Assume x and y are non-zero real numbers. If the algebraic expression \frac{(3x^2y^4)^3}{9… — Armed Services Vocational Aptitude Battery (ASVAB)

Assume

x

and

y

are non-zero real numbers. If the algebraic expression

\frac{(3x^2y^4)^3}{9x^ky^7}

simplifies completely to

3xy^5

, what is the value of the exponent

k

Answer: 5 / k=5 / k = 5

The correct value is 5. By expanding the numerator using the power of a product rule,

(3x^2y^4)^3

becomes

27x^6y^{12}

. Dividing this by the denominator

9x^ky^7

yields

\left(\frac{27}{9}\right) \cdot x^{6-k} \cdot y^{12-7}

, which simplifies to

3x^{6-k}y^5

. Equating this result to the given simplified expression

3x^1y^5

means the exponent of

x

must be 1. Solving the simple equation

6 - k = 1

gives

k = 5

Expand the numerator using the power of a product and power of a power rules.

(3x^2y^4)^3 = 3^3 \cdot (x^2)^3 \cdot (y^4)^3 = 27x^6y^{12}

The outer exponent must be applied to the numerical coefficient and multiplied by the inner exponents of each variable.

Substitute the expanded numerator back into the fractional expression.

\frac{27x^6y^{12}}{9x^ky^7}

This sets up the expression for division.

Simplify the numerical coefficients and apply the quotient rule for exponents (

x^a / x^b = x^{a-b}

\left(\frac{27}{9}\right) \cdot x^{6-k} \cdot y^{12-7} = 3x^{6-k}y^5

To combine the numerical terms and the variable terms with identical bases.

Set the simplified expression equal to the target expression provided in the problem.

3x^{6-k}y^5 = 3x^1y^5

The problem states that the original expression simplifies completely to

3xy^5

Equate the exponents of the base

x

to solve for

k

6 - k = 1 \Rightarrow k = 5

Since the expressions are equal and the bases are the same, their corresponding exponents must be equal.

Applying power and quotient rules of exponents to solve for an unknown variable exponent.

Question