Here's the solution to the problem: The problem asks to simplify the expression $\frac{p^{1/2} q^{2}}{p^{1/4} q^{1/2} r^{1/6}}$ and then evaluate it when $p=16$, $q=9$, and $r=4$, taking only positive roots. Step 1: Simplify the expression using the laws of exponents. 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: $$ \frac{p^{1/4} q^{3/2}}{r^{1/6}} $$ Step 2: Evaluate the simplified expression using the given values $p=16$, $q=9$, and $r=4$. 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: $$ \frac{2 \times 27}{\sqrt[3]{2}} $$ $$ \frac{54}{\sqrt[3]{2}} $$ The final evaluated expression is $\boxed{\frac{54}{\sqrt[3]{2}}}$.