Answer
$y = 4e^{-((1/5)\ln{8})t}$
Work Step by Step
At $t = 0, y = 4$, so $4 = Ce^{k\cdot 0} = C$, or $C = 4$. At $t = 5, y = 1/2$ so
$\frac{1}{2} = 4e^{5k}\implies\ln{\left(\frac{1}{8}\right)}=5k\implies k=-\frac{1}{5}\ln{8}$
So
$y = 4e^{-((1/5)\ln{8})t}$
