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The six cultivars were grouped in three pairs in which each had a similar commercial harvest date.

Each pair comprised by a non-acid cultivar and an acid cultivar, except the pair of ‘Big Top’, which included two non-acid cultivars.

In order to do this for the example of potassium-40, we know that when time is 1.25 billion years, that the amount we have left is half of our initial amount. So let's say we start with N0, whatever that might be. We know, after that long, that half of the sample will be left. Whatever we started with, we're going to have half left after 1.25 billion years. And then to solve for k, we can take the natural log of both sides.

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.

In the last video, we give a bit of an overview of potassium-argon dating.

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

Acid cultivars showed a significantly faster decrease in fruit firmness, especially evident in the case of early season cultivars as ‘Alice’.

If you can on one side of this equation through algebra separate out the Ys and the DYs and on the other side have all the Xs and DXs, and then integrate. But now that we did this we can integrate both sides. So this is going to be Y squared over two and we could put some constant there. And if you're integrating that that's going to be equal to. I haven't used this initial condition yet we could call it. So if I just subtract C one from both sides I have an arbitrary so this is gonna cancel, and I have C two, sorry. These are both constants, arbitrary constants and we don't know what they are yet.

And we know that there's a generalized way to describe that.

And we go into more depth and kind of prove it in other Khan Academy videos.

But we know that the amount as a function of time-- so if we say N is the amount of a radioactive sample we have at some time-- we know that's equal to the initial amount we have.

We'll call that N sub 0, times e to the negative kt-- where this constant is particular to that thing's half-life.