If the equation \frac{(3x^4 y^k)^2}{x^3 y^5} = 9x^5 y^{11} is true for all non-zero values… — Armed Services Vocational Aptitude Battery (ASVAB)

If the equation

\frac{(3x^4 y^k)^2}{x^3 y^5} = 9x^5 y^{11}

is true for all non-zero values of

x

and

y

, what is the value of

k

Cevap: 8 / k=8 / k = 8

By applying the power rule to expand the numerator, the expression becomes

9x^8 y^{2k}

. Using the quotient rule to divide these terms by the denominator

x^3 y^5

, the expression simplifies to

9x^5 y^{2k-5}

. Equating this final simplified form to the given result

9x^5 y^{11}

, we can isolate the exponent of

y

to set up the equation

2k - 5 = 11

. Solving this equation gives

k = 8

Apply the power of a product and power of a power rules to the numerator.

(3x^4 y^k)^2 = 3^2 (x^4)^2 (y^k)^2 = 9x^8 y^{2k}

When a product is raised to a power, each factor inside the parentheses must be individually raised to that power by multiplying the exponents.

Substitute the expanded numerator back into the fraction and apply the quotient rule.

\frac{9x^8 y^{2k}}{x^3 y^5} = 9 x^{8-3} y^{2k-5} = 9x^5 y^{2k-5}

When dividing algebraic terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.

Equate the simplified exponent of

y

to the exponent of

y

on the right side of the original equation.

2k - 5 = 11

Because the expressions are equivalent and their coefficients and

x

-terms already match, their

y

-exponents must be strictly equal.

Solve the resulting linear equation for

k

2k = 16 \rightarrow k = 8

Adding 5 to both sides and then dividing by 2 isolates the variable

k

Operations with Variable Exponents

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