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Example

Simplifying 5y15y^{-1}, (5y)1(5y)^{-1}, and (5y)1(-5y)^{-1} Using the Negative Exponent Definition

Apply the negative exponent definition to three expressions that combine a coefficient with a variable raised to 1-1 — contrasting how parentheses determine which part of the expression the exponent acts on, and how a negative coefficient introduces an extra sign step. All variables are assumed nonzero.

5y1=5y5y^{-1} = \frac{5}{y}: Without parentheses, the exponent 1-1 applies only to the base yy, not to the coefficient 55. Apply the negative exponent definition to y1y^{-1} alone: y1=1yy^{-1} = \frac{1}{y}. The expression becomes 51y=5y5 \cdot \frac{1}{y} = \frac{5}{y}.

(5y)1=15y(5y)^{-1} = \frac{1}{5y}: The parentheses make the entire product 5y5y the base of the exponent 1-1. Apply the negative exponent definition to the whole base: (5y)1=1(5y)1=15y(5y)^{-1} = \frac{1}{(5y)^1} = \frac{1}{5y}.

(5y)1=15y(-5y)^{-1} = -\frac{1}{5y}: The base is now 5y-5y, including the negative sign. Apply the negative exponent definition: (5y)1=1(5y)1=15y(-5y)^{-1} = \frac{1}{(-5y)^1} = \frac{1}{-5y}. Then apply the Negative Sign Placement in Fractions property ab=ab\frac{a}{-b} = -\frac{a}{b} to rewrite the result as 15y-\frac{1}{5y}.

Comparing the three results highlights that parentheses control which quantity the exponent acts on. In part (a), only the variable is reciprocated and the coefficient stays in the numerator. In part (b), the entire product moves to the denominator. In part (c), the negative base produces a negative fraction after the sign is repositioned.

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Updated 2026-04-21

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