Question

which expression is equivalent to \\(\frac{4x^{3}y^{-4}}{(3x)^{-2}}\\)?
option 1: \\(\frac{4x^{4}}{3y^{4}}\\)
option 2: \\(\frac{24x^{6}}{y^{4}}\\)
option 3: \\(\frac{2x}{3y^{4}}\\)
option 4: \\(\frac{36x^{5}}{y^{4}}\\)

Explanation:

Step1: Simplify denominator negative exponent

A term with a negative exponent moves to the numerator: $(3x)^{-2} = \frac{1}{(3x)^2}$, so the expression becomes $4x^3y^{-4} \times (3x)^2$.

Step2: Expand the squared term

Calculate $(3x)^2$: $(3x)^2 = 3^2 \times x^2 = 9x^2$.

Step3: Multiply the constants and x terms

Multiply constants: $4 \times 9 = 36$. Multiply x terms using exponent rule $x^a \times x^b = x^{a+b}$: $x^3 \times x^2 = x^{3+2} = x^5$.

Step4: Rewrite negative y exponent

$y^{-4} = \frac{1}{y^4}$, so combine all parts: $36x^5 \times \frac{1}{y^4} = \frac{36x^5}{y^4}$.

Answer:

Option 4. $\frac{36x^5}{y^4}$