Integer Exponents - Pre Algebra

1. If $3^{-2} \cdot 9^x = 27^{-1}$, what is the value of $x$?
   Answer: $-\frac{1}{2}$ First, express everything in terms of base 3: $3^{-2} \cdot (3^2)^x = (3^3)^{-1}$, which becomes $3^{-2} \cdot 3^{2x} = 3^{-3}$. Using the product rule: $3^{-2+2x} = 3^{-3}$. Therefore $-2+2x = -3$, so $2x = -1$ and $x = -\frac{1}{2}$. Choice A results from incorrectly setting up $-2+2x = -1$. Choice C comes from sign errors in the exponent rules. Choice D results from misinterpreting the bases or making multiple calculation errors.
2. Which expression is equivalent to $\frac{2^{-3} \cdot 4^2}{8^{-1}}$?
   Answer: $2^6$ Convert all terms to base 2: $\frac{2^{-3} \cdot (2^2)^2}{(2^3)^{-1}} = \frac{2^{-3} \cdot 2^4}{2^{-3}}$. This simplifies to $\frac{2^{-3+4}}{2^{-3}} = \frac{2^1}{2^{-3}} = 2^1 \cdot 2^3 = 2^6$. 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.
3. The expression $\frac{(3^2)^{-1} \cdot 3^{-3}}{3^{-7}}$ simplifies to which of the following?
   Answer: $3^{2}$ 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: $(a^m)^n = a^{mn}$, $a^m \cdot a^n = a^{m+n}$, and $\frac{a^m}{a^n} = a^{m-n}$. Let's work through this systematically. First, simplify $(3^2)^{-1}$ using the power rule: $(3^2)^{-1} = 3^{2 \cdot (-1)} = 3^{-2}$. Now the expression becomes $\frac{3^{-2} \cdot 3^{-3}}{3^{-7}}$. Next, handle the multiplication in the numerator: $3^{-2} \cdot 3^{-3} = 3^{-2 + (-3)} = 3^{-5}$. The expression is now $\frac{3^{-5}}{3^{-7}}$. Finally, use the division rule: $\frac{3^{-5}}{3^{-7}} = 3^{-5 - (-7)} = 3^{-5 + 7} = 3^2$. This matches choice B. Choice A ($3^{-2}$) would result from incorrectly adding exponents during division instead of subtracting. Choice C ($3^0$) might come from miscalculating the final subtraction as $-5 - (-7) = -2$ instead of $+2$. Choice D ($3^4$) 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.
4. If $\frac{1}{4^x} = 2^{3x-5}$, what is the value of $x$?
   Answer: $1$ Rewrite in terms of base 2: $\frac{1}{4^x} = \frac{1}{(2^2)^x} = \frac{1}{2^{2x}} = 2^{-2x}$. So the equation becomes $2^{-2x} = 2^{3x-5}$. Setting exponents equal: $-2x = 3x-5$, which gives $-2x-3x = -5$, so $-5x = -5$ and $x = 1$. Choice A results from sign errors. Choice B comes from incorrectly handling the negative exponent. Choice D results from algebraic manipulation errors.
5. If $a^{-2} = \frac{1}{16}$ and $b^3 = 8$, what is the value of $\frac{a^2}{b^{-3}}$?
   Answer: $128$ From $a^{-2} = \frac{1}{16}$, we get $a^2 = 16$. From $b^3 = 8$, we get $b = 2$, so $b^{-3} = \frac{1}{8}$. Therefore $\frac{a^2}{b^{-3}} = \frac{16}{\frac{1}{8}} = 16 \cdot 8 = 128$. Choice A results from incorrectly finding $a^2 \cdot b^{-3}$. Choice B comes from calculation errors with the reciprocal. Choice C results from using $b^3$ instead of $b^{-3}$ in the denominator.
6. If $5^{2x-1} = 125^{x-2}$, what is the value of $x$?
   Answer: $5$ Rewrite with the same base: $5^{2x-1} = (5^3)^{x-2} = 5^{3(x-2)} = 5^{3x-6}$. Since the bases are equal, the exponents must be equal: $2x-1 = 3x-6$. Solving: $-1+6 = 3x-2x$, so $5 = x$. 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 $x$.
7. For which value of $k$ is the equation $2^{k+1} \cdot 4^{k-2} = 8^2$ true?
   Answer: $k = 3$ 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: $4 = 2^2$ and $8 = 2^3$. This transforms the equation into: $2^{k+1} \cdot (2^2)^{k-2} = (2^3)^2$ Using the power rule $(a^m)^n = a^{mn}$, this becomes: $2^{k+1} \cdot 2^{2(k-2)} = 2^{3 \cdot 2}$ $2^{k+1} \cdot 2^{2k-4} = 2^6$ When multiplying powers with the same base, add the exponents: $2^{(k+1) + (2k-4)} = 2^6$ $2^{3k-3} = 2^6$ Since the bases are equal, the exponents must be equal: $3k - 3 = 6$ $3k = 9$ $k = 3$ Let's check why the other answers don't work. Answer B ($k = 2$) gives us $3(2) - 3 = 3 \neq 6$. Answer C ($k = 1$) yields $3(1) - 3 = 0 \neq 6$. Answer D ($k = 4$) produces $3(4) - 3 = 9 \neq 6$. Only answer A ($k = 3$) 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: $4 = 2^2$, $8 = 2^3$, $16 = 2^4$.
8. The expression $\left(\frac{x^{-2}y^3}{x^3y^{-1}}\right)^{-2}$ is equivalent to which of the following?
   Answer: $\frac{x^{10}}{y^8}$ First simplify inside the parentheses: $\frac{x^{-2}y^3}{x^3y^{-1}} = x^{-2-3} \cdot y^{3-(-1)} = x^{-5}y^4$. Then apply the outer exponent: $(x^{-5}y^4)^{-2} = x^{(-5)(-2)} \cdot y^{4(-2)} = x^{10}y^{-8} = \frac{x^{10}}{y^8}$. 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.

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