If y=kx^{1/4} and y=3√2 at x=81, what is the value of y at x=4?
y = k x1/4
3 sqrt 2 = k (81)1/4
k = sqrt 2
y = sqrt 2 ( 4)1/4 = sqrt 2 (22)1/4 = sqrt 2 ( 21/2 ) = sqrt2 sqrt 2 = 2
+36916 
