00:01
All right, this is a really nice half -life problem.
00:04
So carbon 14 is something, a way to measure the age of an item.
00:13
So we need the exponential equation y equals a e to the kt power.
00:19
And a is going to represent that initial amount.
00:22
K is going to be that decay rate and t is your time.
00:27
So they tell you that a piece of ancient charcoal contains only 15 % as much radioactive carbon as a piece of modern charcoal.
00:36
And they want to go, how long ago was the tree burned to make the ancient charcoal? if the half -life of carbon 14 is 5 ,715 years.
00:45
Well, we don't know how much original charcoal we have, right? so we're just going to start with one gram.
00:53
And we know that we want that to be cut in half.
00:57
That's the whole concept of half -life, right? how long it takes for half of the substance to disappear.
01:03
We don't know what rate that is occurring at.
01:06
That's what we have to find first, but we do know that it takes 5 ,100, i lied, 5 ,715 years to do that.
01:15
So what i need to be able to do is undo this exponential function.
01:22
So because i have a natural exponential base, i have to undo that with a natural logarithm.
01:29
So i'll take the natural log of each side.
01:33
I do that so that i can apply some log properties here that i'm allowed to bring the exponent down in front and multiply by the natural log of e, but the natural log of e is one.
01:46
So really all i need to do on my calculator is evaluate the natural log of 0 .5 divided by the half -life of 5 ,715 years.
01:58
So the natural log, let me punch it in the calculator here, the natural log of 0 .5 divided by 5 ,715 is, i'll write it, it's negative 1 .21285596 times 10 to the negative fourth power...