Let's simplify the expression step by step:

$\left( \frac{x^{4} y^{3} z}{16 x^{-16} y z^{5}} \right)^{\frac{1}{2}}$

Step 1: Combine like terms inside the fraction.

For $x$-terms:

$\frac{x^{4}}{x^{-16}} = x^{4 - (-16)} = x^{4 + 16} = x^{20}$

For $y$-terms:

$\frac{y^{3}}{y} = y^{3 - 1} = y^{2}$

For $z$-terms:

$\frac{z}{z^{5}} = z^{1 - 5} = z^{-4}$

Now the expression becomes:

$\left( \frac{x^{20} y^{2} z^{-4}}{16} \right)^{\frac{1}{2}}$

Step 2: Apply the exponent $\frac{1}{2}$ (square root) to each term.

$\left( \frac{x^{20} y^{2} z^{-4}}{16} \right)^{\frac{1}{2}} = \frac{x^{20 \cdot \frac{1}{2}} y^{2 \cdot \frac{1}{2}} z^{-4 \cdot \frac{1}{2}}}{16^{\frac{1}{2}}}$

This simplifies to:

$\frac{x^{10} y^{1} z^{-2}}{4}$

Final Answer:

$\frac{x^{10} y z^{-2}}{4}$ or equivalently:

$\frac{x^{10} y}{4 z^{2}}$

Would you like further explanation or have any questions?

Here are 5 related questions:

1. How does the exponent rule apply when dividing powers of the same base?
2. Can the square root be applied individually to numerator and denominator in all cases?
3. How would the expression change if the exponent was $\frac{1}{3}$ instead of $\frac{1}{2}$?
4. What happens when there are negative exponents in both the numerator and denominator?
5. How would you simplify an expression with multiple variables under a cube root?

Tip: When simplifying expressions with exponents, always apply the exponent rules carefully (multiplying, dividing, or raising powers).