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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.Une toute nouvelle exprience linguistique s'offre vous.Grce des millions de traductions toujours porte de main, dcouvrir de nouvelles cultures n'a jamais t aussi simple. Vous pouvez compter sur Linguee, avec ou sans connexion Internet.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 let's say you were able to figure out the potassium is 1 milligram.Accdez, par exemple, au vocabulaire juridique employ par les plus grands cabinets d'avocats dans de nombreuses langues.Many of the settlers learned to grow certain crops without any watering at all.

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.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.So we could actually generalize this if we were talking about some other radioactive substance.Or I could write it as negative 1.25-- let me write times-- 10 to the ninth k. Or you could view it as multiplying the numerator and the denominator by a negative so that a negative shows up at the top. The negative natural log-- well, I could just write it this way.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.

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