Exponents Drill. Warm-up Problems. Problem 1 If (x 3 y 3 ) -3 = (xy) -z, what is z? A) -6 B) 0 C) 1 D) 6 E) 9. Problem 2 36 =?

Transcription

1 Exponents Drill Warm-up Problems Problem 1 If (x 3 y 3 ) -3 = (xy) -z, what is z? A) -6 B) 0 C) 1 D) 6 E) 9 Problem =? A) 0 B) 1/36 C) 1/6 D) 6 E) 36 Problem 3 3 ( xy) =? 6 6 x y A) (xy) -9 B) (xy) -3 C) xy 3 D) x 3 y E) (xy) 3 Problem =? A) B) 4 20 C) 2 20 D) 2 12

2 E) 2 10 Problem =? A) -80 B) 0 C) 1 D) 9-2 E) 9-4 Problem =? A) 1 B) 6 C) 6 3 D) 6 5 /7 E) 6 10 /7 Problem =? A) 1/96 B) 1/8 C) 1 D) 3 E) 6 Problem =? A) 5/9

3 B) 1/2 C) 5/2 D) 5/4 E) 25/2 Problem (.3 )(.5 ) 8 3 (.25 )(1.2 ) =? A) 1/4 B) 1/2 C) 1 D) 4 E) 6 Problem 10 If (3 10 )(6 5 ) = 2 x 3 y, and x and y are integers, what is x + y? A) 30 B) 20 C) 15 D) 10 E) 5 Problem 11 If 4 x 5 y = 400, where x and y are positive integers, what is 4 x+1 5 y-1? A) 360 B) 320 C) 32 D) 20 E) 16 Problem 12 If = 30 2x, what is x? A) 64 B) 32

4 C) 8 D) 4 E) 2 Problem =? A) 1/4 B) 1/2 C) 1 D) 5/4 E) 5/2 Problem =? A) 1/5 B) 2/5 C) 1 D) 5/2 E) 20 Problem =? A) 1/6 B) 1/5 C) 1/3 D) 2/3 E) 3/2 Problem 16 If x is a positive integer and (2 x )(2 x )(2 x ) = 16 6, then what is the value of x?

5 A) 2 B) 3 C) 4 D) 6 E) 8 Problem =? A) 3 B) 18 C) 19 D) 27 E) 54 Problem 18 If m and n are integers and (4 m )(3 n ) = 6 6, what is m? A) 1 B) 2 C) 3 D) 6 E) 12 Problem =? A) 1/5 B) 2/5 C) 1 D) 2 E) 4 Problem = x, what is x?

6 A) (9 6 )/10 B) 81 C) 9 D) 1 E) 9/10

7 Medium Problems Problem 21 (1/4) -2 + (1/4) -2 + (1/4) -2 + (1/4) -2 =? A) (1/4) -16 B) (1/4) -8 C) 1 D) 4 E) 64 Problem 22 ( )(5 5 ) = 20 x ; what is x? A) 1 B) 5 C) 6 D) 9 E) 10 Problem 23 If 5 2x = 2500, what is the value of 5 x+1? A) 2500 B) 1250 C) 500 D) 250 E) 100 Problem 24 (1/2) -8 (1/2) -6 + (1/2) -8 (1/2) -6 =? A) 0 B) C) 2 4 D) E) 2 28

8 Problem =? A) 1 B) 2 C) 6 D) 10 E) 16 Problem 26 If = 10 3x, what is x+2? A) 0 B) 2 C) 4 D) 6 E) 8 Problem =? A) 4 B) 2 C) 1 D) 1/2 E) 1/4 Problem 28 2x 3x 42 7 =? 3x x A) 1 B) 6 -x C) 7 -x D) 36 -x E) 49 -x

9 Problem 29 y x+ 2y 2x If = 32, what is x? x+ 2y 2x y A) 5 B) 2 C) 1 D) 0 E) -1 Problem =? A) 1/10 B) 1/5 C) 1 D) 4 E) 12

10 Challenge Problems Problem = A) -6 B) -3 C) 1 D) 3 E) 6 x ; what is x? Problem 32 If 5 n is a factor of 78,126, where n is an integer, what is the value of 5 n+1 5 n? A) 100 B) 25 C) 16 D) 5 E) 4 Problem 33 What is z if 6 2z z 48 z z = 1? A) -2 B) -1/2 C) 0 D) 1/2 E) 2 Problem 34 If 5 2n+1 5 2n = , where n is an integer, what is the value of n? A) 8 B) 6 C) 4 D) 2 E) 0

11 Problem 35 If = z, what is the value of z? A) 53 B) 52 C) 25 D) 12.5 E) 6 Problem =? A) 1 B) 2 C) 10 D) 2 10 E) 2 14 Problem 37 If x > 1 and A) 0 B) 1 C) y D) 2y E) y 2 x (2 y) 2y 2x 2 x (4( xy) ) = 1, what is x in terms of y? Problem 38 x+ y 100 If > 1, where x and y are integers, then which of the following MUST be 3x 2y 10 true? I. y > 0 II. x + y > 0 III. x 4y < 0

12 A) II only B) III only C) I and III D) II and III E) I, II, and III Problem x y+ x 3x y 2y =? A) 1 B) 5 C) 7 D) 5 x E) 7 x Problem =? A) 1/6 B) 1/4 C) 1 D) 3 E) 4

13 Problem 1 Solution: E The problem gives us a clue regarding how to proceed by setting the right hand side of the equation to (xy) -z. Because both sides of the equation feature some combination of xy, we must manipulate the left hand side of the equation to come up with a base that matches xy. (x 3 y 3 ) -3 = ((xy) 3 ) -3 = (xy) -9 Thus (xy) -9 = (xy) -z. Because the bases, (xy), are the same, the only way the two sides of the equation will equal one another is when the exponents are equal. So -9 = -z, or z = 9. Problem 2 Solution: B This problem appears very difficult at first glance, but its difficulty is reduced by recognizing that one must identify a common base to proceed. The common base is uncovered by recognizing the common exponent of -4 in the denominator. Because both 3 and 2 are raised to -4 we can group the two together as we do below: = = 3 2 (3 2) There is no longer a common exponent to manipulate but there is a concealed hidden base of 6. That is because 36 = 6 2. In this case the equation can be rewritten as follows: 36 (6 ) 6 6 = = = Now because the bases are the same the numerator and denominator can be combined as follows: ( 4) 2 = 6 = 6 4 The answer is reached by recognizing that a negative exponent is equivalent to rewriting the number as a fraction as follows: = =. 2 Problem 3 Solution: E

14 This problem is attacked by recognizing that one must identify a common base or exponent to proceed. The common base is uncovered by recognizing the common exponent of -6 in the denominator. Because both x and y are raised to -6 we can group the two together as a product as we do below: ( xy) ( xy) = x y ( xy) There is no longer a common exponent to manipulate the problem but there is a common base of xy. Because the bases are the same the numerator and denominator can be combined as follows: ( xy) ( xy) 3 3 ( 6) 3 = ( xy) = ( xy). 6 Problem 4 Solution: D This problem is solved by recognizing that the terms are common and can be combined. To see this, imagine 4 5 were x. In that case the expression could be rewritten as x + x + x + x = 4x. Substituting for x yields 44, which can be rewritten as follows: This can then be combined as follows because 4 is a common base: 4 4 = 4 = A quick scan of the answer choices reveals that 4 6 is not present. To finish the problem off, 4 6 is rewritten as (2 2 ) 6. In this case the external exponent 6 is combined with the internal exponent of 2 via multiplication as follows: (2 ) = 2 = 2. Problem 5 Solution: A In this problem we must deal with a fraction. The key to exponent problems involving fractions is to look for common elements. An element is common when all of the numbers can be divided by that number. In this case, 9 2 is certainly common to 9 2

15 and 9 4, because = 9 9. We can thus start the problem by factoring out a 9 2 from the numerator: (1 9) = Now we have a 9 2 in the numerator and the denominator so the 9 2 s will cancel leaving: (1 9 2 ) = 1 81 = -80 As a takeaway you must recognize that both simple and very difficult problems on the GMAT hinge upon this ability to recognize and factor out common elements. Problem 6 Solution: B In this problem we must deal with a fraction. The key to exponent problems involving fractions is to look for common elements. An element is common when all of the numbers can be divided by that number. In this case, 6 2 is common to 6 3 and 6 4 and of course to the expression in the denominator. If we factor 6 2 out of the numerator we have the following equation: (6 + 6) = = At this point we could finish the problem by computing the value of and dividing by 7. But there is another way to complete the problem that may help in future problems. 6 is common to 6 2 and 6 so we can divide the numerator by 6 and get the following: 6(6 + 1) 6 7 = = To make the problem easier to manage we factored out a 6 from the numerator and were left with 6(6+1). This was helpful because the (6+1) canceled out the 7 in the denominator. In this case the step may not have saved much time but in future problems this step of factoring to uncover a hidden common element will be crucial for solving very difficult problems. Problem 7 Solution: D This problem is solved by breaking the numerator and denominator into the smallest prime factors possible:

16 96 = 8 12 = (2 2 2) (2 2 3) = = 2 3 Rewriting the expression: (2 3) = = = = = = (2 3) The correct answer is 3. Problem 8 Solution: D This problem is solved by breaking the numerator and denominator into the smallest prime factors possible. Starting with the denominator: 90 5 = ( ) 5 = ( ) 5 = (2 5 )(3 10 )(5 5 ) Then breaking down the numerator: (50 3 )(3 10 ) = (2 5 5) 3 (3 10 ) = (2 5 2 ) 3 (3 10 ) = (2 3 )(5 6 )(3 10 ) Combining the numerator and denominator yields that appear in the numerator and denominator, which then yields 5/4.. To solve, cancel any factors =. The answer is 2 4 Problem 9 Solution: D This problem is solved by combining like terms. Here the exponents 3 and 8 are common in the numerator and denominator. The fraction can be rewritten by grouping the like terms:

17 3.3.3 To continue the problem, one must recognize that = 3. Rewriting the equation yields: = = = = (2-2 ) 3 (2 8 ) = (2-6 )(2 8 ) = = 2 2 = The correct answer is 4. Problem 10 Solution: B Because the bases 2 and 3 are raised to the unknown exponents of x and y, respectively, we must start this problem by simplifying the left hand side of the equation to match the bases on right. To do so, break down the 6 5 into its prime factors of 2 and 3 as follows: = 3 10 (2 3) 5 = (3 10 )(2 5 )(3 5 ) = (2 5 )( ) = (2 5 )(3 15 ) = (2 x )(3 y ) Because the bases on either side of the equation are the same, the only way the two sides of the equation will equal one another is when the exponents are equal. So, x = 5 and y = 15. Be careful to answer the correct question: x + y = = 20. Problem 11 Solution: B There are two major approaches to solving this problem. In the first, each side is broken down into prime factors, and then the unknown variables are calculated. Starting with the left hand side of the equation: 4 x 5 y = (2 2 ) x 5 y = 2 2x 5 y On the right hand side, 400 is broken down through prime factorization: 400 = = = 2 2x 5 y Because the bases on either side of the equation are the same, the only way the two sides of the equation will equal one another is when the exponents are equal. So, x = 2 and y = 2. Thus, 4 x+1 5 y-1 = = 320. The other way to look at the problem is to compare 4 x+1 5 y-1 and 4 x 5 y. 4 x+1 is 4 1 times as much as 4 x because x = 4 x+1. 5 y-1 is 5-1 times as much as 5 y since y = 5 y-1.

18 We can then use this information to calculate the correct answer by multiplying 400 by 4/5: 400 4/5 = 320. Problem 12 Solution: D Because the base 30 is raised to 2x on the right hand side we must start this problem by adjusting the left hand side of the equation to match the right. To do so, notice that each term on the right is raised to a power of 8. As such, these terms can be combined through multiplication because they share a common exponent: = (2 3 5) 8 = 30 8 = 30 2x Because the bases on either side of the equation are the same, the only way the two sides of the equation will equal one another is when the exponents are equal. So 8 = 2x, and x = 4. Problem 13 Solution: A As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, there are some obvious common bases, but there are in fact several hidden common bases as well as hidden common exponents. Here two different solutions are provided to demonstrate the flexibility of exponent problem solutions: Solution 1 Find the prime bases, simplify, and solve: (2)(5) = = = = 2 = (2 5) This method is labor intensive, but thorough. Solution 2 Find common exponents and quickly eliminate: (2)(5) = = = 2 = = = 2 = 2 = = 2 = This solution could have saved some time and is scalable to more difficult problems involving common exponents. The answer is 1/4.

19 Problem 14 Solution: B As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, there is a 10 in the numerator and denominator but otherwise there do not appear to be any common bases or exponents worth examining. As such, one may be tempted to break down each base into its prime bases. In this case, however, it is more efficient first to attack the common base of 10 and then to factor the bases of the remaining terms until a common base is discovered, thus saving valuable time. This is possible because there are several multiples of 10 in the numerator and denominator. To simplify, it may be productive to simplify the terms in the numerator to match the base 10 in the denominator = 20 = Now the 20 2 can be factored to expose a hidden common base of 10: = (2 10) = = = = 4/10 = 2/5 The answer is 2/5. Problem 15 Solution: A As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, there are some obvious common bases, and others can be found by breaking down larger numbers such as 4, 8, 25, and 27 into their prime bases. For example, = and = = Now the external exponents need to be distributed as in = = =

20 Finally, the expression is simplified by grouping and canceling like terms: = = = 1 = The answer is 1/6 Problem 16 Solution: E This problem is solved by combining like terms. Here, we need to work with the common bases of 2 on the left hand side of the equation. Because 2 x is multiplied by itself 3 times, the left side of the equation can be simplified by adding the exponents: (2 x )(2 x )(2 x ) = 2 x+x+x = 2 3x = 16 6 To solve for the unknown exponent, the base on the left hand side of the equation must match that of the right. In this case, 2 is a prime number and also a factor of 16, so 16 should be broken down to a power of = ( ) 6 = (2 4 ) 6 The external exponent is combined by multiplying by the internal exponent yielding: 2 3x = (2 4 ) 6 = 2 24 For the left hand side of the equation to equal the right the exponents must be equal, therefore 2 3x = 2 24, or 3x = 24, and x = 8. The correct answer is 8. Problem 17 Solution: E This problem is completed by recognizing and grouping the common terms 3 3 and 3 2 by summing the terms together. To see this, imagine 3 3 were x and 3 2 were y. In that case the equation could be rewritten as x + x + x = 3x and y + y + y = 3y. Substituting 3 3 for x and 3 2 for y yields 3(3 3 ) 3(3 2 ), which can be rewritten as follows: 3(3 3 ) 3(3 2 ) = = At this point the answer can be calculated by subtracting the value 3 3, or 27, from 3 4, or 81. Another way to calculate the value is by factoring out the common term of 3 3 as follows: = 3 3 (3 1) = = 27 2 = 54

21 The factoring step is optional but it is a step that is frequently required for solving more difficult GMAT exponent problems so it should be practiced often. Problem 18 Solution: C As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem there do not appear to be any common bases or exponents worth examining. As such, we must break down the left and right hand side of the equation into prime bases. Starting with left hand side (4 m )(3 n ) = ((2 2 ) m )(3 n ) = (2 2m )(3 n ). On the right hand side 6 6 = (2 3) 6 = (2 6 )(3 6 ). To find the value of m, recognize that for the left hand side of the equation to equal the right, the exponent to which 2 is raised on both sides of the equation must be equal. Thus 2 6 = 2 2m or 6 = 2m, or m =3. The answer is 3. Problem 19 Solution: D This problem is solved by combining like terms. Here the exponents 2 and 3 are common in the numerator and denominator. The fraction can be rewritten by grouping the like terms: = To continue the problem, recognize that =. Rewriting the equation yields: = = 2 = (2-2 ) (2 3 ) = = 2 1 The correct answer is 2. Problem 20 Solution: B In this problem, the fraction needs to be simplified. The key to simplifying exponent problems involving fractions is to look for numbers that are common to both the numerator and denominator. In this case, it is not readily apparent what might be common to both ( ) and 10. From this point, the problem can be solved either by using order of operations or by factoring. Both solutions are examined below:

22 Solution 1 Order of operations: = = = This solution is computationally intensive and the labor involved isn t scalable to larger exponents. If, for example, the numerator were 9 5 or higher, it would not be easy to calculate the value of the numerator within a two minute time frame. Solution 2 discusses a method that allows for greater speed no matter how large the exponent. Solution 2 Factoring the numerator In some exponent problems it is helpful to factor a polynomial such as to expose common elements. In this case 9 2 is a factor of both 9 3 and 9 2. In this case the numerator can be factored by dividing by 9 2 : (9 + 1) = = = The answer is 81. Factoring is a powerful strategy for dealing with problems with high exponents or problems where there is not a clear connection between bases in the numerator and denominator. As a takeaway, it is helpful to recognize that factoring is often signaled by the presence of a polynomial, which in this case was , over a another number or polynomial. Problem 21 Solution: E The first step of this problem is to eliminate the negative exponent and the fraction (1/4). This is completed by recognizing that (1/4) could be rewritten as 4-1 as follows: (1/4) -2 + (1/4) -2 + (1/4) -2 + (1/4) -2 = (4-1 ) -2 + (4-1 ) -2 + (4-1 ) -2 + (4-1 ) -2 = This problem is completed then by recognizing that the term 4 2 is common and the terms can be summed together. To see this, imagine 4 2 were x. In that case the equation could be rewritten as x + x + x + x = 4x. Substituting 4 2 for x yields 4 4 2, which can be rewritten as follows: (4 1 ) (4 2 ) This can then be combined as follows because 4 is a common base:

23 4 4 2 = = 4 3 = 64. Problem 22 Solution: C The key to exponent problems is to look for common elements such as a common base or exponent. The fact that the problem is seeking a value for an unknown exponent means that the base of the left hand side must match the base of the right hand side, otherwise there is no way to equate the unknown variable on the right hand side, which is an exponent, with the known values on the left. ( )(5 5 ) = 20 x so ( ) (5 5 ) = (4 5) x = (4 x )(5 x ) This problem is completed then by recognizing that the term 4 6 is common and can be factored out. To see this, imagine 4 6 were x. In that case the expression in parentheses could be rewritten as 4x + x. Substituting 4 6 for x yields , which means the equation can be rewritten as follows: 4 6 (4+1)(5 5 ) = (4 6 )5(5 5 ) = = 4 x 5 x For the two sides of the equation to be equal the exponents of the 4 and 5 must be equivalent on both sides so x must equal 6. x = 6. Problem 23 Solution: D At first glance, 5 2x = 2500 appears to be an equation we can solve because 25 is 5 2. Unfortunately, this is not the case and a brief examination of powers of 5 will bear this out: 5, 25, 125, 625, 3125, etc. Therefore it is wholly unproductive to attempt to solve for x and then use this value to calculate 5 x+1. Instead, this problem is solved by solving for 5 x first. To see this, examine the expression that must be solved for: 5 x+1 = (5 x )(5 1 ) = 5(5 x ) The question tells us that 5 2x = This expression can be modified in a manner similar to what was done above: 5 2x = (5 x ) 2 = 2500 Now both sides of the equation are perfect squares and 5 x can be isolated by taking the square root of both sides:

24 x ( ) 2 5 = 2500, so 5 x = 50 To solve the equation, substitute 50 in for 5 x to yield: 5(5 x ) = 5(50) = 250. Problem 24 Solution: D The first step of this problem is to eliminate the negative exponent and the fraction (1/2). This is completed by recognizing that (1/2) could be rewritten as 2-1 as follows: (1/2) -8 (1/2) -6 + (1/2) -8 (1/2) -6 = (2-1 ) -8 (2-1 ) -6 + (2-1 ) -8 (2-1 ) -6 = The next step is to group the common terms 2 8 and 2 6 by summing like terms together. To see this, imagine 2 8 were x and 2 6 were y. In that case the equation could be rewritten as x + x = 2x and y + y = 2y. Substituting 2 8 for x and 2 6 for y yields , which can be rewritten as follows: = At this point the answer can be calculated by factoring out 2 7 because it is common to both 2 9 and can be factored out of the expression as follows: = (2 7 ) (2 2 ) (2 7 )(1) = (2 7 )(2 2 1) = (2 7 )(4 1) = (2 7 )(3) = Problem 25 Solution: E As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, many terms share an exponent while none obviously shares a base. To group terms according to a shared exponent notice the 10 5 in the numerator and the 5 5 in the denominator. In this case 10 5 /5 5 is equivalent to 2 5. This method of grouping like exponents can be repeated for the entire expression: = = 2 2 = = 2 = 2 =

25 Problem 26 Solution: C For the left hand side of the equation to equal the right hand side, all of the bases must be equal. In this case, the base 10 is not common to the other bases so the first goal of the problem is to eliminate the 10 somehow. How to do so is not immediately obvious, but factoring both sides of the equation helps reveal the path to elimination. (2 2 ) = (2 5) 3x = 2 3x 5 3x The exponents on the left side must match the exponents on the right side, so 3x = 6 and x = 2. Therefore the answer to the problem is x+2 = 2+2 = 4. Problem 27 Solution: E This problem is solved by combining like terms. In this problem there are both common bases and common exponents so a choice must be made as to which to attack. In this case, it is more efficient to group terms according to a shared exponent. To group terms according to a shared exponent notice the 20 5 in the numerator and the 10 5 in the denominator. In this case 20 5 /10 5 is equivalent to 2 5. This method of grouping like exponents can be repeated for the entire expression: = = = 2 = 2 = The expression could also be simplified by combining like terms according to shared bases: = = = = = The answer is 1/4. Problem 28 Solution: C This problem is solved by breaking down the numerator and denominator until we have common bases or exponents. In this case, some combination of 7 s and 6 s are common to all of the bases in this problem and no exponents are common. Thus, the bases 42, 49, and 36 should be factored into these common bases:

26 2x 3x 2x 3x 2x 2x 3x 2x 5x 42 7 (6 7) = = = = 3x x 3x x 2 3x 2 x 6x 2x (7 7) (6 6) (7 ) (6 ) x-2x 7 5x-6x = 7 -x In this problem, one could break down 36 x to ( ) x but this work is excessive because 6 is common to both the numerator and denominator. In general, one should not necessarily break bases down to their lowest prime factors in exponent problems unless it is necessary to do so. Scanning the answer choices in this problem provides a hint that answers have large bases in them, so from that information alone it is probably not a good choice to break all the bases down to their prime factors. Problem 29 Solution: A As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, many terms share an exponent while none obviously shares a base. To group terms according to a shared exponent notice the 8 y in the numerator and the 2 y in the denominator. In this case 8 y /2 y is equivalent to (8/2) y = 4 y = (2 2 ) y = 2 2y. This method of grouping like exponents can be repeated for the entire expression: y x+ 2y 2x y x+ 2y 2x y 1 x+ 2y 2x y y x+ 2y 2x x+ 2 y = = x = 2 2y 2 -x-2y 2 2x = 2 2y-x-2y+2x = 2 x = 32 To solve the problem then we must convert the right hand side of the equation to have the same base as the left so 32 must be expressed in terms of a power of 2, or 2 5. Thus 2 x = 2 5. For the left hand side to equal the right, the exponents must be equal, thus x = 5. The answer is 5. Problem 30 Solution: E As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, there do not appear to be any common bases or exponents worth examining. As such, one may be tempted to break down each base into its prime bases. In this case, however, it is more productive to factor the bases until a common base is discovered, thus saving valuable time. This is possible because there are several multiples of 10 in the numerator and denominator. To simplify them, it may be productive to simplify the terms in the numerator to match the base 10 in the denominator (2 10) (3 10) (4 10) = = = = =

27 The answer is 12. Problem 31 Solution: A The key to exponent problems is to look for common elements such as a common base or exponent. The fact that the problem is seeking a value for an unknown exponent means that the base of the left hand side must match the base of the right hand side, otherwise there is no way to equate the unknown variable on the right hand side, which is an exponent, with the known values on the left = (2 5) = x x x The next step is to recognize that the negative exponent and fraction need to be eliminated. This is done by recognizing that (1/2) and (1/25) can be rewritten as 2-1 and 25-1 respectively or in the case of the equation, they can be rewritten as follows: = ( ) This problem is completed then by recognizing that the 2 4 terms are common and can be summed together. To see this, imagine 2 4 were x. In that case the equation could be rewritten as x+x+x+x = 4x. Substituting 2 4 for x yields 4 2 4, which can be rewritten as 2 6. Working with the original equation we have = 2 -x 5 -x. To finish the problem the bases on the left hand side must match those on the right so the 25 3 must match 5 -x. To fix this situation then 25 3 can be rewritten as (5 2 ) 3 = 5 6. Thus the final equation is = 2 -x 5 -x. For the two sides of the equation to be equivalent the exponents of each base much match so 6 = -x. x = -6. Problem 32 Solution: E Given the constraint that n must be an integer and 5 n must be a factor of 78,126, it is reasonable to consider that not only could n be a positive integer but it could also possibly be zero. Using long division, we can see that 5 will not divide into 78,126 without a remainder. Though this is a thorough method for reaching such a conclusion, the application of this method takes a lot of time. Instead, one should rely on knowledge from number properties to recall that a number is only divisible by 5 if that number ends in 5 or 0. In this case, 78,126 ends in 6 and is thus not divisible by 5. For this reason, n must be zero because 5 0 = 1 is the largest power of 5 that is a factor of 78,126. Substituting in 0 for n into 5 n+1 5 n yields = 5 1 = 4. The correct answer is 4.

28 Problem 33 Solution: D This problem is solved by breaking each number down into the smallest prime factors possible. 6 2z z 48 z z = (2 3) 2z+3 (2 3 ) 5-z (2 4 3) z-3 (2 5 3) -3z = 1 Now the exterior exponents should be distributed across the prime factors by multiplying the exterior exponents by the interior exponents: 2 2z+3 3 2z z 2 4z-12 3 z z 3-3z = 1 The exponents can be added because the powers are multiplied by each other: 2 2z z+4z-12-15z 3 2z+3+z-3-3z = z 3 0 = 1 A number raised to an exponent equals 1 when that exponent = 0. Thus this problem can be solved by understanding that 6 12z must equal z = 0 so 6 = 12z, z = 6/12 = 1/2. The answer is z = 1/2. Note that this problem could be solved by plugging in the answer choices for z, but this is a time intensive strategy that is very likely to exceed our time budget of 2 minutes. Problem 34 Solution: B For the left hand side of the equation to equal the right hand side, all of the bases must be equal. In this case, the base 10 is not common to the other bases so the first goal of the problem is to eliminate the 10 somehow. How to do so is not immediately obvious, but factoring the left hand size of the equation helps reveal the path to elimination. 5 2n+1 5 2n = 5 2n (5 1 1) = 5 2n 4 = 5 2n 2 2 Breaking each number on the right hand side down into the smallest prime factors possible yields: 5 2n 2 2 = (2 5) = = Now the 2 2 on each side cancels because it is a common to both sides. For the left hand side of the equation to be equal to the right hand side, the exponents must be equal or 2n = 12, so n = 6.

29 Problem 35 Solution: D For the left hand side of the equation to equal the right hand side, we need to make all of the bases equal because the exponents clearly are not. In this case, the bases 26 and 25 on the right are not common to 5 on the left, so the first goal of the problem is to eliminate the 26 and 25 somehow. How to do so is not immediately obvious, but factoring the left hand size of the equation helps reveal the path to elimination = 5 25 ( ) = = z 5 25 = 25 z Breaking each number on the right hand side down into the smallest prime factors possible yields 25 z = (5 2 ) z = 5 2z, so 5 25 = 5 2z. For the left hand side of the equation to be equal to the right hand side, the exponents must be equal: 2z = 25, or z = Problem 36 Solution: E One could solve the problem by multiplying out the numerator and dividing that by the product in the denominator. That solution would take far too much time and this problem is better solved by combining like terms, which are the shared exponents in the numerator and denominator in this case. To group terms according to a shared exponent notice the 10 2 in the numerator and the in the denominator. In this case is equivalent to 5 = 22. This method of grouping like exponents can be repeated for the entire expression: = = = = Problem 37 Solution: C This problem is solved by combining like terms. Here, we need to work with the common bases of x, y, and 2 in the numerator and denominator. x (2 y) x 2 y x 2 y = = (4( xy) ) (2 x y ) 2 x y 2y 2x 2y 2x 2x 2y 2x 2x 2 x x 2x 2x 2x

30 Now we can eliminate common terms from the numerator and denominator. x 2 y x = 2 x y x 2y 2x 2x 2y 2x 2x 2x 2x Now x is the only base that remains so we can combine the terms and set the expression equal to the right side of the equation: x x 2 y 2y 2x = x = 1 2x For the left hand side to equal the right hand side of the equation, either x must be 1 or the exponent that x is raised to must equal 0. We know x is larger than 1, so the exponent must be 0. 2y 2x = 0 or 2y = 2x and x = y. So the answer is y. Problem 38 Solution: B The inequality should be simplified before attempting to process any information from the three Roman numeral statements. Because we are dealing with an exponent problem we need to identify either a common base or a common exponent to simplify. Here, the exponent can be ruled out though, so we must work with the bases. 100 = 10 2, so we can use that as a starting point. x+ y 2 x+ y 2x+ 2y 100 (10 ) 10 = = = 10 = 10 3x 2y 3x 2y 3x 2y x+ 2y 3x+ 2y 4y x Now for 10 4y-x to be greater than 1, we know that our exponent must be positive, or 4y-x > 0. If 4y-x were 0 then 10 0 would be 1, and if 4y-x were less than 1 then the expression would be a fraction and less than one. Now, we turn our attention to the Roman Numerals: I. Not necessarily true. We can plug in numbers to test, and the easiest might be zero. With y = 0, we get 100 x /10 3x = 10 2x /10 3x = 10 -x. So long as x < 0 then 10 -x > 1 when y = 0, so I is not necessarily true. II. Not necessarily true. If x = -4 and y = 1, then 10 4y-x = = 10 8, which is greater than 1 even though x + y is not greater than 0. III. True. This is another way of stating that 4y x > 0 and from our above work we saw that this means the exponent will be positive and the entire expression will be greater than 1.

31 III only is the correct statement. Problem 39 Solution: E This problem is solved by breaking down the numerator and denominator until we have common bases or exponents. In this case, some combination of 7 s and 5 s are common to all of the bases in this problem and no exponents are obviously common. Thus, the bases 35 and 25 should be factored to highlight these common bases: 2x 2y+ x 2x 2y+ x 2x 2y+ x 2y+ x 3x+ 2y 2y+ x (5 7) = = = = 5 (3x+2y)-(3x+2y) 7 (2y+x)-(2y) = x 3x y 2y 3x 2 y 2y 3x 2y 2y 3x+ 2y 2y (5 ) The answer is 7 x. Problem 40 Solution: E As with other exponent problems, we must look for either a shared base or a shared exponent. In this problem, there are some obvious common bases, but there are in fact several hidden common bases as well as hidden common exponents. Further, it is awkward to deal with fractional bases, so before simplifying it is more effective to convert the fractional bases to integers by using negative exponents. For example, 1/4 = 4-1. Applying this to the expression yields: = Now the problem is solved by finding common bases and simplifying (2 ) (3 ) = = = = 2 (-10)-(-12) = 2 2 = (2 3) The answer is 4.