If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra 2>Unit 7

Lesson 2: Constructing exponential models according to rate of change

# Constructing exponential models: half life

Sal models the decay of a Carbon-14 sample using an exponential function.

## Want to join the conversation?

• If carbon-14's mass gets halved every 5730 years, then wouldn't carbon-14 never disappear? If the current mass gets halved, there would always be something to half for the next 5730 years.
• When we say that a mass is "halved", we don't mean that, after 5730 years, half suddenly disappears. We're saying that atoms of carbon-14 (let's say there are 1000) are disappearing at a certain rate (we don't know the exact rate) which is decelerating (so we lose less every 5730 years). So, after 5730 years, we will have in total lost 500 atoms of carbon-14, leaving us at 500. But this happened gradually. Maybe we lost a couple atoms every 10 years or so.

Eventually, we get down to the point where we have very few atoms, say 3. If atoms are gradually disappearing, can you really say that after 5730 years we'll have 1.5 atoms (which doesn't make sense)? No, you'd say that maybe we lost 2 atoms over all that time. 1/2 is not a perfect rate. It's just really, really close.

it's also worth noting that, eventually, when you reach only 1 atom, you can't really divide that anymore, so you're stuck at 1 atom. It would be silly to keep halving that, because then we wouldn't have carbon-14 anymore.
• On Sal wrote 1⁄2^2 but shouldn't he do it with parentheses?
• Yes, Sal should be using parentheses. Technically, his version only applies the exponent to the 1, not the whole fraction.
• I was so curious how can we use Exponential function in our real life?
• One common use for exponential models is calculating compounding interest. When you leave money in a bank, the bank may give you interest, which is additional money based on the amount you already have. When the amount of interest you get increases as you gain interest, then we call it compounding interest.

CI = P * (1 - r/n) ^ (n * t) - P
where CI is the compound interest, P is the original amount of money (principal amount), r is the interest rate, n is the number of times the interest is calculated (compounded), and t is the length of time the money is left in the bank.
• Sal's answer is 741*(1/2)^t/5730. I got 741*(2)^-t/5730. They are the same thing, right?
(1 vote)
• Yes! If you factor the -1 out of your exponent, you would get 741*((2)^-1)^t/5730. 2^-1 simplifies to 1/2, so your whole expression would then be the same as Sal's: 741*(1/2)^t/5730
• Just saying that at Sal should have put parentheses around the 1/2.
• Hi! At Sal said that 1/2 can be called common ratio. But the "step" for the common ratio should be 1. Here t goes from 0 to 5730. Shouldn't the common ration be (1/2)^(1/5730)? As always thanks for the grate job. Greeting from Greece!
• Yes. He doesn't stress this, in the interest, maybe, of making this easy to learn. Every year the C14 is reduced by a factor of "the 5,730th root of 1/2". But a step of 5730 yrs is just as valid.
(1 vote)
• At why didn't you multiply 741 by 1.5 instead of 0.5?
• Read the problem carefully. It tells you that the carbon-14 loses half its mass. Your method of multiplying by 1.5 would mean that its mass is increasing by 1/2.
Hope this helps.