which expression is equivalent to $(4x^{3})(2x)^{-4}$?\n$\frac{1}{4x}$\n$\frac{4}{x}$\n$\frac{8}{x}$\n$\frac{…

Explanation:

Step1: Expand $(2x)^{-4}$

According to the power - of - a - product rule $(ab)^n=a^n b^n$ and negative exponent rule $a^{-n}=\frac{1}{a^n}$, we have $(2x)^{-4}=2^{-4}x^{-4}=\frac{1}{16x^{4}}$.

Step2: Multiply with $4x^{3}$

$(4x^{3})(2x)^{-4}=4x^{3}\times\frac{1}{16x^{4}}$.

Step3: Simplify the product

Using the rule $\frac{a^m}{a^n}=a^{m - n}$ and $\frac{4}{16}=\frac{1}{4}$, we get $\frac{4x^{3}}{16x^{4}}=\frac{1}{4x^{4 - 3}}=\frac{1}{4x}$.

Answer:

$\frac{1}{4x}$