Answer
$y=0.45e^{0.602t}$
Work Step by Step
$(2, \frac{3}{2})$ lies on the curve.
Thus, $$\frac{3}{2} = Ce^{2k}$$
$(4, 5)$ lies on the curve.
Thus, $$5 = Ce^{4k}$$
Dividing eq 1. by eq 2. $$\frac{10}{3} = e^{2k}$$ $$k=\frac{ln(10/3)}{2}=0.602$$
Putting this in eq 1 gives,
$$\frac{3}{2}=C(\frac{10}{3})$$ Or $$C=\frac{9}{20}=0.45$$
So, the function is $$y=0.45e^{0.602t}$$