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  1. 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.