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So maybe I could say k initial-- the potassium-40 initial-- is going to be equal to the amount of potassium 40 we have today-- 1 milligram-- plus the amount of potassium-40 we needed to get this amount of argon-40. And that number of milligrams there, it's really just 11% of the original potassium-40 that it had to come from. And so our initial-- which is really this thing right over here. This is going to be equal to-- and I won't do any of the math-- so we have 1 milligram we have left is equal to 1 milligram-- which is what we found-- plus 0.01 milligram over 0.11. And what you see here is, when we want to solve for t-- assuming we know k, and we do know k now-- that really, the absolute amount doesn't matter. Because if we're solving for t, you want to divide both sides of this equation by this quantity right over here. We're going to divide that by the negative-- I'll use parentheses carefully-- the negative natural log of 2-- that's that there-- divided by 1.25 times 10 to the ninth. So the whole point of this-- I know the math was a little bit involved, but it's something that you would actually see in a pre-calculus class or an algebra 2 class when you're studying exponential growth and decay.

So you get this side-- the left-hand side-- divide both sides. So it's negative natural log of 2 divided by 1.25. But the whole point I wanted to do this is to show you that it's not some crazy voodoo here.

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.

the dating math problem-73the dating math problem-81the dating math problem-58

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.

And let's say that the argon-- actually, I'm going to say the potassium-40 found, and let's say the argon-40 found-- let's say it is 0.01 milligram. And to figure out our initial amount, we just have to remember that for every argon-40 we see, that must have decayed from-- when you have potassium-40, when it decays, 11% decays into argon-40 and the rest-- 89%-- decays into calcium-40. So however much argon-40, that is 11% of the decay product.

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.

Students should be guided to recognize the use of the logarithm when the exponential function has the given base of $e$, as in this problem.

Note that the purpose of this task is algebraic in nature -- closely related tasks exist which approach similar problems from numerical or graphical stances.




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