ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Exponent Rules
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Exponent RulesQuestion 1 of 20

Simplify the expression (k3n+1)2k2n\frac{(k^{3n+1})^2}{k^{2n}}.

k4n+1k^{4n+1}
k4n+2k^{4n+2}
kn+3k^{n+3}
k8n+2k^{8n+2}
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Exponent Rules

Practice Exponent Rules in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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 Exponent Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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

Simplify the expression (k3n+1)2k2n\frac{(k^{3n+1})^2}{k^{2n}}.

  1. k4n+1k^{4n+1}
  2. k4n+2k^{4n+2} (correct answer)
  3. kn+3k^{n+3}
  4. k8n+2k^{8n+2}
Explanation: First, simplify the numerator using the power rule, (am)n=amn(a^m)^n = a^{mn}. We must distribute the 2 to both terms in the exponent: (k3n+1)2=k2(3n+1)=k6n+2(k^{3n+1})^2 = k^{2(3n+1)} = k^{6n+2}. The expression now becomes k6n+2k2n\frac{k^{6n+2}}{k^{2n}}. Now, use the quotient rule, aman=am−n\frac{a^m}{a^n} = a^{m-n}: k(6n+2)−2n=k4n+2k^{(6n+2) - 2n} = k^{4n+2}.

Question 2

Which of the following expressions is NOT equivalent to x12y6\frac{x^{12}}{y^6}?

  1. (x2y−1)6(x^2y^{-1})^6
  2. (x6y3)2\left(\frac{x^6}{y^3}\right)^2
  3. x6⋅(x2y−2)3x^6 \cdot (x^2y^{-2})^3
  4. (x3y−1)4(x^3y^{-1})^4 (correct answer)
Explanation: Evaluate each choice to see if it simplifies to x12y6\frac{x^{12}}{y^6}.\A) (x2y−1)6=x2⋅6y−1⋅6=x12y−6=x12y6(x^2y^{-1})^6 = x^{2 \cdot 6}y^{-1 \cdot 6} = x^{12}y^{-6} = \frac{x^{12}}{y^6}. This is equivalent.\B) (x6y3)2=(x6)2(y3)2=x12y6\left(\frac{x^6}{y^3}\right)^2 = \frac{(x^6)^2}{(y^3)^2} = \frac{x^{12}}{y^6}. This is equivalent.\C) x6⋅(x2y−2)3=x6⋅x6y−6=x6+6y−6=x12y−6=x12y6x^6 \cdot (x^2y^{-2})^3 = x^6 \cdot x^{6}y^{-6} = x^{6+6}y^{-6} = x^{12}y^{-6} = \frac{x^{12}}{y^6}. This is equivalent.\D) (x3y−1)4=x3⋅4y−1⋅4=x12y−4=x12y4(x^3y^{-1})^4 = x^{3 \cdot 4}y^{-1 \cdot 4} = x^{12}y^{-4} = \frac{x^{12}}{y^4}. This is NOT equivalent.

Question 3

The side length of a square is 5x3y25x^3y^2. What is the area of the square?

  1. 10x6y410x^6y^4
  2. 25x5y425x^5y^4
  3. 25x6y425x^6y^4 (correct answer)
  4. 10x5y410x^5y^4
Explanation: The area of a square is given by the formula A=s2A = s^2, where ss is the side length. In this case, s=5x3y2s = 5x^3y^2. So, the area is (5x3y2)2(5x^3y^2)^2. To simplify, apply the exponent 2 to each factor inside the parentheses: 52(x3)2(y2)25^2(x^3)^2(y^2)^2. This simplifies to 25x3⋅2y2⋅2=25x6y425x^{3 \cdot 2}y^{2 \cdot 2} = 25x^6y^4.

Question 4

The number of bacteria in a lab sample is modeled by the expression 500⋅(2h)500 \cdot (2^h), where hh is the number of hours. Which expression represents the number of bacteria after 3h3h hours?

  1. 500⋅(6h)500 \cdot (6^h)
  2. 1500⋅(6h)1500 \cdot (6^h)
  3. 1500⋅(2h)1500 \cdot (2^h)
  4. 500⋅(8h)500 \cdot (8^h) (correct answer)
Explanation: To find the number of bacteria after 3h3h hours, substitute 3h3h for hh in the given expression: 500⋅(23h)500 \cdot (2^{3h}). Using the rules of exponents, 23h2^{3h} can be rewritten as (23)h(2^3)^h. Since 23=82^3 = 8, the expression is equivalent to 500⋅(8h)500 \cdot (8^h).

Question 5

Which of the following is equivalent to (3x−2)−2⋅x3(3x^{-2})^{-2} \cdot x^3?

  1. −9x7-9x^7
  2. x9\frac{x}{9}
  3. −6x-6x
  4. x79\frac{x^7}{9} (correct answer)
Explanation: First, apply the power rule to (3x−2)−2(3x^{-2})^{-2}. The exponent -2 applies to both the 3 and the x−2x^{-2}. This gives 3−2⋅(x−2)−2=3−2⋅x43^{-2} \cdot (x^{-2})^{-2} = 3^{-2} \cdot x^{4}. Since 3−2=132=193^{-2} = \frac{1}{3^2} = \frac{1}{9}, this part simplifies to 19x4\frac{1}{9}x^4. Now, multiply this by the second part of the expression: (19x4)⋅x3(\frac{1}{9}x^4) \cdot x^3. Using the product rule, add the exponents of x: x4⋅x3=x4+3=x7x^4 \cdot x^3 = x^{4+3} = x^7. The final result is 19x7\frac{1}{9}x^7, or x79\frac{x^7}{9}.

Question 6

The volume of a rectangular prism is found by multiplying its length, width, and height. If a prism has a length of 3x23x^2, a width of 2x52x^5, and a height of xx, which expression represents its volume?

  1. 6x76x^7
  2. 5x85x^8
  3. 6x86x^8 (correct answer)
  4. 6x106x^{10}
Explanation: To find the volume, multiply the three dimensions: (3x2)(2x5)(x)(3x^2)(2x^5)(x). First, multiply the numerical coefficients: 3⋅2=63 \cdot 2 = 6. Next, multiply the variable terms using the product rule for exponents, remembering that x=x1x = x^1: x2⋅x5⋅x1=x2+5+1=x8x^2 \cdot x^5 \cdot x^1 = x^{2+5+1} = x^8. Combining the coefficient and the variable gives the volume as 6x86x^8.

Question 7

Which of the following expressions is equivalent to (2x4y−1)3(2x^4y^{-1})^3?

  1. 6x12y3\frac{6x^{12}}{y^3}
  2. 8x7y−2\frac{8x^7}{y^{-2}}
  3. 8x12y3\frac{8x^{12}}{y^3} (correct answer)
  4. 6x12y−36x^{12}y^{-3}
Explanation: To simplify the expression (2x4y−1)3(2x^4y^{-1})^3, the exponent 3 must be applied to each factor inside the parentheses. This is the power of a product rule. For the coefficient, 23=82^3 = 8. For the variables, we use the power of a power rule, which means multiplying the exponents: (x4)3=x4⋅3=x12(x^4)^3 = x^{4 \cdot 3} = x^{12} and (y−1)3=y−1⋅3=y−3(y^{-1})^3 = y^{-1 \cdot 3} = y^{-3}. Combining these gives 8x12y−38x^{12}y^{-3}. To write the expression with positive exponents, we use the rule a−n=1ana^{-n} = \frac{1}{a^n}, so y−3=1y3y^{-3} = \frac{1}{y^3}. The final simplified expression is 8x12y3\frac{8x^{12}}{y^3}.

Question 8

If a2x⋅a3x−1=a13a^{2x} \cdot a^{3x-1} = a^{13}, what is the value of xx?

  1. x=145x = \frac{14}{5} (correct answer)
  2. x=135x = \frac{13}{5}
  3. x=3x = 3
  4. x=125x = \frac{12}{5}
Explanation: Using the product rule for exponents, a2x⋅a3x−1=a2x+3x−1=a5x−1a^{2x} \cdot a^{3x-1} = a^{2x + 3x - 1} = a^{5x-1}. Setting this equal to a13a^{13}: 5x−1=135x - 1 = 13, so 5x=145x = 14, therefore x=145x = \frac{14}{5}. Choice B incorrectly sets 5x=135x = 13. Choice C assumes 2x+3x=132x + 3x = 13 without considering the −1-1. Choice D results from the error 5x−1=125x - 1 = 12.

Question 9

If (xayb)c=x12y15\left(\frac{x^a}{y^b}\right)^c = \frac{x^{12}}{y^{15}} and a=3a = 3, b=5b = 5, what is the value of cc?

  1. c=4c = 4 (correct answer)
  2. c=3c = 3
  3. c=5c = 5
  4. c=125c = \frac{12}{5}
Explanation: Using the power rule, (xayb)c=xacybc\left(\frac{x^a}{y^b}\right)^c = \frac{x^{ac}}{y^{bc}}. Substituting a=3a = 3 and b=5b = 5: x3cy5c=x12y15\frac{x^{3c}}{y^{5c}} = \frac{x^{12}}{y^{15}}. This gives us 3c=123c = 12 and 5c=155c = 15. Both equations yield c=4c = 4. Choice B incorrectly uses c=ac = a. Choice C incorrectly uses c=bc = b. Choice D results from incorrectly solving 123=15c\frac{12}{3} = \frac{15}{c}.

Question 10

Simplify the expression m−5n6m2n−3\frac{m^{-5}n^6}{m^2n^{-3}}.

  1. n9m7\frac{n^9}{m^7} (correct answer)
  2. n3m3\frac{n^3}{m^3}
  3. m7n9m^7n^9
  4. m7n9\frac{m^7}{n^9}
Explanation: To simplify the expression, use the quotient rule for exponents, which states axay=ax−y\frac{a^x}{a^y} = a^{x-y}. For the variable mm, the new exponent is −5−2=−7-5 - 2 = -7. For the variable nn, the new exponent is 6−(−3)=6+3=96 - (-3) = 6 + 3 = 9. This gives the expression m−7n9m^{-7}n^9. To express this with positive exponents, move the term with the negative exponent to the denominator, resulting in n9m7\frac{n^9}{m^7}.

Question 11

If (xk)2⋅x5=x17(x^k)^2 \cdot x^5 = x^{17}, what is the value of kk?

  1. 3
  2. 6 (correct answer)
  3. 10
  4. 24
Explanation: First, simplify the left side of the equation. Using the power rule, (xk)2=x2k(x^k)^2 = x^{2k}. The equation becomes x2k⋅x5=x17x^{2k} \cdot x^5 = x^{17}. Using the product rule, combine the terms on the left: x2k+5=x17x^{2k+5} = x^{17}. Since the bases are the same, the exponents must be equal: 2k+5=172k+5 = 17. Solving for kk: 2k=122k = 12, so k=6k = 6.

Question 12

What is the value of the expression 28⋅2−322\frac{2^8 \cdot 2^{-3}}{2^2}?

  1. 1
  2. 8 (correct answer)
  3. 32
  4. 128
Explanation: First, simplify the numerator using the product rule for exponents (ax⋅ay=ax+ya^x \cdot a^y = a^{x+y}): 28⋅2−3=28+(−3)=252^8 \cdot 2^{-3} = 2^{8+(-3)} = 2^5. Now the expression is 2522\frac{2^5}{2^2}. Next, use the quotient rule for exponents (axay=ax−y\frac{a^x}{a^y} = a^{x-y}): 2522=25−2=23\frac{2^5}{2^2} = 2^{5-2} = 2^3. Finally, calculate the value: 23=2⋅2⋅2=82^3 = 2 \cdot 2 \cdot 2 = 8.

Question 13

The expression (2x5y)2x3y3\frac{(2x^5y)^2}{x^3y^3} simplifies to the form axbycax^by^c. What is the value of bb?

  1. -1
  2. 4
  3. 7 (correct answer)
  4. 13
Explanation: First, simplify the numerator by applying the exponent 2 to each factor inside the parentheses: (2x5y)2=22(x5)2y2=4x10y2(2x^5y)^2 = 2^2(x^5)^2y^2 = 4x^{10}y^2. The expression becomes 4x10y2x3y3\frac{4x^{10}y^2}{x^3y^3}. Now, simplify the fraction. The coefficient is 4. For the variable xx, use the quotient rule: x10−3=x7x^{10-3} = x^7. For the variable yy, use the quotient rule: y2−3=y−1y^{2-3} = y^{-1}. The simplified expression is 4x7y−14x^7y^{-1}. In the form axbycax^by^c, we have a=4a=4, b=7b=7, and c=−1c=-1. The question asks for the value of bb, which is 7.

Question 14

Simplify the expression x−5⋅x8x−2x^{-5} \cdot \frac{x^8}{x^{-2}}.

  1. xx
  2. x5x^5 (correct answer)
  3. x6x^6
  4. x−15x^{-15}
Explanation: First, simplify the fraction x8x−2\frac{x^8}{x^{-2}} using the quotient rule for exponents: x8−(−2)=x8+2=x10x^{8 - (-2)} = x^{8+2} = x^{10}. Now the expression is x−5⋅x10x^{-5} \cdot x^{10}. Next, use the product rule for exponents to combine these terms: x−5+10=x5x^{-5+10} = x^5.

Question 15

If p=(23)2p = (2^3)^2 and q=2(32)q = 2^{(3^2)}, what is the relationship between pp and qq?

  1. p=qp = q
  2. p>qp > q
  3. p<qp < q (correct answer)
  4. p=−qp = -q
Explanation: First, evaluate pp. Using the power of a power rule, p=(23)2=23⋅2=26p = (2^3)^2 = 2^{3 \cdot 2} = 2^6. We know 26=642^6 = 64. Next, evaluate qq. The exponent is 323^2, which must be calculated first: 32=93^2 = 9. So, q=29q = 2^9. We know 29=5122^9 = 512. Since 64<51264 < 512, it follows that p<qp < q.

Question 16

The expression (4×107)(2×10−3)\frac{(4 \times 10^7)}{(2 \times 10^{-3})} is equivalent to which of the following?

  1. 2×1042 \times 10^4
  2. 2×10−212 \times 10^{-21}
  3. 2×10−102 \times 10^{-10}
  4. 2×10102 \times 10^{10} (correct answer)
Explanation: To simplify this expression, handle the coefficients and the powers of 10 separately. First, divide the coefficients: 4÷2=24 \div 2 = 2. Next, divide the powers of 10 using the quotient rule for exponents: 10710−3=107−(−3)=107+3=1010\frac{10^7}{10^{-3}} = 10^{7 - (-3)} = 10^{7+3} = 10^{10}. Combining the two parts gives the final answer: 2×10102 \times 10^{10}.

Question 17

The expression (a−2b4)(a3b−1)−2(a^{-2}b^4)(a^3b^{-1})^{-2} is equivalent to which of the following?

  1. ba\frac{b}{a}
  2. a8b6a^8b^6
  3. b6a8\frac{b^6}{a^8} (correct answer)
  4. a4b2\frac{a^4}{b^2}
Explanation: First, apply the power rule to the second part of the expression: (a3b−1)−2=a3⋅−2b−1⋅−2=a−6b2(a^3b^{-1})^{-2} = a^{3 \cdot -2}b^{-1 \cdot -2} = a^{-6}b^2. Now, multiply this result by the first part of the expression using the product rule (add exponents): (a−2b4)(a−6b2)=a−2+(−6)b4+2=a−8b6(a^{-2}b^4)(a^{-6}b^2) = a^{-2 + (-6)}b^{4 + 2} = a^{-8}b^6. Finally, write the expression with positive exponents: a−8b6=b6a8a^{-8}b^6 = \frac{b^6}{a^8}.

Question 18

Which of the following expressions is equivalent to (x9)1/3⋅x−2(x^9)^{1/3} \cdot x^{-2}?

  1. xx (correct answer)
  2. x5x^5
  3. x−6x^{-6}
  4. x25x^{25}
Explanation: First, simplify (x9)1/3(x^9)^{1/3} using the power rule for exponents, which means multiplying the exponents: x9⋅(1/3)=x3x^{9 \cdot (1/3)} = x^3. Now the expression is x3⋅x−2x^3 \cdot x^{-2}. Using the product rule, add the exponents: x3+(−2)=x1=xx^{3 + (-2)} = x^1 = x.

Question 19

Which of the following is equivalent to (4p5q0)22p3\frac{(4p^5q^0)^2}{2p^3}?

  1. 2p72p^7
  2. 8p78p^7 (correct answer)
  3. 4p7q24p^7q^2
  4. 8p128p^{12}
Explanation: First, simplify the term q0q^0, which is equal to 1. The expression becomes (4p5)22p3\frac{(4p^5)^2}{2p^3}. Next, apply the exponent 2 to the terms in the numerator: (4p5)2=42(p5)2=16p10(4p^5)^2 = 4^2(p^5)^2 = 16p^{10}. The expression is now 16p102p3\frac{16p^{10}}{2p^3}. Finally, simplify the fraction. Divide the coefficients: 16÷2=816 \div 2 = 8. Use the quotient rule for the variable: p10÷p3=p10−3=p7p^{10} \div p^3 = p^{10-3} = p^7. The final simplified expression is 8p78p^7.

Question 20

The expression 2x0y5(2y)3\frac{2x^0y^5}{(2y)^3} is equivalent to which of the following?

  1. y24\frac{y^2}{4} (correct answer)
  2. xy24\frac{x y^2}{4}
  3. 4y24y^2
  4. 14y2\frac{1}{4y^2}
Explanation: First, simplify terms with zero exponents: x0=1x^0 = 1. The expression becomes 2(1)y5(2y)3=2y5(2y)3\frac{2(1)y^5}{(2y)^3} = \frac{2y^5}{(2y)^3}. Next, simplify the denominator by applying the exponent 3 to both factors: (2y)3=23y3=8y3(2y)^3 = 2^3y^3 = 8y^3. The expression is now 2y58y3\frac{2y^5}{8y^3}. Simplify the coefficients: 28=14\frac{2}{8} = \frac{1}{4}. Simplify the variables using the quotient rule: y5y3=y5−3=y2\frac{y^5}{y^3} = y^{5-3} = y^2. Combining these results gives 14y2\frac{1}{4}y^2, or y24\frac{y^2}{4}.