Constructing exponential models: half life | Mathematics II | High School Math | Khan Academy

We're told carbon 14 is an element which loses exactly half of its mass every 5,730 years. The mass of a sample of carbon 14 can be modeled by a function m which depends on its age t in years. We measure that the initial mass of a sample of carbon 14 is 741 grams. Write a function that models the mass of the carbon-14 sample remaining t years since the initial measurement.

All right, so like always, pause the video and see if you can come up with this function m that is going to be a function of t, the years since the initial measurement.

All right, let's work through it together. What I like to do is I always like to start up with a little bit of a table to get a sense of things. So let's think about t: how much time, how many years have passed since the initial measurement and what the amount of mass we're going to have. Well, we know that the initial mass of a sample of carbon 14 is 741 grams. So at t equals zero, our mass is 741.

Now what's another interesting t that we could think about? Well, we know every 5,730 years we lose exactly half of our mass of carbon 14. Every 5,730 years. So let's think about what happens when t is 5730. Well, we're going to lose half of our mass, so we're going to multiply this times one-half. So this is going to be 741 times one-half. I'm not even going to calculate what that is right now.

And then let's say we have another hundred 5,730 years take place. So that's going to be, and I'm just gonna write two times five thousand seven hundred thirty. I could calculate it was going to be ten thousand, eleven thousand four hundred and sixty or something like that. All right, but let's just go at two times five thousand: is it ten? Yeah, ten thousand plus one thousand four hundred, so eleven thousand four hundred plus sixty. Yeah, so eleven thousand four hundred sixty, but let's just leave it like this.

Well, then it's going to be this times one-half. So it's gonna be 741 times one-half times one-half. So we're going to multiply by one-half again, and so this is the same thing as 741 times one-half squared.

And then let's just think about if we wait another five thousand seven hundred thirty years. So three times five thousand seven hundred thirty, well then it's going to be one-half times this. So it's going to be 741; this times one-half is going to be one-half to the third power.

So you might notice a little bit of a pattern here. However many half-lives we have, we're going to multiply, we're going to raise one-half to that power and then multiply it times our initial mass. This is one half-life has gone by, two half-lives we have an exponent of two, and three half-lives we multiply by three. Sorry, we multiply by one-half three times.

So what's going to be a general way to express m of t? Well, m of t is going to be our initial value 741 times, and you might already be identifying this as an exponential function. We're going to multiply times this number which we could call our common ratio, as many half-lives have passed by.

So how do we know how many half-lives have passed by? Well, we could take t and we can divide it by the half-life and try to test this out. When t equals zero, it's going to be one-half to the zero power, which is just one, and we're just going to have 741. When t is equal to 5730, this exponent is going to be one, which we want it to be.

We’re going to multiply our initial value by one-half once. When this exponent is two, times five thousand, when t is two times five thousand seven hundred thirty, well then the exponent is going to be two, and we're going to multiply by one-half twice. It's gonna be one-half to the second power, and it's going to work for everything in between.

When we are a fraction of a half-life along, we're gonna get a non-integer exponent, and that too will work out. And so this is our function. We are done. We have written our function m that models the mass of carbon 14 remaining t years since the initial measurement.