So how can we use this information-- in what we just figured out here, which is derived from the half-life-- to figure out how old this sample right over here? So we need to figure out what our initial amount is. So if you want to think about the total number of potassium-40s that have decayed since this was kind of stuck in the lava.

It might be 1 gram, kilogram, 5 grams-- whatever it might be-- whatever we start with, we take e to the negative k times 1.25 billion years. So you get the natural log of 1/2-- we don't have that N0 there anymore-- is equal to the natural log of this thing.

The natural log is just saying-- to what power do I have to raise e to get e to the negative k times 1.25 billion? k is equal to the natural log of 1/2 times negative 1.25 times 10 to the ninth power.

So you get 1 over this quantity, which is 1 plus 0.01 over the 11%. And then, if you want to solve for t, you want to take the natural log of both sides. And then, to solve for t, you divide both sides by negative k. And you can see, this a little bit cumbersome mathematically, but we're getting to the answer. The mathematics really is something that you would see in high school.

So we got the natural log of 1 over 1 plus 0.01 over 0.11 over negative k. We're just dividing both sides of this equation by negative k. So let's take the natural log of our previous answer. If you saw a sample that had this ratio of argon-40 to potassium-40, you would actually be able to do that high school mathematics.

So we could actually generalize this if we were talking about some other radioactive substance.

And now let's think about a situation-- now that we've figured out a k-- let's think about a situation where we find in some sample-- so let's say the potassium that we find is 1 milligram. And usually, these aren't measured directly, and you really care about the relative amounts.

But since floods jumble materials of different origins and ages together, that meant the scientists had to date dozens of different minerals.

The youngest crystal in the footprint layer would represent the oldest possible age for the prints; the oldest crystal in the layer above it would represent the youngest they could be.

And so we could make this as over 1.25 times 10 to the ninth. If I have a natural log of b-- we know from our logarithm properties, this is the same thing as the natural log of b to the a power.

So the negative natural log of 1/2 is the same thing as the natural log of 1/2 to the negative 1 power. Anything to the negative power is just its multiplicative inverse. So negative natural log of 1 half is just the natural log of 2 over here. It's essentially the natural log of 2 over the half-life of the substance.

In this video, I want to go through a concrete example.