Can someone help me with this calculus problem about decay?

Question

A radioactive substance decays at a rate proportional to the amount present. If 20 percent of such a substance decays in 19 years, what is the half-life of the substance?

Answer 1

Let N(t) be the amount present at time t.

We are told that dN/dt = -k*N where k is a constant. I have written -k to remind us that this is decay (so k is positive).

If you know how to solve this, do so. You should get N(t) = C e^(-kt) where C is a constant -- in fact, if you evaluate this at time t=0, you will see that C is N(0), that is, the amount present at time t=0. So let's call this amount N0. Then
N(t) = N0 e^(-kt)

You are told that 20% of the substance decays in 19 years. That means that N(19) = 0.8*N(0) (because if 20% has decayed, 80% remains). Therefore

N(19) = N0 e^(-19k) = 0.8 N0

Solve this for k. Note that it does not depend on N0.

Lastly, to find the half-life, solve

N0 e^(-kt) = 0.5 N0 for t (since you will now know k).