Solve the Exponential Expression: 4^(2x) × (1/4) × 4^(-2)

Apply the laws of exponents for negative exponents, in the opposite direction:

$\frac{1}{a^n} = a^{-n}$

to the middle term in the multiplication in the problem:

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2}=4^{2x}\cdot4^{-1}\cdot4^{-2}$

Next, we'll recall the law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and we'll apply this law to the last expression that we obtained :

$4^{2x}\cdot4^{-1}\cdot4^{-2}=4^{2x+(-1)+(-2)}=4^{2x-1-2}=4^{2x-3}$

We obtained the most simplified expression,

Let's summarize the steps so far, as follows:

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2}=4^{2x}\cdot4^{-1}\cdot4^{-2} =4^{2x-3}$

A quick look at the options will reveal that there isn't such an answer among the options and another check of what we've done so far will show that there are no calculation errors,

This means that another mathematical manipulation is needed on the expression we got, a hint for the required manipulation could be the fact that answer D is similar to our expression but the exponent has a minus sign compared to the exponent we got in the final expression and the expression itself is in a fraction where the numerator is 1, which reminds us of the negative exponent law, let's check this suspicion and handle the expression we got in the following way:

$4^{2x-3}=4^{-(-2x+3)}=4^{-(3+(-2x))}=4^{-(3-2x)}$

The goal is to present the expression that we obtained in the form of a term with a negative exponent. We did this by taking the minus sign outside the parentheses in the exponent and rearranging the expression inside the parentheses using the commutative law of addition and then simplified the expression in parentheses,

Now let's use the negative exponent law again:

$a^{-n} =\frac{1}{a^n}$

And apply it to the expression that we obtained:

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

Therefore the expression that we obtained earlier can be written as:

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2} =4^{2x-3} = 4^{-(3-2x)}=\frac{1}{4^{3-2x}}$

The correct answer is indeed answer D.