Question

which values of a and b make the equation true?\\(\frac{(2xy)^{4}}{4xy} = 4x^{a}y^{b}\\)\\(\circ\\) \\(a = 0, b = 0\\)\\(\circ\\) \\(a = 3, b = 3\\)\\(\circ\\) \\(a = 4, b = 4\\)\\(\circ\\) \\(a = 5, b = 5\\)

Explanation:

Step1: Simplify the numerator

Using the power of a product rule \((ab)^n = a^n b^n\), we have \((2xy)^4 = 2^4 x^4 y^4 = 16x^4 y^4\).

Step2: Divide by the denominator

Now, divide \(\frac{16x^4 y^4}{4xy}\). For the coefficients, \(\frac{16}{4} = 4\). For the \(x\) terms, using the quotient rule \(\frac{x^m}{x^n}=x^{m - n}\), we get \(x^{4 - 1}=x^3\). For the \(y\) terms, \(\frac{y^4}{y^1}=y^{4 - 1}=y^3\). So the left - hand side simplifies to \(4x^3y^3\).

Step3: Compare with the right - hand side

The right - hand side is \(4x^a y^b\). By comparing the exponents of \(x\) and \(y\) on both sides, we have \(a = 3\) and \(b = 3\).

Answer:

a = 3, b = 3