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Difficulty: Very hardProperties of Exponents in Algebraic Expressions

For all non-zero real numbers xx and yy, which of the following is equivalent to the expression (2x−1y2+12x−1y2)−3(x2y−3)−2\frac{\left( 2x^{-1} y^2 + \frac{1}{2} x^{-1} y^2 \right)^{-3}}{(x^2 y^{-3})^{-2}}?

  1. A
    5x72y12\frac{5x^7}{2y^{12}}
  2. 8x7125y12\frac{8x^7}{125y^{12}}Answer
  3. C
    65x78y12\frac{65x^7}{8y^{12}}
  4. D
    8125x12y36\frac{8}{125x^{12}y^{36}}
  5. E
    x7y12\frac{x^7}{y^{12}}

Answer

8x7125y12\frac{8x^7}{125y^{12}}
To find the equivalent expression, we first combine the like terms inside the parentheses in the numerator to get 52x−1y2\frac{5}{2} x^{-1} y^2. Raising this product to the power of −3-3 yields (52)−3(x−1)−3(y2)−3=8125x3y−6\left(\frac{5}{2}\right)^{-3} (x^{-1})^{-3} (y^2)^{-3} = \frac{8}{125} x^3 y^{-6}. Next, the denominator simplifies to (x2y−3)−2=x−4y6(x^2 y^{-3})^{-2} = x^{-4} y^6. Dividing the numerator by the denominator requires subtracting the exponents of like bases: for xx, we have 3−(−4)=73 - (-4) = 7, and for yy, we have −6−6=−12-6 - 6 = -12. This results in 8125x7y−12\frac{8}{125} x^7 y^{-12}, which is equivalent to the correct expression 8x7125y12\frac{8x^7}{125y^{12}}.

Step-by-Step Solution

1
Combine the like terms inside the parentheses in the numerator.
2x−1y2+12x−1y2=(2+12)x−1y2=52x−1y22x^{-1} y^2 + \frac{1}{2} x^{-1} y^2 = \left(2 + \frac{1}{2}\right) x^{-1} y^2 = \frac{5}{2} x^{-1} y^2
Before applying the outer negative exponent, it is mathematically simpler to combine the like terms inside the grouping.
2
Apply the power of −3-3 to the term in the numerator.
(52x−1y2)−3=(52)−3(x−1)−3(y2)−3=8125x3y−6\left(\frac{5}{2} x^{-1} y^2\right)^{-3} = \left(\frac{5}{2}\right)^{-3} (x^{-1})^{-3} (y^2)^{-3} = \frac{8}{125} x^3 y^{-6}
The power of a product rule (ab)n=anbn(ab)^n = a^n b^n and the power of a power rule (am)n=amn(a^m)^n = a^{mn} are applied to expand the term.
3
Simplify the denominator by applying the power of −2-2.
(x2y−3)−2=(x2)−2(y−3)−2=x−4y6(x^2 y^{-3})^{-2} = (x^2)^{-2} (y^{-3})^{-2} = x^{-4} y^6
The power of a product rule is applied to the denominator to resolve the outer exponent.
4
Divide the simplified numerator by the simplified denominator using the quotient rule for exponents.
8125x3y−6x−4y6=8125x3−(−4)y−6−6=8125x7y−12=8x7125y12\frac{\frac{8}{125} x^3 y^{-6}}{x^{-4} y^6} = \frac{8}{125} x^{3 - (-4)} y^{-6 - 6} = \frac{8}{125} x^7 y^{-12} = \frac{8x^7}{125y^{12}}
The quotient rule am/an=am−na^m / a^n = a^{m-n} is used to subtract the exponents of the corresponding variables, and negative exponents are rewritten in the denominator.

Key Concept

Properties of Exponents in Algebraic Expressions

Alternative Method

Alternatively, you can write out all variables with positive exponents before simplifying. Rewrite the term in the numerator as 2y2x+y22x=5y22x\frac{2y^2}{x} + \frac{y^2}{2x} = \frac{5y^2}{2x}. Raising this to the −3-3 power flips the fraction and cubes it, yielding (2x5y2)3=8x3125y6\left(\frac{2x}{5y^2}\right)^3 = \frac{8x^3}{125y^6}. Simplifying the denominator yields 1(x2y−3)2=1x4y−6=y6x4\frac{1}{(x^2 y^{-3})^2} = \frac{1}{x^4 y^{-6}} = \frac{y^6}{x^4}. Dividing the numerator by the denominator yields 8x3125y6÷y6x4=8x3125y6⋅x4y6=8x7125y12\frac{8x^3}{125y^6} \div \frac{y^6}{x^4} = \frac{8x^3}{125y^6} \cdot \frac{x^4}{y^6} = \frac{8x^7}{125y^{12}}.
Estimated Time:2m 0s
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Topics

Properties of Exponents in Algebraic Expressions

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