{"id":783,"date":"2016-06-01T20:49:10","date_gmt":"2016-06-01T20:49:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=783"},"modified":"2023-11-08T13:18:43","modified_gmt":"2023-11-08T13:18:43","slug":"read-terms-and-expressions-with-exponents-2","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/read-terms-and-expressions-with-exponents-2\/","title":{"raw":"2.1 - Exponent Rules","rendered":"2.1 &#8211; Exponent Rules"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(2.1.1) - Notation and terminology for exponents\r\n<ul>\r\n \t<li>Identify the components of a term containing integer exponents<\/li>\r\n \t<li>Evaluate expressions containing integer exponents<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(2.1.2) - The product and quotient rules\r\n<ul>\r\n \t<li>Use the product rule to multiply exponential expressions<\/li>\r\n \t<li>Use the quotient rule to divide exponential expressions<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(2.1.3) - The power rule\r\n<ul>\r\n \t<li>Use the power rule to simplify expressions with exponents raised to powers<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(2.1.4) - Negative and zero exponent rules\r\n<ul>\r\n \t<li>Define and use the zero exponent rule<\/li>\r\n \t<li>Define and use the negative exponent rule<\/li>\r\n \t<li>Combine all the exponents rules to simplify expressions<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(2.1.5) - Power of a product and a quotient\r\n<ul>\r\n \t<li>Simplify an expression with a product raised to a power<\/li>\r\n \t<li>Simplify an expression with a quotient raised to a power<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(2.1.6) - Definition of a simplified exponential expression<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\n<\/div>\r\n\r\n[caption id=\"attachment_4445\" align=\"aligncenter\" width=\"282\"]<img class=\"wp-image-4445\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183306\/Screen-Shot-2016-05-31-at-10.38.21-AM-150x150.png\" alt=\"Image of a woman taking a picture with a camera repeated five times in different colors.\" width=\"282\" height=\"282\" \/> Repeated Image[\/caption]\r\n<h1>(2.1.1) - Notation and terminology for exponents<\/h1>\r\nWe use exponential notation to write repeated multiplication. For example [latex]10\\cdot10\\cdot10[\/latex] can be written more succinctly as [latex]10^{3}[\/latex]. The 10 in [latex]10^{3}[\/latex]<sup>\u00a0<\/sup>is called the <b>base<\/b>. The 3 in [latex]10^{3}[\/latex]<sup>\u00a0<\/sup>is called the <b>exponent<\/b>. The expression [latex]10^{3}[\/latex] is called the exponential expression. Knowing the names for the parts of an exponential expression or term will help you learn how to perform mathematical operations on them.\r\n<p style=\"text-align: center\">[latex]\\text{base}\\rightarrow10^{3\\leftarrow\\text{exponent}}[\/latex]<\/p>\r\n[latex]10^{3}[\/latex] is read as \u201c10 to the third power\u201d or \u201c10 cubed.\u201d It means [latex]10\\cdot10\\cdot10[\/latex], or 1,000.\r\n\r\n[latex]8^{2}[\/latex]\u00a0is read as \u201c8 to the second power\u201d or \u201c8 squared.\u201d It means [latex]8\\cdot8[\/latex], or 64.\r\n\r\n[latex]5^{4}[\/latex]\u00a0is read as \u201c5 to the fourth power.\u201d It means [latex]5\\cdot5\\cdot5\\cdot5[\/latex], or 625.\r\n\r\n[latex]b^{5}[\/latex]\u00a0is read as \u201c<i>b<\/i> to the fifth power.\u201d It means [latex]{b}\\cdot{b}\\cdot{b}\\cdot{b}\\cdot{b}[\/latex]. Its value will depend on the value of <i>b<\/i>.\r\n\r\nThe exponent applies only to the number that it is next to. Therefore, in the expression [latex]xy^{4}[\/latex],\u00a0only the <i>y<\/i> is affected by the 4. [latex]xy^{4}[\/latex]\u00a0means [latex]{x}\\cdot{y}\\cdot{y}\\cdot{y}\\cdot{y}[\/latex]. The <em>x<\/em> in this term is a <strong>coefficient<\/strong> of <em>y<\/em>.\r\n\r\nIf the exponential expression is negative, such as [latex]\u22123^{4}[\/latex], it means [latex]\u2013\\left(3\\cdot3\\cdot3\\cdot3\\right)[\/latex] or [latex]\u221281[\/latex].\r\n\r\nIf [latex]\u22123[\/latex] is to be the base, it must be written as [latex]\\left(\u22123\\right)^{4}[\/latex], which means [latex]\u22123\\cdot\u22123\\cdot\u22123\\cdot\u22123[\/latex], or 81.\r\n\r\nLikewise,\u00a0[latex]\\left(\u2212x\\right)^{4}=\\left(\u2212x\\right)\\cdot\\left(\u2212x\\right)\\cdot\\left(\u2212x\\right)\\cdot\\left(\u2212x\\right)=x^{4}[\/latex], while [latex]\u2212x^{4}=\u2013\\left(x\\cdot x\\cdot x\\cdot x\\right)[\/latex].\r\n\r\nYou can see that there is quite a difference, so you have to be very careful! The following examples show how to identify the base and the exponent, as well as how to identify the expanded and exponential format of writing repeated multiplication.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIdentify the exponent and the base in the following terms, then simplify:\r\n<ol>\r\n \t<li>[latex]7^{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\frac{1}{2}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]2x^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-5\\right)^{2}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"211363\"]Show Solution[\/reveal-answer]\r\n<p style=\"text-align: left\">[hidden-answer a=\"211363\"]<\/p>\r\n1) [latex]7^{2}[\/latex]\r\n\r\nThe exponent in this term is 2 and the base is 7. To simplify, expand the term: [latex]7^{2}=7\\cdot{7}=49[\/latex]\r\n\r\n2) [latex]{\\left(\\frac{1}{2}\\right)}^{3}[\/latex]\r\n\r\nThe exponent on this term is 3, and the base is [latex]\\frac{1}{2}[\/latex]. To simplify, expand the multiplication and remember how to multiply fractions: [latex]{\\left(\\frac{1}{2}\\right)}^{3}=\\frac{1}{2}\\cdot{\\frac{1}{2}}\\cdot{\\frac{1}{2}}=\\frac{1}{16}[\/latex]\r\n\r\n3) \u00a0[latex]2x^{3}[\/latex]\r\n\r\nThe exponent on this term is 3, and the base is x, the 2 is not getting the exponent because there are no parentheses that tell us it is. \u00a0This term is in its most simplified form.\r\n\r\n4)\u00a0[latex]\\left(-5\\right)^{2}[\/latex]\r\n\r\nThe exponent on this terms is 2 and the base is [latex]-5[\/latex]. To simplify, expand the multiplication: [latex]\\left(-5\\right)^{2}=-5\\cdot{-5}=25[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p class=\"no-indent\">In the following video you are provided more examples of applying exponents to various bases.<\/p>\r\nhttps:\/\/youtu.be\/ocedY91LHKU\r\n<h3>Evaluate expressions<\/h3>\r\nEvaluating expressions containing exponents is the same as evaluating the linear expressions from earlier in the course. You substitute the value of the variable into the expression and simplify.\r\n\r\nYou can use the order of operations\u00a0to evaluate the expressions containing exponents. First, evaluate anything in Parentheses or grouping symbols. Next, look for Exponents, followed by Multiplication and Division (reading from left to right), and lastly, Addition and Subtraction (again, reading from left to right).\r\n\r\nSo, when you evaluate the expression [latex]5x^{3}[\/latex]\u00a0if [latex]x=4[\/latex], first substitute the value 4 for the variable <i>x<\/i>. Then evaluate, using order of operations.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nEvaluate the following expressions for the given value.\r\n<ol>\r\n \t<li>[latex]5x^{3}[\/latex]<i>\u00a0<\/i>if [latex]x=4[\/latex]<\/li>\r\n \t<li>[latex]\\left(5x\\right)^{3}[\/latex]\u00a0if [latex]x=4[\/latex]<\/li>\r\n \t<li>[latex]x^{3}[\/latex] if [latex]x=\u22124[\/latex]<\/li>\r\n \t<li>[latex]3^x[\/latex] if [latex]x = 4[\/latex]<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"411363\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"411363\"]\r\n\r\n1) Substitute 4 for the variable <i>x<\/i>.\r\n<p style=\"text-align: center\">[latex]5\\cdot4^{3}[\/latex]<\/p>\r\nEvaluate [latex]4^{3}[\/latex]. Multiply.\r\n<p style=\"text-align: center\">[latex]5\\left(4\\cdot4\\cdot4\\right)=5\\cdot64=320[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]5x^{3}=320[\/latex]\u00a0when [latex]x=4[\/latex]\r\n\r\n&nbsp;\r\n\r\n2) [latex]\\left(5x\\right)^{3}[\/latex]\u00a0if [latex]x=4[\/latex]\r\n\r\nSubstitute 4 for the variable <i>x<\/i>.\u00a0notice the how adding parentheses can change the outcome when you are simplifying terms with exponents.\r\n<p style=\"text-align: center\">[latex]\\left(5\\cdot4\\right)3[\/latex]<\/p>\r\nMultiply inside the parentheses, then apply the exponent\u2014following the rules of PEMDAS.\r\n<p style=\"text-align: center\">[latex]20^{3}[\/latex]<\/p>\r\nEvaluate [latex]20^{3}[\/latex].\r\n<p style=\"text-align: center\">[latex]20\\cdot20\\cdot20=8,000[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(5x\\right)3=8,000[\/latex] when [latex]x=4[\/latex]\r\n\r\n&nbsp;\r\n\r\n3) \u00a0[latex]x^{3}[\/latex] if [latex]x=\u22124[\/latex].\r\n\r\nSubstitute [latex]\u22124[\/latex] for the variable <i>x<\/i>.\r\n<p style=\"text-align: center\">[latex]\\left(\u22124\\right)^{3}[\/latex]<\/p>\r\nEvaluate. Note how placing parentheses around the [latex]\u22124[\/latex] means the negative sign also gets multiplied.\r\n<p style=\"text-align: center\">[latex]\u22124\\cdot\u22124\\cdot\u22124[\/latex]<\/p>\r\nMultiply.\r\n<p style=\"text-align: center\">[latex]\u22124\\cdot\u22124\\cdot\u22124=\u221264[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x^{3}=\u221264[\/latex] when [latex]x=\u22124[\/latex]\r\n\r\n4) \u00a0[latex]3^x[\/latex] if [latex]x = 4[\/latex]\r\n\r\nSubstitute x = 4 into the exponent. \u00a0[latex]3^x=3^4=3\\cdot3\\cdot3\\cdot3=81[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]3^{x}=81[\/latex] when [latex]x=4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"75\" height=\"66\" \/>\r\n\r\nCaution! Whether to include a negative sign as part of a base or not often leads to confusion. To clarify\u00a0whether a negative sign is applied before or after the exponent, here is an example.\r\n\r\nWhat is the difference in the way you would evaluate these two terms?\r\n<ol>\r\n \t<li style=\"text-align: left\">[latex]-{3}^{2}[\/latex]<\/li>\r\n \t<li style=\"text-align: left\">[latex]{\\left(-3\\right)}^{2}[\/latex]<\/li>\r\n<\/ol>\r\nTo evaluate 1), you would apply the exponent to the three first, then apply the negative sign last, like this:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}-\\left({3}^{2}\\right)\\\\=-\\left(9\\right) = -9\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">To evaluate 2), you would apply the exponent to the 3 and the negative sign:<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{\\left(-3\\right)}^{2}\\\\=\\left(-3\\right)\\cdot\\left(-3\\right)\\\\={ 9}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">The key to remembering this is to follow the order of operations. The first expression does not include parentheses so you would apply the exponent to the integer 3 first, then apply the negative sign. The second expression includes parentheses, so hopefully you will remember that the negative sign also gets squared.<\/p>\r\n\r\n<\/div>\r\n<p id=\"video0\" class=\"no-indent\" style=\"text-align: left\">In the next sections, you will learn how to simplify expressions that contain exponents. Come back to this page if you forget how to apply the order of operations to a term with exponents, or forget which is the base and which is the exponent!<\/p>\r\n<p class=\"no-indent\" style=\"text-align: left\">In the following video you are provided with examples of evaluating exponential expressions for a given number.<\/p>\r\nhttps:\/\/youtu.be\/pQNz8IpVVg0\r\n<h1>(2.1.2) - The product and quotient rules<\/h1>\r\n<h3>The product rule<\/h3>\r\n<b>Exponential notation<\/b> was developed to write repeated multiplication more efficiently. There are times when it is easier or faster to leave the expressions in exponential notation when multiplying or dividing. Let\u2019s look at rules that will allow you to do this.\r\n\r\nFor example, the notation [latex]5^{4}[\/latex]\u00a0can be expanded and written as [latex]5\\cdot5\\cdot5\\cdot5[\/latex], or 625. And don\u2019t forget, the exponent only applies to the number immediately to its left, unless there are parentheses.\r\n\r\nWhat happens if you multiply two numbers in exponential form with the same base? Consider the expression [latex]{2}^{3}{2}^{4}[\/latex]. Expanding each exponent, this can be rewritten as [latex]\\left(2\\cdot2\\cdot2\\right)\\left(2\\cdot2\\cdot2\\cdot2\\right)[\/latex] or [latex]2\\cdot2\\cdot2\\cdot2\\cdot2\\cdot2\\cdot2[\/latex]. In exponential form, you would write the product as [latex]2^{7}[\/latex]. Notice that 7 is the sum of the original two exponents, 3 and 4.\r\n\r\nWhat about [latex]{x}^{2}{x}^{6}[\/latex]? This can be written as [latex]\\left(x\\cdot{x}\\right)\\left(x\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\right)=x\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}[\/latex] or [latex]x^{8}[\/latex]. And, once again, 8 is the sum of the original two exponents. This concept can be generalized in the following way:\r\n<div class=\"textbox shaded\">\r\n<h3>The Product Rule for Exponents<\/h3>\r\nFor any number <i>x<\/i> and any integers <i>a<\/i> and <i>b<\/i>,\u00a0[latex]\\left(x^{a}\\right)\\left(x^{b}\\right) = x^{a+b}[\/latex].\r\n\r\nTo multiply exponential terms with the same base, add the exponents.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"772709\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"772709\"]Solution\r\n\r\nUse the product rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\nAt first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.\r\n<div style=\"text-align: center\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\r\nNotice we get the same result by adding the three exponents in one step.\r\n<div style=\"text-align: center\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\r\n<div style=\"text-align: center\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center\"><img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183308\/traffic-sign-160659.png\" alt=\"traffic-sign-160659\" width=\"96\" height=\"83\" \/><\/h3>\r\nCaution! Do not try to apply this rule to sums.\r\n\r\nThink about the expression\u00a0[latex]\\left(2+3\\right)^{2}[\/latex]\r\n<p style=\"text-align: center\">Does [latex]\\left(2+3\\right)^{2}[\/latex] equal [latex]2^{2}+3^{2}[\/latex]?<\/p>\r\nNo, it does not because of the order of operations!\r\n<p style=\"text-align: center\">[latex]\\left(2+3\\right)^{2}=5^{2}=25[\/latex]<\/p>\r\n<p style=\"text-align: center\">and<\/p>\r\n<p style=\"text-align: center\">[latex]2^{2}+3^{2}=4+9=13[\/latex]<\/p>\r\nTherefore, you can only use this rule when the numbers inside the parentheses are being multiplied (or divided, as we will see next).\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">In the following video you will see more examples of using the product rule for exponents to simplify expressions.<\/span>\r\n\r\nhttps:\/\/youtu.be\/P0UVIMy2nuI\r\n\r\nIn our last product rule example we will show that an exponent can be an algebraic expression. \u00a0We can use the product rule for exponents no matter what the exponent looks like, as long as the base is the same.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nMultiply.\u00a0[latex]x^{a+2}\\cdot{x^{3a-9}}[\/latex]\r\n[reveal-answer q=\"740329\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"740329\"]\r\n\r\nWe have two exponentiated terms with the same base, so we can multiply them together. The product rule for exponents says that we can add the exponents.\r\n\r\n[latex]x^{a+2}\\cdot{x^{3a-9}}=x^{(a+2)+(3a-9)}=x^{4a-7}[\/latex]\r\n\r\nThe expression can't be simplified any further.\r\n<h4>Answer<\/h4>\r\n[latex]x^{a+2}\\cdot{x^{3a-9}}=x^{4a-7}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>The quotient rule<\/h3>\r\nLet\u2019s look at dividing terms containing exponential expressions. What happens if you divide two numbers in exponential form with the same base? Consider the following expression.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\frac{{{4}^{5}}}{{{4}^{2}}}[\/latex]<\/p>\r\nYou can rewrite the expression as: [latex] \\displaystyle \\frac{4\\cdot 4\\cdot 4\\cdot 4\\cdot 4}{4\\cdot 4}[\/latex]. Then you can cancel the common factors of 4 in the numerator and denominator: [latex] \\displaystyle [\/latex]\r\n\r\nFinally, this expression can be rewritten as [latex]4^{3}[\/latex]\u00a0using exponential notation. Notice that the exponent, 3, is the difference between the two exponents in the original expression, 5 and 2.\r\n\r\nSo,\u00a0[latex] \\displaystyle \\frac{{{4}^{5}}}{{{4}^{2}}}=4^{5-2}=4^{3}[\/latex].\r\n\r\nBe careful that you subtract the exponent in the denominator from the exponent in the numerator.\r\n\r\nSo, to divide two exponential terms with the same base, subtract the exponents.\r\n<div class=\"textbox shaded\">\r\n<h3>The Quotient (Division) Rule for Exponents<\/h3>\r\nFor any non-zero number [latex]x[\/latex] and any integers [latex]a[\/latex] and [latex]b[\/latex]: [latex]\\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]\\displaystyle \\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle \\frac{{t}^{23}}{{t}^{15}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"978732\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"978732\"]\r\n\r\nUse the quotient rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]\\displaystyle \\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}={\\left(-2\\right)}^{14 - 9}={\\left(-2\\right)}^{5}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle \\frac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs we showed with the product rule, you may be given a quotient with an exponent that is an algebraic expression to simplify. \u00a0As long as the bases agree, you may use the quotient rule for exponents.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\displaystyle \\frac{y^{x-3}}{y^{9-x}}[\/latex]\r\n[reveal-answer q=\"836863\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"836863\"]\r\n\r\nWe have a quotient whose terms have the same base so we can use the quotient rule for exponents.\r\n\r\n[latex]\\displaystyle \\frac{y^{x-3}}{y^{9-x}}=y^{(x-3)-(9-x)}=y^{2x-12}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">In the following video, you will we more examples of using the quotient rule for exponents.<\/span>\r\n\r\nhttps:\/\/youtu.be\/xy6WW7y_GcU\r\n<h1>(2.1.3) - The power rule<\/h1>\r\nAnother word for exponent is power. \u00a0You have likely seen or heard an example such as [latex]3^5[\/latex] can be described as 3 raised to the 5th power. In this section we will further expand our capabilities with exponents. We will learn what to do when a term with a\u00a0power\u00a0is raised to another power, and what to do when two numbers or variables are multiplied and both are raised to an exponent. \u00a0We will also learn what to do when numbers or variables that are divided are raised to a power. \u00a0We will begin by raising powers to powers.\r\n\r\nLet\u2019s simplify [latex]\\left(5^{2}\\right)^{4}[\/latex]. In this case, the base is [latex]5^2[\/latex]<sup>\u00a0<\/sup>and the exponent is 4, so you multiply [latex]5^{2}[\/latex]<sup>\u00a0<\/sup>four times: [latex]\\left(5^{2}\\right)^{4}=5^{2}\\cdot5^{2}\\cdot5^{2}\\cdot5^{2}=5^{8}[\/latex]<sup>\u00a0<\/sup>(using the Product Rule\u2014add the exponents).\r\n\r\n[latex]\\left(5^{2}\\right)^{4}[\/latex]<sup>\u00a0<\/sup>is a power of a power. It is the fourth power of 5 to the second power. And we saw above that the answer is [latex]5^{8}[\/latex]. Notice that the new exponent is the same as the product of the original exponents: [latex]2\\cdot4=8[\/latex].\r\n\r\nSo, [latex]\\left(5^{2}\\right)^{4}=5^{2\\cdot4}=5^{8}[\/latex]\u00a0(which equals 390,625, if you do the multiplication).\r\n\r\nLikewise, [latex]\\left(x^{4}\\right)^{3}=x^{4\\cdot3}=x^{12}[\/latex]\r\n\r\nThis leads to another rule for exponents\u2014the <b>Power Rule for Exponents<\/b>. To simplify a power of a power, you multiply the exponents, keeping the base the same. For example, [latex]\\left(2^{3}\\right)^{5}=2^{15}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>The Power Rule for Exponents<\/h3>\r\nFor any positive number <i>x<\/i> and integers <i>a<\/i> and <i>b<\/i>: [latex]\\left(x^{a}\\right)^{b}=x^{a\\cdot{b}}[\/latex].\r\n\r\nTake a moment to contrast how this is different from the product rule for exponents found on the previous page.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{\\left({x}^{2}\\right)}^{7}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"841688\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"841688\"]\r\n\r\nUse the power rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{\\left({x}^{2}\\right)}^{7}={x}^{2\\cdot 7}={x}^{14}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}={\\left(2t\\right)}^{5\\cdot 3}={\\left(2t\\right)}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}={\\left(-3\\right)}^{5\\cdot 11}={\\left(-3\\right)}^{55}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">\u00a0In the following video you will see more examples of using the power rule to simplify expressions with exponents.<\/span>\r\n\r\nhttps:\/\/youtu.be\/VjcKU5rA7F8\r\n\r\nBe careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.\r\n<table style=\"width: 20%\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\" colspan=\"5\">Product Rule<\/th>\r\n<th style=\"text-align: center\" colspan=\"6\">Power Rule<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]5^{3}\\cdot5^{4}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>\u00a0[latex]5^{3+4}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]5^{7}[\/latex]<\/td>\r\n<td>but<\/td>\r\n<td>[latex]\\left(5^{3}\\right)^{4}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]5^{3\\cdot4}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]5^{12}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x^{5}\\cdot x^{2}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]x^{5+2}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]x^{7}[\/latex]<\/td>\r\n<td>but<\/td>\r\n<td>[latex]\\left(x^{5}\\right)^{2}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>\u00a0[latex]x^{5\\cdot2}[\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]x^{10}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\left(3a\\right)^{7}\\cdot\\left(3a\\right)^{10} [\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]\\left(3a\\right)^{7+10} [\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]\\left(3a\\right)^{17}[\/latex]<\/td>\r\n<td>but<\/td>\r\n<td>[latex]\\left(\\left(3a\\right)^{7}\\right)^{10} [\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]\\left(3a\\right)^{7\\cdot10} [\/latex]<\/td>\r\n<td>=<\/td>\r\n<td>[latex]\\left(3a\\right)^{70}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h1>(2.1.4) - Negative and zero exponent rules<\/h1>\r\n<h3>Zero exponent rule<\/h3>\r\nReturn to the quotient rule. We worked with expressions for which\u00a0\u00a0[latex]a&gt;b[\/latex] so that the difference [latex]a-b[\/latex] would never be zero or negative.\r\n<div class=\"textbox shaded\">\r\n<h3>The Quotient (Division) Rule for Exponents<\/h3>\r\nFor any non-zero number [latex]x[\/latex]\u00a0and any integers [latex]a[\/latex] and [latex]b[\/latex]: [latex]\\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]\r\n\r\n<\/div>\r\nWhat would happen if [latex]a=b[\/latex]? In this case, we would use the <em>zero exponent rule of exponents<\/em> to simplify the expression to 1. To see how this is done, let us begin with an example.\r\n<p style=\"text-align: center\">[latex]\\frac{t^{8}}{t^{8}}=\\frac{\\cancel{t^{8}}}{\\cancel{t^{8}}}=1[\/latex]<\/p>\r\nIf we were to simplify the original expression using the quotient rule, we would have\r\n<div style=\"text-align: center\">[latex]\\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[\/latex]<\/div>\r\nIf we equate the two answers, the result is [latex]{t}^{0}=1[\/latex]. This is true for any nonzero real number, or any variable representing a real number.\r\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\r\nThe 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.\r\n<div class=\"textbox shaded\">\r\n<h3>The Zero Exponent Rule of Exponents<\/h3>\r\nFor any nonzero real number [latex]a[\/latex], the zero exponent rule of exponents states that\r\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify each expression using the zero exponent rule of exponents.\r\n<ol>\r\n \t<li>[latex]\\Large\\frac{{c}^{3}}{{c}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large\\frac{-3{x}^{5}}{{x}^{5}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large\\frac{{\\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]\\Large\\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"375469\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"375469\"]\r\n\r\nUse the zero exponent and other rules to simplify each expression.\r\n\r\n1.\r\n<p style=\"text-align: center\">[latex]\\large\\begin{array}\\text{ }\\frac{c^{3}}{c^{3}} \\hfill&amp; =c^{3-3} \\\\ \\hfill&amp; =c^{0} \\\\ \\hfill&amp; =1\\end{array}[\/latex]<\/p>\r\n2.\r\n<p style=\"text-align: center\">[latex]\\large\\begin{array}{ccc}\\hfill \\frac{-3{x}^{5}}{{x}^{5}}&amp; =&amp; -3\\cdot \\frac{{x}^{5}}{{x}^{5}}\\hfill \\\\ &amp; =&amp; -3\\cdot {x}^{5 - 5}\\hfill \\\\ &amp; =&amp; -3\\cdot {x}^{0}\\hfill \\\\ &amp; =&amp; -3\\cdot 1\\hfill \\\\ &amp; =&amp; -3\\hfill \\end{array}[\/latex]<\/p>\r\n3.\r\n<p style=\"text-align: center\">[latex]\\large\\begin{array}{cccc}\\hfill \\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}&amp; =&amp; \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{1+3}}\\hfill &amp;\\text{Use the product rule in the denominator}.\\hfill \\\\ &amp; =&amp;\\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}}\\hfill &amp; \\text{Simplify}.\\hfill \\\\ &amp; =&amp;{\\left({j}^{2}k\\right)}^{4 - 4}\\hfill &amp;\\text{Use the quotient rule}.\\hfill \\\\ &amp; =&amp; {\\left({j}^{2}k\\right)}^{0}\\hfill&amp;\\text{Simplify}.\\hfill \\\\ &amp; =&amp; 1&amp; \\end{array}[\/latex]<\/p>\r\n4.\r\n<p style=\"text-align: center\">[latex]\\large\\begin{array}{cccc}\\hfill \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}&amp; =&amp;5{\\left(r{s}^{2}\\right)}^{2 - 2}\\hfill &amp;\\text{Use the quotient rule}.\\hfill \\\\ &amp; =&amp; 5{\\left(r{s}^{2}\\right)}^{0}\\hfill &amp;\\text{Simplify}.\\hfill \\\\ &amp; =&amp; 5\\cdot 1\\hfill &amp;\\text{Use the zero exponent rule}.\\hfill \\\\ &amp; =&amp; 5\\hfill &amp;\\text{Simplify}.\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">In the following video you will see more examples of simplifying expressions whose exponents may be zero.<\/span>\r\n\r\nhttps:\/\/youtu.be\/rpoUg32utlc\r\n<h3>Negative exponent rule<\/h3>\r\nAnother useful result occurs if we relax the condition that [latex]a&gt;b[\/latex] in the quotient rule even further. For example, can we simplify [latex]\\displaystyle \\frac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]a&lt;b[\/latex]\u2014that is, where the difference [latex]a-b[\/latex] is negative\u2014we can use the <em>negative rule of exponents<\/em> to simplify the expression to its reciprocal.\r\n\r\nDivide one exponential expression by another with a larger exponent. Use our example, [latex]\\displaystyle \\frac{{h}^{3}}{{h}^{5}}[\/latex].\r\n<div style=\"text-align: center\">[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}&amp; =&amp; \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h}\\hfill \\\\ &amp; =&amp; \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h}\\hfill \\\\ &amp; =&amp; \\frac{1}{h\\cdot h}\\hfill \\\\ &amp; =&amp; \\frac{1}{{h}^{2}}\\hfill \\end{array}[\/latex]<\/div>\r\nIf we were to simplify the original expression using the quotient rule, we would have\r\n<div style=\"text-align: center\">[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}&amp; =&amp; {h}^{3 - 5}\\hfill \\\\ &amp; =&amp; \\text{ }{h}^{-2}\\hfill \\end{array}[\/latex]<\/div>\r\nPutting the answers together, we have [latex]\\displaystyle {h}^{-2}=\\frac{1}{{h}^{2}}[\/latex]. This is true for any nonzero real number, or any variable representing a nonzero real number.\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 or vice versa.\r\n<div style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{ccc}{a}^{-n}=\\frac{1}{{a}^{n}}&amp; \\text{and}&amp; {a}^{n}=\\frac{1}{{a}^{-n}}\\end{array}[\/latex]<\/div>\r\nWe have shown that the exponential expression [latex]{a}^{n}[\/latex] is defined when [latex]n[\/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[\/latex] is defined for any integer [latex]n[\/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>The Negative Rule of Exponents<\/h3>\r\nFor any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that\r\n<div style=\"text-align: center\">[latex]\\displaystyle {a}^{-n}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]\\Large\\frac{{(2b) }^{3}}{{(2b) }^{10}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large\\frac{{z}^{2}\\cdot z}{{z}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large\\frac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"552758\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"552758\"]\r\n<h4>Solution<\/h4>\r\n<ol>\r\n \t<li>[latex]\\Large\\frac{{(2b) }^{3}}{{(2b)}^{10}}={(2b)}^{3 - 10}={(2b) }^{-7}=\\frac{1}{{(2b)}^{7}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large\\frac{{z}^{2}\\cdot z}{{z}^{4}}=\\frac{{z}^{2+1}}{{z}^{4}}=\\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\\frac{1}{z}[\/latex]<\/li>\r\n \t<li>[latex]\\Large\\frac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}={\\left(-5{t}^{3}\\right)}^{4 - 8}={\\left(-5{t}^{3}\\right)}^{-4}=\\frac{1}{{\\left(-5{t}^{3}\\right)}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">In the following video you will see examples of simplifying expressions with negative exponents.<\/span>\r\n\r\nhttps:\/\/youtu.be\/Gssi4dBtAEI\r\n<h3>Combine exponent rules to simplify expressions<\/h3>\r\nIn the next examples we will combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/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]\\displaystyle \\frac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"163692\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"163692\"]\r\n<h4>Solution<\/h4>\r\n<ol>\r\n \t<li>[latex]{b}^{2}\\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\\frac{1}{{b}^{6}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}={\\left(-x\\right)}^{5 - 5}={\\left(-x\\right)}^{0}=1[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle \\frac{-7z}{{\\left(-7z\\right)}^{5}}=\\frac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}={\\left(-7z\\right)}^{1 - 5}={\\left(-7z\\right)}^{-4}=\\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">The following video shows more examples of how to combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.<\/span>\r\n\r\nhttps:\/\/youtu.be\/EkvN5tcp4Cc\r\n<h1>(2.1.5) - Power of a product and a quotient<\/h1>\r\n<h3>Finding the Power of a Product<\/h3>\r\nTo simplify the power of a product of two exponential expressions, we can use the <em>power of a product rule of exponents,<\/em> which breaks up the power of a product of factors into the product of the powers of the factors. For instance, 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{array}{ccc}\\hfill {\\left(pq\\right)}^{3}&amp; =&amp; \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}}\\hfill \\\\ &amp; =&amp; p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q\\hfill \\\\ &amp; =&amp; \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}}\\hfill \\\\ &amp; =&amp; {p}^{3}\\cdot {q}^{3}\\hfill \\end{array}[\/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 of Exponents<\/h3>\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a product rule of exponents states that\r\n<div style=\"text-align: center\">[latex]{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/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(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(2^a{t}\\right)}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle \\frac{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[reveal-answer q=\"788982\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"788982\"]\r\n\r\nUse the product and quotient rules and the new definitions to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{\\left(a{b}^{2}\\right)}^{3}={\\left(a\\right)}^{3}\\cdot {\\left({b}^{2}\\right)}^{3}={a}^{1\\cdot 3}\\cdot {b}^{2\\cdot 3}={a}^{3}{b}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(2^a{t}\\right)}^{15}={\\left(2^a\\right)}^{15}\\cdot {\\left(t\\right)}^{15}={2}^{a\\cdot15}\\cdot{t}^{15}=2^{15a}\\cdot{t}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}={\\left(-2\\right)}^{3}\\cdot {\\left({w}^{3}\\right)}^{3}=-8\\cdot {w}^{3\\cdot 3}=-8{w}^{9}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle \\frac{1}{{\\left(-7z\\right)}^{4}}=\\frac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}=\\frac{1}{2,401{z}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}={\\left({e}^{-2}\\right)}^{7}\\cdot {\\left({f}^{2}\\right)}^{7}={e}^{-2\\cdot 7}\\cdot {f}^{2\\cdot 7}={e}^{-14}{f}^{14}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe may even encounter\r\n\r\n<span style=\"color: #000000\">In the following video, \u00a0we provide more examples of how to find the power of a product.<\/span>\r\n\r\nhttps:\/\/youtu.be\/p-2UkpJQWpo\r\n<h3>Finding the Power of a Quotient<\/h3>\r\nTo simplify the power of a quotient of two expressions, we can use the <em>power of a quotient rule,<\/em> which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let\u2019s look at the following example.\r\n<div style=\"text-align: center\">[latex]\\displaystyle {\\left({e}^{-2}{f}^{2}\\right)}^{7}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/div>\r\nLet\u2019s rewrite the original problem differently and look at the result.\r\n<div style=\"text-align: center\">[latex]\\Large \\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}&amp; =&amp; {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ &amp; =&amp; \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\r\nIt appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.\r\n<div style=\"text-align: center\">[latex]\\Large \\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}&amp; =&amp; {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ &amp; =&amp; \\frac{{\\left({f}^{2}\\right)}^{7}}{{\\left({e}^{2}\\right)}^{7}}\\hfill \\\\ &amp; =&amp; \\frac{{f}^{2\\cdot 7}}{{e}^{2\\cdot 7}}\\hfill \\\\ &amp; =&amp; \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\r\n<div style=\"text-align: left\">\r\n<div class=\"textbox shaded\">\r\n<h3>The Power of a Quotient Rule of Exponents<\/h3>\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a quotient rule of exponents states that\r\n<div style=\"text-align: center\">[latex]\\displaystyle {\\left(\\frac{a}{b}\\right)}^{n}=\\frac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]\\displaystyle {\\left(\\frac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle {\\left(\\frac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle {\\left(\\frac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle {\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle {\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"660878\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"660878\"]\r\n<ol>\r\n \t<li>[latex]\\Large{\\left(\\frac{4}{{z}^{11}}\\right)}^{3}=\\frac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}=\\frac{64}{{z}^{11\\cdot 3}}=\\frac{64}{{z}^{33}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large{\\left(\\frac{p}{{q}^{3}}\\right)}^{6}=\\frac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}=\\frac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}=\\frac{{p}^{6}}{{q}^{18}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large{\\left(\\frac{-1}{{t}^{2}}\\right)}^{27}=\\frac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\\frac{-1}{{t}^{2\\cdot 27}}=\\frac{-1}{{t}^{54}}=-\\frac{1}{{t}^{54}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large{\\left({j}^{3}{k}^{-2}\\right)}^{4}={\\left(\\frac{{j}^{3}}{{k}^{2}}\\right)}^{4}=\\frac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}=\\frac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}=\\frac{{j}^{12}}{{k}^{8}}[\/latex]<\/li>\r\n \t<li>[latex]\\Large{\\left({m}^{-2}{n}^{-2}\\right)}^{3}={\\left(\\frac{1}{{m}^{2}{n}^{2}}\\right)}^{3}=\\frac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}=\\frac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}=\\frac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}=\\frac{1}{{m}^{6}{n}^{6}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000\">The following video provides more examples of simplifying expressions using the power of a quotient and other exponent rules.<\/span>\r\n\r\nhttps:\/\/youtu.be\/BoBe31pRxFM\r\n<h1>(2.1.6) - Definition of a simplified exponential expression<\/h1>\r\n<div class=\"textbox shaded\">\r\n<h3><strong>Definition:<\/strong> An exponential expression is <span style=\"text-decoration: underline\">\"<em><strong>simplified\"<\/strong><\/em><\/span> when:<\/h3>\r\n<ul>\r\n \t<li>No parenthesis appear<\/li>\r\n \t<li>No powers are raised to powers<\/li>\r\n \t<li>Each base occurs only once<\/li>\r\n \t<li>No negative or zero exponents appear<\/li>\r\n<\/ul>\r\n<\/div>\r\n&nbsp;\r\n<h2>Summary<\/h2>\r\n<ul>\r\n \t<li>Evaluating expressions containing exponents is the same as evaluating any expression. You substitute the value of the variable into the expression and simplify.<\/li>\r\n \t<li>The product rule for exponents:\u00a0For any number [latex]x[\/latex] and any integers [latex]a[\/latex] and [latex]b[\/latex],\u00a0[latex]\\left(x^{a}\\right)\\left(x^{b}\\right) = x^{a+b}[\/latex].<\/li>\r\n \t<li>The quotient rule for exponents:\u00a0For any non-zero number [latex]x[\/latex] and any integers [latex]a[\/latex] and [latex]b[\/latex]: [latex] \\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]<\/li>\r\n \t<li>The power rule for exponents:\r\n<ol>\r\n \t<li>For any nonzero numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]x[\/latex], [latex]\\left(ab\\right)^{x}=a^{x}\\cdot{b^{x}}[\/latex].<\/li>\r\n \t<li>For any number [latex]a[\/latex], any non-zero number [latex]b[\/latex], and any integer [latex]x[\/latex], [latex] \\displaystyle {\\left(\\frac{a}{b}\\right)}^{x}=\\frac{a^{x}}{b^{x}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<h2><\/h2>\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(2.1.1) &#8211; Notation and terminology for exponents\n<ul>\n<li>Identify the components of a term containing integer exponents<\/li>\n<li>Evaluate expressions containing integer exponents<\/li>\n<\/ul>\n<\/li>\n<li>(2.1.2) &#8211; The product and quotient rules\n<ul>\n<li>Use the product rule to multiply exponential expressions<\/li>\n<li>Use the quotient rule to divide exponential expressions<\/li>\n<\/ul>\n<\/li>\n<li>(2.1.3) &#8211; The power rule\n<ul>\n<li>Use the power rule to simplify expressions with exponents raised to powers<\/li>\n<\/ul>\n<\/li>\n<li>(2.1.4) &#8211; Negative and zero exponent rules\n<ul>\n<li>Define and use the zero exponent rule<\/li>\n<li>Define and use the negative exponent rule<\/li>\n<li>Combine all the exponents rules to simplify expressions<\/li>\n<\/ul>\n<\/li>\n<li>(2.1.5) &#8211; Power of a product and a quotient\n<ul>\n<li>Simplify an expression with a product raised to a power<\/li>\n<li>Simplify an expression with a quotient raised to a power<\/li>\n<\/ul>\n<\/li>\n<li>(2.1.6) &#8211; Definition of a simplified exponential expression<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"attachment_4445\" style=\"width: 292px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4445\" class=\"wp-image-4445\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183306\/Screen-Shot-2016-05-31-at-10.38.21-AM-150x150.png\" alt=\"Image of a woman taking a picture with a camera repeated five times in different colors.\" width=\"282\" height=\"282\" \/><\/p>\n<p id=\"caption-attachment-4445\" class=\"wp-caption-text\">Repeated Image<\/p>\n<\/div>\n<h1>(2.1.1) &#8211; Notation and terminology for exponents<\/h1>\n<p>We use exponential notation to write repeated multiplication. For example [latex]10\\cdot10\\cdot10[\/latex] can be written more succinctly as [latex]10^{3}[\/latex]. The 10 in [latex]10^{3}[\/latex]<sup>\u00a0<\/sup>is called the <b>base<\/b>. The 3 in [latex]10^{3}[\/latex]<sup>\u00a0<\/sup>is called the <b>exponent<\/b>. The expression [latex]10^{3}[\/latex] is called the exponential expression. Knowing the names for the parts of an exponential expression or term will help you learn how to perform mathematical operations on them.<\/p>\n<p style=\"text-align: center\">[latex]\\text{base}\\rightarrow10^{3\\leftarrow\\text{exponent}}[\/latex]<\/p>\n<p>[latex]10^{3}[\/latex] is read as \u201c10 to the third power\u201d or \u201c10 cubed.\u201d It means [latex]10\\cdot10\\cdot10[\/latex], or 1,000.<\/p>\n<p>[latex]8^{2}[\/latex]\u00a0is read as \u201c8 to the second power\u201d or \u201c8 squared.\u201d It means [latex]8\\cdot8[\/latex], or 64.<\/p>\n<p>[latex]5^{4}[\/latex]\u00a0is read as \u201c5 to the fourth power.\u201d It means [latex]5\\cdot5\\cdot5\\cdot5[\/latex], or 625.<\/p>\n<p>[latex]b^{5}[\/latex]\u00a0is read as \u201c<i>b<\/i> to the fifth power.\u201d It means [latex]{b}\\cdot{b}\\cdot{b}\\cdot{b}\\cdot{b}[\/latex]. Its value will depend on the value of <i>b<\/i>.<\/p>\n<p>The exponent applies only to the number that it is next to. Therefore, in the expression [latex]xy^{4}[\/latex],\u00a0only the <i>y<\/i> is affected by the 4. [latex]xy^{4}[\/latex]\u00a0means [latex]{x}\\cdot{y}\\cdot{y}\\cdot{y}\\cdot{y}[\/latex]. The <em>x<\/em> in this term is a <strong>coefficient<\/strong> of <em>y<\/em>.<\/p>\n<p>If the exponential expression is negative, such as [latex]\u22123^{4}[\/latex], it means [latex]\u2013\\left(3\\cdot3\\cdot3\\cdot3\\right)[\/latex] or [latex]\u221281[\/latex].<\/p>\n<p>If [latex]\u22123[\/latex] is to be the base, it must be written as [latex]\\left(\u22123\\right)^{4}[\/latex], which means [latex]\u22123\\cdot\u22123\\cdot\u22123\\cdot\u22123[\/latex], or 81.<\/p>\n<p>Likewise,\u00a0[latex]\\left(\u2212x\\right)^{4}=\\left(\u2212x\\right)\\cdot\\left(\u2212x\\right)\\cdot\\left(\u2212x\\right)\\cdot\\left(\u2212x\\right)=x^{4}[\/latex], while [latex]\u2212x^{4}=\u2013\\left(x\\cdot x\\cdot x\\cdot x\\right)[\/latex].<\/p>\n<p>You can see that there is quite a difference, so you have to be very careful! The following examples show how to identify the base and the exponent, as well as how to identify the expanded and exponential format of writing repeated multiplication.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Identify the exponent and the base in the following terms, then simplify:<\/p>\n<ol>\n<li>[latex]7^{2}[\/latex]<\/li>\n<li>[latex]{\\left(\\frac{1}{2}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]2x^{3}[\/latex]<\/li>\n<li>[latex]\\left(-5\\right)^{2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q211363\">Show Solution<\/span><\/p>\n<p style=\"text-align: left\">\n<div id=\"q211363\" class=\"hidden-answer\" style=\"display: none\">\n<p>1) [latex]7^{2}[\/latex]<\/p>\n<p>The exponent in this term is 2 and the base is 7. To simplify, expand the term: [latex]7^{2}=7\\cdot{7}=49[\/latex]<\/p>\n<p>2) [latex]{\\left(\\frac{1}{2}\\right)}^{3}[\/latex]<\/p>\n<p>The exponent on this term is 3, and the base is [latex]\\frac{1}{2}[\/latex]. To simplify, expand the multiplication and remember how to multiply fractions: [latex]{\\left(\\frac{1}{2}\\right)}^{3}=\\frac{1}{2}\\cdot{\\frac{1}{2}}\\cdot{\\frac{1}{2}}=\\frac{1}{16}[\/latex]<\/p>\n<p>3) \u00a0[latex]2x^{3}[\/latex]<\/p>\n<p>The exponent on this term is 3, and the base is x, the 2 is not getting the exponent because there are no parentheses that tell us it is. \u00a0This term is in its most simplified form.<\/p>\n<p>4)\u00a0[latex]\\left(-5\\right)^{2}[\/latex]<\/p>\n<p>The exponent on this terms is 2 and the base is [latex]-5[\/latex]. To simplify, expand the multiplication: [latex]\\left(-5\\right)^{2}=-5\\cdot{-5}=25[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p class=\"no-indent\">In the following video you are provided more examples of applying exponents to various bases.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Simplify  Basic Exponential Expressions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ocedY91LHKU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Evaluate expressions<\/h3>\n<p>Evaluating expressions containing exponents is the same as evaluating the linear expressions from earlier in the course. You substitute the value of the variable into the expression and simplify.<\/p>\n<p>You can use the order of operations\u00a0to evaluate the expressions containing exponents. First, evaluate anything in Parentheses or grouping symbols. Next, look for Exponents, followed by Multiplication and Division (reading from left to right), and lastly, Addition and Subtraction (again, reading from left to right).<\/p>\n<p>So, when you evaluate the expression [latex]5x^{3}[\/latex]\u00a0if [latex]x=4[\/latex], first substitute the value 4 for the variable <i>x<\/i>. Then evaluate, using order of operations.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Evaluate the following expressions for the given value.<\/p>\n<ol>\n<li>[latex]5x^{3}[\/latex]<i>\u00a0<\/i>if [latex]x=4[\/latex]<\/li>\n<li>[latex]\\left(5x\\right)^{3}[\/latex]\u00a0if [latex]x=4[\/latex]<\/li>\n<li>[latex]x^{3}[\/latex] if [latex]x=\u22124[\/latex]<\/li>\n<li>[latex]3^x[\/latex] if [latex]x = 4[\/latex]<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q411363\">Show Solution<\/span><\/p>\n<div id=\"q411363\" class=\"hidden-answer\" style=\"display: none\">\n<p>1) Substitute 4 for the variable <i>x<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]5\\cdot4^{3}[\/latex]<\/p>\n<p>Evaluate [latex]4^{3}[\/latex]. Multiply.<\/p>\n<p style=\"text-align: center\">[latex]5\\left(4\\cdot4\\cdot4\\right)=5\\cdot64=320[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]5x^{3}=320[\/latex]\u00a0when [latex]x=4[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>2) [latex]\\left(5x\\right)^{3}[\/latex]\u00a0if [latex]x=4[\/latex]<\/p>\n<p>Substitute 4 for the variable <i>x<\/i>.\u00a0notice the how adding parentheses can change the outcome when you are simplifying terms with exponents.<\/p>\n<p style=\"text-align: center\">[latex]\\left(5\\cdot4\\right)3[\/latex]<\/p>\n<p>Multiply inside the parentheses, then apply the exponent\u2014following the rules of PEMDAS.<\/p>\n<p style=\"text-align: center\">[latex]20^{3}[\/latex]<\/p>\n<p>Evaluate [latex]20^{3}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]20\\cdot20\\cdot20=8,000[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(5x\\right)3=8,000[\/latex] when [latex]x=4[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>3) \u00a0[latex]x^{3}[\/latex] if [latex]x=\u22124[\/latex].<\/p>\n<p>Substitute [latex]\u22124[\/latex] for the variable <i>x<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]\\left(\u22124\\right)^{3}[\/latex]<\/p>\n<p>Evaluate. Note how placing parentheses around the [latex]\u22124[\/latex] means the negative sign also gets multiplied.<\/p>\n<p style=\"text-align: center\">[latex]\u22124\\cdot\u22124\\cdot\u22124[\/latex]<\/p>\n<p>Multiply.<\/p>\n<p style=\"text-align: center\">[latex]\u22124\\cdot\u22124\\cdot\u22124=\u221264[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x^{3}=\u221264[\/latex] when [latex]x=\u22124[\/latex]<\/p>\n<p>4) \u00a0[latex]3^x[\/latex] if [latex]x = 4[\/latex]<\/p>\n<p>Substitute x = 4 into the exponent. \u00a0[latex]3^x=3^4=3\\cdot3\\cdot3\\cdot3=81[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3^{x}=81[\/latex] when [latex]x=4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"75\" height=\"66\" \/><\/p>\n<p>Caution! Whether to include a negative sign as part of a base or not often leads to confusion. To clarify\u00a0whether a negative sign is applied before or after the exponent, here is an example.<\/p>\n<p>What is the difference in the way you would evaluate these two terms?<\/p>\n<ol>\n<li style=\"text-align: left\">[latex]-{3}^{2}[\/latex]<\/li>\n<li style=\"text-align: left\">[latex]{\\left(-3\\right)}^{2}[\/latex]<\/li>\n<\/ol>\n<p>To evaluate 1), you would apply the exponent to the three first, then apply the negative sign last, like this:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}-\\left({3}^{2}\\right)\\\\=-\\left(9\\right) = -9\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">To evaluate 2), you would apply the exponent to the 3 and the negative sign:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{\\left(-3\\right)}^{2}\\\\=\\left(-3\\right)\\cdot\\left(-3\\right)\\\\={ 9}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">The key to remembering this is to follow the order of operations. The first expression does not include parentheses so you would apply the exponent to the integer 3 first, then apply the negative sign. The second expression includes parentheses, so hopefully you will remember that the negative sign also gets squared.<\/p>\n<\/div>\n<p id=\"video0\" class=\"no-indent\" style=\"text-align: left\">In the next sections, you will learn how to simplify expressions that contain exponents. Come back to this page if you forget how to apply the order of operations to a term with exponents, or forget which is the base and which is the exponent!<\/p>\n<p class=\"no-indent\" style=\"text-align: left\">In the following video you are provided with examples of evaluating exponential expressions for a given number.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Evaluate Basic Exponential Expressions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pQNz8IpVVg0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(2.1.2) &#8211; The product and quotient rules<\/h1>\n<h3>The product rule<\/h3>\n<p><b>Exponential notation<\/b> was developed to write repeated multiplication more efficiently. There are times when it is easier or faster to leave the expressions in exponential notation when multiplying or dividing. Let\u2019s look at rules that will allow you to do this.<\/p>\n<p>For example, the notation [latex]5^{4}[\/latex]\u00a0can be expanded and written as [latex]5\\cdot5\\cdot5\\cdot5[\/latex], or 625. And don\u2019t forget, the exponent only applies to the number immediately to its left, unless there are parentheses.<\/p>\n<p>What happens if you multiply two numbers in exponential form with the same base? Consider the expression [latex]{2}^{3}{2}^{4}[\/latex]. Expanding each exponent, this can be rewritten as [latex]\\left(2\\cdot2\\cdot2\\right)\\left(2\\cdot2\\cdot2\\cdot2\\right)[\/latex] or [latex]2\\cdot2\\cdot2\\cdot2\\cdot2\\cdot2\\cdot2[\/latex]. In exponential form, you would write the product as [latex]2^{7}[\/latex]. Notice that 7 is the sum of the original two exponents, 3 and 4.<\/p>\n<p>What about [latex]{x}^{2}{x}^{6}[\/latex]? This can be written as [latex]\\left(x\\cdot{x}\\right)\\left(x\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\right)=x\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}\\cdot{x}[\/latex] or [latex]x^{8}[\/latex]. And, once again, 8 is the sum of the original two exponents. This concept can be generalized in the following way:<\/p>\n<div class=\"textbox shaded\">\n<h3>The Product Rule for Exponents<\/h3>\n<p>For any number <i>x<\/i> and any integers <i>a<\/i> and <i>b<\/i>,\u00a0[latex]\\left(x^{a}\\right)\\left(x^{b}\\right) = x^{a+b}[\/latex].<\/p>\n<p>To multiply exponential terms with the same base, add the exponents.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\n<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q772709\">Show Answer<\/span><\/p>\n<div id=\"q772709\" class=\"hidden-answer\" style=\"display: none\">Solution<\/p>\n<p>Use the product rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\n<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<p>At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.<\/p>\n<div style=\"text-align: center\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\n<p>Notice we get the same result by adding the three exponents in one step.<\/p>\n<div style=\"text-align: center\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\n<div style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183308\/traffic-sign-160659.png\" alt=\"traffic-sign-160659\" width=\"96\" height=\"83\" \/><\/h3>\n<p>Caution! Do not try to apply this rule to sums.<\/p>\n<p>Think about the expression\u00a0[latex]\\left(2+3\\right)^{2}[\/latex]<\/p>\n<p style=\"text-align: center\">Does [latex]\\left(2+3\\right)^{2}[\/latex] equal [latex]2^{2}+3^{2}[\/latex]?<\/p>\n<p>No, it does not because of the order of operations!<\/p>\n<p style=\"text-align: center\">[latex]\\left(2+3\\right)^{2}=5^{2}=25[\/latex]<\/p>\n<p style=\"text-align: center\">and<\/p>\n<p style=\"text-align: center\">[latex]2^{2}+3^{2}=4+9=13[\/latex]<\/p>\n<p>Therefore, you can only use this rule when the numbers inside the parentheses are being multiplied (or divided, as we will see next).<\/p>\n<\/div>\n<p><span style=\"color: #000000\">In the following video you will see more examples of using the product rule for exponents to simplify expressions.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Simplify Expressions Using the Product Rule of Exponents (Basic)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/P0UVIMy2nuI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our last product rule example we will show that an exponent can be an algebraic expression. \u00a0We can use the product rule for exponents no matter what the exponent looks like, as long as the base is the same.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Multiply.\u00a0[latex]x^{a+2}\\cdot{x^{3a-9}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q740329\">Show Answer<\/span><\/p>\n<div id=\"q740329\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have two exponentiated terms with the same base, so we can multiply them together. The product rule for exponents says that we can add the exponents.<\/p>\n<p>[latex]x^{a+2}\\cdot{x^{3a-9}}=x^{(a+2)+(3a-9)}=x^{4a-7}[\/latex]<\/p>\n<p>The expression can&#8217;t be simplified any further.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x^{a+2}\\cdot{x^{3a-9}}=x^{4a-7}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>The quotient rule<\/h3>\n<p>Let\u2019s look at dividing terms containing exponential expressions. What happens if you divide two numbers in exponential form with the same base? Consider the following expression.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{{{4}^{5}}}{{{4}^{2}}}[\/latex]<\/p>\n<p>You can rewrite the expression as: [latex]\\displaystyle \\frac{4\\cdot 4\\cdot 4\\cdot 4\\cdot 4}{4\\cdot 4}[\/latex]. Then you can cancel the common factors of 4 in the numerator and denominator: [latex]\\displaystyle[\/latex]<\/p>\n<p>Finally, this expression can be rewritten as [latex]4^{3}[\/latex]\u00a0using exponential notation. Notice that the exponent, 3, is the difference between the two exponents in the original expression, 5 and 2.<\/p>\n<p>So,\u00a0[latex]\\displaystyle \\frac{{{4}^{5}}}{{{4}^{2}}}=4^{5-2}=4^{3}[\/latex].<\/p>\n<p>Be careful that you subtract the exponent in the denominator from the exponent in the numerator.<\/p>\n<p>So, to divide two exponential terms with the same base, subtract the exponents.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Quotient (Division) Rule for Exponents<\/h3>\n<p>For any non-zero number [latex]x[\/latex] and any integers [latex]a[\/latex] and [latex]b[\/latex]: [latex]\\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]\\displaystyle \\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}[\/latex]<\/li>\n<li>[latex]\\displaystyle \\frac{{t}^{23}}{{t}^{15}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q978732\">Show Answer<\/span><\/p>\n<div id=\"q978732\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the quotient rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]\\displaystyle \\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}={\\left(-2\\right)}^{14 - 9}={\\left(-2\\right)}^{5}[\/latex]<\/li>\n<li>[latex]\\displaystyle \\frac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>As we showed with the product rule, you may be given a quotient with an exponent that is an algebraic expression to simplify. \u00a0As long as the bases agree, you may use the quotient rule for exponents.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\displaystyle \\frac{y^{x-3}}{y^{9-x}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836863\">Show Answer<\/span><\/p>\n<div id=\"q836863\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a quotient whose terms have the same base so we can use the quotient rule for exponents.<\/p>\n<p>[latex]\\displaystyle \\frac{y^{x-3}}{y^{9-x}}=y^{(x-3)-(9-x)}=y^{2x-12}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">In the following video, you will we more examples of using the quotient rule for exponents.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Simplify Expressions Using the Quotient Rule of Exponents (Basic)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xy6WW7y_GcU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(2.1.3) &#8211; The power rule<\/h1>\n<p>Another word for exponent is power. \u00a0You have likely seen or heard an example such as [latex]3^5[\/latex] can be described as 3 raised to the 5th power. In this section we will further expand our capabilities with exponents. We will learn what to do when a term with a\u00a0power\u00a0is raised to another power, and what to do when two numbers or variables are multiplied and both are raised to an exponent. \u00a0We will also learn what to do when numbers or variables that are divided are raised to a power. \u00a0We will begin by raising powers to powers.<\/p>\n<p>Let\u2019s simplify [latex]\\left(5^{2}\\right)^{4}[\/latex]. In this case, the base is [latex]5^2[\/latex]<sup>\u00a0<\/sup>and the exponent is 4, so you multiply [latex]5^{2}[\/latex]<sup>\u00a0<\/sup>four times: [latex]\\left(5^{2}\\right)^{4}=5^{2}\\cdot5^{2}\\cdot5^{2}\\cdot5^{2}=5^{8}[\/latex]<sup>\u00a0<\/sup>(using the Product Rule\u2014add the exponents).<\/p>\n<p>[latex]\\left(5^{2}\\right)^{4}[\/latex]<sup>\u00a0<\/sup>is a power of a power. It is the fourth power of 5 to the second power. And we saw above that the answer is [latex]5^{8}[\/latex]. Notice that the new exponent is the same as the product of the original exponents: [latex]2\\cdot4=8[\/latex].<\/p>\n<p>So, [latex]\\left(5^{2}\\right)^{4}=5^{2\\cdot4}=5^{8}[\/latex]\u00a0(which equals 390,625, if you do the multiplication).<\/p>\n<p>Likewise, [latex]\\left(x^{4}\\right)^{3}=x^{4\\cdot3}=x^{12}[\/latex]<\/p>\n<p>This leads to another rule for exponents\u2014the <b>Power Rule for Exponents<\/b>. To simplify a power of a power, you multiply the exponents, keeping the base the same. For example, [latex]\\left(2^{3}\\right)^{5}=2^{15}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>The Power Rule for Exponents<\/h3>\n<p>For any positive number <i>x<\/i> and integers <i>a<\/i> and <i>b<\/i>: [latex]\\left(x^{a}\\right)^{b}=x^{a\\cdot{b}}[\/latex].<\/p>\n<p>Take a moment to contrast how this is different from the product rule for exponents found on the previous page.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{\\left({x}^{2}\\right)}^{7}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q841688\">Show Solution<\/span><\/p>\n<div id=\"q841688\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the power rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{\\left({x}^{2}\\right)}^{7}={x}^{2\\cdot 7}={x}^{14}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}={\\left(2t\\right)}^{5\\cdot 3}={\\left(2t\\right)}^{15}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}={\\left(-3\\right)}^{5\\cdot 11}={\\left(-3\\right)}^{55}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">\u00a0In the following video you will see more examples of using the power rule to simplify expressions with exponents.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Simplify Expressions Using the Power Rule of Exponents (Basic)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VjcKU5rA7F8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.<\/p>\n<table style=\"width: 20%\">\n<thead>\n<tr>\n<th style=\"text-align: center\" colspan=\"5\">Product Rule<\/th>\n<th style=\"text-align: center\" colspan=\"6\">Power Rule<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]5^{3}\\cdot5^{4}[\/latex]<\/td>\n<td>=<\/td>\n<td>\u00a0[latex]5^{3+4}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]5^{7}[\/latex]<\/td>\n<td>but<\/td>\n<td>[latex]\\left(5^{3}\\right)^{4}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]5^{3\\cdot4}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]5^{12}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x^{5}\\cdot x^{2}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]x^{5+2}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]x^{7}[\/latex]<\/td>\n<td>but<\/td>\n<td>[latex]\\left(x^{5}\\right)^{2}[\/latex]<\/td>\n<td>=<\/td>\n<td>\u00a0[latex]x^{5\\cdot2}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]x^{10}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\left(3a\\right)^{7}\\cdot\\left(3a\\right)^{10}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]\\left(3a\\right)^{7+10}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]\\left(3a\\right)^{17}[\/latex]<\/td>\n<td>but<\/td>\n<td>[latex]\\left(\\left(3a\\right)^{7}\\right)^{10}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]\\left(3a\\right)^{7\\cdot10}[\/latex]<\/td>\n<td>=<\/td>\n<td>[latex]\\left(3a\\right)^{70}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h1>(2.1.4) &#8211; Negative and zero exponent rules<\/h1>\n<h3>Zero exponent rule<\/h3>\n<p>Return to the quotient rule. We worked with expressions for which\u00a0\u00a0[latex]a>b[\/latex] so that the difference [latex]a-b[\/latex] would never be zero or negative.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Quotient (Division) Rule for Exponents<\/h3>\n<p>For any non-zero number [latex]x[\/latex]\u00a0and any integers [latex]a[\/latex] and [latex]b[\/latex]: [latex]\\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]<\/p>\n<\/div>\n<p>What would happen if [latex]a=b[\/latex]? In this case, we would use the <em>zero exponent rule of exponents<\/em> to simplify the expression to 1. To see how this is done, let us begin with an example.<\/p>\n<p style=\"text-align: center\">[latex]\\frac{t^{8}}{t^{8}}=\\frac{\\cancel{t^{8}}}{\\cancel{t^{8}}}=1[\/latex]<\/p>\n<p>If we were to simplify the original expression using the quotient rule, we would have<\/p>\n<div style=\"text-align: center\">[latex]\\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[\/latex]<\/div>\n<p>If we equate the two answers, the result is [latex]{t}^{0}=1[\/latex]. This is true for any nonzero real number, or any variable representing a real number.<\/p>\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\n<p>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.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Zero Exponent Rule of Exponents<\/h3>\n<p>For any nonzero real number [latex]a[\/latex], the zero exponent rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify each expression using the zero exponent rule of exponents.<\/p>\n<ol>\n<li>[latex]\\Large\\frac{{c}^{3}}{{c}^{3}}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{-3{x}^{5}}{{x}^{5}}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{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=\"q375469\">Show Answer<\/span><\/p>\n<div id=\"q375469\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the zero exponent and other rules to simplify each expression.<\/p>\n<p>1.<\/p>\n<p style=\"text-align: center\">[latex]\\large\\begin{array}\\text{ }\\frac{c^{3}}{c^{3}} \\hfill& =c^{3-3} \\\\ \\hfill& =c^{0} \\\\ \\hfill& =1\\end{array}[\/latex]<\/p>\n<p>2.<\/p>\n<p style=\"text-align: center\">[latex]\\large\\begin{array}{ccc}\\hfill \\frac{-3{x}^{5}}{{x}^{5}}& =& -3\\cdot \\frac{{x}^{5}}{{x}^{5}}\\hfill \\\\ & =& -3\\cdot {x}^{5 - 5}\\hfill \\\\ & =& -3\\cdot {x}^{0}\\hfill \\\\ & =& -3\\cdot 1\\hfill \\\\ & =& -3\\hfill \\end{array}[\/latex]<\/p>\n<p>3.<\/p>\n<p style=\"text-align: center\">[latex]\\large\\begin{array}{cccc}\\hfill \\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}}\\hfill &\\text{Use the product rule in the denominator}.\\hfill \\\\ & =&\\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}}\\hfill & \\text{Simplify}.\\hfill \\\\ & =&{\\left({j}^{2}k\\right)}^{4 - 4}\\hfill &\\text{Use the quotient rule}.\\hfill \\\\ & =& {\\left({j}^{2}k\\right)}^{0}\\hfill&\\text{Simplify}.\\hfill \\\\ & =& 1& \\end{array}[\/latex]<\/p>\n<p>4.<\/p>\n<p style=\"text-align: center\">[latex]\\large\\begin{array}{cccc}\\hfill \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}& =&5{\\left(r{s}^{2}\\right)}^{2 - 2}\\hfill &\\text{Use the quotient rule}.\\hfill \\\\ & =& 5{\\left(r{s}^{2}\\right)}^{0}\\hfill &\\text{Simplify}.\\hfill \\\\ & =& 5\\cdot 1\\hfill &\\text{Use the zero exponent rule}.\\hfill \\\\ & =& 5\\hfill &\\text{Simplify}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">In the following video you will see more examples of simplifying expressions whose exponents may be zero.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Simplify Expressions Using the Quotient and Zero Exponent Rules\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/rpoUg32utlc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Negative exponent rule<\/h3>\n<p>Another useful result occurs if we relax the condition that [latex]a>b[\/latex] in the quotient rule even further. For example, can we simplify [latex]\\displaystyle \\frac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]a<b[\/latex]\u2014that is, where the difference [latex]a-b[\/latex] is negative\u2014we can use the <em>negative rule of exponents<\/em> to simplify the expression to its reciprocal.<\/p>\n<p>Divide one exponential expression by another with a larger exponent. Use our example, [latex]\\displaystyle \\frac{{h}^{3}}{{h}^{5}}[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}& =& \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h}\\hfill \\\\ & =& \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h}\\hfill \\\\ & =& \\frac{1}{h\\cdot h}\\hfill \\\\ & =& \\frac{1}{{h}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n<p>If we were to simplify the original expression using the quotient rule, we would have<\/p>\n<div style=\"text-align: center\">[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\\hfill \\\\ & =& \\text{ }{h}^{-2}\\hfill \\end{array}[\/latex]<\/div>\n<p>Putting the answers together, we have [latex]\\displaystyle {h}^{-2}=\\frac{1}{{h}^{2}}[\/latex]. This is true for any nonzero real number, or any variable representing a nonzero real number.<\/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 or vice versa.<\/p>\n<div style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{ccc}{a}^{-n}=\\frac{1}{{a}^{n}}& \\text{and}& {a}^{n}=\\frac{1}{{a}^{-n}}\\end{array}[\/latex]<\/div>\n<p>We have shown that the exponential expression [latex]{a}^{n}[\/latex] is defined when [latex]n[\/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[\/latex] is defined for any integer [latex]n[\/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>The Negative Rule of Exponents<\/h3>\n<p>For any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]\\displaystyle {a}^{-n}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]\\Large\\frac{{(2b) }^{3}}{{(2b) }^{10}}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{{z}^{2}\\cdot z}{{z}^{4}}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q552758\">Show Answer<\/span><\/p>\n<div id=\"q552758\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Solution<\/h4>\n<ol>\n<li>[latex]\\Large\\frac{{(2b) }^{3}}{{(2b)}^{10}}={(2b)}^{3 - 10}={(2b) }^{-7}=\\frac{1}{{(2b)}^{7}}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{{z}^{2}\\cdot z}{{z}^{4}}=\\frac{{z}^{2+1}}{{z}^{4}}=\\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\\frac{1}{z}[\/latex]<\/li>\n<li>[latex]\\Large\\frac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}={\\left(-5{t}^{3}\\right)}^{4 - 8}={\\left(-5{t}^{3}\\right)}^{-4}=\\frac{1}{{\\left(-5{t}^{3}\\right)}^{4}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">In the following video you will see examples of simplifying expressions with negative exponents.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Simplify Expressions Using the Quotient and Negative Exponent Rules\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Gssi4dBtAEI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Combine exponent rules to simplify expressions<\/h3>\n<p>In the next examples we will combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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]\\displaystyle \\frac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q163692\">Show Answer<\/span><\/p>\n<div id=\"q163692\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Solution<\/h4>\n<ol>\n<li>[latex]{b}^{2}\\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\\frac{1}{{b}^{6}}[\/latex]<\/li>\n<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}={\\left(-x\\right)}^{5 - 5}={\\left(-x\\right)}^{0}=1[\/latex]<\/li>\n<li>[latex]\\displaystyle \\frac{-7z}{{\\left(-7z\\right)}^{5}}=\\frac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}={\\left(-7z\\right)}^{1 - 5}={\\left(-7z\\right)}^{-4}=\\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">The following video shows more examples of how to combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Simplify Expressions Using Exponent Rules (Product, Quotient, Zero Exponent)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/EkvN5tcp4Cc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(2.1.5) &#8211; Power of a product and a quotient<\/h1>\n<h3>Finding the Power of a Product<\/h3>\n<p>To simplify the power of a product of two exponential expressions, we can use the <em>power of a product rule of exponents,<\/em> which breaks up the power of a product of factors into the product of the powers of the factors. For instance, 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{array}{ccc}\\hfill {\\left(pq\\right)}^{3}& =& \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}}\\hfill \\\\ & =& p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q\\hfill \\\\ & =& \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}}\\hfill \\\\ & =& {p}^{3}\\cdot {q}^{3}\\hfill \\end{array}[\/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 of Exponents<\/h3>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a product rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/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(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(2^a{t}\\right)}^{15}[\/latex]<\/li>\n<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\displaystyle \\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788982\">Show Answer<\/span><\/p>\n<div id=\"q788982\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the product and quotient rules and the new definitions to simplify each expression.<\/p>\n<ol>\n<li>[latex]{\\left(a{b}^{2}\\right)}^{3}={\\left(a\\right)}^{3}\\cdot {\\left({b}^{2}\\right)}^{3}={a}^{1\\cdot 3}\\cdot {b}^{2\\cdot 3}={a}^{3}{b}^{6}[\/latex]<\/li>\n<li>[latex]{\\left(2^a{t}\\right)}^{15}={\\left(2^a\\right)}^{15}\\cdot {\\left(t\\right)}^{15}={2}^{a\\cdot15}\\cdot{t}^{15}=2^{15a}\\cdot{t}^{15}[\/latex]<\/li>\n<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}={\\left(-2\\right)}^{3}\\cdot {\\left({w}^{3}\\right)}^{3}=-8\\cdot {w}^{3\\cdot 3}=-8{w}^{9}[\/latex]<\/li>\n<li>[latex]\\displaystyle \\frac{1}{{\\left(-7z\\right)}^{4}}=\\frac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}=\\frac{1}{2,401{z}^{4}}[\/latex]<\/li>\n<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}={\\left({e}^{-2}\\right)}^{7}\\cdot {\\left({f}^{2}\\right)}^{7}={e}^{-2\\cdot 7}\\cdot {f}^{2\\cdot 7}={e}^{-14}{f}^{14}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>We may even encounter<\/p>\n<p><span style=\"color: #000000\">In the following video, \u00a0we provide more examples of how to find the power of a product.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Simplify Expressions Using Exponent Rules (Power of a Product)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/p-2UkpJQWpo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Finding the Power of a Quotient<\/h3>\n<p>To simplify the power of a quotient of two expressions, we can use the <em>power of a quotient rule,<\/em> which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let\u2019s look at the following example.<\/p>\n<div style=\"text-align: center\">[latex]\\displaystyle {\\left({e}^{-2}{f}^{2}\\right)}^{7}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/div>\n<p>Let\u2019s rewrite the original problem differently and look at the result.<\/p>\n<div style=\"text-align: center\">[latex]\\Large \\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}& =& {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ & =& \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\n<p>It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.<\/p>\n<div style=\"text-align: center\">[latex]\\Large \\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}& =& {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ & =& \\frac{{\\left({f}^{2}\\right)}^{7}}{{\\left({e}^{2}\\right)}^{7}}\\hfill \\\\ & =& \\frac{{f}^{2\\cdot 7}}{{e}^{2\\cdot 7}}\\hfill \\\\ & =& \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"text-align: left\">\n<div class=\"textbox shaded\">\n<h3>The Power of a Quotient Rule of Exponents<\/h3>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a quotient rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]\\displaystyle {\\left(\\frac{a}{b}\\right)}^{n}=\\frac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]\\displaystyle {\\left(\\frac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\displaystyle {\\left(\\frac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\n<li>[latex]\\displaystyle {\\left(\\frac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\n<li>[latex]\\displaystyle {\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\n<li>[latex]\\displaystyle {\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q660878\">Show Answer<\/span><\/p>\n<div id=\"q660878\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\Large{\\left(\\frac{4}{{z}^{11}}\\right)}^{3}=\\frac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}=\\frac{64}{{z}^{11\\cdot 3}}=\\frac{64}{{z}^{33}}[\/latex]<\/li>\n<li>[latex]\\Large{\\left(\\frac{p}{{q}^{3}}\\right)}^{6}=\\frac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}=\\frac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}=\\frac{{p}^{6}}{{q}^{18}}[\/latex]<\/li>\n<li>[latex]\\Large{\\left(\\frac{-1}{{t}^{2}}\\right)}^{27}=\\frac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\\frac{-1}{{t}^{2\\cdot 27}}=\\frac{-1}{{t}^{54}}=-\\frac{1}{{t}^{54}}[\/latex]<\/li>\n<li>[latex]\\Large{\\left({j}^{3}{k}^{-2}\\right)}^{4}={\\left(\\frac{{j}^{3}}{{k}^{2}}\\right)}^{4}=\\frac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}=\\frac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}=\\frac{{j}^{12}}{{k}^{8}}[\/latex]<\/li>\n<li>[latex]\\Large{\\left({m}^{-2}{n}^{-2}\\right)}^{3}={\\left(\\frac{1}{{m}^{2}{n}^{2}}\\right)}^{3}=\\frac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}=\\frac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}=\\frac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}=\\frac{1}{{m}^{6}{n}^{6}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000\">The following video provides more examples of simplifying expressions using the power of a quotient and other exponent rules.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Simplify Expressions Using Exponent Rules (Power of a Quotient)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BoBe31pRxFM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(2.1.6) &#8211; Definition of a simplified exponential expression<\/h1>\n<div class=\"textbox shaded\">\n<h3><strong>Definition:<\/strong> An exponential expression is <span style=\"text-decoration: underline\">&#8220;<em><strong>simplified&#8221;<\/strong><\/em><\/span> when:<\/h3>\n<ul>\n<li>No parenthesis appear<\/li>\n<li>No powers are raised to powers<\/li>\n<li>Each base occurs only once<\/li>\n<li>No negative or zero exponents appear<\/li>\n<\/ul>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Summary<\/h2>\n<ul>\n<li>Evaluating expressions containing exponents is the same as evaluating any expression. You substitute the value of the variable into the expression and simplify.<\/li>\n<li>The product rule for exponents:\u00a0For any number [latex]x[\/latex] and any integers [latex]a[\/latex] and [latex]b[\/latex],\u00a0[latex]\\left(x^{a}\\right)\\left(x^{b}\\right) = x^{a+b}[\/latex].<\/li>\n<li>The quotient rule for exponents:\u00a0For any non-zero number [latex]x[\/latex] and any integers [latex]a[\/latex] and [latex]b[\/latex]: [latex]\\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]<\/li>\n<li>The power rule for exponents:\n<ol>\n<li>For any nonzero numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]x[\/latex], [latex]\\left(ab\\right)^{x}=a^{x}\\cdot{b^{x}}[\/latex].<\/li>\n<li>For any number [latex]a[\/latex], any non-zero number [latex]b[\/latex], and any integer [latex]x[\/latex], [latex]\\displaystyle {\\left(\\frac{a}{b}\\right)}^{x}=\\frac{a^{x}}{b^{x}}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ul>\n<h2><\/h2>\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-783\">\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>Simplify Basic Exponential Expressions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ocedY91LHKU\">https:\/\/youtu.be\/ocedY91LHKU<\/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>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>Repeated Image. <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>Evaluate Basic Exponential Expressions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/pQNz8IpVVg0\">https:\/\/youtu.be\/pQNz8IpVVg0<\/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>Ex: Expanding and Evaluating Exponential Notation . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/KOnQpKSpVRo\">https:\/\/youtu.be\/KOnQpKSpVRo<\/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>Ex: Simplify Exponential Expressions Using the Product Property of Exponents . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hA9AT7QsXWo\">https:\/\/youtu.be\/hA9AT7QsXWo<\/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>Simplify Expressions Using the Quotient Rule of Exponents (Basic). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xy6WW7y_GcU\">https:\/\/youtu.be\/xy6WW7y_GcU<\/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>Simplify Expressions Using the Power Rule of Exponents (Basic). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/VjcKU5rA7F8\">https:\/\/youtu.be\/VjcKU5rA7F8<\/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>Simplify Expressions Using the Quotient and Zero Exponent Rules. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Simplify Expressions Using the Quotient and Negative Exponent Rules . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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>Simplify Expressions Using Exponent Rules (Product, Quotient, Zero Exponent). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/EkvN5tcp4Cc\">https:\/\/youtu.be\/EkvN5tcp4Cc<\/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>Simplify Expressions Using Exponent Rules (Power of a Product). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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>Simplify Expressions Using Exponent Rules (Power of a Quotient). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/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>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@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><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\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/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":21,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Simplify Basic Exponential Expressions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ocedY91LHKU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Repeated Image\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Evaluate Basic Exponential Expressions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/pQNz8IpVVg0\",\"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@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex: Expanding and Evaluating Exponential Notation \",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/KOnQpKSpVRo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex: Simplify Exponential Expressions Using the Product Property of Exponents \",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/hA9AT7QsXWo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using the Quotient Rule of Exponents (Basic)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/xy6WW7y_GcU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using the Power Rule of Exponents (Basic)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/VjcKU5rA7F8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et. al\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using the Quotient and Zero Exponent Rules\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/rpoUg32utlc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using the Quotient and Negative Exponent Rules \",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Gssi4dBtAEI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using Exponent Rules (Product, Quotient, Zero Exponent)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/EkvN5tcp4Cc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using Exponent Rules (Power of a Product)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/p-2UkpJQWpo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Expressions Using Exponent Rules (Power of a Quotient)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/BoBe31pRxFM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"d04dde00-5811-4c62-abb6-6c8154312732","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-783","chapter","type-chapter","status-web-only","hentry"],"part":774,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/783","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":29,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/783\/revisions"}],"predecessor-version":[{"id":5724,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/783\/revisions\/5724"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/774"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/783\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/media?parent=783"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=783"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/contributor?post=783"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/license?post=783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}