Ordinary Differential Equations

Assume that the half life of the radium in a piece of lead is 1600 years. How much radium will be lost in 100 years?

A = P * (1/2)^(t/k)

k = 1600
t = 100

A = P * (1/2)^(100/1600)
A = P * (1/2)^(1/16)

Now, compare P * (1/2)^(1/16) to P

(Initial - Final) / Initial = % change

(P - P * (1/2)^(1/16)) / P =>
1 - (1/2)^(1/16) =>
1 - 0.95760328069857364693630563514792 =>
0.04239671930142635306369436485208

4.2397%