Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Exponential growth in DK. The formula A equal a sub naughty of E to the KTA. Sub naughty is the initial amount, K is a constant, T is time, and just A is the ending amount. Now, if our constant is greater than 0, then we're talking about growth, and if our constant's less than 0, we're talking about decay. So for exponential growth, it's just going to be a graph that's an exponential graph that's increasing. Now in story problems, our time is never going to be 0. So we're really talking about this portion of the graph here where our time is positive in exponential decay. It's going to be an exponential graph that's decreasing. And once again, due to the time factor, we're going to be having it be only from the time being positive and exponential growth, our amounts are going to be increasing and exponential decay our amounts are going to be decreasing. Let's look at some examples. If we had $25,000 invested at 4% continuously, we're going to use a equal A, not E to the RT. So we want to find out how much we have. If we invested 25,000 compounded continuously is the E portion, the rate is going to be .04 and we're going to do it for one year. So if we have E equal 25,000 E to the .04 T, we're going to get, Dang it, If we look at a $25,000 investment at 4% compounded continuously, our final amount is going to equal the initial investment times E to the rate, which in this case is .04. And let's say we want to know what happens after one year. So a time is going to be 1. If we then pick up our calculator and plug this in, we're going to get a equal $26,020.20 thirty cents, sorry, at the and at the end of approximately 1 year, we've made $1000 by not doing anything other than investing our money at 4%. Let's look at doubling time. To do doubling time, we want to figure out how long it will take that 25,000 to turn into 50,000. So we started with 25,000 and we wanted to end with 50,000 E .04 T. So in this case, we're going to divide each side by 25,000, IE there's a double, there's a 2 right there. So twenty 50,000 / 25,000 is 2. So you could think of every first dollar invested, you get out $2.00. So it really doesn't matter how much the 25,000 and the 50,000 is, if we knew that we started with P amount and we're going to end with 2P amount, the P and the 2P will cancel and it will always be one and two for a doubling time. Now to solve this, we're going to put this into logarithmic form. We're going to get .04 T equal the natural log OF2SO2T is going to equal the natural log of 2 / .04. We plugged that into our calculator and I don't know, I think it's something like 17.7 years. Looking at some more examples, what if we had population growth? If we started with a population that was 203.3 million in 1970 and it grows to 300.9 million in 2007. So in 2007 we have 300.9, that's our ending amount. In 1970 we have 203.3. That's going to be our starting amount, E the K, we don't know in this in this problem, we're going to find it. But the T, how long was it? It was 37 years. So now we're going to take the 300.9, divide it by 203.3. That's going to equal E to the 37 K taking the natural logarithm to get rid of that E So natural log of 300.9 divided by 203.3 would equal 37 K or K is going to equal 137th of that natural log of 300.9 / 203.3. Now I personally never round until the very, very last step. And the reason is that we don't want to have any rounding errors as we're going. Errors will be compounded if we round as we go. So we're going to get K is approximately .011. Now when will it be 315,000,000? So we're going to have 315. We start with the 203.3 E. We now know that K, that K was that 137th natural log 300.9 / 203.3, but we don't know our time this time. So we're going to do the same steps. We're going to divide each side by the 203.3 and we're going to get E to the 137th natural log 300 and .9 / 203.3 times T Then if we take the natural log to get rid of the base E and then we're going to divide each side by that K value to get T by itself. So T is going to be approximately 40 years. Now it asks for when, so we need to take the original 1970 and add 40 years to it. So in approximately 2010 we will have 315 million. If we look at a carbon dating problem, we want to know after 5715 years, amount of carbon present is half what was there originally. So if we had originally P amount after 5715 years, we are having half that amount present, E or K we're going to find and our T is 5715. Now some sources state that that number could be as 5730 ± 40. So it could have been 5690 up to 5770 depending on which source you're using. So the PS here are going to cancel and we're going to take the natural log to get rid of the base E. So the natural log of 1/2 is going to equal K * 5715 or K is going to be 1 / 5715 times the natural log of 1/2. Well, that's going to be approximately negative. Why is it negative? Because it's a decay .00012. If this 505,715 change by a little bit, that K is not going to change by very much. It's still very small. What happens if we found, say, something that had 76% of its original carbon 14 and we want to know what year it is? Well, if it's got 76% left, we would have .76 of the original. So if we called, the original PP would equal P * E to that K, which is approximately -.00012 * t The PS on each side are going to cancel. We could have used other variables, we could have used A's. So we're going to take the natural log of .76 and that's going to equal -.00012 T So to get T by itself, lane of .76 divided by that K and we're going to get T being approximately 2268 years old. Thank you and have a wonderful day.