 This video will talk about compound interest. Alright, so now we have compound interest. That means it's going to gain interest more than just once a year, or it might only be once a year. We're going to have this equation where A is equal to P, again the principle, times 1 plus the rate over N to the NT. And N is the number of times it's compounded per year. That's the one thing that's different. And then since we're compounding it so many times per year, then we do this thing over and over and over again per year. So we're going to take it to that compounding value times our T. So, find T when A is 30,146 equal to my P times 1 plus my rate, which is 0.053. And it didn't tell me what my N is, so we're going to assume the N is 1 to the 1 times my T. It's annual. Alright, we're ready to solve now. And to solve, we want to get rid of what's in front of the parentheses, because our unknown has to do with that parentheses. It's actually the exponent, but we're going to take our 30,146 and divide that by the 21,057, equal to our 1 plus 0.053, or 5.3%. That's annual, so I really don't need that fraction there to the T. Okay, now I have a base. This might be better written as 1.053 to the T. So now I have a base raised to an exponent. So if you remember, I have a number over here and a number over here with an exponent. So we can either take the log of that base like that, or let's just take the log. We can use any log we want. So let's take the log of this 30,146 over that 21, blah, blah, blah, blah, and take the log of this side. I'm taking the log. I could take a natural log, but I don't really want to do anything else because I need to use my calculator to figure this out. I don't want an exact answer. I want the decimal that I might round. Now we have that property that says I have a log with an exponent in my argument, so it can come out front. And we have the log of that fraction divided by the 21,057, equal to T times the log of my 1.053. And so to finish the problem off, I just have to divide both sides by the log of 1.053. This technically would be what my answer was, but that doesn't really say anything to me. So I really need to say, okay, this thing I'm going to take to my calculator and get a time that makes more sense to me than just some fractions and logs. So again, clear everything out. I have the log of in parentheses. I have 30,146 divided by the 21,057, or 57 cents. And now I can close the parentheses because that's my first argument divided by the log of my bottom argument, which is 1.053. That definitely rounds to seven years. So we will say seven years. How long will it take $10,000 to double if it is deposited in an account that is compounded monthly at a rate of 8.375? There should be a percent in there. So let's see what we have. $10,000, that's going to be our original amount. So that's our P and doubled. We'll come back to that one. The deposited account compounded monthly. So this is our N and actually there are 12 months in a year, so N is going to be 12. And then this is our R. But remember, we have to put the decimal in there and move the decimal two places. So it's 0.08375. And then this double is actually our A because that's the amount that we want to get out. So if we started with $10,000, we would have two times that $10,000. And that's equal to our $10,000 times the one plus the rate, which is 0.08375 divided by our N, which is 12, raised to the 12 times T, which we are trying to find. So here we have our equation. And we want to get to this base by itself. So I need to divide off $10,000. And when I divide off $10,000, it's going to divide off both sides. I'm just going to be left with 2 equal to that nice parentheses. And then we have to get the exponent here. It has my variable in it. So we need to take a log. And we can take a log or a natural log. And that will get rid of the exponent because of our property that we can use. And in fact, I'm going to use it all in one step. I'm going to take the natural log of both sides to the natural log of 2 is equal to, but then I'm going to bring down my exponent first. And then I'm going to write my natural log of my argument that is in the parentheses. And my 12T again is already out front. So I took the natural log of both sides and I applied the property that said the exponent can come out front all in one step. So I'm ready to divide off the ln, the natural log. So that will give me the natural log of 2 divided by the natural log of that big number. And that's going to equal to 12T. So T is going to be equal to the ln 2 divided by the ln 1 plus .08375 divided by 12. And then all of that, which would be in brackets, would be divided by 12 to get that T all by itself. So let's call up the calculator so we can see exactly how long that is. This is just some number, but it doesn't mean anything to me. So if I put it in my calculator, I have to put it in exactly like I see it. So I'm going to put in parentheses ln 2, close the parentheses for that argument, divided by natural log of 1 plus and then my decimal. And I should be able to just leave it like that and not have to worry about any parentheses because it will divide before it adds. And that closes that argument. Now I need to close the bracket. I need a double parentheses there and then divide it by 12. So that'll be approximately eight years. All right, so now we have our final kind of compounding interest and that's our continuous. And this gives us the A. It's that same A that we've always had and the amount that we're going to accumulate. And then the P is the same thing, the beginning amount times E. Remember that's that natural number raised to the R. And we've got to remember here that R is going to be a decimal and times T. So let's see what we have. A is, we're going to put our A in here, equal to our P times E to the R. But this is 2.62. So we're going to do the decimal two places. So 0.0262 times T. And again, we want to get to the point where we have this E all by itself. So if I divide it, I should be able to divide the zeros right off. And so I'm going to write this as 325 over 250. I'm sure I can reduce it further, but I'm happy with that. And then 0.0262 to the T. And then here, remember that we have, we need a log, but we need U's LN because of base E. So the natural log of 325 over 250 is equal to, and I'm going to bring my exponent down again. So the 0.0262 T is my exponent. And then times the natural log of E. And come over here and remind yourself again that the natural log of E, the exponent on E, that's assumed based on here, that will get me E would be 1. So when I see L and E, I really could just say that I'm multiplying by 1 here. So all I have to do finally then is take my natural log of that 325 over 250 and divide that whole thing by my decimal, 0.0262. And I should find out what T is. So again, call it the calculator. And this time, let's move it over a little bit. We're going to clear all this out. And we have the natural log of 325 divided by 250. Close the argument divided by my 0.0262. And we find out that T will take approximately 10 years.