Radioactivity from Chernobyl After the release of radioactive material into the atmosphere from a nuclear power plant at Chernobyl (Ukraine) in $1986,$ the hay in Austria was contaminated by iodine 131 (half-life 8 days). If it is safe to feed the hay to cows when $10 \%$ of the iodine 131 remains, how long did the farmers need to wait to use this hay?
So they tell us that after the nuclear power plant, Chernobyl released radioactive material into the atmosphere. The hay in Austria contained Mayadeen 1 31 which they tell us has 1/2 life of eight dates. And then they also tell us that 10% of dieting 13 will be safe for the cows to eat. So when will the farmers be able to feed those cows? The hay? So this half life of eight days is really just a fancy way of saying that if we're using the inhibited decay equation, which is that top one that we have up there, that means a of eight is going to be equal to 1/2 of whatever are starting amount eight, not ISS. So we don't know necessarily how much we're starting with other than just 100% so we can have this year. And if we want to figure out when is going to be equal to 10% well, we would want to just set this equal to, because this is really just saying when is 0.1 of a not left. So first thing we're gonna need to do is let's set up our equation in general. So the A T is equal to a not times e to the Katie. So we don't know what our constant K is yet. So let's go ahead and do that. So we're going to use this right here to help us. So when we plug in eight days, we should have won half a knot is equal to a not times e to the k t. And now we just need to solve four cable action. I forgot. We need to replace this t here with eight. Right now. We can. So the first thing we would do is divide each side by a Not they can't out. We have 1/2 is equal to e to the eight k. And then we would take the natural log on each side, the EU natural cats out and we have a K left. Then we were just dividing side. I won eighth or dividing side by eight, which would give us K is equal to 1/8 natural log of 1/2 and then we can pull out or we can rewrite natural log of 1/2. That's just negative. 18 natural log of to And now with this here, we can plug it into our original equation again. So it be a of T is equal to a not e to the so negative 18 natural log of two times teeth and we're going to do is replace 80 with 800.1 A. Not because that's when it would be 10% is left. So 0.1 a not is equal to a not e raise the negative 1/8 natural log of two times t. And so this time, instead of solving for tea, we want to solve for air instead of solid for K. We want to solve for teeth this time. So same steps essentially first, divided side by a note, a knots cancel out. And then we would want to take the natural log on each side, just like what we did before. So it would be natural log of 0.1 than natural log of this. Those counts out and we'd have natural r of 0.1 is equal to negative 1/8 natural law. Go to Times T, So that's going to imply that T is equal to so we'd multiply over by negative eight would be negative eight natural log of 0.1 and that we would divide by natural log of. So that would be our exact answer. And if we were to plug this into a calculator, it should give us something around, um, 26.6 and then our units remember this here was in days. Sorry. Units over here should also be in days. So after about 26 27 days, the hay will be safe or the cows to eat.