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

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

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

Answer

Explanation:

Step1: Simplify the left - hand side numerator

Use the power - of - a - product rule ((ab)^n=a^n b^n). So, ((2xy)^4 = 2^4x^4y^4=16x^4y^4).

Step2: Simplify the left - hand side fraction

(\frac{(2xy)^4}{4xy}=\frac{16x^4y^4}{4xy}). Then, use the quotient rule (\frac{a^m}{a^n}=a^{m - n}). We have (\frac{16}{4}x^{4 - 1}y^{4 - 1}=4x^{3}y^{3}).

Step3: Compare with the right - hand side

Since (4x^{3}y^{3}=4x^{a}y^{b}), then (a = 3) and (b = 3).

Answer:

a = 3, b = 3