Simplify the Exponential Expression: (a^2x/a^y) × (a^2y/a^-y)

The following problem involves a multiplication operation between two fractions.

Let's begin by applying the rule for multiplying fractions. It states that multiplication of two fractions is calculated by placing one fraction over a line and then proceeding to multiply the numerators together and multiplying the denominators together.

$\frac{x}{y}\cdot\frac{w}{m}=\frac{x\cdot w}{y\cdot m}$

Let's apply it to the problem:

$\frac{a^{2x}}{a^{y}}\cdot\frac{a^{2y}}{a^{-y}}=\frac{a^{2x}\cdot a^{2y}}{a^y\cdot a^{-y}}$

Note that in both the numerator and denominator separately there exists a multiplication operation between terms with identical bases. Hence we'll apply the power law for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Let's apply it separately to the numerator and denominator in the problem:

$\frac{a^{2x}\cdot a^{2y}}{a^y\cdot a^{-y}}=\frac{a^{2x+2y}}{a^{y+(-y)}}=\frac{a^{2x+2y}}{a^0}$

Next, remember that any number to the power of 0 mathematically equals 1 :

$a^0=1$

So let's return to the problem:

$\frac{a^{2x+2y}}{a^{0}}=\frac{a^{2x+2y}}{1}=a^{2x+2y}$

Remember that division by 1 doesn't change the value of the number, as shown below:

$X:1=\frac{X}{1}=X$

Let's proceed to examine the result that we obtained above:

$a^{2x+2y}\text{ }$

In terms of simplification using the laws of exponents we have indeed finished given that this is the most simplified form of the expression.

However it's worth noting that in the exponent we obtained an expression that can be factored using common factor extraction:

$2x+2y=2(x+y)$

In this case the common factor is the number 2,

Let's return to the result of the expression simplification, as follows:

$a^{2x+2y}=a^{2(x+y)}$Therefore the correct answer is C.