If $y=kx^{\frac{1}{4}}$ and $y=3\sqrt{2}$ at $x=81$, what is the value of $y$ at $x=4$?
I notice that this is a problem involving direct variation, where $y$ is proportional to a power of $x$. To find the constant of proportionality $k$, I can plug in the given values of $y$ and $x$ and solve for $k$. I get $3\sqrt{2}=k(81)^{\frac{1}{4}}$, which simplifies to $3\sqrt{2}=k\cdot 3$. Dividing both sides by $3$, I get $k=\sqrt{2}$. Now that I have $k$, I can plug in the other value of $x$ and find $y$. I get $y=\sqrt{2}(4)^{\frac{1}{4}}$, which simplifies to $y=\sqrt{2}\cdot \sqrt{2}$. Multiplying these radicals, I get $y=2$. 