+4 If $y=kx^{\frac{1}{4}}$ and $y=3\sqrt{2}$ at $x=81$, what is the value of $y$ at $x=4$?
We can find k by substituting the known values of x and y into the equation.
3\sqrt{2} = k(81)^(1/4)
3\sqrt{2} = k(3)
k = 3\sqrt{2}/3 = \sqrt{2}
Now we can find y at x=4 by substituting this value of k into the equation.
y = \sqrt{2}(4)^(1/4) = \sqrt{2}(2) = 2\sqrt{2}
Therefore, the value of y at x=4 is 2\sqrt{2}.
