If $y=kx^{\frac{1}{4}}$ and $y=3\sqrt{2}$ at $x=81$, what is the value of $y$ at $x=4$?
First we have to solve for $k$ where $3\sqrt{2}=k\cdot81^{\frac{1}{4}}$.  Since $81^{\frac{1}{4}}=3$, we have $3\sqrt{2}=k\cdot3$, so $k = \sqrt{2}$.

When $x=4$, we have$$y=k\cdot4^{\frac{1}{4}}=k\cdot\sqrt{2}.$$Since $k=\sqrt{2}$, we have $$y=\sqrt{2}\cdot\sqrt{2}=\boxed{2}.$$
2
Level 3
Algebra
Problem: If $y=kx^{\frac{1}{4}}$ and $y=3\sqrt{2}$ at $x=81$, what is the value of $y$ at $x=4$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $y$ at $x=4$, we first need to determine the constant $k$ using the information given at $x=81$. 

1. Substitute $x=81$ into the equation $y=kx^{\frac{1}{4}}$:
   - $y = k(81)^{\frac{1}{4}}$.
   
2. Calculate $(81)^{\frac{1}{4}}$:
   - $81 = 3^4$, so $(81)^{\frac{1}{4}} = 3$.

3. Now, substitute this back into the equation:
   - $y = k \cdot 3$.

4. We know that $y=3\sqrt{2}$ when $x=81$, so set the equations equal:
   - $3k = 3\sqrt{2}$.

5. Solve for $k$:
   - $k = \sqrt{2}$.

Now that we have $k$, we can find $y$ at $x=4$.

6. Substitute $x=4$ into the equation $y=kx^{\frac{1}{4}}$:
   - $y = \sqrt{2}(4)^{\frac{1}{4}}$.

7. Calculate $(4)^{\frac{1}{4}}$:
   - $4 = 2^2$, so $(4)^{\frac{1}{4}} = 2^{\frac{1}{2}} = \sqrt{2}$.

8. Substitute this back into the equation:
   - $y = \sqrt{2} \cdot \sqrt{2}$.

9. Simplify the expression:
   - $y = 2$.

Thus, the value of $y$ at $x=4$ is $2$.<|END_OF_LATENT|> The final answer is $2$.