To simplify the expression p^(1/2 q²)p^(1/4 q¹/2 r¹/6) and then evaluate it when p=16, q=9, and r=4, taking only positive roots.

Recall the exponent rules: $\frac{a^m}{a^n} = a^{m-n}$ and $a^{-n} = \frac{1}{a^n}$.

$$\frac{p^{1/2} q^{2}}{p^{1/4} q^{1/2} r^{1/6}}$$

Combine the terms with the same base:
For $p$: $p^{1/2 - 1/4} = p^{2/4 - 1/4} = p^{1/4}$
For $q$: $q^{2 - 1/2} = q^{4/2 - 1/2} = q^{3/2}$
For $r$: The $r^{1/6}$ term is in the denominator, so it becomes $r^{-1/6}$ when moved to the numerator.
The simplified expression is:

$$p^{1/4} q^{3/2} r^{-1/6}$$

This can also be written as:

$$\frac{p^{1/4} q^{3/2}}{r^{1/6}}$$

Substitute these values into the simplified expression:

$$\frac{(16)^{1/4} (9)^{3/2}}{(4)^{1/6}}$$

Calculate each part:

• $(16)^{1/4}$: This is the fourth root of 16. Since $2^4 = 16$, $(16)^{1/4} = 2$. (Taking the positive root as specified).
• $(9)^{3/2}$: This can be written as $(\sqrt{9})^3$. Since $\sqrt{9} = 3$ (taking the positive root), we have $3^3 = 27$.
• $(4)^{1/6}$: This can be written as $(2^2)^{1/6}$. Using the rule $(a^m)^n = a^{mn}$, we get $2^{2 \times 1/6} = 2^{2/6} = 2^{1/3}$. This is the cube root of 2, or $\sqrt[3]{2}$.

Now substitute these calculated values back into the expression:

$$\frac{2 \times 27}{\sqrt[3]{2}}$$ $$\frac{54}{\sqrt[3]{2}}$$