 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Exponential Growth and Decay - Problem 2

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Share

So an application of exponential decay is when we are actually given a scenario about something that's increasing in quantity. For this particular problem we're going to be looking at we're dealing with the amount of a drug in a system and how it's going to decrease over time.

So we have this drug and it's decreasing according to this formula right here and we're given the amount of time that this is occurring is in hours. And what this question is asking for is, it's two-part question, first part is how much is remaining after two hours?

So this is just a function, the amount of drug is a function of time. So what we really have to do is plug in 2 in for time and see what we come up with. So this just tells us we're dealing with base of 2 is equal to 10e to the negative point 2 times 2. We plug this into our calculator and we're left with 10 times e to the -.2 times 2 and that gives us around 6.7. It's a problem that doesn't tell us what our unit of measurement is if it's grams or whatever it is, so we can't have a unit, but we're left with 6.7 whatever it is.

Second part is when will there be half the original amount left in the system? For this one we actually have to do two steps. The first thing is we have to find what the original amount is, so we need to find the original amount and the first thing we need to think about is when does that occur? That occurs at time zero.

So just like we did up here where we plugged in t2 for t to find that two hours, we plug in t is equals to 0 to find the original amount, so we had -2 times 0 is 0, e to the 0 anything with a 0 is 1 so our original amount is 10. So that's our original amount.

We're now asked to figure out when we have half of that, half of 10 is 5 and so that's going to mean that we want to have left, so this is, so studying our equation 5 is equal to 10e to the -.2 times t and we now have a exponential equation we need to solve. Get our exponential by itself, so we need to divide by 10 leaving us with 1/2 is equal to e to the negative point 2t. We need to get our exponent down, we have to do natural log in order to do that natural log because we already have an e here, natural log, natural log and I have natural log of 1/2is equal to -.2t natural log of e.

The natural log of e becomes 1 so we can drop that out in order to finish this up divide by our coefficient on y, natural log of 1/2 over -.2 is equal to 4.

We can plug this into our calculator let's figure out what our answer is, natural log of 1/2 divided by -.2 and we end up with about 3 point which one is it? 3.46 and out units for time is hours.

So a application of exponential decay in this case we're dealing with concentration of drugs in one's bloodstream, but it could be a variety of different recipes or ingredients if you will and the process is always about the same.