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    Hard GRE Exponents Questions Practice Questions

    July 8, 20269 min read64 views
    Hard GRE Exponents Questions Practice Questions

    Exponents represent one of the most frequent topics on the GRE Quantitative Reasoning section, appearing in roughly 15% of all math problems. Solving Hard GRE Exponents Questions requires a deep understanding of base manipulation, negative powers, and the relationship between roots and fractional exponents. While basic arithmetic focuses on multiplication, advanced exponentiation tests your ability to recognize patterns and simplify complex expressions into manageable parts.

    To succeed, students should utilize a comprehensive GRE Prep strategy that combines conceptual knowledge with rigorous practice. This article provides a deep dive into the mechanics of exponent rules and offers challenging practice sets to sharpen your skills.

    Concept Explanation

    Exponents are mathematical notations that indicate how many times a base number is multiplied by itself, governed by specific algebraic laws. At the advanced level, the GRE tests your proficiency with non-integer exponents and the manipulation of bases that are not identical. The fundamental rules you must internalize include the Product Rule (amΓ—an=am+n)(a^m \times a^n = a^{m+n}), the Quotient Rule (am/an=amβˆ’n)(a^m / a^n = a^{m-n}), and the Power of a Power Rule (am)n=amn(a^m)^n = a^{mn}.

    Harder questions often involve bases that can be rewritten as powers of a common prime. For example, if an equation involves both 8 and 16, you should immediately recognize them as 232^3 and 242^4. Additionally, negative exponents signify the reciprocal of the base: aβˆ’n=1ana^{-n} = \frac{1}{a^n}. Fractional exponents represent roots, where am/n=amna^{m/n} = \sqrt[n]{a^m}. On the GRE, you will frequently encounter problems where you must factor out the greatest common factor from an exponential expression, such as simplifying 310+393^{10} + 3^9 into 39(3+1)=4Γ—393^9(3 + 1) = 4 \times 3^9. Mastery of these GRE practice questions with explanations is essential for achieving a high score in the Quantitative section.

    Solved Examples

    1. Example 1: Base Manipulation
      Solve for xx if 27xβˆ’1=9x+427^{x-1} = 9^{x+4}.
      1. Identify a common base for 27 and 9. Both are powers of 3: 27=3327 = 3^3 and 9=329 = 3^2.
      2. Substitute these into the equation: (33)xβˆ’1=(32)x+4(3^3)^{x-1} = (3^2)^{x+4}.
      3. Apply the Power of a Power Rule: 33(xβˆ’1)=32(x+4)3^{3(x-1)} = 3^{2(x+4)}.
      4. Set the exponents equal to each other: 3xβˆ’3=2x+83x - 3 = 2x + 8.
      5. Solve for xx: x=11x = 11.
    2. Example 2: Factoring Exponential Sums
      What is the value of 512βˆ’510511+510\frac{5^{12} - 5^{10}}{5^{11} + 5^{10}}?
      1. Identify the smallest power in the numerator and denominator, which is 5105^{10}.
      2. Factor 5105^{10} out of both: 510(52βˆ’1)510(51+1)\frac{5^{10}(5^2 - 1)}{5^{10}(5^1 + 1)}.
      3. Cancel the common term 5105^{10}: 52βˆ’15+1\frac{5^2 - 1}{5 + 1}.
      4. Simplify the remaining values: 25βˆ’16=246=4\frac{25 - 1}{6} = \frac{24}{6} = 4.
    3. Example 3: Negative and Fractional Exponents
      Simplify (16βˆ’3/4Γ—82/3)βˆ’1(16^{-3/4} \times 8^{2/3})^{-1}.
      1. Express bases as powers of 2: 16=2416 = 2^4 and 8=238 = 2^3.
      2. Substitute: ((24)βˆ’3/4Γ—(23)2/3)βˆ’1((2^4)^{-3/4} \times (2^3)^{2/3})^{-1}.
      3. Multiply exponents: (2βˆ’3Γ—22)βˆ’1(2^{-3} \times 2^2)^{-1}.
      4. Combine inside the parentheses: (2βˆ’1)βˆ’1(2^{-1})^{-1}.
      5. Multiply the final exponents: 21=22^1 = 2.

    Practice Questions

    Test your knowledge with these Hard GRE Exponents Questions. For more variety, check out unlimited GRE practice questions online.

    1. If 4x=8yβˆ’14^x = 8^{y-1} and 9x+y=2439^{x+y} = 243, what is the value of xx?
    2. Which is greater: 21002^{100} or 3753^{75}?
    3. Simplify the expression: 2n+4βˆ’2(2n)2(2n+3)\frac{2^{n+4} - 2(2^n)}{2(2^{n+3})}.

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    Practice GRE Questions
    1. If xx and yy are positive integers and 2xβ‹…5y=20,0002^x \cdot 5^y = 20,000, what is the value of x+yx + y?
    2. Find the units digit of 742+3217^{42} + 3^{21}.
    3. If nn is an integer, is (βˆ’1)n2+n(-1)^{n^2 + n} always positive?
    4. Solve for zz: (0.25)zβˆ’2=64(0.25)^{z-2} = 64.
    5. If x=210x = 2^{10}, what is the value of xxx^x in terms of base 2?

    Answers & Explanations

    1. Answer: 1.5 (or 3/2)
      Rewrite both equations with common bases. (22)x=(23)yβˆ’1β†’2x=3yβˆ’3(2^2)^x = (2^3)^{y-1} \rightarrow 2x = 3y - 3. For the second, (32)x+y=35β†’2x+2y=5(3^2)^{x+y} = 3^5 \rightarrow 2x + 2y = 5. Solving the system: from the first, 3y=2x+33y = 2x + 3, so y=(2x+3)/3y = (2x+3)/3. Substitute into the second: 2x+2((2x+3)/3)=52x + 2((2x+3)/3) = 5. Multiply by 3: 6x+4x+6=15β†’10x=9β†’x=0.96x + 4x + 6 = 15 \rightarrow 10x = 9 \rightarrow x = 0.9. (Re-checking: if 2x=3yβˆ’32x = 3y-3 and 2x+2y=52x+2y=5, then 3yβˆ’3+2y=5β†’5y=8β†’y=1.63y-3+2y=5 \rightarrow 5y=8 \rightarrow y=1.6. Then 2x=3(1.6)βˆ’3=1.8β†’x=0.92x = 3(1.6)-3 = 1.8 \rightarrow x=0.9).
    2. Answer: 3753^{75}
      To compare exponents with different bases, find a common exponent. The GCD of 100 and 75 is 25. Rewrite: 2100=(24)25=16252^{100} = (2^4)^{25} = 16^{25}. Rewrite: 375=(33)25=27253^{75} = (3^3)^{25} = 27^{25}. Since 27>1627 > 16, 3753^{75} is greater.
    3. Answer: 7/8
      Factor out 2n2^n from the numerator: 2n(24βˆ’2)=2n(16βˆ’2)=14β‹…2n2^n(2^4 - 2) = 2^n(16 - 2) = 14 \cdot 2^n. The denominator is 2β‹…2nβ‹…23=2nβ‹…162 \cdot 2^n \cdot 2^3 = 2^n \cdot 16. The expression becomes 14β‹…2n16β‹…2n=1416=78\frac{14 \cdot 2^n}{16 \cdot 2^n} = \frac{14}{16} = \frac{7}{8}.
    4. Answer: 9
      Prime factorize 20,000. 20,000=2Γ—10,000=2Γ—104=2Γ—(2Γ—5)4=2Γ—24Γ—54=25Γ—5420,000 = 2 \times 10,000 = 2 \times 10^4 = 2 \times (2 \times 5)^4 = 2 \times 2^4 \times 5^4 = 2^5 \times 5^4. Thus x=5x = 5 and y=4y = 4. x+y=9x + y = 9.
    5. Answer: 2
      Units digits of 7 follow a cycle of 4: 7, 9, 3, 1. 42Γ·442 \div 4 leaves a remainder of 2, so the units digit of 7427^{42} is 9. Units digits of 3 follow a cycle of 4: 3, 9, 7, 1. 21Γ·421 \div 4 leaves a remainder of 1, so the units digit of 3213^{21} is 3. 9+3=129 + 3 = 12. The units digit is 2.
    6. Answer: Yes
      The exponent is n2+n=n(n+1)n^2 + n = n(n+1). Since nn and n+1n+1 are consecutive integers, one must be even. The product of an even and odd number is always even. (βˆ’1)even=1(-1)^{ \text{even}} = 1, which is always positive.
    7. Answer: -1
      Rewrite bases to 4 or 2. 0.25=1/4=4βˆ’10.25 = 1/4 = 4^{-1}. 64=4364 = 4^3. Equation: (4βˆ’1)zβˆ’2=43β†’βˆ’z+2=3β†’βˆ’z=1β†’z=βˆ’1(4^{-1})^{z-2} = 4^3 \rightarrow -z + 2 = 3 \rightarrow -z = 1 \rightarrow z = -1.
    8. Answer: 2102402^{10240}
      Substitute x=210x = 2^{10} into xxx^x: (210)(210)(2^{10})^{(2^{10})}. Since 210=10242^{10} = 1024, this is (210)1024(2^{10})^{1024}. Multiply the exponents: 210Γ—1024=2102402^{10 \times 1024} = 2^{10240}.
    Interactive quizQuestion 1 of 5

    1. If \( 3^a \times 3^b = 3^{12} \) and \( a = 2b \), what is the value of \( b \)?

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    Frequently Asked Questions

    How do I handle negative exponents on the GRE?

    Negative exponents should be treated as the reciprocal of the positive exponent. For example, xβˆ’nx^{-n} is equal to 1/xn1/x^n, which allows you to move terms between the numerator and denominator to simplify the expression.

    What is the best way to compare large exponential values?

    To compare large exponents, try to manipulate the expressions so they either have the same base or the same exponent. Once one part is identical, you can directly compare the remaining values to determine which is larger.

    Can I add exponents if the bases are being added?

    No, exponent rules like the Product Rule only apply when bases are being multiplied. When bases are added, such as 34+343^4 + 3^4, you must use factoring to simplify the expression into 2(34)2(3^4).

    How do fractional exponents work?

    Fractional exponents represent roots where the denominator of the fraction is the index of the root and the numerator is the power. For instance, x2/3x^{2/3} is the cube root of xx squared, or x23\sqrt[3]{x^2}.

    Why is (βˆ’1)n(-1)^n sometimes positive and sometimes negative?

    The result of (βˆ’1)n(-1)^n depends entirely on whether nn is even or odd. If nn is an even integer, the result is 1; if nn is an odd integer, the result is -1.

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