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Exponent Rules - Math 1

  1. Which expression is equivalent to x−3⋅x7x−2⋅x3\frac{x^{-3} \cdot x^7}{x^{-2} \cdot x^3}?
    Answer: x3x^{3} First simplify the numerator: x−3⋅x7=x−3+7=x4x^{-3} \cdot x^7 = x^{-3+7} = x^4. Then simplify the denominator: x−2⋅x3=x−2+3=x1=xx^{-2} \cdot x^3 = x^{-2+3} = x^1 = x. Finally: x4x=x4−1=x3\frac{x^4}{x} = x^{4-1} = x^3. Choice A results from incorrectly subtracting exponents in multiplication. Choice C comes from adding all exponents without considering the fraction. Choice D results from multiplying all exponents.
  2. Which expression is equivalent to (2−325)−1\left(\frac{2^{-3}}{2^5}\right)^{-1}?
    Answer: 282^{8} First simplify inside the parentheses: 2−325=2−3−5=2−8\frac{2^{-3}}{2^5} = 2^{-3-5} = 2^{-8}. Then apply the outer exponent: (2−8)−1=2(−8)⋅(−1)=28(2^{-8})^{-1} = 2^{(-8) \cdot (-1)} = 2^8. Choice A results from incorrectly calculating the quotient as 2−3−5=2−22^{-3-5} = 2^{-2} and then applying the power rule. Choice C forgets to apply the outer negative exponent. Choice D comes from multiple errors in exponent manipulation.
  3. A financial model uses the expression 52x−15x−3⋅5−2\frac{5^{2x-1}}{5^{x-3}} \cdot 5^{-2} to calculate compound interest. What is this expression simplified?
    Answer: 5x5^{x} When you encounter expressions with exponents that have the same base, you're working with the laws of exponents. This problem tests your ability to simplify expressions involving division and multiplication of exponential terms. To simplify 52x−15x−3⋅5−2\frac{5^{2x-1}}{5^{x-3}} \cdot 5^{-2}, apply the exponent rules step by step. First, handle the division: when dividing powers with the same base, subtract the exponents. So 52x−15x−3=5(2x−1)−(x−3)=52x−1−x+3=5x+2\frac{5^{2x-1}}{5^{x-3}} = 5^{(2x-1)-(x-3)} = 5^{2x-1-x+3} = 5^{x+2}. Next, multiply this result by 5−25^{-2}. When multiplying powers with the same base, add the exponents: 5x+2⋅5−2=5(x+2)+(−2)=5x+2−2=5x5^{x+2} \cdot 5^{-2} = 5^{(x+2)+(-2)} = 5^{x+2-2} = 5^x. Looking at the wrong answers: Choice A (53x−65^{3x-6}) likely comes from incorrectly adding all the original exponents rather than following proper order of operations. Choice B (5x+25^{x+2}) represents stopping after the division step and forgetting to multiply by 5−25^{-2}. Choice C (5x−45^{x-4}) results from sign errors when combining the exponents, particularly mishandling the subtraction in the division step. Remember this sequence for exponent problems: division means subtract exponents, multiplication means add exponents, and always work left to right while carefully tracking positive and negative signs. These rules are fundamental building blocks you'll use throughout algebra and beyond.
  4. Which expression is equivalent to (3−2⋅533−4⋅5−1)−2\left(\frac{3^{-2} \cdot 5^3}{3^{-4} \cdot 5^{-1}}\right)^{-2}?
    Answer: 134⋅58\frac{1}{3^4 \cdot 5^8} First, simplify the expression inside the parentheses: 3−2⋅533−4⋅5−1=3−2−(−4)⋅53−(−1)=3−2+4⋅53+1=32⋅54\frac{3^{-2} \cdot 5^3}{3^{-4} \cdot 5^{-1}} = 3^{-2-(-4)} \cdot 5^{3-(-1)} = 3^{-2+4} \cdot 5^{3+1} = 3^2 \cdot 5^4. Now apply the outer exponent: (32⋅54)−2=32⋅(−2)⋅54⋅(−2)=3−4⋅5−8=134⋅58\left(3^2 \cdot 5^4\right)^{-2} = 3^{2 \cdot (-2)} \cdot 5^{4 \cdot (-2)} = 3^{-4} \cdot 5^{-8} = \frac{1}{3^4 \cdot 5^8}. Choice B results from making an error with the negative exponent on 5, getting 3−4⋅583^{-4} \cdot 5^8 instead of 3−4⋅5−83^{-4} \cdot 5^{-8}. Choice C results from making an error with the negative exponent on 3, getting 34⋅5−83^4 \cdot 5^{-8} instead of 3−4⋅5−83^{-4} \cdot 5^{-8}. Choice D results from forgetting the negative sign on the outer exponent entirely, giving 34⋅583^4 \cdot 5^8.
  5. Which of the following expressions is equivalent to (a3b−2a−1b4)−3\left(\frac{a^3b^{-2}}{a^{-1}b^4}\right)^{-3}?
    Answer: b18a12\frac{b^{18}}{a^{12}} First simplify inside parentheses: a3b−2a−1b4=a3−(−1)⋅b−2−4=a3+1⋅b−6=a4b−6\frac{a^3b^{-2}}{a^{-1}b^4} = a^{3-(-1)} \cdot b^{-2-4} = a^{3+1} \cdot b^{-6} = a^4b^{-6}. Then apply the outer exponent: (a4b−6)−3=a4⋅(−3)⋅b−6⋅(−3)=a−12b18=b18a12(a^4b^{-6})^{-3} = a^{4 \cdot (-3)} \cdot b^{-6 \cdot (-3)} = a^{-12}b^{18} = \frac{b^{18}}{a^{12}}. Choice A incorrectly handles the signs of the exponents. Choice C forgets that negative exponents create reciprocals. Choice D incorrectly places both variables in the denominator.
  6. If x−3=8x^{-3} = 8, what is the value of x6x^6?
    Answer: 164\frac{1}{64} From x−3=8x^{-3} = 8, we get 1x3=8\frac{1}{x^3} = 8, so x3=18x^3 = \frac{1}{8}. Therefore x6=(x3)2=(18)2=164x^6 = (x^3)^2 = \left(\frac{1}{8}\right)^2 = \frac{1}{64}. Choice B results from incorrectly calculating (18)3\left(\frac{1}{8}\right)^3. Choice C comes from taking the reciprocal incorrectly. Choice D results from calculating 8−2=1648^{-2} = \frac{1}{64} but then making an additional error.
  7. A radioactive substance decays according to the formula N(t)=N0⋅(12)t/hN(t) = N_0 \cdot (\frac{1}{2})^{t/h}, where hh is the half-life. If a sample has a half-life of 6 hours and contains 80 grams after 18 hours, how many grams were present initially?
    Answer: 640640 grams Given N(t)=N0⋅(12)t/hN(t) = N_0 \cdot (\frac{1}{2})^{t/h} with h=6h = 6 hours, t=18t = 18 hours, and N(18)=80N(18) = 80 grams. Substituting: 80=N0⋅(12)18/6=N0⋅(12)3=N0⋅1880 = N_0 \cdot (\frac{1}{2})^{18/6} = N_0 \cdot (\frac{1}{2})^3 = N_0 \cdot \frac{1}{8}. Therefore N0=80⋅8=640N_0 = 80 \cdot 8 = 640 grams. Choice A (160) results from computing N0=80⋅2N_0 = 80 \cdot 2 (using (12)1(\frac{1}{2})^1 instead of (12)3(\frac{1}{2})^3). Choice B (320) results from computing N0=80⋅4N_0 = 80 \cdot 4 (using (12)2(\frac{1}{2})^2 instead of (12)3(\frac{1}{2})^3). Choice C (480) results from computing N0=80⋅6N_0 = 80 \cdot 6 (incorrectly using the half-life value as the multiplier).
  8. What is the simplified form of (a−2b3a−4b−1)−2\left(\frac{a^{-2}b^3}{a^{-4}b^{-1}}\right)^{-2}?
    Answer: 1a4b8\frac{1}{a^{4}b^{8}} First simplify inside the parentheses: a−2b3a−4b−1=a−2−(−4)⋅b3−(−1)=a−2+4⋅b3+1=a2b4\frac{a^{-2}b^3}{a^{-4}b^{-1}} = a^{-2-(-4)} \cdot b^{3-(-1)} = a^{-2+4} \cdot b^{3+1} = a^2b^4. Now apply the outer exponent: (a2b4)−2=a2⋅(−2)⋅b4⋅(−2)=a−4b−8=1a4b8(a^2b^4)^{-2} = a^{2 \cdot (-2)} \cdot b^{4 \cdot (-2)} = a^{-4}b^{-8} = \frac{1}{a^4b^8}. Choice A would result from incorrectly making the exponents positive. Choice B comes from incorrectly placing aa in the denominator and bb in the numerator. Choice C ignores the negative outer exponent entirely.
  9. If a−2=19a^{-2} = \frac{1}{9}, what is the value of a4a^4?
    Answer: 8181 From a−2=19a^{-2} = \frac{1}{9}, we get 1a2=19\frac{1}{a^2} = \frac{1}{9}, so a2=9a^2 = 9. Therefore a4=(a2)2=92=81a^4 = (a^2)^2 = 9^2 = 81. Choice B incorrectly inverts the final answer. Choice C comes from cubing instead of squaring (93=7299^3 = 729). Choice D incorrectly applies negative exponents to the final result.
  10. Which of the following expressions is equivalent to x−4y2x−7y−3\frac{x^{-4} y^2}{x^{-7} y^{-3}} when x≠0x \neq 0 and y≠0y \neq 0?
    Answer: x3y5x^3 y^5 Using the quotient rule for exponents: x−4y2x−7y−3=x−4−(−7)⋅y2−(−3)=x−4+7⋅y2+3=x3y5\frac{x^{-4} y^2}{x^{-7} y^{-3}} = x^{-4-(-7)} \cdot y^{2-(-3)} = x^{-4+7} \cdot y^{2+3} = x^3 y^5. Choice A results from incorrectly computing y2−(−3)=y2−3=y−1y^{2-(-3)} = y^{2-3} = y^{-1} instead of y5y^5, then possibly making another error. Choice C results from incorrectly computing x−4−(−7)=x−4−7=x−11x^{-4-(-7)} = x^{-4-7} = x^{-11}, then perhaps correcting to x−3x^{-3}. Choice D results from computing x−4−(−7)=x−11x^{-4-(-7)} = x^{-11} instead of x3x^3, then writing x−11=1x11x^{-11} = \frac{1}{x^{11}}.
  11. What is the simplified form of x−2y−1x−5y3⋅(xy)−2\frac{x^{-2}y^{-1}}{x^{-5}y^{3}} \cdot (xy)^{-2}?
    Answer: xy6\frac{x}{y^{6}} When you encounter expressions with negative exponents and multiplication, you need to systematically apply the rules of exponents to simplify step by step. Start by simplifying the fraction x−2y−1x−5y3\frac{x^{-2}y^{-1}}{x^{-5}y^{3}}. When dividing powers with the same base, subtract the exponents: x−2−(−5)⋅y−1−3=x3⋅y−4=x3y4x^{-2-(-5)} \cdot y^{-1-3} = x^{3} \cdot y^{-4} = \frac{x^3}{y^4}. Next, handle (xy)−2(xy)^{-2}. This equals x−2y−2=1x2y2x^{-2}y^{-2} = \frac{1}{x^2y^2}. Now multiply these results: x3y4⋅1x2y2=x3x2y4y2=x3−2y4+2=x1y6=xy6\frac{x^3}{y^4} \cdot \frac{1}{x^2y^2} = \frac{x^3}{x^2y^4y^2} = \frac{x^{3-2}}{y^{4+2}} = \frac{x^1}{y^6} = \frac{x}{y^6}, which is answer C. Let's examine why the other options are wrong. Answer A (x5y2\frac{x^{5}}{y^{2}}) likely results from incorrectly adding exponents instead of subtracting when dividing, or mishandling the negative exponents. Answer B (1xy6\frac{1}{xy^{6}}) suggests an error in the x-exponent calculation, possibly forgetting that x−2÷x−5=x3x^{-2} \div x^{-5} = x^3, not x−1x^{-1}. Answer D (1x5y2\frac{1}{x^{5}y^{2}}) appears to come from multiple sign errors and incorrect exponent arithmetic. Remember: negative exponents mean reciprocals, and when dividing powers with the same base, subtract the bottom exponent from the top exponent. Work systematically through each step to avoid sign errors.
  12. A scientist observes that a bacterial population decreases according to the formula P(t)=P0⋅2−t/3P(t) = P_0 \cdot 2^{-t/3}, where tt is time in hours. If the population is 800 bacteria at t=6t = 6 hours, how many bacteria were present initially (at t=0t = 0)?
    Answer: 32003200 bacteria Given P(t)=P0⋅2−t/3P(t) = P_0 \cdot 2^{-t/3} and P(6)=800P(6) = 800. Substituting: 800=P0⋅2−6/3=P0⋅2−2=P0⋅14800 = P_0 \cdot 2^{-6/3} = P_0 \cdot 2^{-2} = P_0 \cdot \frac{1}{4}. Therefore P0=800⋅4=3200P_0 = 800 \cdot 4 = 3200. Choice A (200) would result if student incorrectly computed P0=8004P_0 = \frac{800}{4} instead of P0=800⋅4P_0 = 800 \cdot 4. Choice B (1600) would result from calculating 2−6/3=2−2=122^{-6/3} = 2^{-2} = \frac{1}{2} instead of 14\frac{1}{4}. Choice D (6400) would result from using 2−6/3=2−2=182^{-6/3} = 2^{-2} = \frac{1}{8} instead of 14\frac{1}{4}.
  13. A researcher models population decay using P(t)=P0⋅(12)3t⋅2t−1P(t) = P_0 \cdot \left(\frac{1}{2}\right)^{3t} \cdot 2^{t-1}. Which expression represents the simplified form of the decay factor (12)3t⋅2t−1\left(\frac{1}{2}\right)^{3t} \cdot 2^{t-1}?
    Answer: 2−2t−12^{-2t-1} When you encounter exponential expressions with different bases, your goal is to rewrite everything using the same base to simplify the calculation. Here, you need to express both terms using base 2. Start by recognizing that 12=2−1\frac{1}{2} = 2^{-1}. This means (12)3t=(2−1)3t=2−3t\left(\frac{1}{2}\right)^{3t} = \left(2^{-1}\right)^{3t} = 2^{-3t} using the power rule (am)n=amn(a^m)^n = a^{mn}. Next, rewrite 2t−12^{t-1} using the property that am−n=am⋅a−na^{m-n} = a^m \cdot a^{-n}: 2t−1=2t⋅2−12^{t-1} = 2^t \cdot 2^{-1}. Now multiply the expressions: 2−3t⋅2t⋅2−12^{-3t} \cdot 2^t \cdot 2^{-1}. When multiplying powers with the same base, add the exponents: 2−3t+t+(−1)=2−2t−12^{-3t + t + (-1)} = 2^{-2t - 1}. Looking at the wrong answers: Choice A (24t−12^{4t-1}) incorrectly treats 12\frac{1}{2} as positive 2, leading to 23t2^{3t} instead of 2−3t2^{-3t}. Choice B (2−2t+12^{-2t+1}) makes an error in the constant term, writing +1+1 instead of −1-1. Choice D (122t+1\frac{1}{2^{2t+1}}) is mathematically equivalent to choice C since 122t+1=2−(2t+1)=2−2t−1\frac{1}{2^{2t+1}} = 2^{-(2t+1)} = 2^{-2t-1}, but it's not simplified to the same form as the other exponential expressions. The answer is C: 2−2t−12^{-2t-1}. Strategy tip: Always convert fractions with exponential bases to negative exponents first—it makes combining terms much cleaner and reduces arithmetic errors.

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