During a physics calculation, a student models a system's energy output using the followin… — Armed Services Vocational Aptitude Battery (ASVAB)

During a physics calculation, a student models a system's energy output using the following expression, where $x$ and $y$ are non-zero variables:

\frac{(2x^{-2}y^3)^3}{4x^{-4}y^5}

What is the fully simplified form of this expression using only positive exponents?

\frac{2y^4}{x^2}

Applying the power of a product rule to the numerator yields

8x^{-6}y^9

. Dividing this by the denominator

4x^{-4}y^5

involves dividing the coefficients (

8/4 = 2

) and subtracting the exponents of like bases. For

x

-6 - (-4) = -2

, and for

y

9 - 5 = 4

. The resulting expression is

2x^{-2}y^4

. To represent this with positive exponents, the

x^{-2}

term is moved to the denominator, giving

\frac{2y^4}{x^2}

Apply the power of a product rule to the numerator.

(2x^{-2}y^3)^3 = 2^3 \cdot (x^{-2})^3 \cdot (y^3)^3 = 8x^{-6}y^9

The exponent outside the parentheses applies to every individual factor inside.

Divide the numerical coefficients.

8 \div 4 = 2

Numerical coefficients are handled separately from variables with exponents.

Apply the quotient of powers rule to the variables.

x^{-6 - (-4)} = x^{-2}

and

y^{9 - 5} = y^4

When dividing like bases, you subtract the denominator's exponent from the numerator's exponent.

Combine the results and convert the negative exponent to a positive exponent.

2x^{-2}y^4 = \frac{2y^4}{x^2}

Moving a term with a negative exponent to the denominator changes its exponent to positive.

Simplifying rational algebraic expressions using the power of a product rule, the quotient of powers rule, and negative exponents.

Question