 Okay, so let's try this half-life equation. So remember, half-lives are first-order rate equations, okay? So if you subject this to the first-order rate law, you should be able to get the right answer. So this one says, a half-life of cobalt-60 is 5.3 years. How much of a 1.00 mixed sample of cobalt-60 is left after a 10-year period? Okay, so the first thing we're going to have to do is figure out what is the k of this equation, okay? So, well, I guess remember the first-order integrated rate law is the ln of the concentration of A0 divided by concentration of At equals kT. So in this case, concentration of A0 is going to be the original mass, okay? And the mass at time t is actually what we're looking for. So this would be integrated rate law for the rate equation with concentrations. We're going to use this for the half-life of one of these radioactive isotopes. So we'll just change this to M0 and Mt there, okay? So we have t, we have M0. We're looking for Mt, you don't know what that is. So what do we have to find k first, okay? So for first-order rate equation, remember k is the ln of 2 divided by, in the case of half-life, it's going to be the half-life. So that's going to give us a value that equals 0.131 per year, okay? Is everybody okay with what we've done so far? Yes. Okay. So we're looking for Mt. This is M0, okay, so M at time t. So we can change this around to make it easier for us to solve. So remember ln of M0 divided by Mt is going to change to ln, if we're going to expand it, of M0 minus the ln of Mt equals kt. So we again want to solve for Mt, so let's isolate the ln of Mt now, okay? So ln of Mt is going to equal the ln of M0 minus kt, look like that, is everybody okay with that? So now, how do we get rid of ln? We've got to take it to the e, everything to the e, okay? So we'll take both sides to the e, both sides. So when we do e to the ln that cancels out like that, so that's going to give us a new equation. I'll just write it up here, because I'm too tall to do it. So that's Mt equals e ln of M0 minus kt, can I erase this portion down here? Yes. Is everybody okay with that? You get this? Yes. Okay, so that's what we're looking for is Mt. So all we need to do is really plug in values for this, okay? So Mt equals e to the ln of, well, 1.000 minus k, 0.131 per year times t, which is 10.6 years. Is everybody okay with what we've done there? So if you solve for that, you should get the answer. So that number times 10.6, and then say, well, the ln of 1, of course, is 0. So it's going to be negative 1.38, and we're going to say e to that number. So that's the long way to do this, like you're doing it, like a first-order rate equation, okay? Integrated first-order rate equation. Any questions on this one? No. Okay. Let's do it faster right here, okay? We'll do the same problem really quick though, okay?