 Episode 17 of Math 1050, College Algebra. I'm Dennis Allison and I teach mathematics here at Utah Valley State College. If you remember in the previous episode, we were talking about exponential functions, that is functions of the form f of x equals a to the x, where a is a constant, bigger than zero. And we looked at a special function, e to the x, where e has a very specific value. Let's see, who can remind us, what is the value of e approximately? 2.718? 2.718. And you know, if you needed to know e to a further, to a longer decimal expansion, you could find it on your calculator. But let me just tell you what the first few digits are. It's not that difficult to remember. If you expand e further, it's 2.7, and then it's 1828, 1828, whoops, 1828. And then 459045. I don't know why you would need to know this, I don't know the number out that far, but the hard parts remember the 2.7. After that, we have the number 1828 twice. That's the year Andrew Jackson was elected president. And then 459045, you can remember that by an isosceles right triangle, where the angles are 45 degrees, 90 and 45 degrees. So now you know e to 16 significant digits. And the graph of e to the x, while we're looking at this here, the graph of e to the x, which we're going to be using today looks like this, the graph of e to the x has three target points. And what we do is we start off from the origin, we go up one, and if I go one to the right from the origin, I go up e, or 2.7, roughly. And if I go to the left one unit, I go up one over e, and you remember how much one over e is approximately? One over e, if you calculate it on your calculator, it's about 0.36, or roughly a third. So you go up about a third. And then this is the graph of the function e to the x, f of x equals e to the x. And this function is called the natural exponential function. And sometimes you see it abbreviated as exp of x, the so-called natural exponential function. Well, let's look at the list of objectives for today's class. We want to look at several applications of exponential functions, and these are all involved with compound interest. Now, we want to talk about compounding over n periods per year, like per month or per day or per quarter. And then we also want to look at something called continuous compounding, where interest is being added in a continuous fashion into the account. We'll also look at population growth, which is sort of a variation of the continuous compounding. And then finally in today's lesson, we're going to look at an introduction to logarithms, and the ideas for logarithms continue into the next episode. Okay, well, let's begin with compound interest. You know, if you have a certain amount of money that you put into an account, that's referred to as the principal. So I'm going to let p represent the principal in a banking account. And let's say r represents the rate of interest. That would be the annual rate of interest. And then if I were to put a principal of p dollars into an account that has a percentage rate of r, then the interest that I would earn in a year would be r times p. So for example, class, if you were to put $100 in a bank account that paid 5% interest, and the interest was put in at the end of the year, how much money would be in your account at the end of the year? $100, 5% interest? $105? $105, exactly. You see, what you do is you take the amount of principal that you started with, and you add on the interest that's added to that account. And that would be the principal plus r times the principal which is p times 1 plus r. Okay, now with that idea, let me derive a few formulas that we want to use to talk about compounding interest. Suppose that a at t represents the amount in the account after, let's say, t years. Okay, so if I say a at zero, that would be the amount in the account initially before you have any compounding that takes place. So a evaluated at zero would just be p, whatever the principal is that you put in the account. Excuse me. Okay, now after one year, after one year, I'll call that a at one, the amount in the account would be p plus rp, rp being the interest that's added to the account, and that would be p times 1 plus r. Okay, so if the interest is added at the end of the year, then you would have p times 1 plus r dollars in the account. After two years, well, let's see, at the beginning of that second year, you would begin with the amount that you carried over, p times 1 plus r, plus you would earn interest on that at a rate of r, p times 1 plus r. So this is your interest rate, and this is the new principal that you're collecting interest on. And if I factor out a common factor there, let's see, I could factor out a p and 1 plus r. And if I factor that out, let's see, if I factor that out, there'd be a 1 left over here, and there would be an r left over there. So I have 1 plus r twice, so I'll write this as p times 1 plus r squared. At the end of three years, how much money would be in the account? Well, I would begin the third year with this amount of money, p times 1 plus r carried over from the previous year, plus I would get an interest rate of r times that, p times 1 plus r to the second power. So if I factor out the common factor, just like I did before, this would be p times 1 plus r squared. And when I factor that out, there'd be a 1 left over here, and there'll be an r left over there. And this reduces to be 1 plus r to the third power. Now, I think you can see then, that as a general rule, if I were to leave money in this account for, let's say t years, rather than zero, one, two, or three, you notice there's a pattern in these answers, p, p times 1 plus r, p times 1 plus r squared, p times 1 plus r cubed. And you notice that power is the same as the number of years. So after t years, this would be p times 1 plus r to the t power. Now, let me just take an example of how I would use that formula in a specific problem. Suppose that we were to deposit, let's say $1,000 in an account that pays 5% interest, and this money is, this interest rate is compounded annually. That is, the interest is added into the account at the end of each year. That's not the way most banks would normally compound money, but suppose that were the case. And so we might ask the question, how much is in the account at the end of five years? At the end of five years. Well, I can use the formula that I just arrived to work this problem. The formula said p of t is, oops, a of t is p times 1 plus r to the t power. So in this case, a at five would be 1,000, because that was the initial principle, times 1 plus 0.05, because that's the interest rate to the fifth power, because we're talking about five years. So this is gonna be 1,000 times 1.05. Maybe somebody can get out their calculator and be calculating this for us, so I don't have to stop to do this. This will be 1,000 times 1.05 to the fifth power. Stephen, it looks like maybe you have an answer for that. What did you get? $1,276 and 28 cents. And 28 cents. Now, did you round up or round down when you said 28 cents? I rounded down. You rounded down, okay? Now, you know what? At a bank, I don't believe they would ever round up, because they would say you haven't earned that 28th cent there, so it's really 27 point something cents. So I think they would round that down, and they would call that a 27. So at a bank that's always rounded down because technically you haven't earned that extra full penny, you'd have to ask your individual bank to see, but we're talking about small potatoes here, but it looks like at the end of five years, you'd have $1,276 and 27 cents in that case. Okay, well now from this formula, I wanna look at some other formulas that go with it that will be useful when we compound numbers. Suppose that rather than adding the interest at the end of the year, suppose we divide up the year like in the months, for example, and we add interest at the end of every month so that your interest gets in there a little bit earlier and then the interest begins to draw interest, so that would be called compounding monthly. Most savings accounts that you would see at a savings loan or at a bank are compounded daily, so actually the interest is added at the end of the day and that way the sooner the interest gets in, the faster the money compounds and so the more money you could actually earn. Now the formula we just had said, whoops, I keep writing a P there, A of T, the amount at time T is P times one plus R to the T power. Now this was for money that's being compounded annually, but suppose we have in compounding periods in a year. For example, what if the money's being compounded per month? There would be 12 compounding periods per year. Here's how we adjust the formula. The amount in the account now will be the initial principal times the interest rate per period. Well, if there were in periods in a year, you're getting an interest rate of R over in per period. So for example, if it were 5% per year, how much would that be per month? It would be 5% over 12. But then on the other hand, the power changes because instead of having T years, now we wanna look at how many compounding periods there are and in T years there would be in times T compounding periods. So for example, in T years there would be 12 times T months. So you divide by 12 or divide by N inside, but you multiply by N in the exponent. Now this actually tends to give a slightly larger answer. Suppose that we were to take that same principle of $1,000, Stephen, maybe you could get your calculator out again for us here. And suppose we were going to have our money compounded at 5% per year, 5% per year, but it's compounded monthly and we're gonna leave the money in the account for five years. I think you'll find this ends up being a little bit more than what it was before. The way I'll calculate the amount in the account after five years will be 1,000 times one plus 0.05 divided by 12 because this money is being compounded monthly. And then in five years there are 12 times five months. There are that many compounding periods. So this gives me 1,000 times one plus 0.05 over 12 to the 60th power. And Stephen, how much do you get this time? $1283 and since we're not always rounding down 35 cents. And 35 cents. Okay, so we're gonna round down there. And before I think we had $1,270 something dollars or was it 81, I've forgotten, but it was slightly less than this, not much, not much. Now how would we have done this differently if we had been compounding daily, which is actually what most banks do? How would I change this expression up here? You changed the 12 to 365. Right, we'd put 365 there, which makes this interest rate per day even smaller than what it was per month. But then what would we put up here? 365. 365 times five. So now we have a bigger period and you would find out that this ends up being slightly higher, like it might go to $1,284, $1,285, wouldn't be much more, but slightly more. Now let's go to the next graphic and we'll see two of these three formulas listed. There are three formulas for computing interest. First of all, the interest compounded annually and our formula is A equals P times one plus R to the T power. Interest compounded N times per year and A equals P times one plus R over N to the NT power. And now finally, here's a new formula for continuously compounded interest. Now let's come back to the green screen and let me explain what that means. Rather than being compounded per month, per week, per day, even per hour or even per minute, what if money was being continuously added to your account? It would be very small, but that's called continuous compounding. And then the formula changes and the formula becomes this. A equals P times E to the RT. Now this is where E is that same number, 2.718 approximately. We gave a longer decimal expansion for that and you notice that the interest rate is now moved into the exponent. Now to explain how this happens, you really have to know a little bit about calculus. So what we're gonna do is take this formula for granted and let's see how much would be $1,000 placed in an account that pays 5% interest annually, but it's compounded continuously and we leave it for five years. Now last time we had $1,283.35, that's what we got. Let me just write that down, $1,283.35. That's what we saw when we were compounding it daily. So this time the amount in the account after five years will be 1,000 times E to the 0.05, that's the interest rate, times five, that's the value of T. And this is 1,000 times E to the 0.25 power. That's the 1 fourth power. And when you multiply that out, Stephen, what did you get? $1,284.02. And 2 cents. So by continuous compounding, we have increased this by a total of 67 cents. Now you might say, well Dennis, of all these compounding schemes, it sounds like continuous compounding is the fastest growing scheme of all, but most banks don't do this because you see if you had an account where money was compounded continuously, then you'd have to have a time clock so that if you went to close your account, they'd have to know at what time of day you actually closed it because every moment there's interest being added to the account. So what they do is they just run all the files at the end of the day and upgrade them with the newly added interest at the end of the day. And you see it only makes a 67 cents difference over a period of five years. So there's a very slight increase in going to continuous compounding and it would be too much bother, I think, for a bank to have to determine at what time you actually made a deposit or made a withdrawal. Let's go to the next graphic and we'll see the results of three different schemes for some other numbers. I just computed these before class and I'll show you the answers and you could check these out on your own. The example says verify the value of a $1,000 investment that should be singular there at 4% after three years. Now that's 4% annually but the money may be compounded by these three different schemes. Then it also says find the equivalent annual interest rate. I'll show you how to do that in just a moment. Now in part A, if the money is compounded annually you get $1,124.86 and the equivalent annual interest rate is 4% because this is compounded annually. If the money were compounded monthly you would get $1,127.27 and it says the equivalent annual interest rate is 4.074. Let me show you how that number was just derived. If we go back to our original formula that says A equals P times one plus R to the T power this was for an annual compounding. Now under this scheme where we were compounding monthly our $1,000 grew to $1,127.27 but it started off at $1,000. Now what would be the annually compounded interest rate that would have produced this amount after three years? So what I'm gonna do is solve this equation for R and that's the percentage interest rate that you saw on the screen. It's gonna be slightly more than 4% because I'm acting now as if this were just annual compounding. So I would divide by 1,000 and I would get 1.127.27 equals one plus R to the third power. And then to solve for R I'd take the cube root of both sides. So I'd take the cube root of 1.127.27 equals one plus R and then if I solve for R I would subtract one from this answer. So on my calculator I'd compute the cube root of 1.127.27 and then I would subtract one. Now if you do that you'll find out that R is approximately 0.04074 R 4.074% Now what that means is rather than having your money invested at 4% compounded monthly for three years if you invested it annually, excuse me, if you invested it in a scheme that compounded it annually it would, the interest rate would have to go up to 4.074% in order for it to grow to that same amount. So this is the equivalent annual interest rate if it were compounded annually. Now let's go back to that graphic one more time. And it says in part C that if you were compounding continuously the money would grow to $1,127.49 that's what, only a gain of 22 cents after three years. What would be the equivalent annual interest rate if it were being compounded annually rather than continuously and it would be slightly higher 4.081%. Now in order to get that figure we have to know something about logarithms that will come up later on in this episode in the next episode. So we'll just have to take that number as it's given and move on. Okay, let's go on to the next graphic and look at an example. In this problem it says assume that the population of a city grows exponentially at an annual rate of 2.4%. Now by the way, when I say it grows exponentially what I mean by that, that sort of a label growing exponentially means that the population of a city grows continuously. That is, people are added to the population. It could be any moment of the day. I mean people aren't born at the end of the day or people aren't born at noon only and the same way for people who die within the population or move away. People may move into the town at any given time or they move out of town theoretically at any given time. So we say it grows exponentially meaning I'm gonna use my natural exponential function to model this. So as a continuously compounded problem and it says its population in the year 2003 is 286,000. Now if this continues, predict its population in the year 2025. Okay, well let's take that information and put it up here. We're assuming the population grows exponentially. Now what that means mathematically for us is that the population at time t, and here I will call it p because it stands for population, equals the initial population, which I'll call p sub zero. That's the population in the year 2003 times e to the r t power. You see this looks like a continuously compounded interest problem except now we're talking about money growing, we're talking about population growing rather than money in an account growing. So the initial population was 286,000. So I'm gonna substitute 286,000 right here for p sub zero times e to the, let's see now the interest rate or the rate of growth was 2.4%. That'll be 0.024% times t because we haven't been given the t yet. Now the question was if this is the population in the year 2003, what's the population in the year 2025 if this continues? Well how many years lapses between those two times? 22 years, so what I'm gonna do is calculate p at 22. So if I calculate p after 22 years that'll be 286,000 times e to the 0.024 times 22 power. Now let me just show you how I would calculate that now on a calculator. Okay now if we multiply that out that's 286,000 times the exponential function and I can find the exponential function it's written right above this key that says ln on my calculator. So if I push second then you'll see it's e raised to the power of and I'll open parentheses 0.024 times 22. Close parentheses and I'll enter. And this looks like it's around 485,000. I don't think we should take it so literally that we should round it off to the nearest person. I'll say it's 285,000. So this is approximately, whoops, 485,000 it should be. 485,000 and you know to be really more reasonable about it I think we should say it's about half a million is what it is. So it's about half a million people. So we have a town here with a population of 286,000 people and it's growing at a rate of 2.4% per year it's growing exponentially. And so at the end of 22 years we'll have about half a million people in this town. Okay, let me go to a graphic, not to a graphic but to a newspaper article that was in the Salt Lake Tribune and let me show you what this looks like. If you can read this, can you zoom down on this? I want to show in particular the graph that's over on the right-hand side. Now this is a newspaper article that was printed on Thursday, December 28th. The year was 2000 so this is a rather old cutout from the newspaper that I made. And you know it gives the population of Utah back here in the year 1900 and the entire state had a population of about 277,000. Not too far off from what that example had in it initially that we just looked at. By the year 1910 the population was about 377,000 and now you notice that it just keeps climbing every year, there's a big jump between 1940 and 1950 from 552,000 to 696,000. Anybody have an explanation of why it would jump so much during the 1940s? Baby boomers, yeah, right after World War II there were probably a lot of children born right then and you notice from that point on this begins to rise rather dramatically and if you look at this, shall we call this graph overall, it looks a little bit like an exponential function. The population is a little bit different from an exponential function. The population is growing more or less exponentially. I don't think we should expect that an exponential function would calculate all of these numbers precisely but it does look like the state has been growing exponentially more or less. And in the year 2000 the population of the state this was estimated at about 2,160,000. Now you see this is in a newspaper article from December 28th the year 2000 so they didn't have an accurate estimate of the population at that time. Okay, so it does look like populations may grow more or less exponentially. Okay, let's go to another example. This one also comes from the newspaper and this was in the sports section of the Salt Lake Tribune just this spring. Now I'm talking to you in the spring of 2003 and this is during spring training and this is an article about the New York Yankees. You know the title of the article says the Yankees reach another milestone this time for salary. Let me just read a little bit to you and there's a mathematics problem here about continuously compounded interest. Now the article says even before the start of the season the New York Yankees were smashing barriers. New York set a record with $138 million payroll last year that would have been 2002 according to final tabulation by the commissioner's office and is on the verge of becoming the first team to reach $750 million. Now it says the Yankees 2003 payroll stands at $149.2 million that's this year for 22 signed players likely to be on the opening day roster plus an injured pitcher, John Lieber according to contract. Okay, well let's put some of this information down on the green screen and let's predict what their team salary will be next year. It says in the year 2002 the team salary was $138 million and this year in 2003, now the season hasn't actually started yet the team salary is $149.5 no, .2 it was million dollars. Now let's assume exponential growth in the team salaries and the question is will the team salary next year in 2004 for the New York Yankees? So we want to make a prediction on this. Well, I'm going to take 2002 as my base year I'm going to call that t equals zero 2003 I'll call that t equals one one year later and 2004 I'll call that t equals two. Now if salaries are growing exponentially that would say that the salary at time t should be s of zero that would be the starting salary in the year zero times e to the rt power. Trouble is we don't know what r is. So I'm going to say that s at t is equal to s at zero now in the year zero that was 138 million I'll just put 138 there we'll have to remember this is in millions times e to the rt power and I'm wanting to calculate the team salary in the year t equals two so s at two equals well what I do know is that s at one is equal to 149.2 million dollars and that would be 138 times e to the r times one power I'm putting in a one for t there and so e to the r is 149.2 over 138 that's what e to the r is so what is s at two let's see I think maybe I can squeeze this in right here what is s at two that is an s there well that will be 138 times e to the r times two power which is 138 times e to the r squared you notice I'm using properties of exponents here I'm putting r on the inside and the two on the outside and if I raise e to the r and square it then that's e to the two r power and I just found out that e to the r is this ratio so this is 138 times 149.2 over 138 squared now what I'll need to do is to calculate that on my calculator Stephen you have your calculator out already have you calculated that many chance okay and I'll be calculating here those of you at home try calculating it and I'll work it over here on the side and I get 161.3 approximately 161.3 is that what you got Stephen okay so of course this is measured in millions so the New York Yankees have not topped 150 million dollar team salary but they're right on the verge of it and we would predict that next year if things continue to grow exponentially it'll be 161.3 now of course in the real world there are a number of factors that affect salaries and what the team salary will be you know there may be a major player who retires an expensive player who retires or is traded or it may be that they pick up a player with a really big contract and so this figure could be a little bit higher or lower but this is our prediction based on the evidence that we have here okay so much for exponential functions and compound interest let's move now to another topic and this is this is a completely different topic it's an introduction to logarithms let me begin on the green screen and tell you that if you write an expression such as 5 squared equals 25 I would say this is written in what I'll call exponential form that's because I'm using an exponent namely the 2 and I put the exponent up in the air we're all accustomed to this and so 5 squared is 25 now there is another form of writing this called logarithmic form and what you do is you write log base 5 that's the base that was over here so now I write that as a subscript and I put a 25 after it the log base 5 of 25 equals 2 now this is referred to as logarithmic form and it turns out that in this notation some problems are much easier to solve this idea was invented by a guy named Napierre in the 17th century and you see what's happened is what used to be the exponent is now written on the ground here on the right-hand side the logarithm is equal to the exponent the number that was the old base becomes a subscript and the number which used to be the exponential value 25 is now written inside the logarithm so we read this as the log base 5 of 25 equals 2 let me just write one more like this suppose I were to say 2 to the 5th power is 32 I'm using the same numbers but I've reversed their positions 2 to the 5th power is 32 this is written in exponential form how would I write that in logarithmic form well I write the log base now the base is whatever the base was over here base 2 but I write that as a subscript and the exponent is going to be put over here on the right-hand side the logarithm always equals the exponent so the 32 has to go in here now what if I reverse this process this time I'm going to write something in logarithmic form I want you to tell me what is this exponential equivalent suppose we say the log base 4 of 1 fourth is negative 1 how would I write that in exponential form 4 to the negative 1 power 4 to the negative 1 power is 1 fourth and that's true 4 to the negative 1 power is 1 fourth you notice that whatever is the subscript I call that the base, log base 4 that becomes the base of the exponential and the negative 1 the logarithm equals the negative 1 that becomes the exponent so 4 to the negative 1 power equals the quantity inside 1 fourth so the number inside here couldn't be anything other than 1 fourth or this would not have been true let's go to the next graphic and we'll see some other examples of this expressed on the screen on the left-hand side we have the exponential form of an expression 4 to the third power 64 and on the right-hand side we have the logarithmic equivalent of that the log base 4 of 64 equals 3 and then as a second example we have 2 to the negative 1 is a half and on the right-hand side the logarithmic equivalent of that says log base 2 of 1 half equals the exponent and so the exponent is negative 1 now, right below this is a very important statement that we will refer to frequently during this material for this next exam you want to remember that a logarithm is an exponent you see the logarithms on the right-hand side always equal to the exponent they equal to 3 the second one equals negative 1 so the logarithm is an exponent because it's equal to the exponent so whatever in doubt what the logarithmic value is you set it equal to the exponent okay, in this next example it asks us to do the same thing again with a few more problems write the following exponential expressions in logarithmic form and this gives us some practice in recognizing this notation and knowing where the individual pieces go let me write those three problems down right here on my green board and then I'm going to ask people here in the classroom how to write each of these in logarithmic form part A was that 5 cubed is 125 part B is that 0.2 to the zero power is 1 and the last one is 9 to the negative 1 half power is 1 third now first of all would you agree with all three of these statements 5 cubed is 125 I think so 5 times 5 times 5 is 125 any non-zero number to the zero power is 1 so 0.2 to the zero power is 1 and 9 to the negative a half well now see what does the one half power mean the square root of 1 over the square root of yeah exactly so the one half means take the square root and the negative means flip it over so we're supposed to take the square root of 9 and then invert it so the square root of 9 is 3 and we invert it and get a third well let's see Matt what is the logarithmic equivalent of 5 to the third power is 125 a log base 5 of 125 equals 3 that's exactly right now these two things mean the same thing Sam what would be the equivalent of 0.2 to the zero power is 1 it would be log base 0.2 to the 1 equals 0 okay I would say log base 0.2 of 1 equals 0 so if you take this base raise it to that power then you'll get 1 last case Susan what would you say is the logarithmic statement for C log base 9 of 1 third log base 9 of 1 third I'll put parenthesis around that equals negative one half equals negative one half that's exactly right okay we have another example after this that's a little bit different so let's go to the next graphic okay this time let's define the value of the variable in each one of the following logarithmic expressions let me write each one of these on the screen here for the class to see as well as the people at home log base 4 of x equals 2 the question is what is x and then in the next one it says the log base 0.01 of 100 equals y so the question is what is y and in part C of the log base B of 6 equals one half this is a B looks sort of like a 6 doesn't it so let me try to emphasize that's the letter B okay coming back to the classroom to figure out what x is here what I would do is I would convert this to exponential form because you and I are much more comfortable with exponential form because we've just introduced the logarithmic notation can anyone tell me what this says in exponential form 4 squared equals x exactly you see here's the base 4 the logarithm is equal to the exponent so 2 is the exponent and that has to equal the third number x so x is 4 squared so what that tells me is that x is equal to 16 yeah that's the answer to this logarithmic expression or actually equation it's a logarithmic equation okay in the next case let's see Matt can you tell me what is the what is the exponential equivalent of this 0.01 yeah rate to the y equals 100 equals 100 now let's just go over that you remember y is the exponent because the logarithm equals the exponent so the y goes in the air this is the base so 0.01 is the base and this is equal to 100 now this may look rather complicated but think of a way to a simpler way to write this so we can solve for y 1 over 100 to the y power equals 100 right 0.01 means 1 over 100 to the y power equals 100 now what do you do to 1 over 100 to get 100 well you have to flip it over and to flip it over what should the exponent be negative 1 so y is going to be negative 1 yeah in the next case you didn't think of that let me show you another route you could have gone taken to solve this we could have written the inside as 100 to the negative 1 power yeah because 1 over 100 means 100 to the negative 1 power now if I multiply exponents together this says 100 to the negative y power equals 100 and by the way this is 100 to the first power and therefore these exponents should be alike so y should be 1 and therefore y is negative 1 now this is a little bit longer tack to get to the answer but if you find this more comfortable or more agreeable to you you're welcome to follow that approach what we did back here the shorter way was to just say what we need to do is invert the 1 over 100 to get 100 so the exponent that will do that is the exponent negative 1 and y was negative 1 okay so either way you look at it this answer is y equals negative 1 okay now we have a third problem here log base B of 6 is a half Steven how would you write this exponentially B to the 1 half power equals 6 B to the 1 half power equals 6 exactly because this is the base B we always take that subscript to be the base this is the exponent over here so it's B to the 1 half power and the number inside is the value of the exponential expression so this says the square root of B is 6 and therefore B must be what must be 36 must be 36 so B equals 36 is the answer by the way the base that I put here for the logarithm obviously it doesn't have to be an integer because we have a case here where I used a decimal but the base always has to be a positive number so that's the only restriction that we place on that logarithm okay now you can find some logarithmic values on your calculator and let's go to the next graphic to describe the two that are available on your calculator the two logarithms available on your calculator call common logarithms and natural logarithms now the logarithm log base 10 of a number say log base 10 of x this sort of a logarithm is called a common logarithm and it's actually the older of the two logarithms that we're going to discuss here log base 10 or common logarithms go back for several hundred years and because they're so quote common this is normally abbreviated as just log x so if you don't see a subscript on the log then that's normally taken to be log base 10 so in that first sentence it says the logarithm log base 10 of x is a common logarithm and is abbreviated as log x with no subscript at all now the other logarithm that's rather common is the log base e you remember e is the number 2.718 approximately and the logarithm log base e of x is called a natural logarithm and it's abbreviated as ln of x for natural logarithm now you may say Dennis, I don't see anything natural at all about the number e but if you recall when we had the function e to the x we call that the natural exponential function and now in this case log base e we call the natural logarithmic function and they both have abbreviated names and this one has an abbreviated name because it's used so often now if you look on your calculator the last sentence here says your calculator is designated for logarithms and you'll have a key that says L-O-G and a key that says ln now let me just show you on this on this graphing calculator come back to the green screen if you can zoom in close enough over in the left hand column I have two keys one of them says L-O-G one of them says ln and let me just show you how these work suppose I want to take the log of 100 and I think if I put it over here on the side I can write just beside it the log of 100 now what this means is log base 10 but if there's no base shown there we just assume it's a 10 even if it's not expressed there now if I were to put a 10 here what exponent would you put on 10 to get 100 right so that answers 2 so the log of 100 should be 2 let's try this on the calculator and see what happens I'm going to press the log button and I'm going to enter 100 and then push enter and we get a 2 over here on the right hand side okay let's take another example what if I were to take the log of 0.1 now let's see to figure out what this answer would be remember this means base 10 what exponent would I put on base 10 to get 1 tenth for the answer negative 1, yeah you see if you think of this as a fraction this is the fraction 1 over 10 10 to the negative 1 power is 1 tenth or 0.1 let's try that here on the calculator and I'm going to take the log of 0.1 and when I enter it it's exactly right okay now the other key on this calculator I want you to notice is the natural log key now what if I take ln ln of 1 I'm wondering what that would be well let's see now you know there's actually a base here the base is e 2.718 approximately I'm going to put on base e to get 1 for the answer 0 yeah I think that would be 0 does everybody agree with that because see what we're thinking is an exponential form e to the 0 power is equal to 1 that's what I'm thinking now of course the e isn't shown but when you see ln it's assumed to be base e let's try computing this I'm going to push the natural log button the natural log of 1 equals so if I push enter and I get a zero right over there and let's take one more okay this one's kind of tricky what would be ln of the quantity e to the 3rd power does anyone have a guess as to what that answer would be 3 that would be 3 okay Stephen tell us how you came up with that answer well you want to find out what what power you raise e to to get e to the 3rd and well you're kind of given the answer there so the question is what were you going to say well to get e to the 3rd from e simply raise it to the 3rd power so the answer is 3 so we have the answer right there in front of us so the question is going back to the green screen if we were to take this base e which generally of course is not shown if you were to take that base e what would the power be well the power should be 3 now I'm going to try that here on the green screen and I'm actually going to substitute in the approximate value of e so I should get an approximate value for 3 so I'm going to take the natural log of the quantity see how much was e again who can tell us 2.7 2.7 okay approximately that's the number we've been using and I'm going to raise that to the 3rd power so I have 2.718 raise to the 3rd power close parenthesis now do you think I'm going to get an answer of exactly 3 no no because this isn't exactly e is this number a little bit too big or a little bit too small for e a little bit too small so anyone want to take a wild guess as to what the answer we're actually going to see on the screen when I press enter 2.97 2.97 okay so Sam's going out of limb 2.97 anybody want to go for 2.98 2.99 okay Matt says 2.99 but I notice he kind of glanced at his calculator so we'll see if he's already worked this let's see 2.99688 et cetera okay now let's do this one more time natural log of the quantity I'm going to put in a more accurate decimal expansion of e 2.718 now if you remember then it was 281828 and then 459045 you might say Dennis are we really supposed to remember all that no not at all but at the beginning of this episode I told you more digits in that number I'm going to raise that to the 3rd power close parenthesis and now I should get I hope 2.9999 I'm going to get a bunch of 9s I still won't get 3 because I think that number is a little bit still too small so I press enter and I got 3 why do you think it said exactly 3? calculator rounded the calculator rounded off you see this only goes out so far and it rounds off the last digit so it rounded off to 3 so that is our best estimate for that number okay let's go to the next graphic and look at this example it says evaluate the following logarithms well I'm going to do this first of all without a calculator and then I want to show you one more graphic before we leave about using your calculator okay without a calculator I'm wondering if someone can tell me I'm going to write this on the screen here what is the log of 100,000 and while we're at it let me write down the other problems what is the natural log of eq oh my goodness all I just happen to make up the same problem I had on the on my graphic so I think you'll know what that one is and for part c what is the log base 12 of 1 over 144 that's kind of a ringer it's not a natural log or a common log okay coming back to the green screen in part a when I say the log of 100,000 what base are we assuming here base 10 okay so 10 to some power is 100,000 does anyone know what power that would be it'd be the fifth power what's a quick way you can tell the power on 10 that will give you 100,000 just count the zeros yeah 1, 2, 3, 4, 5 that's 10 to the 5th so for example if this had been 100 just 100 100 has 2 zeros so 10 to the 2nd power is 100 so that answer is 5 this problem is way too familiar for us because I just happen to make it up again for the second time this answer is how much it could be 3 it could be 3 it could be if you were lucky this is your lucky day and okay now in the last case log base 12 of 1 over 144 let's see Susan do you want to take a guess at that one negative 2 is exactly right you are a winner okay negative 2 now which one of these would I have not been able to have computed on my calculator C because we don't have a button that says log base 12 we have a log base 10 we have a natural log now you might say well gee what's wrong with these cheap calculators they don't even give us a log base 12 button what are we supposed to do if we encounter a log base 12 and we don't know the answer otherwise well we will find a way in the next episode to convert any logarithm base to base 10 or base E so we can actually use these two buttons to work a base 12 problem but we haven't seen how to do this yet okay now in the last minute or so let's go to the last graphic we have one more graphic here and this title logarithms and calculators I wanted to give you an example of two different calculators and the order of the buttons that you have to push in order to do these first of all I'm going to use my graphing calculator once again and if you come back to the green screen the question was to find the log of 481 now on this TI-82 and other TI calculators similar to it if you want to take the log of 481 what you do is you enter it in that order you push the log button and then 481 and the answer is 2.6821 and back on that graphic you would have seen this is the answer I had written out I would have guessed there was going to be a 2 in front because this number is more than 100 481 is more than 100 and the log of 100 is 2 so this would be a little bit more than 2 2.68 now if I go to a different TI calculator just happens this is another TI calculator this is a little TI-30 that I brought from home and on this calculator I have a log button and a natural log button written not in the left-hand column but on the third row and if I want to take the log of 481 on this calculator I enter 481 and after I push the logarithm and I get 2.6821 et cetera now you notice there was a slight pause before that number came up let me go back and do that again I'm going to enter 481 and when I push the log button there's a slight pause before the number appears there you go it was only a fraction of a second but that's because the calculator has in it a little sub-program where it's actually making some arithmetic computations to come up with that number it doesn't have this answer memorized it actually has to compute it and that little pause is telling you it's sort of like thinking just for a moment you might say thinking in quotes before it can give you that answer now on the other hand if I say 2 times 2 look how fast the answer comes up 2 times 2 equals bingo 4 because there's only one calculation there 2 times 2 is 4 but for the logarithm value there was a slight pause because there's more going on than meets the eye and you'll hear about that in a calculus class I'll see you next time for episode 18 where we will continue talking about logarithms see you then