$\displaystyle 16^4 = (2^4)^4 = 2^1^6$. Thus, $\displaystyle z$ is equal to 16.

Begin by distributing the exponent through the parentheses. The power rule dictates that an exponent raised to another exponent means that the two exponents are multiplied:

$\displaystyle (3x^{4}y^2)(4xy^2)^{-3}= (3xy^2)(4^{-3}x^{-3}y^{-6})$

Any negative exponents can be converted to positive exponents in the denominator of a fraction:

$\displaystyle \frac{3x^4y^2}{64x^3y^6}$

The like terms can be simplified by subtracting the power of the denominator from the power of the numerator:

$\displaystyle \frac{3x^4y^2}{64x^3y^6}=\frac{3}{64}x^{4-3}y^{2-6}=\frac{3}{64}xy^{-4}$

$\displaystyle \frac{3x}{64y^4}$

In order to solve this problem, each of the answer choices needs to be simplified.

Instead of simplifying completely, make all terms into a form such that they have 100 as the exponent.  Then they can be easily compared.

$\displaystyle 2^3^0^0 = (2^3)^1^0^0 = 8^1^0^0$, $\displaystyle 3^5^0^0 = (3^5)^1^0^0 = 243^1^0^0$, $\displaystyle 4^4^0^0 = (4^4)^1^0^0 = 256^1^0^0$, and $\displaystyle 2^6^0^0 = (2^6)^1^0^0 = 64^1^0^0$.

Thus, ordering from least to greatest: $\displaystyle 8^1^0^0, 25^1^0^0, 64^1^0^0, 243^1^0^0, 256^1^0^0$.

$\displaystyle \frac{(x^{4})^{3}(5x)^{-5}}{xy}$

Step 1: Distribute the exponents in the numberator.

$\displaystyle \frac{x^{12}5^{-5}x^{-5}}{xy}$

Step 2: Represent the negative exponents in the demoninator.

$\displaystyle \frac{x^{12}}{5^{5}x^{5}xy}$

Step 3: Simplify by combining terms.

$\displaystyle \frac{x^{12}}{3125x^{6}y}$

$\displaystyle \frac{x^6}{3125y}$

Use the power rule to distribute the exponent:

$\displaystyle (4x^3)^4\rightarrow 4^4\cdot x^{3\cdot 4}\rightarrow 256x^{12}$

Step 1: Distribute the exponent through the terms in parentheses:

$\displaystyle \left ( \frac{a^{-8}b^{16}}{a^{-12}b^{-24}c^{28}} \right )$

Step 2: Use the division of exponents rule.  Subtract the exponents in the numerator from the exponents in the denominator:

$\displaystyle {\color{Red} \left ( \frac{a^4b^{40}}{c^{28}} \right )}$

When a power applies to an exponent, it acts as a multiplier, so 2a becomes 4a and -b becomes -2b. The negative exponent is moved to the denominator.

When faced with a problem that has an exponent raised to another exponent, the powers are multiplied: $\displaystyle x^{2\cdot2}y^{4\cdot2}$ then simplify: $\displaystyle x^4y^8$.

Solve each term separately.  A number to the zeroth power is equal to 1, but be careful to apply the signs after the terms have been simplified.

$\displaystyle -25^0-(5^2)^0 = -1-1 =-2$

$\displaystyle 64x^{16}y^4$ is the correct answer because the order of operations were followed and the multiplication and power rules of exponents were obeyed. These rules are as follows: PEMDAS (parentheses,exponents, multiplication, division, addition, subtraction), for multiplication of exponents follow the format $\displaystyle a^m*a^n=a^{m+n}$, and $\displaystyle (a^m)^n=a^{m*n}$.

$\displaystyle [8x^5x^3y^2]^2$

First we simplify terms within the parenthesis because of the order of operations and the multiplication rule of exponents:

$\displaystyle [8x^8y^2]^2$ 

Next we use the power rule to distribute the outer power:

$\displaystyle 8^2(x^8)^2(y^2)^2$=$\displaystyle 64x^{16}y^4$

**note that in the first step it isn't necessary to combine the two x powers because the individuals terms will still add to x^16 at the end if you use the power rule correctly. However, following the order of operations is a great way to avoid simple math errors and is relevant in many problems.