First, we can solve the differential equation.

After years, After years, we still have But after years, however, almost half of the carbon has decayed. Give the answer to three significant figures.

The time period calculated in this example is called the half-life of carbon In fact over any period of years, the amount of carbon in an isolated sample will decay by half. Roughly speaking, while an organism is alive, its interactions with its environment maintain a constant ratio of carbon to carbon in the organism; but after it dies, the carbon is no longer replenished, and the ratio of carbon to carbon decays in a predictable way. See Wikipedia for more on radiocarbon dating. More exponential decay examples.

Exponential decay and semi-log plots. Two videos ago we learned about half-lives. And we saw that they're good if we are trying to figure out how much of a compound we have left after one half-life, or two half-lives, or three half-lives.

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And to address that issue in the last video, I proved that it involved a little bit of sophisticated math. And if you haven't taken calculus, you can really just skip that video.

## Content - Radioactive decay and half-life

You don't have to watch it for an intro math class. But if you're curious, that's where we proved the following formula. That at any given point of time, if you have some decaying atom, some element, it can be described as the amount of element you have at any period of time is equal to the amount you started off with, times e to some constant-- in the last video I use lambda. I could use k this time-- minus k times t. And then for a particular element with a particular half-life you can just solve for the k, and then apply it to your problem.

So let's do that in this video, just so that all of these variables can become a little bit more concrete. So let's figure out the general formula for carbon. Carbon, that's the one that we addressed in the half-life.

## Radioactive decay and exponential laws

We saw that carbon has a half-life of 5, years. So let's see if we can somehow take this information and apply it to this equation.

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So this tells us that after one half-life-- so t is equal to 5, N of 5, is equal to the amount we start off with. So we're starting off with, well, we're starting off with N sub 0 times e to the minus-- wherever you see the t you put the minus 5, so minus k, times 5, That's how many years have gone by. So if we try to solve this equation for k, what do we get? Divide both sides by N naught.

If we take the natural log of both sides, what do we get? The natural log of e to anything, the natural log of e to the a is just a. I just took the natural log of both sides. The natural log and natural log of both sides of that. But let's see if we can do that again here, to avoid-- for those who might have skipped it. So it equals 1. So now we have the general formula for carbon, given its half-life. At any given point in time, after our starting point-- so this is for, let's call this for carbon, for c the amount of carbon we're going to have left is going to be the amount that we started with times e to the minus k.

This is our formula for carbon, for carbon If we were doing this for some other element, we would use that element's half-life to figure out how much we're going to have at any given period of time to figure out the k value. So let's use this to solve a problem. Let's say that I start off with, I don't know, say I start off with grams of carbon, carbon And I want to know, how much do I have after, I don't know, after years? How much do I have? Well I just plug into the formula. N of is equal to the amount that I started off with, grams, times e to the minus 1.

So what is that? So I already have that 1.

### Your Answer

So let me say, times equals-- and of course, this throws a negative out there, so let me put the negative number out there. So there's a negative. And I have to raise e to this power.