To solve for the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that make the given equation true:
[tex]\[
\frac{(2xy)^4}{4xy} = 4x^a y^b
\][/tex]
we begin by simplifying the left-hand side of the equation step-by-step.
1. Simplify [tex]\((2xy)^4\)[/tex]:
[tex]\[\begin{align*}
(2xy)^4 & = (2^4)(x^4)(y^4) \\
& = 16x^4y^4.
\end{align*}\][/tex]
2. Substitute this back into the original equation:
[tex]\[
\frac{16x^4y^4}{4xy}.
\][/tex]
3. Next, simplify the fraction [tex]\(\frac{16x^4y^4}{4xy}\)[/tex]:
[tex]\[\begin{align*}
\frac{16x^4y^4}{4xy} & = \frac{16}{4} \cdot \frac{x^4}{x} \cdot \frac{y^4}{y} \\
& = 4 \cdot x^{4-1} \cdot y^{4-1} \\
& = 4x^3y^3.
\end{align*}\][/tex]
So now we have:
[tex]\[
4x^3y^3 = 4x^a y^b.
\][/tex]
4. By comparing the exponents of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] on both sides of the equation:
[tex]\[\begin{align*}
x^3 & = x^a, \quad \text{which implies that} \; a = 3, \\
y^3 & = y^b, \quad \text{which implies that} \; b = 3.
\end{align*}\][/tex]
Therefore, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that make the equation true are:
[tex]\[
a = 3 \quad \text{and} \quad b = 3.
\][/tex]
Among the given options:
- [tex]\(a = 0, b = 0\)[/tex]
- [tex]\(a = 3, b = 3\)[/tex]
- [tex]\(a = 4, b = 4\)[/tex]
- [tex]\(a = 5, b = 5\)[/tex]
The correct answer is:
[tex]\[ a = 3, b = 3. \][/tex]
