Knowing the fundamental rules of exponents will help you simplify\u00a0mathematical\u00a0equations and statements so you can arrive at a\u00a0solution with relative ease. As with everything, it may take some practice to get the hang of working with exponents. But if you take your time, don't allow yourself to be overwhelmed by variables, and memorize each rule, you'll be a master in no time.<\/p>\n
Before getting into the theory of exponents, let's first look at\u00a0a few notations and related terminology.<\/p>\n
Let's take the term We read this as \"x<\/em>\u00a0raised to the power of\u00a0n<\/em>,\" which means we multiply\u00a0x<\/em>\u00a0n number of times.<\/p>\n Here,\u00a0x<\/em>\u00a0is any real number and\u00a0n<\/em>\u00a0is a positive integer.\u00a0We refer to\u00a0x<\/em>\u00a0as\u00a0\"base,\" and\u00a0n<\/em>\u00a0as\u00a0\"power\"\u00a0or \"exponent.\"<\/p>\n Another example, we read There are eight\u00a0fundamental rules of exponents.<\/p>\n A\u00a0number to the\u00a01 power is equal to the number itself.<\/p>\n We multiply\u00a0x<\/em>\u00a0one time, which is equal to the number itself.<\/p>\n Examples:<\/strong><\/p>\n Any\u00a0number raised to the power of 0 (except zero) is equal to 1.<\/p>\n Examples:<\/strong><\/p>\n Note: When multiplying, you can add two exponents with the same base.<\/p>\n As the bases of both the terms\u00a0are the same, we can add\u00a0their powers.<\/p>\n Examples:<\/strong><\/p>\n In special cases, we can apply this rule even when there are the same bases but with different signs (positive\/negative).\u00a0We just need to rewrite the terms in accordance with\u00a0our rule.<\/p>\n For example:<\/strong><\/p>\n Simplify: We can write This implies, By subtracting the exponents, we can divide two powers with the same base.<\/p>\n Here the bases of both the terms under division are the same, so we subtract\u00a0the\u00a0exponents.<\/p>\n Examples:<\/strong><\/p>\n To raise a power to another power, we multiply the exponents.<\/p>\n We raise the inner term to an exponent and so we multiply\u00a0both the exponents.<\/p>\n Examples:<\/strong><\/p>\n Any non-zero number raised to a negative power equals its reciprocal raised to the opposite positive power.<\/p>\n x<\/em>\u00a0is raised to a negative exponent and we can write it as\u00a0x<\/em>'s reciprocal with the opposite sign of\u00a0n<\/em>.<\/p>\n Examples:<\/strong><\/p>\n When there is a power to a fraction, we can rewrite the term as both the numerator and denominator raised to the power of the exponent.<\/p>\n Examples:<\/strong><\/p>\n When you have a fractional exponent, the numerator is the power and the denominator is the root (square root for denominator = 2, cube root for denominator = 3, and so on).<\/p>\n As we studied the fundamental laws of exponents, we shall now implement\u00a0these rules in the following examples.\u00a0Observe the statement or equation carefully before rushing in to apply the rules. You may be able to solve what seems like a\u00a0five-step\u00a0problem\u00a0in just two steps. This is\u00a0especially true when\u00a0dealing with exponents.<\/p>\n Simplify: Solution\u00a0<\/strong><\/p>\n From the power rule, we get Simplify: Solution<\/strong><\/p>\n From the negative exponent rule, we get From the quotient rule, we get From the product rule, we get From the power rule, we get Hence, Simplify: Solution<\/strong><\/p>\n Look carefully. We can simplify this in two steps without even simplifying the individual terms under multiplication. Upon observation, we find that the bases of the powers are equal and we can apply the rule of product.<\/p>\n Applying the zero rule we get, Our aim is not just to solve problems or equations. Our aim is to solve them\u00a0efficiently\u00a0and effectively. It is true that we can solve or handle the statements without using these rules but that process is highly ineffective, complex, and time-consuming. So, we\u00a0use these rules of exponents to solve and simplify equations and statements efficiently.<\/p>\nx^n<\/code>.<\/p>\nx^n<\/code>=x\u22c5x\u22c5x<\/em>...n times.<\/p>\n9^5<\/code> as \"9 raised to the power of 5\" and it is equal to 9\u22c59\u22c59\u22c59\u22c59.<\/p>\n9^5<\/code> = 9\u22c59\u22c59\u22c59\u22c59 = 59,049<\/em><\/p>\nRules of Exponents<\/h2>\n
1. The One Rule<\/h3>\n
x^n<\/code> = n<\/em>\u00a0where\u00a0n = 1\u00a0<\/em><\/p>\nx^1<\/code> = x<\/em><\/p>\n9^1<\/code> = 9<\/em><\/p>\n01 = 0<\/code><\/p>\n-1^1 = -1<\/code><\/p>\n2. The Zero Rule<\/h3>\n
x^0 = 1<\/code><\/p>\n9^0 = 1<\/code><\/p>\n1^0 = 1<\/code><\/p>\n0^0<\/code> is undefined.<\/p>\n3. The Product Rule<\/h3>\n
x^a\u22c5x^b = x^a+b<\/code><\/p>\n9^5\u22c59^7 = 9^(5+7) = 9^12<\/code><\/p>\n8^-2\u22c58^10 = 8^(-2+10) = 8^8<\/code><\/em><\/p>\n(-9)^5\u22c5(9)^7<\/code><\/p>\n(-9)^5<\/code> as (-1\u22c59)^5<\/code> which is equal to (-1)^5\u22c5(9)^5<\/code>.<\/em><\/p>\n(-9)^5\u22c5(9)^7 = (-1)^5\u22c5(9)^5\u22c5(9)^7 = (-1)^5\u22c5(9)^5+7 = (-1)^5\u22c5(9)^12 = -9^12<\/code>.<\/em><\/p>\n4. The Quotient Rule<\/h3>\n
x^a\/x^b = x^(a-b)<\/code><\/em><\/p>\n9^5\/9^7 = 9^(5-7) = 9^-2<\/code><\/p>\n2^2\/2^2 = 2^(2-2) = 2^0 = 1<\/code><\/p>\n5. The Power Rule<\/h3>\n
(x^m)^n = x^(m\u22c5n)<\/code><\/p>\n(9^5)^7 = 9^(5\u22c57) = 9^35<\/code><\/p>\n(8^2)^2 = 8^(2\u22c52) = 8^4<\/code><\/p>\n6. The Negative Exponent<\/h3>\n
x^-n = 1\/x^n<\/code><\/p>\n9^-5 = 1\/9^5<\/code><\/p>\n1^-1 = 1\/1^1 = 1<\/code><\/p>\n7. The Power of a Fraction Rule<\/h3>\n
(x\/y)^n = x^n\/x^y<\/code><\/p>\n(2\/5)^5 = 2^5\/5^5<\/code><\/em><\/p>\n(1\/2)^2 = 1^2\/2^2 = 1\/2^2<\/code><\/em><\/p>\n8. A Fractional Exponent<\/h3>\n
b^n\/m = (^m \u221ab)^n<\/code><\/p>\nExamples<\/h2>\n
Problem 1<\/strong><\/h3>\n
((x\/y)^n)^(1\/n)<\/code><\/p>\n(x\/y)^(n\u22c5(1\/n))<\/code> which is equal to (x\/y)^1 = x\/y<\/code>.<\/p>\nProblem 2<\/h3>\n
((x^a\/x^b)x^c)^-d<\/code><\/p>\n1\/((x^a\/x^b)x^c)^d<\/code>.<\/p>\n1\/((x^a-b)x^c)^d<\/code>.<\/p>\n1\/(x^a-b+c)^d<\/code>.<\/p>\n1\/x^((a-b+c)^d)<\/code> which equals to 1\/x^(ad-bd+cd)<\/code>.<\/p>\n1\/((x^a\/x^b)x^c)^-d = 1\/(x^(ad-bd+cd))<\/code><\/p>\nProblem 3<\/h3>\n
(((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^-1)\u22c5(((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^1)<\/code><\/p>\n(((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^-1)\u22c5(((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^1) = ((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^-1+1 = ((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^0<\/code><\/p>\n((x\u22c5x\u22c5x\u22c5x\u22c5x)^y)^0 = 1<\/code><\/p>\nConclusion<\/h2>\n