Calculus 10th Edition

$y=0.45e^{0.602t}$
$(2, \frac{3}{2})$ 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 $$y=0.45e^{0.602t}$$