{"id":1986,"date":"2019-08-26T11:12:28","date_gmt":"2019-08-26T16:12:28","guid":{"rendered":"http:\/\/www.kaptest.com\/blog\/prep\/?p=1986"},"modified":"2023-08-29T15:11:40","modified_gmt":"2023-08-29T15:11:40","slug":"psat-math-exponents-and-radicals","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/psat\/psat-math-exponents-and-radicals\/","title":{"rendered":"PSAT Math: Exponents and Radicals"},"content":{"rendered":"

We often turn to our calculators to solve difficult radical and exponent problems, especially in math-intensive classes. However, being too calculator dependent can cost you time and points on the PSAT. Further, on the PSAT, many radical and exponent problems are structured in such a way that your calculator can\u2019t help you, even if it is allowed.<\/p>\n

This chapter will review algebra and arithmetic rules that you may have learned at some point but likely haven\u2019t used in a while. It will reacquaint you with the formulas and procedures you\u2019ll need to simplify even the toughest expressions and equations on the PSAT.<\/p>\n

We\u2019ll start with exponents.<\/p>\n

Exponents on the PSAT<\/h3>
<\/div><\/div><\/div>
\nQuestions involving exponents often look intimidating, but when you know the rules governing them, you\u2019ll see that there are plenty of shortcuts. First, it\u2019s important to understand the anatomy of a term that has an exponent. This term is comprised of two pieces: a base and an exponent (also called a power). The base is the number in larger type and is the value being multiplied by itself. The exponent, written as a superscript, shows you how many times the base is being multiplied by itself.
\nBase<\/i> \u21d2 34 <\/sup>\u21d0<\/sup> Exponent<\/sup><\/i> is the same as 3 \u00d7 3 \u00d7 3 \u00d7 3<\/p>\n

The following table lists the rules you\u2019ll need to handle any exponent question you\u2019ll see on the PSAT.<\/p>\n

\n\n\n\n\n\n\n\n\n\n\n
Rule<\/th>\nExample<\/th>\n<\/tr>\n<\/thead>\n
When multiplying two terms with the same base, add the exponents.<\/td>\n\n
\n
ab<\/sup><\/i> \u00d7 ac<\/sup><\/i> = a<\/i>(b+c<\/i>)<\/sup> \u2192<\/span> 42<\/sup> \u00d7 43<\/sup> = 42+3<\/sup> = 45<\/sup><\/div>\n<\/div>\n<\/td>\n<\/tr>\n
When dividing two terms with the same base, subtract the exponents.<\/td>\n\n
ab<\/sup>ac<\/sup>=a(b\u2212c)<\/sup>\u2192\u200943<\/sup>42<\/sup>=4(3\u22122)<\/sup>=41<\/sup><\/div>\n<\/td>\n<\/tr>\n
When raising a power to another power, multiply the exponents.<\/td>\n(ab<\/sup><\/i>)c<\/i><\/sup> = a<\/i>(bc<\/i>)<\/sup> \u2192<\/span> (43<\/sup>)2<\/sup> = 43\u00d72<\/sup> = 46<\/sup>; (2x<\/i>\u200a2<\/sup>)3<\/sup> = 21\u00d73<\/sup> x<\/i>2\u00d73<\/sup> = 8x<\/i>6<\/sup><\/span><\/td>\n<\/tr>\n
When raising a product to a power, apply the power to all factors in the product.<\/td>\n\n
(ab<\/i>)c<\/i><\/sup> = a<\/i>c<\/i><\/sup> \u00d7 b<\/i>c<\/i><\/sup> \u2192<\/span> (2m<\/i>)3<\/sup> = 23<\/sup> \u00d7 m<\/i>3<\/sup> = 8m<\/i>3<\/sup><\/div>\n<\/td>\n<\/tr>\n
Any term raised to the zero power equals 1.<\/td>\na<\/i>0<\/sup> = 1 \u2192<\/span> 40<\/sup> = 1<\/span><\/td>\n<\/tr>\n
A base raised to a negative exponent can be rewritten as the reciprocal raised to the positive of the original exponent.<\/td>\n\n
a\u2013b<\/sup>=1\/ab<\/sup>;\u20091\/a\u2013b<\/sup> = ab<\/sup>\u2009\u2192\u2009\u20094\u20132<\/sup> = 1\/42<\/sup>;\u20091\/4\u20132<\/sup>=42<\/sup><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Different things happen to different kinds of numbers when they are raised to powers. Compare the locations and values of the variables and numbers on the following number line to the results in the table for a summary.<\/p>\n

<\/div><\/div><\/div>\n<\/div>\n
\n\n\n\n\n\n\n\n\n\n\n\n
Quantity<\/th>\nEven Exponent Result<\/th>\nOdd Exponent Result<\/th>\nExample<\/th>\n<\/tr>\n<\/thead>\n
w<\/i><\/td>\npositive, absolute value increases<\/td>\nnegative, absolute value increases<\/td>\n(\u20135)2<\/sup> = 25; (\u20135)3<\/sup> = \u2013125<\/span><\/td>\n<\/tr>\n
\u20131<\/td>\nalways 1<\/td>\nalways \u20131<\/td>\nn\/a<\/td>\n<\/tr>\n
x<\/i><\/td>\npositive, absolute value decreases<\/td>\nnegative, absolute value decreases<\/td>\n\n
(\u20131\/2)2<\/sup> =\u20091\/4;\u2009(\u20131\/2)3<\/sup> =\u2009\u20131\/8<\/div>\n<\/td>\n<\/tr>\n
0<\/td>\nalways 0<\/td>\nalways 0<\/td>\nn\/a<\/td>\n<\/tr>\n
y<\/i><\/td>\npositive, absolute value decreases<\/td>\npositive, absolute value decreases<\/td>\n\n
(1\/4)2<\/sup>=\u20091\/16;\u2009(1\/4)3<\/sup> =\u20091\/64<\/div>\n<\/td>\n<\/tr>\n
1<\/td>\nalways 1<\/td>\nalways 1<\/td>\nn\/a<\/td>\n<\/tr>\n
z<\/i><\/td>\npositive, absolute value increases<\/td>\npositive, absolute value increases<\/td>\n32<\/sup> = 9; 33<\/sup> = 27<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

PSAT Math Practice Question: Exponents<\/h3>
<\/div><\/div><\/div>\n<\/div>\n
Which of the following has the same value as\u00a0 64<\/sup> \u00d7 363<\/sup> \/ 45<\/sup>?<\/section>\n
    \n
  1. 39<\/sup>\/2<\/li>\n
  2. 310<\/sup><\/li>\n
  3. 22<\/sup> \u00d7 39<\/sup><\/li>\n
  4. 312<\/sup>\/22<\/sup><\/li>\n<\/ol>\n

    Use the Kaplan Method for Math to solve this question, working through it step-by-step. The following table shows Kaplan\u2019s strategic thinking on the left, along with suggested math scratchwork on the right.<\/p>\n

    \n\n\n\n\n\n\n\n
    Strategic Thinking<\/th>\nMath Scratchwork<\/th>\n<\/tr>\n<\/thead>\n
    Step 1: Read the question, identifying and organizing important information as you go<\/b> You\u2019re asked to identify the expression that has the same value as the one presented; this means you need to simplify it.<\/td>\n<\/td>\n<\/tr>\n
    Step 2: Choose the best strategy to answer the question<\/b> As written, you can\u2019t combine the bases or the exponents. However, 36 = 62<\/sup>,<\/span> so rewrite the numerator to reflect this relationship. Then combine the bases in the numerator by adding the exponents. The bases are now being raised to the same power, 5, so rewrite the expression using a single exponent. Then simplify by dividing.<\/td>\n\u00a0 64<\/sup> \u00d7 363<\/sup> \/ 45<\/sup> = (62<\/sup>)2<\/sup> \u00d7 363<\/sup> \/ 45<\/sup> = 362<\/sup> \u00d7 363<\/sup> \/ 45<\/sup> \u00a0 365<\/sup>\/45<\/sup> = (36\/4)5<\/sup> = 95<\/sup><\/td>\n<\/tr>\n
    Step 3: Check that you answered the\u00a0<\/b>right\u00a0<\/i><\/b>question<\/b> Although 95<\/sup> is correct, it’s not one of the answer choices, so you’ll need to simplify even further. Rewrite 9 as 32<\/sup>\u00a0and then use exponent rules to simplify. The result is 310<\/sup>, which is (B).<\/td>\n\u00a0 \u00a0 95<\/sup> = (32<\/sup>)5<\/sup> =310<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Radicals on the PSAT<\/h3>
    <\/div><\/div><\/div>
    \nA radical can be written using a fractional exponent. You can think of addition and subtraction (and multiplication and division) as opposites; similarly, raising a number to a power and taking the root of the number are another opposite pair.<\/p>\n<\/div>\n
    Specifically, when you raise a term to the n<\/i>th power, taking the n<\/i>th root will return the original term. Consider for example\u00a034<\/sup> = 3 \u00d7 3 \u00d7 3 \u00d7 3 =81.<\/span>\u00a0If you take the fourth root of 81 (that is, determine the number that can be multiplied by itself four times to get 81), you will arrive at the original term:\u00a04<\/sup>\u221a81 = 4<\/sup>\u221a(3\u00d73\u00d73\u00d73) = 3 .<\/span><\/div>\n
    Radicals can be intimidating at first, but remembering the basic rules for radicals can make them much easier to tackle. The following table contains all the formulas you\u2019ll need to know to achieve \u201cradical\u201d success on the PSAT<\/span>.<\/div>\n
    <\/div>\n
    \n\n\n\n\n\n\n\n\n\n
    Rule<\/th>\nExample<\/th>\n<\/tr>\n<\/thead>\n
    When a fraction is under a radical, you can rewrite it using two radicals: one containing the numerator and the other containing the denominator.<\/td>\n\n
    \u221a(a\/b) = \u221aa \/ \u221ab\u2009\u2192\u2009\u221a(4\/9) = \u221a4\/\u221a9 = 2\/3<\/div>\n<\/td>\n<\/tr>\n
    Two factors under a single radical can be rewritten as separate radicals multiplied together.<\/td>\n\n
    \u221a(ab) = \u221aa \u00d7 \u221ab\u2009\u2192\u2009\u221a75 = \u221a25 \u00d7 \u221a3 = 5\u221a3<\/div>\n<\/td>\n<\/tr>\n
    A radical can be written using a fractional exponent.<\/td>\n\n
    \u221aa= a(1\/2)<\/sup>,3<\/sup>\u221aa = a(1\/3)<\/sup>\u2009\u2192\u2009\u221a289 = 289(1\/2)<\/sup><\/div>\n<\/td>\n<\/tr>\n
    <\/td>\n\n

    a(b\/c)<\/sup> = c<\/sup>\u221aab<\/sup>\u2009\u2192\u20095(2\/3)<\/sup>\u21923<\/sup>\u221a52<\/sup><\/p>\n<\/td>\n<\/tr>\n

    When a number is squared, the original number can be positive or negative, but the square root of a number can only be positive.<\/td>\nIf a<\/i>2<\/sup> = 81<\/span>, then a<\/i> = \u00b19<\/span>, BUT \u221a81 = 9 only.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    It is not considered proper notation to leave a radical in the denominator of a fraction. However, it\u2019s sometimes better to keep them through intermediate steps to make the math easier (and sometimes the radical is eliminated along the way). Once all manipulations are complete, the denominator can be rationalized to remove a remaining radical by multiplying both the numerator and denominator by that same radical.<\/p>\n<\/div>\n

    \n\n\n\n\n\n\n
    1. Original Fraction<\/th>\n2. Rationalization<\/th>\n3. Intermediate Math<\/th>\n4. Resulting Fraction<\/th>\n<\/tr>\n<\/thead>\n
    x\/\u221a5<\/td>\nx\/\u221a5 \u00d7 \u221a5\/\u221a5<\/td>\n\n
    x\u221a5\/\u221a(5\u00d75) = x\u221a5\/\u221a25 = x\u221a5\/5<\/div>\n<\/td>\n
    		x\u221a5\/5<\/td>\n<\/tr>\n
    14\/\u221a(x2<\/sup>+2)<\/td>\n\n
    14\/\u221a(x2<\/sup>+2) \u00d7 \u221a(x2<\/sup>+2)\/\u221a(x2<\/sup>+2)<\/div>\n<\/td>\n
    		\n
    14\u221a(x2<\/sup>+2) \/ \u221a(x2<\/sup>+2)(x2<\/sup>+2) = 14\u221a(x2<\/sup>+2)\/\u221a(x2<\/sup>+2)2<\/sup><\/div>\n<\/td>\n
    		14\u221a(x2<\/sup>+2)\/x2<\/sup>+2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    Sometimes, you\u2019ll have an expression such as 2 + \u221a5 in the denominator. To rationalize this, multiply by its conjugate, which is found by negating the second term; in this case, the conjugate is 2 \u2212 \u221a5 \u00a0. As a general rule of thumb, you are not likely to see a radical in the denominator of the answer choices on the PSAT<\/span>, so you\u2019ll need to be comfortable with rationalizing expressions that contain radicals.<\/p>\n