}{a^mz}[\/latex]\u00a0using exponent rules, it may be helpful to break the fraction up into a product of fractions, then simplify:<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{5 a^m z^2}{a^mz}&=5\\cdot\\dfrac{a^m}{a^m}\\cdot\\dfrac{z^2}{z}&&\\text{Separate into fractions}\\\\&=5 \\cdot a^{m-m}\\cdot z^{2-1}&&\\text{Subtract the exponents}\\\\&=5\\cdot a^0 \\cdot z^1 &&\\text{Simplify }a^0=1\\text{ and }z^1=z\\\\&=5z\\end{aligned}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Simplify each expression.<\/p>\n<ol>\n<li>[latex]\\dfrac{{c}^{3}}{{c}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{-3{x}^{5}}{{x}^{5}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>We can apply the zero exponent rule and other rules to simplify each expression:<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 100%;\">\n<p style=\"text-align: initial; font-size: 1rem;\">Applying the rules of exponents<\/p>\n<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\">\n<p style=\"text-align: initial; font-size: 1rem;\">1.<\/p>\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align}\\frac{c^{3}}{c^{3}}&=c^{3-3}&&\\text{Apply the quotient rule: subtract exponents} \\\\ & =c^{0}&&\\text{Apply the zero exponent rule} \\\\ & =1\\end{align}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\">\n<p style=\"text-align: initial; font-size: 1rem;\">2.<\/p>\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align} \\frac{-3{x}^{5}}{{x}^{5}}& = -3\\cdot \\frac{{x}^{5}}{{x}^{5}} \\\\ & = -3\\cdot {x}^{5 - 5}&&\\text{Apply the quotient rule: subtract exponents} \\\\ & = -3\\cdot {x}^{0}&&\\text{Apply the zero exponent rule}\\\\ & = -3\\cdot 1 \\\\ & = -3 \\end{align}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\">\n<p style=\"text-align: initial; font-size: 1rem;\">3.<\/p>\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align} \\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}& = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{1+3}} && \\text{Use the product rule in the denominator. The base is }(j^2k). \\\\ & = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}} && \\text{Use the quotient rule}. \\\\ & = {\\left({j}^{2}k\\right)}^{4 - 4} \\\\ & = {\\left({j}^{2}k\\right)}^{0} && \\text{Use the zero exponent rule}. \\\\ & = 1 \\end{align}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\">\n<p style=\"text-align: initial; font-size: 1rem;\">4.<\/p>\n<p style=\"text-align: initial; font-size: 1rem;\">[latex]\\begin{align} \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}& = 5{\\left(r{s}^{2}\\right)}^{2 - 2}&&\\text{Use the quotient rule}. \\\\ & = 5{\\left(r{s}^{2}\\right)}^{0}&&\\text{Use the zero exponent rule}. \\\\ & = 5\\cdot 1 \\\\ & = 5 \\end{align}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Simplify each expression using the zero exponent rule of exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{{t}^{7}}{{t}^{7}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(d{e}^{2}\\right)}^{11}}{2{\\left(d{e}^{2}\\right)}^{11}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{w}^{4}\\cdot {w}^{2}}{{w}^{6}}[\/latex]<\/li>\n<li>[latex]\\dfrac{-5{t}^{3}\\cdot {t}^{4}}{{t}^{2}\\cdot {t}^{5}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qq703483\">Show Answer<\/span><\/p>\n<div id=\"qq703483\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{2}[\/latex]<\/li>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]-5[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Negative Exponents<\/span><\/p>\n<p>The quotient rule for exponents can also be used to determine what it means to have a negative exponent [latex]x^{-n}[\/latex]. If\u00a0[latex]m<n,\\;m-n<0[\/latex], so\u00a0[latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex] will have a negative exponent.\n\nConsider, for example,\u00a0[latex]\\dfrac{{x}^{2}}{{x}^{4}}={x}^{2-4}=x^{-2}[\/latex]. We have a negative exponent, but what does that mean?\n\nAnother way to simplify\u00a0[latex]\\dfrac{{x}^{2}}{{x}^{4}}[\/latex] is to expand the numerator and denominator then cancel common factors:\n\n\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x^{2}}{x^{4}} &=\\frac{x\\cdot x}{x\\cdot x\\cdot x\\cdot x} \\\\[1mm] &=\\frac{\\cancel{x}\\cdot\\cancel{x}}{\\cancel{x}\\cdot\\cancel{x}\\cdot x\\cdot x} \\\\[1mm] & =\\frac{1}{x^2}\\end{align}[\/latex]<\/p>\n<p>Consequently, [latex]x^{-2}=\\dfrac{1}{x^2}[\/latex].<\/p>\n<p>A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar\u2014from numerator to denominator. This can be generalized to [latex]a^{-n}=\\dfrac{1}{a^n}[\/latex].<\/p>\n<p>If the negative exponent is on the denominator, [latex]\\dfrac{1}{x^{-n}}[\/latex], we can use division of fractions to simplify it:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{1}{x^{-n}}&=1\\div x^{-n}\\\\&=1\\div \\dfrac{1}{x^n}\\\\&=1\\times \\dfrac{x^n}{1}\\\\&=x^n\\end{aligned}[\/latex]<\/p>\n<div style=\"text-align: left;\">A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar\u2014from denominator to\u00a0<span style=\"font-size: 1em;\">numerator<\/span><span style=\"font-size: 1em;\">.<\/span><\/div>\n<div class=\"textbox shaded\">\n<h3>NEGATIVE EXPONENTS<\/h3>\n<p>For any real numbers [latex]a\\neq0[\/latex] and [latex]n[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]a^{-n}=\\dfrac{1}{a^n}[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{{a}^{-n}}=a^n[\/latex]<\/p>\n<p>A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Simplify the expressions. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{x^3}{x^{10}}[\/latex]<\/li>\n<li>[latex]\\dfrac{z^2\\cdot z}{z^4}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(-5t^3\\right)}^4}{\\left(-5t^3\\right)^8}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\"><span style=\"font-size: 16px; orphans: 1;\">Applying the rules for exponents<\/span><br \/>\n<span style=\"font-size: 16px; orphans: 1;\"><br \/>\n<\/span><\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\">[latex]1.\\\\ \\begin{aligned}\\dfrac{x^3}{x^{10}}&=x^{3 - 10}&&\\text{Quotient rule}\\\\&=x^{-7}&&\\text{Negative exponent rule}\\\\&=\\dfrac{1}{x^7}\\end{aligned}[\/latex]<\/span><\/td>\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\">[latex]2.\\\\ \\begin{aligned}\\dfrac{z^2\\cdot z}{z^4}&=\\dfrac{z^{2+1}}{z^4}&&\\text{Product rule}\\\\&=\\dfrac{z^3}{z^4}&&\\text{Quotient rule}\\\\&={z}^{3 - 4}\\\\&={z}^{-1}&&\\text{Negative exponent rule}\\\\&=\\dfrac{1}{z}\\end{aligned}[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\"><span style=\"font-size: 16px; orphans: 1;\">[latex]3.\\\\ \\begin{aligned}\\dfrac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}&={\\left(-5{t}^{3}\\right)}^{4 - 8}&&\\text{Product rule: the base is }-5t^3\\\\&={\\left(-5{t}^{3}\\right)}^{-4}&&\\text{Negative exponent rule}\\\\&=\\dfrac{1}{{\\left(-5{t}^{3}\\right)}^{4}}\\end{aligned}[\/latex]<\/span><\/span><\/td>\n<td style=\"width: 50%; vertical-align: top;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Simplify the expressions. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{{\\left(-3t\\right)}^{2}}{{\\left(-3t\\right)}^{8}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{f}^{47}}{{f}^{49}\\cdot f}[\/latex]<\/li>\n<li>[latex]\\dfrac{2{k}^{4}}{5{k}^{7}}[\/latex]<\/li>\n<li>[latex]\\dfrac{\\left(-3x^4\\right)^5}{\\left(-3x^4\\right)^{12}}[\/latex]<\/li>\n<li>[latex]\\dfrac{5y^{-8}}{y^{-6}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qq85205\">Show Answer<\/span><\/p>\n<div id=\"qq85205\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{1}{{\\left(-3t\\right)}^{6}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{f}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{2}{5{k}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{\\left(-3x^4\\right)^7}[\/latex]<\/li>\n<li>[latex]\\dfrac{5}{y^2}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{b}^{2}\\cdot {b}^{-8}[\/latex]<\/li>\n<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}[\/latex]<\/li>\n<li>[latex]\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\"><span style=\"font-size: 16px; orphans: 1;\">Applying the rules for exponents<\/span><\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\"><span style=\"font-size: 16px; orphans: 1;\">1.<\/span><\/p>\n<p>[latex]\\begin{aligned}{b}^{2}\\cdot {b}^{-8}&={b}^{2 + (-8)}&&\\text{Product rule}\\\\&={b}^{-6}\\\\&=\\frac{1}{{b}^{6}}&&\\text{Negative exponent rule}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%; vertical-align: top;\">2.<\/p>\n<p>[latex]\\begin{aligned}{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}&={\\left(-x\\right)}^{5+(-5)}&&\\text{Product rule}\\\\&={\\left(-x\\right)}^{0}\\\\&=1&&\\text{Zero exponent rule}\\end{aligned}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">3.<\/p>\n<p>[latex]\\begin{aligned}\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}&=\\dfrac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}\\\\&={\\left(-7z\\right)}^{1 - 5}&&\\text{Quotient rule}\\\\&={\\left(-7z\\right)}^{-4}\\\\&=\\dfrac{1}{{\\left(-7z\\right)}^{4}}&&\\text{Negative exponent rule}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%; vertical-align: top;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Simplify. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{t}^{-11}\\cdot {t}^{6}[\/latex]<\/li>\n<li>[latex]\\dfrac{{25}^{12}}{{25}^{13}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qq592096\">Show Answer<\/span><\/p>\n<div id=\"qq592096\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{t}^{-5}=\\dfrac{1}{{t}^{5}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{25}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Power of a Product Rule<\/h2>\n<p>To simplify the power of a product of two exponential expressions, we can use the\u00a0<em><strong>power of a product rule of exponents<\/strong>,\u00a0<\/em>which breaks up the power of a product of factors into the product of the powers of the factors. For example, consider [latex]{\\left(pq\\right)}^{3}[\/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align} {\\left(pq\\right)}^{3}& = \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}} \\\\ & = p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q \\\\& = \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}} \\\\ & = {p}^{3}\\cdot {q}^{3} \\end{align}[\/latex]<\/div>\n<p>In other words, [latex]{\\left(pq\\right)}^{3}={p}^{3}\\cdot {q}^{3}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>The Power of a Product Rule<\/h3>\n<p>For any real numbers [latex]a,\\;b[\/latex] and [latex]n[\/latex], the power of a product rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Simplify each of the following products as much as possible. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(2t\\right)}^{15}[\/latex]<\/li>\n<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>We can use the product and quotient rules and the new definitions to simplify each expression. If a number is raised to a power, we can evaluate it.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%;\" colspan=\"2\">Applying the rules for exponents<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">1.<\/p>\n<p>[latex]\\begin{aligned}{\\left(a{b}^{2}\\right)}^{3}&={\\left(a\\right)}^{3}\\cdot {\\left({b}^{2}\\right)}^{3}&&\\text{Power of a product rule}\\\\&={a}^{1\\cdot 3}\\cdot {b}^{2\\cdot 3}\\\\&={a}^{3}{b}^{6}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%;\">2.<\/p>\n<p>[latex]\\begin{aligned}(2{t})^{15}&={\\left(2\\right)}^{15}\\cdot {\\left(t\\right)}^{15}&&\\text{Power of a product rule}\\\\&={2}^{15}{t}^{15}&&\\text{Evaluate }2^{15}\\text{ using a calculator}\\\\&=32,768{t}^{15}\\end{aligned}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">3.<\/p>\n<p>[latex]\\begin{aligned}{\\left(-2{w}^{3}\\right)}^{3}&={\\left(-2\\right)}^{3}\\cdot {\\left({w}^{3}\\right)}^{3}&&\\text{Power of a product rule}\\\\&=-8\\cdot {w}^{3\\cdot 3}&&{(-2)}^3=-8\\\\&=-8{w}^{9}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%;\">4.<\/p>\n<p>[latex]\\begin{aligned}\\dfrac{1}{{\\left(-7z\\right)}^{4}}&=\\dfrac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}&&\\text{Power of a product rule}\\\\&=\\dfrac{1}{2,401{z}^{4}}&&{(-7)}^4\\text{ is evaluated}\\end{aligned}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">5.<\/p>\n<p>[latex]\\begin{aligned}{\\left({e}^{-2}{f}^{2}\\right)}^{7}&={\\left({e}^{-2}\\right)}^{7}\\cdot {\\left({f}^{2}\\right)}^{7}&&\\text{Power of a product rule}\\\\&={e}^{-2\\cdot 7}\\cdot {f}^{2\\cdot 7}\\\\&={e}^{-14}{f}^{14}&&\\text{Negative exponent rule}\\\\&=\\dfrac{{f}^{14}}{{e}^{14}}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left({g}^{2}{h}^{3}\\right)}^{5}[\/latex]<\/li>\n<li>[latex]{\\left(5t\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(-3{y}^{5}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{\\left({a}^{6}{b}^{7}\\right)}^{3}}[\/latex]<\/li>\n<li>[latex]{\\left({r}^{3}{s}^{-2}\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qq540817\">Show Answer<\/span><\/p>\n<div id=\"qq540817\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{g}^{10}{h}^{15}[\/latex]<\/li>\n<li>[latex]125{t}^{3}[\/latex]<\/li>\n<li>[latex]-27{y}^{15}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{a}^{18}{b}^{21}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{r}^{12}}{{s}^{8}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Power of a Quotient Rule<\/h2>\n<p>To simplify the power of a quotient of two expressions, we can use the\u00a0<em><strong>power of a quotient rule<\/strong>,\u00a0<\/em>which states that the power of a quotient of factors is the quotient of the powers of the factors.The power of a quotient rule is an extension of the power of a product rule since a quotient can be written as a product:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(\\dfrac{a}{b}\\right)^n&=\\left(a\\times\\dfrac{1}{b}\\right)^n\\\\&=a^n\\times \\left(b^{-1}\\right)^n\\\\&=a^n\\times b^{-n}\\\\&=a^n\\times \\dfrac{1}{b^n}\\\\&=\\dfrac{a^n}{b^n}\\end{aligned}[\/latex].<\/p>\n<div style=\"text-align: center;\"><\/div>\n<div class=\"textbox shaded\">\n<h3>The Power of a Quotient Rule<\/h3>\n<p>For any real numbers [latex]a,\\;b[\/latex] and [latex]n[\/latex], provided [latex]b\\neq0[\/latex], the power of a quotient rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\left(\\dfrac{a}{b}\\right)}^{n}=\\dfrac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Simplify each of the following quotients as much as possible. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\n<li>[latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\n<li>[latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%;\" colspan=\"2\">Applying the rules for exponents<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">1.<\/p>\n<p>[latex]\\begin{aligned}{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}&=\\dfrac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}&&\\text{Quotient to a power rule}\\\\&=\\dfrac{64}{{z}^{11\\cdot 3}}&&\\text{Evaluate }4^3=64\\text{. Power to a power rule.} \\\\&=\\dfrac{64}{{z}^{33}}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%;\">2.<\/p>\n<p>[latex]\\begin{aligned}{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}&=\\dfrac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}&&\\text{Power of a quotient rule}\\\\&=\\dfrac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}&&\\text{Power to a power rule}\\\\&=\\dfrac{{p}^{6}}{{q}^{18}}\\end{aligned}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">3.<\/p>\n<p>[latex]\\begin{aligned}{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}&=\\dfrac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}&&\\text{Power of a quotient rule}\\\\&=\\dfrac{-1}{{t}^{2\\cdot 27}}&&\\text{Power of a power rule}\\\\&=\\dfrac{-1}{{t}^{54}}&&\\text{Put the negative sign in front of the fraction}\\\\&=-\\dfrac{1}{{t}^{54}}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%;\">4.<\/p>\n<p>[latex]\\begin{aligned}{\\left({j}^{3}{k}^{-2}\\right)}^{4}&={\\left(\\dfrac{{j}^{3}}{{k}^{2}}\\right)}^{4}&&\\text{Negative exponent rule}\\\\&=\\dfrac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}&&\\text{Power of a quotient rule}\\\\&=\\dfrac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}&&\\text{Power to a power rule}\\\\&=\\dfrac{{j}^{12}}{{k}^{8}}\\end{aligned}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">5.<\/p>\n<p>[latex]\\begin{aligned}{\\left({m}^{-2}{n}^{-2}\\right)}^{3}&={\\left(\\dfrac{1}{{m}^{2}{n}^{2}}\\right)}^{3}&&\\text{Negative exponent rule}\\\\&=\\dfrac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}&&\\text{Power of a quotient rule}\\\\&=\\dfrac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}&&\\text{Evaluate }1^3=1\\\\&=\\dfrac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}&&\\text{Power to a power rule}\\\\&=\\dfrac{1}{{m}^{6}{n}^{6}}\\end{aligned}[\/latex]<\/td>\n<td style=\"width: 50%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Simplify each of the following quotients as much as possible. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left(\\dfrac{{b}^{5}}{c}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{5}{{u}^{8}}\\right)}^{4}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{-1}{{w}^{3}}\\right)}^{35}[\/latex]<\/li>\n<li>[latex]{\\left({p}^{-4}{q}^{3}\\right)}^{8}[\/latex]<\/li>\n<li>[latex]{\\left({c}^{-5}{d}^{-3}\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qq815731\">Show Answer<\/span><\/p>\n<div id=\"qq815731\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{{b}^{15}}{{c}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{625}{{u}^{32}}[\/latex]<\/li>\n<li>[latex]\\dfrac{-1}{{w}^{105}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{q}^{24}}{{p}^{32}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{c}^{20}{d}^{12}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Simplifying Exponential Expressions<\/h2>\n<p>Recall that to simplify an expression means to rewrite it by combining terms or exponents. Evaluating an expression means to get a numerical answer. The rules for exponents can be combined to simplify expressions.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Simplify each expression and write the answer with positive exponents only.<\/p>\n<ol>\n<li>[latex]{\\left(6{m}^{2}{n}^{-1}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}[\/latex]<\/li>\n<li>[latex]\\left(-2{a}^{3}{b}^{-1}\\right)\\left(5{a}^{-2}{b}^{2}\\right)[\/latex]<\/li>\n<li>[latex]{\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<table style=\"border-collapse: collapse; width: 99.0613%; height: 850px;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; vertical-align: top;\" colspan=\"2\">\n<p style=\"text-align: initial;\">Applying the rules for exponents<\/p>\n<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">\n<p style=\"text-align: initial;\">1.<\/p>\n<p style=\"text-align: initial;\">[latex]\\begin{align} {\\left(6{m}^{2}{n}^{-1}\\right)}^{3}& = {\\left(6\\right)}^{3}{\\left({m}^{2}\\right)}^{3}{\\left({n}^{-1}\\right)}^{3}&& \\text{Power of a product rule} \\\\ & = {6}^{3}{m}^{2\\cdot 3}{n}^{-1\\cdot 3}&& \\text{Power rule} \\\\ & = 216{m}^{6}{n}^{-3}&&\\text{Evaluate: }2^6=216\\text{ and simplify}. \\\\ & = \\frac{216{m}^{6}}{{n}^{3}}&& \\text{Negative exponent rule} \\end{align}[\/latex]<\/p>\n<\/td>\n<td style=\"width: 50%; vertical-align: top;\">\n<p style=\"text-align: initial;\">2.<\/p>\n<p style=\"text-align: initial;\">[latex]\\begin{align} {17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}&=17^{5 +(-4)+(-3)}&& \\text{Product rule} \\\\ & = {17}^{-2}&&\\text{Negative exponent rule}\\\\&=\\dfrac{1}{{17}^{2}}&&\\text{Evaluate}\\\\&=\\dfrac{1}{289}\\end{align}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">\n<p style=\"text-align: initial;\">3.<\/p>\n<p style=\"text-align: initial;\">[latex]\\begin{align} {\\left(\\dfrac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}&=\\dfrac{{\\left({u}^{-1}v\\right)}^{2}}{{\\left({v}^{-1}\\right)}^{2}}&&\\text{Power of a quotient rule}\\\\&=\\dfrac{{u}^{-2}{v}^{2}}{{v}^{-2}}&&\\text{Power of a product rule}\\\\&={u}^{-2}{v}^{2-(-2)}&&\\text{Quotient rule}\\\\&={u}^{-2}{v}^{4}&&\\text{Evaluate: }2-(-2)=2+2=4\\\\&=\\dfrac{{v}^{4}}{{u}^{2}}&&\\text{Negative exponent rule}\\end{align}[\/latex]<\/p>\n<\/td>\n<td style=\"width: 50%; vertical-align: top;\">\n<p style=\"text-align: initial;\">4.<\/p>\n<p style=\"text-align: initial;\">[latex]\\begin{align}\\left(-2a^3b^{-1}\\right)\\left(5a^{-2}b^2 \\right)&=\\left(-2\\cdot 5\\right)\\left({a}^{3}\\cdot {a}^{-2}\\right)\\left( {b}^{-1}\\cdot {b}^{2}\\right)&& \\text{Commutative\/associative properties} \\\\ & = -10{a}^{3+(-2)}{b}^{-1+2}&& \\text{Product rule} \\\\ & = -10a^1b^1\\\\&=-10ab \\end{align}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">\n<p style=\"text-align: initial;\">5.<\/p>\n<p style=\"text-align: initial;\">[latex]\\begin{align} {\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}& = {\\left({x}^{2}\\sqrt{2}\\right)}^{4+(-4)} && \\text{Product rule: base is }(x^2\\sqrt{2}) \\\\ & = {\\left({x}^{2}\\sqrt{2}\\right)}^{0}&& \\text{Zero exponent rule}\\\\ & = 1 \\end{align}[\/latex]<\/p>\n<\/td>\n<td style=\"width: 50%; vertical-align: top;\">\n<p style=\"text-align: initial;\">6.<\/p>\n<p><span style=\"text-align: initial;\">[latex]\\begin{align} \\frac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}& = \\frac{{\\left(3\\right)}^{5}\\cdot {\\left({w}^{2}\\right)}^{5}}{{\\left(6\\right)}^{2}\\cdot {\\left({w}^{-2}\\right)}^{2}}&& \\text{Power of a product rule} \\\\ & = \\frac{{3}^{5}{w}^{2\\cdot 5}}{{6}^{2}{w}^{-2\\cdot 2}}&& \\text{Power rule} \\\\ & = \\frac{243{w}^{10}}{36{w}^{-4}} && \\text{Evaluate: }3^5=243\\text{ and }6^2=36 \\\\ & = \\frac{27{w}^{10-\\left(-4\\right)}}{4}&& \\text{Quotient rule and simplify fraction} \\\\ & = \\frac{27{w}^{14}}{4}\\end{align}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 6<\/h3>\n<p>Simplify each expression and write the answer with positive exponents only.<\/p>\n<ol>\n<li>[latex]{\\left(2u{v}^{-2}\\right)}^{-3}[\/latex]<\/li>\n<li>[latex]{x}^{8}\\cdot {x}^{-12}\\cdot x[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{{e}^{2}{f}^{-3}}{{f}^{-1}}\\right)}^{2}[\/latex]<\/li>\n<li>[latex]\\left(9{r}^{-5}{s}^{3}\\right)\\left(3{r}^{6}{s}^{-4}\\right)[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{4}{9}t{w}^{-2}\\right)}^{-3}{\\left(\\dfrac{4}{9}t{w}^{-2}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(2{h}^{2}k\\right)}^{4}}{{\\left(7{h}^{-1}{k}^{2}\\right)}^{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q31244\">Show Answer<\/span><\/p>\n<div id=\"q31244\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{{v}^{6}}{8{u}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{x}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{e}^{4}}{{f}^{4}}[\/latex]<\/li>\n<li>[latex]\\dfrac{27r}{s}[\/latex]<\/li>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]\\dfrac{16{h}^{10}}{49}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>TRY IT 7<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm43896\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43896&#38;theme=oea&#38;iframe_resize_id=ohm43896&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe> title= &#8220;Interactive problem&#8221;<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 8<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14060\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14060&#38;theme=oea&#38;iframe_resize_id=ohm14060&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe> title= &#8220;Interactive problem&#8221;<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 9<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14056\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14056&#38;theme=oea&#38;iframe_resize_id=ohm14056&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe> title= &#8220;Interactive problem&#8221;<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 10<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm14057\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14057&#38;theme=oea&#38;iframe_resize_id=ohm14057&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe> title= &#8220;Interactive problem&#8221;<\/p>\n<\/div>\n<h2>Simplifying Exponential Functions<\/h2>\n<p>Each of the properties of exponents can be used to simplify and write equivalent exponential functions. For example, the function [latex]f(x)=2^{x+3}[\/latex] can be written as the equivalent exponential function [latex]f(x)=8\\left(2^x\\right)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=2^{x+3}\\\\&=2^x\\cdot 2^3\\\\&=2^x\\cdot 8\\\\&=8\\left(2^x\\right)\\end{aligned}[\/latex]<\/p>\n<p>Being able to simplify an exponential function into the standard form [latex]f(x)=ar^{x-h}+k[\/latex] makes the function easier to graph and allows us to determine the transformations that were made from the parent function [latex]f(x)=r^x[\/latex]. In the case of\u00a0[latex]f(x)=2^{x+3}=8\\left(2^x\\right)[\/latex], the 8 tells us that the graph of [latex]f(x)=2^x[\/latex] has been stretched by a factor of 8.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 7<\/h3>\n<p><span class=\"TextRun SCXW130104365 BCX9\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"auto\"><span class=\"NormalTextRun SCXW130104365 BCX9\">Use exponential rules to <\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">write equivalent<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\"> exponential functions<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">.<\/span>\u00a0<\/span><\/p>\n<ol>\n<li>[latex]f(x)=3^{x+2}[\/latex]<\/li>\n<li>[latex]g(x)=5^{2x+1}[\/latex]<\/li>\n<li>[latex]h(x)=4^{3x-2}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>1.<\/p>\n<p>[latex]\\begin{aligned}f(x)&=3^{x+2}\\\\&=3^x\\cdot 3^2&&\\text{Product rule (in reverse)}\\\\&=3^x\\cdot 9&&\\text{Evaluate }3^2=9\\\\&=9\\left(3^x\\right)&&\\text{Write in standard form}\\end{aligned}[\/latex]<\/p>\n<p>2.<\/p>\n<p>[latex]\\begin{aligned}g(x)&=5^{2x+1}\\\\&=5^{2x}\\cdot 5^1&&\\text{Product rule (in reverse)}\\\\&=\\left(5^2\\right)^x\\cdot 5&&\\text{Evaluate }5^1=5\\\\&=5\\left(25^x\\right)&&\\text{Write in standard form}\\end{aligned}[\/latex]<\/p>\n<p>3.<\/p>\n<p>[latex]\\begin{aligned}h(x)&=4^{3x-2}\\\\&=4^{3x}\\cdot 4^{-2}&&\\text{Product rule (in reverse)}\\\\&=\\left(4^3\\right)^x\\cdot \\dfrac{1}{4^2}&&\\text{Power to a power rule (in reverse) and negative exponent rule}\\\\&=64^x\\cdot \\dfrac{1}{16}&&\\text{Evaluate: }4^3=64\\text{ and }4^2=16\\\\&=\\dfrac{1}{16}\\left(64^x\\right)&&\\text{Write in standard form}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 11<\/h3>\n<p><span class=\"TextRun SCXW130104365 BCX9\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"auto\"><span class=\"NormalTextRun SCXW130104365 BCX9\">Use exponential rules to <\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">write equivalent<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\"> exponential functions<\/span><span class=\"NormalTextRun SCXW130104365 BCX9\">.<\/span>\u00a0<\/span><\/p>\n<ol>\n<li>[latex]f(x)=2^{x+4}[\/latex]<\/li>\n<li>[latex]g(x)=3^{x-2}[\/latex]<\/li>\n<li>[latex]h(x)=5^{2x-1}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm292\">Show Answer<\/span><\/p>\n<div id=\"qhjm292\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]f(x)=16\\left(2^x\\right)[\/latex]<\/li>\n<li>[latex]g(x)=\\dfrac{1}{9}\\left(3^x\\right)[\/latex]<\/li>\n<li>[latex]h(x)=\\dfrac{1}{5}\\left(25^{x}\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2209\">\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>Adaptation and Revision. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><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><li>Simplifying Exponential Functions. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Try It: hjm292. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <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>Simplify Expressions With Zero Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/rpoUg32utlc\">https:\/\/youtu.be\/rpoUg32utlc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Simplify Expressions With Negative Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Gssi4dBtAEI\">https:\/\/youtu.be\/Gssi4dBtAEI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Power of a Product. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/p-2UkpJQWpo\">https:\/\/youtu.be\/p-2UkpJQWpo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Power of a Quotient. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BoBe31pRxFM\">https:\/\/youtu.be\/BoBe31pRxFM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 44120, 43231. <strong>Authored by<\/strong>: Brenda Gardner. <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 7833, 14060. <strong>Authored by<\/strong>: Tyler Wallace. <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 109762, 109765. <strong>Authored 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>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 51959. <strong>Authored by<\/strong>: Roy Shahbazian. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 93393. <strong>Authored by<\/strong>: Michael Jenck. <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 14047, 14058, 14059, 14046, 14051, 14056, 14057. <strong>Authored by<\/strong>: James Souza. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 43896. <strong>Authored by<\/strong>: Carla Kulinsky. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Adaptation and 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http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Simplify Expressions With Zero Exponents\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/rpoUg32utlc\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Simplify Expressions With Negative Exponents\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Gssi4dBtAEI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Power of a Product\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/p-2UkpJQWpo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Power of a Quotient\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/BoBe31pRxFM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 44120, 43231\",\"author\":\"Brenda Gardner\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 7833, 14060\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 109762, 109765\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 51959\",\"author\":\"Roy 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