Middle School Math Quiz: Integer Exponents
8 questions · exam conditions
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Integer ExponentsQuestion 1 of 8

If 3−2⋅9x=27−13^{-2} \cdot 9^x = 27^{-1}, what is the value of xx?

−52-\frac{5}{2}
−12-\frac{1}{2}
12\frac{1}{2}
52\frac{5}{2}
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Middle School Math Quiz

Middle School Math Quiz: Integer Exponents

Practice Integer Exponents in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integer Exponents, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If 3−2⋅9x=27−13^{-2} \cdot 9^x = 27^{-1}, what is the value of xx?

  1. −52-\frac{5}{2}
  2. −12-\frac{1}{2} (correct answer)
  3. 12\frac{1}{2}
  4. 52\frac{5}{2}
Explanation: First, express everything in terms of base 3: 3−2⋅(32)x=(33)−13^{-2} \cdot (3^2)^x = (3^3)^{-1}, which becomes 3−2⋅32x=3−33^{-2} \cdot 3^{2x} = 3^{-3}. Using the product rule: 3−2+2x=3−33^{-2+2x} = 3^{-3}. Therefore −2+2x=−3-2+2x = -3, so 2x=−12x = -1 and x=−12x = -\frac{1}{2}. Choice A results from incorrectly setting up −2+2x=−1-2+2x = -1. Choice C comes from sign errors in the exponent rules. Choice D results from misinterpreting the bases or making multiple calculation errors.

Question 2

If a−2=116a^{-2} = \frac{1}{16} and b3=8b^3 = 8, what is the value of a2b−3\frac{a^2}{b^{-3}}?

  1. 1128\frac{1}{128}
  2. 132\frac{1}{32}
  3. 3232
  4. 128128 (correct answer)
Explanation: From a−2=116a^{-2} = \frac{1}{16}, we get a2=16a^2 = 16. From b3=8b^3 = 8, we get b=2b = 2, so b−3=18b^{-3} = \frac{1}{8}. Therefore a2b−3=1618=16⋅8=128\frac{a^2}{b^{-3}} = \frac{16}{\frac{1}{8}} = 16 \cdot 8 = 128. Choice A results from incorrectly finding a2⋅b−3a^2 \cdot b^{-3}. Choice B comes from calculation errors with the reciprocal. Choice C results from using b3b^3 instead of b−3b^{-3} in the denominator.

Question 3

The expression (x−2y3x3y−1)−2\left(\frac{x^{-2}y^3}{x^3y^{-1}}\right)^{-2} is equivalent to which of the following?

  1. x10y8\frac{x^{10}}{y^8} (correct answer)
  2. x8y10\frac{x^8}{y^{10}}
  3. y8x10\frac{y^8}{x^{10}}
  4. y10x8\frac{y^{10}}{x^8}
Explanation: First simplify inside the parentheses: x−2y3x3y−1=x−2−3⋅y3−(−1)=x−5y4\frac{x^{-2}y^3}{x^3y^{-1}} = x^{-2-3} \cdot y^{3-(-1)} = x^{-5}y^4. Then apply the outer exponent: (x−5y4)−2=x(−5)(−2)⋅y4(−2)=x10y−8=x10y8(x^{-5}y^4)^{-2} = x^{(-5)(-2)} \cdot y^{4(-2)} = x^{10}y^{-8} = \frac{x^{10}}{y^8}. Choice B results from sign errors in exponent calculations. Choice C comes from incorrectly switching the variables' positions. Choice D results from multiple errors in applying negative exponents.

Question 4

Which expression is equivalent to 2−3⋅428−1\frac{2^{-3} \cdot 4^2}{8^{-1}}?

  1. 222^2
  2. 242^4
  3. 262^6 (correct answer)
  4. 282^8
Explanation: Convert all terms to base 2: 2−3⋅(22)2(23)−1=2−3⋅242−3\frac{2^{-3} \cdot (2^2)^2}{(2^3)^{-1}} = \frac{2^{-3} \cdot 2^4}{2^{-3}}. This simplifies to 2−3+42−3=212−3=21⋅23=26\frac{2^{-3+4}}{2^{-3}} = \frac{2^1}{2^{-3}} = 2^1 \cdot 2^3 = 2^6. Choice A results from incorrectly handling negative exponents. Choice B comes from errors in combining exponents. Choice D results from adding exponents instead of using proper rules for multiplication and division.

Question 5

If 52x−1=125x−25^{2x-1} = 125^{x-2}, what is the value of xx?

  1. −5-5
  2. −2-2
  3. 22
  4. 55 (correct answer)
Explanation: Rewrite with the same base: 52x−1=(53)x−2=53(x−2)=53x−65^{2x-1} = (5^3)^{x-2} = 5^{3(x-2)} = 5^{3x-6}. Since the bases are equal, the exponents must be equal: 2x−1=3x−62x-1 = 3x-6. Solving: −1+6=3x−2x-1+6 = 3x-2x, so 5=x5 = x. Choice A results from sign errors in solving the linear equation. Choice B comes from incorrectly setting up the equation. Choice C results from calculation errors when solving for xx.

Question 6

The expression (32)−1⋅3−33−7\frac{(3^2)^{-1} \cdot 3^{-3}}{3^{-7}} simplifies to which of the following?

  1. 3−23^{-2}
  2. 323^{2} (correct answer)
  3. 303^{0}
  4. 343^{4}
Explanation: When you encounter expressions with multiple exponents like this one, you're working with the laws of exponents. The key is to simplify step-by-step using these rules: (am)n=amn(a^m)^n = a^{mn}, am⋅an=am+na^m \cdot a^n = a^{m+n}, and aman=am−n\frac{a^m}{a^n} = a^{m-n}. Let's work through this systematically. First, simplify (32)−1(3^2)^{-1} using the power rule: (32)−1=32⋅(−1)=3−2(3^2)^{-1} = 3^{2 \cdot (-1)} = 3^{-2}. Now the expression becomes 3−2⋅3−33−7\frac{3^{-2} \cdot 3^{-3}}{3^{-7}}. Next, handle the multiplication in the numerator: 3−2⋅3−3=3−2+(−3)=3−53^{-2} \cdot 3^{-3} = 3^{-2 + (-3)} = 3^{-5}. The expression is now 3−53−7\frac{3^{-5}}{3^{-7}}. Finally, use the division rule: 3−53−7=3−5−(−7)=3−5+7=32\frac{3^{-5}}{3^{-7}} = 3^{-5 - (-7)} = 3^{-5 + 7} = 3^2. This matches choice B. Choice A (3−23^{-2}) would result from incorrectly adding exponents during division instead of subtracting. Choice C (303^0) might come from miscalculating the final subtraction as −5−(−7)=−2-5 - (-7) = -2 instead of +2+2. Choice D (343^4) could result from adding all the exponents together without following proper order of operations. Remember: when dividing powers with the same base, subtract the bottom exponent from the top exponent. Watch your signs carefully, especially when subtracting negative numbers—subtracting a negative becomes addition.

Question 7

For which value of kk is the equation 2k+1⋅4k−2=822^{k+1} \cdot 4^{k-2} = 8^2 true?

  1. k=3k = 3 (correct answer)
  2. k=2k = 2
  3. k=1k = 1
  4. k=4k = 4
Explanation: When you encounter exponential equations with different bases, the key strategy is to rewrite everything using the same base. Since 2, 4, and 8 are all powers of 2, you can express everything in terms of base 2. First, rewrite each term using base 2: 4=224 = 2^2 and 8=238 = 2^3. This transforms the equation into: 2k+1⋅(22)k−2=(23)22^{k+1} \cdot (2^2)^{k-2} = (2^3)^2 Using the power rule (am)n=amn(a^m)^n = a^{mn}, this becomes: 2k+1⋅22(k−2)=23⋅22^{k+1} \cdot 2^{2(k-2)} = 2^{3 \cdot 2} 2k+1⋅22k−4=262^{k+1} \cdot 2^{2k-4} = 2^6 When multiplying powers with the same base, add the exponents: 2(k+1)+(2k−4)=262^{(k+1) + (2k-4)} = 2^6 23k−3=262^{3k-3} = 2^6 Since the bases are equal, the exponents must be equal: 3k−3=63k - 3 = 6 3k=93k = 9 k=3k = 3 Let's check why the other answers don't work. Answer B (k=2k = 2) gives us 3(2)−3=3≠63(2) - 3 = 3 \neq 6. Answer C (k=1k = 1) yields 3(1)−3=0≠63(1) - 3 = 0 \neq 6. Answer D (k=4k = 4) produces 3(4)−3=9≠63(4) - 3 = 9 \neq 6. Only answer A (k=3k = 3) satisfies our equation. Study tip: When solving exponential equations, always look for ways to express all terms using the same base. Powers of 2 (like 4, 8, 16) are especially common on pre-algebra exams, so memorize these relationships: 4=224 = 2^2, 8=238 = 2^3, 16=2416 = 2^4.

Question 8

If 14x=23x−5\frac{1}{4^x} = 2^{3x-5}, what is the value of xx?

  1. −1-1
  2. 00
  3. 11 (correct answer)
  4. 22
Explanation: Rewrite in terms of base 2: 14x=1(22)x=122x=2−2x\frac{1}{4^x} = \frac{1}{(2^2)^x} = \frac{1}{2^{2x}} = 2^{-2x}. So the equation becomes 2−2x=23x−52^{-2x} = 2^{3x-5}. Setting exponents equal: −2x=3x−5-2x = 3x-5, which gives −2x−3x=−5-2x-3x = -5, so −5x=−5-5x = -5 and x=1x = 1. Choice A results from sign errors. Choice B comes from incorrectly handling the negative exponent. Choice D results from algebraic manipulation errors.