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Question

Difficulty: HardOperations with Variable Exponents

Consider the algebraic expression below, where x≠0x \neq 0 and y≠0y \neq 0. Which of the following is its fully simplified form?

(3x−2y32xy−2)−2×(9x24y4)−1 \left( \frac{3x^{-2}y^3}{2xy^{-2}} \right)^{-2} \times \left( \frac{9x^2}{4y^4} \right)^{-1}
  1. A
    1681x12y40 \frac{16}{81x^{12}y^{40}}
  2. B
    27x48y6 \frac{27x^4}{8y^6}
  3. 16x481y6 \frac{16x^4}{81y^6} Answer
  4. D
    16y281 \frac{16y^2}{81}

Answer

The expression fully simplifies to 16x481y6\frac{16x^4}{81y^6}.
The correct expression is derived by sequentially simplifying the contents of the parentheses using the quotient rule, applying the outer negative exponents to both the numerical coefficients and variables, and then multiplying the resulting terms by adding the exponents of like bases.

Step-by-Step Solution

1
Simplify the fraction inside the first set of parentheses using the quotient rule for exponents (am/an=am−na^m / a^n = a^{m-n}).
3x−2y32xy−2=32x−2−1y3−(−2)=32x−3y5 \frac{3x^{-2}y^3}{2xy^{-2}} = \frac{3}{2} x^{-2-1} y^{3-(-2)} = \frac{3}{2} x^{-3} y^5
Simplifying inside the parentheses first follows the order of operations and makes subsequent calculations more manageable.
2
Apply the outer negative exponent (−2-2) to all components of the first simplified term, including the numerical coefficient.
(32x−3y5)−2=(32)−2(x−3)−2(y5)−2=49x6y−10 \left( \frac{3}{2} x^{-3} y^5 \right)^{-2} = \left( \frac{3}{2} \right)^{-2} (x^{-3})^{-2} (y^5)^{-2} = \frac{4}{9} x^6 y^{-10}
The power of a product rule states that (ab)n=anbn(ab)^n = a^n b^n, and applying a negative power to a fraction flips it: (a/b)−n=(b/a)n(a/b)^{-n} = (b/a)^n.
3
Apply the outer negative exponent (−1-1) to the second term.
(9x24y4)−1=4y49x2=49x−2y4 \left( \frac{9x^2}{4y^4} \right)^{-1} = \frac{4y^4}{9x^2} = \frac{4}{9} x^{-2} y^4
A negative exponent of −1-1 simply takes the reciprocal (flips the fraction).
4
Multiply the two resulting expressions by grouping coefficients and using the product rule for exponents (am×an=am+na^m \times a^n = a^{m+n}).
(49x6y−10)×(49x−2y4)=(49×49)(x6×x−2)(y−10×y4)=1681x4y−6 \left( \frac{4}{9} x^6 y^{-10} \right) \times \left( \frac{4}{9} x^{-2} y^4 \right) = \left( \frac{4}{9} \times \frac{4}{9} \right) (x^6 \times x^{-2}) (y^{-10} \times y^4) = \frac{16}{81} x^4 y^{-6}
When multiplying terms with the same base, their exponents are added together.
5
Rewrite the final expression using only positive exponents.
1681x4y−6=16x481y6 \frac{16}{81} x^4 y^{-6} = \frac{16x^4}{81y^6}
Standard mathematical notation requires fully simplified algebraic expressions to be written without negative exponents.

Key Concept

Simplifying rational expressions using the laws of exponents, including the quotient rule, power rule, product rule, and handling negative exponents alongside numerical coefficients.
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Topics

Operations with Variable Exponents

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