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    GRE Exponents Questions Practice Questions with Answers

    June 26, 20267 min read94 views
    GRE Exponents Questions Practice Questions with Answers

    GRE Exponents Questions Practice Questions with Answers

    Seven distinct rules govern how powers interact, and internalizing these is essential for solving any quantitative comparison or problem-solving task on the exam. Understanding GRE exponents questions involves recognizing the relationship between a base and its power, where the exponent dictates how many times the base is multiplied by itself. These concepts form the bedrock of the GRE Prep curriculum, as they frequently appear in algebra, geometry, and data interpretation sections alike.

    Concept Explanation

    Exponents are mathematical notations that represent repeated multiplication of a number by itself. In the expression xnx^n, xx is the base and nn is the exponent. The rules of exponents allow test-takers to simplify complex algebraic expressions efficiently. For instance, when multiplying terms with the same base, you add the exponents: xaΓ—xb=xa+bx^a \times x^b = x^{a+b}. Conversely, when dividing, you subtract them: xaxb=xaβˆ’b\frac{x^a}{x^b} = x^{a-b}. A power raised to another power requires multiplication of the exponents, such as (xa)b=xab(x^a)^b = x^{ab}. It is also vital to remember that any non-zero number raised to the power of zero is 1, and negative exponents indicate the reciprocal of the base, such that xβˆ’n=1xnx^{-n} = \frac{1}{x^n}. According to Wikipedia's entry on exponentiation, these properties are consistent across all real numbers, which is a standard assumption in GRE mathematics.

    Solved Examples

    1. Simplify the expression: 35Γ—3234\frac{3^5 \times 3^2}{3^4}
      1. Apply the product rule to the numerator: 35Γ—32=35+2=373^5 \times 3^2 = 3^{5+2} = 3^7.
      2. Apply the quotient rule: 3734=37βˆ’4=33\frac{3^7}{3^4} = 3^{7-4} = 3^3.
      3. Calculate the final value: 3Γ—3Γ—3=273 \times 3 \times 3 = 27.
    2. Solve for xx: (23)x=64(2^3)^x = 64
      1. Apply the power of a power rule: 23x=642^{3x} = 64.
      2. Express 64 as a power of 2: 64=2664 = 2^6.
      3. Set the exponents equal to each other: 3x=63x = 6.
      4. Divide by 3: x=2x = 2.
    3. Evaluate the expression: 4βˆ’2+204^{-2} + 2^0
      1. Convert the negative exponent to a fraction: 4βˆ’2=142=1164^{-2} = \frac{1}{4^2} = \frac{1}{16}.
      2. Apply the zero exponent rule: 20=12^0 = 1.
      3. Add the terms: 116+1=1716\frac{1}{16} + 1 = \frac{17}{16} or 1.06251.0625.

    Practice Questions

    1. If 5k=1255^k = 125, what is the value of 5kβˆ’25^{k-2}?
    2. Simplify the expression: (x2y3)2Γ—(xyβˆ’1)3(x^2 y^3)^2 \times (x y^{-1})^3.
    3. Which is greater: 2102^{10} or 454^5?
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    5. Solve for nn: 9n+1=27nβˆ’19^{n+1} = 27^{n-1}.
    6. What is the units digit of 7257^{25}?
    7. Simplify: (24Γ—54)103\frac{(2^4 \times 5^4)}{10^3}.
    8. If 2x=82^x = 8 and 3y=813^y = 81, what is the value of x+yx + y?
    9. Evaluate: (12)βˆ’3+(13)βˆ’2(\frac{1}{2})^{-3} + (\frac{1}{3})^{-2}.
    10. If xx is a positive integer, is (βˆ’1)2x(-1)^{2x} greater than (βˆ’1)2x+1(-1)^{2x+1}?
    11. Solve for aa: 4a2a=16\frac{4^a}{2^a} = 16.

    Answers & Explanations

    1. Answer: 5. Since 53=1255^3 = 125, k=3k = 3. Therefore, 53βˆ’2=51=55^{3-2} = 5^1 = 5.
    2. Answer: x7y3x^7 y^3. Expand the first part: x4y6x^4 y^6. Expand the second part: x3yβˆ’3x^3 y^{-3}. Multiply them: x4+3y6βˆ’3=x7y3x^{4+3} y^{6-3} = x^7 y^3.
    3. Answer: They are equal. 454^5 can be written as (22)5(2^2)^5, which simplifies to 2102^{10}.
    4. Answer: 5. Rewrite both sides with base 3: (32)n+1=(33)nβˆ’1(3^2)^{n+1} = (3^3)^{n-1}, so 32n+2=33nβˆ’33^{2n+2} = 3^{3n-3}. Set exponents equal: 2n+2=3nβˆ’32n + 2 = 3n - 3. Solving for nn gives 55.
    5. Answer: 7. The units digits of powers of 7 follow a cycle: 7, 9, 3, 1. Since 25Γ·425 \div 4 leaves a remainder of 1, the units digit is the first in the cycle, which is 7.
    6. Answer: 10. Using the power of a product rule, 24Γ—54=(2Γ—5)4=1042^4 \times 5^4 = (2 \times 5)^4 = 10^4. Then, 104103=101=10\frac{10^4}{10^3} = 10^1 = 10.
    7. Answer: 7. Since 23=82^3 = 8, x=3x = 3. Since 34=813^4 = 81, y=4y = 4. Thus, 3+4=73 + 4 = 7.
    8. Answer: 17. A negative exponent flips the fraction: 23+32=8+9=172^3 + 3^2 = 8 + 9 = 17.
    9. Answer: Yes. For any integer xx, 2x2x is even, so (βˆ’1)even=1(-1)^{ \text{even}} = 1. 2x+12x+1 is odd, so (βˆ’1)odd=βˆ’1(-1)^{ \text{odd}} = -1. Since 1>βˆ’11 > -1, the answer is yes.
    10. Answer: 4. Simplify the fraction: (42)a=2a(\frac{4}{2})^a = 2^a. Since 2a=162^a = 16 and 24=162^4 = 16, a=4a = 4.
    Interactive quizQuestion 1 of 5

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    Frequently Asked Questions

    How do I handle negative exponents in GRE questions?

    To handle negative exponents, simply take the reciprocal of the base and change the exponent to a positive number. For example, 5βˆ’25^{-2} becomes 152\frac{1}{5^2}, which is 125\frac{1}{25}.

    Can I add exponents if the bases are different?

    No, the product rule only applies when the bases are identical. If you have 23Γ—322^3 \times 3^2, you must evaluate each term separately rather than adding the exponents.

    What is the difference between βˆ’32-3^2 and (βˆ’3)2(-3)^2?

    In βˆ’32-3^2, the exponent only applies to the 3, resulting in -9. In (βˆ’3)2(-3)^2, the exponent applies to the negative sign as well, resulting in +9.

    Do I need to memorize high powers for the GRE?

    While you don't need to memorize everything, knowing powers of 2 (up to 2102^{10}), 3 (up to 353^5), and squares up to 25 is highly beneficial for speed. You can also use the AI Question Generator to practice these mental math drills.

    How do I solve equations where the variable is in the exponent?

    The most effective strategy is to rewrite both sides of the equation so they have the same base. Once the bases match, you can set the exponents equal to each other and solve for the variable.

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