 Episode 21 of Math 1050 College Algebra for our television course. I'm Dennis Allison and I teach in the math department. We'll be looking at a review today of the chapter on logarithms and exponential functions. Let's go to our first list of review ideas. Number one, you want to be prepared to solve problems that are similar to those that we've just been discussing in class in the last few episodes and those assigned for homework. Now you know occasionally there have been some problems that we didn't necessarily cover in our television course, but you'll find examples like those in the textbook. You want to make sure you know how to do those. One of the differences between this exam and the first two exams is you can use a calculator to solve certain problems. Those problems will be given to you separately on the test. So you should be able to use your calculator to compute logarithmic values as we as we encounter them. You cannot use notes and of course you cannot use your textbook. You'll be given some extra scratch paper with the exam, but you'll have to turn that in when you turn in your exam. And we encourage you to show all your work to justify your answers. Otherwise if you have a wrong answer you may lose full credit, whereas if you show some work then if we can find the mistake that you might make in that then you can get partial credit. And we try to be fairly generous with the partial credit. Okay let's look at the review ideas for episode 16. Now in episode 16 we introduced exponential values and how to compute those with your calculator. Now if we can come back to the green screen let me just remind you that on your calculator to find an exponential value you'll either have a key that says y to the x or it could say x to the y. This depends on the manufacturer of your calculator. Or another common exponential key is this little little carrot thing pointing up. And you enter the base then you push this button, one of these buttons and then you push your exponent afterwards and then you hit the equal sign or the enter sign and you'll get the exponential. So for example if you want to say 2 to the third power then you hit the equal sign and it should say 8. Now of course the reason you hit the equal sign is because when you hit 3, when you push 3, your calculator doesn't know if you're pausing to enter 31, 2 to the 31 power or if you're finished. So when you hit the equal sign or the enter sign then the calculator knows that you're done. Okay let's go back to episode 16 and look at the other items on that list. You should be able to sketch the fundamental exponential functions and then in connection with that you should be able to sketch transformations of those functions and in each case this should be done by plotting only the target points not making a table of values. Why don't we take a few examples of those right here. Suppose that I wanted to graph the function f of x equals 4 to the x power. Then there were three target points that we associate with this exponential function and let's see let's just get our graph system, our coordinate system set up here and the target points were let's see now at 0, at 0 we go up one unit and if I go over 1 from the origin then I should go up whatever the base is so I go up 4 so I'm going up 4 right there and if I go to the left one I go up the reciprocal of the base 1 fourth. Now you notice I'm doing this in a rather mechanical way but I think all of these values make sense because you see when I choose x to be 0, when x is 0 I get 4 to the 0 or 1, when I choose x to be 1 right here then when x is 1 4 to the first power is 4 so I went up 4 and when x was negative 1 then the value there is 4 to the negative 1 power and 4 to the negative 1 is 1 fourth so we went up the reciprocal of the base. Now on the basis of only those three target points we can draw our exponential graph but we have to know that there's a horizontal asymptote along the x-axis and we have to know that there is no vertical asymptote so I don't draw in any vertical dotted lines. So this is how we'd like for you to graph this function and not make a table of other values but plot only these three these three points. Let's take one more example of an exponential function and I'm going to try to hide it in a sort of a minimal way. Let's say we call this function g of x and let's say this function is 3 to the negative x power. What is the effect of the negative in the exponent? What is that done to hide the exponential function? Jeff what's another way you could write that? 1 over 3 to the x. 1 over 3 to the x. Yeah so this is actually 1 third to the x and you see a lot of times students think that this graph is going to look like the one I just graphed before where it goes up on the right-hand side but actually it's going to be going up on the left-hand side because the base is smaller than 1. The base that we choose either 3 or 1 third we always choose positive bases here so you'll never see a base zero or you'll never see a base that has a negative number in it. Okay well I want to graph the function 1 third to the x power or 3 to the negative x and to do that I'll plot three target points just as before and I'll plot them in the same way that is at zero I'll go up one. If I go over one I go up the base now when I say the base I mean the base in this expression 1 third and if I go to the negative if I go over to negative one I go up the reciprocal of the base and the reciprocal of 1 third is 3 so I get a point at 3 and so this graph comes down and approaches the positive x axis as a horizontal asymptote. Now you might say well Dennis how are we supposed to know if 3 is the base or 1 third is the base? Well you see a function is in standard form when it's written a to the x power and this is where a is bigger than zero. So if there's a negative exponent here I have to get it to the form a to the x and then a is the base that I use when I'm plotting these points. Okay now let's look at transformations of these functions. This is still an episode 16 and this time I'm going to take the function f of x equals 2 times e to the x minus 1 minus 2. Now I have three different transformations in this function three different basic transformations. When I when I subtract 2 on the outside that's going to lower the graph 2 units and when I subtract 1 directly on the x that's going to shift it to the right 1 unit and when I multiply by 2 that's going to be a stretch so it's going to double the altitude of the points or stretch them away from the x axis. Now when I write 2e to the x minus 1 you may be asking whether this means 2 times e all raised to the x minus 1 or 2 times the quantity e to the x minus 1 and it's the understanding in an algebra that if you raise something to a power and you multiply if there are no parentheses the exponent comes in before the 2 so in other words the exponent is applied and then afterwards I multiply by 2 and then I subtract 2 on the outside so so this is the way we interpret this expression. Okay so when I go to draw this graph I have to know something about the number e and can anyone tell me what the value of e is approximately? Well it's about 2.7 we said 2.718 but I think when we get down to locating points I think 2.7 is close enough and by the way 1 over e we said is about 0.36 or what I use is roughly 1.3 this is about 1.3 so you want to keep these two values in mind let me add the 1.8 on there just to be a little bit more precise but I don't think we'll be able to plot points with as much accuracy as to show hundreds and thousands in the decimal expansion and 1 over e I'll be locating as roughly 1.3. Okay so we said we're going to be lowering this graph 2 units so if I lower the graph 2 units I'll put in a horizontal line right here this is the line y equals negative 2 and we said we were going to be shifting the graph 1 unit to the right so my vertical axis is going to be shifted over 1 unit. Let's see that was the x-axis this is the y-axis so this is the line whoops x equals 1 and so this is the point I would consider my new origin is how we refer to it this is the original origin and now we're at the new origin and what I'll be doing now is to double the altitude of every point because there's a stretch of 2. Okay so at zero I'd normally go up 1 rather at the origin I normally go up 1 I'm going to go up 2 units because I've stretched it 2 and if I go to the right one I would normally go up e I'll go up twice that amount let's see now e is about 2.7 so twice that's about 5.4 I'll go up a little bit over 5 between 5 and 6 1 2 3 4 5 6 between 5 and 6 about half or between I'll go up to a point right about there. Now you might say Dennis exactly how high is that well since I started 2 units below the original x axis this is actually at an altitude of 2e minus 2. 2e minus 2 is actually how high that value is right there and going back to the origin if I go over to the left one I go up 1 over e which we said was about one-third so I'll double it and go up about two-thirds. Okay so if I connect this point this point and this point my graph looks like this and it approaches a horizontal asymptote which is which is the x axis after it's been shifted down to units. Okay so this is the graph of f so I might put a name tag on it right there. Okay so that's how we go about sketching transformations of exponential functions. Let's go to episode 17 now and look at some of the review items for that episode. There were two formulas for compound interest that you should know and be able to apply on this exam. You know we will return to the notion of compound interest later on when we get to one of the last episodes of this course and we'll talk about annuities and how you figure the payment on a loan. So these two formulas are useful but we'll see more information about this later on. If we come to the green screen let me remind you what the formulas are. Okay the first formula was a equals p times 1 plus r over n raised to the n t power. This is where p was the initial principle that was invested whereas a is the amount in the account after a certain amount of time. r is the interest rate per year but it's compounded n times per year so you divide by n and that that figures the interest rate per period and then you raise this to the number of periods power. So if t is the number of years n periods in a year then n times t is the number of periods. Okay the other formula is for continuous compounding in which we say a equals p e to the r t power. So this time with continuous compounding you don't have periods as such so there is no n but this is the annual interest rate and this is the number of years that the money is being compounded. So you should know these two formulas and rather than working examples with these I think our time is probably better spent going back to our list of items for episode 17 and seeing what are the things you'll need to know here. So let's go back to episode 17. You should be able to convert between exponential and logarithmic expressions. Okay well let's see let's work an example of that. Suppose we said that 5 to the third power is 125. We say that this is an this is an exponential expression because we're using an exponent and this is what most of us are accustomed to seeing rather than its logarithmic equivalent. Now the equivalent of this is to say log base 5 of 125 is equal to 3. Now if you have trouble remembering which numbers go where remember that a logarithm is equal to an exponent that was a fundamental fact that we've mentioned many times so the exponent goes over here on the right the logarithm is equal to 3 and the base base 5 raised to the third power that's also the subscript on the log and that leaves the numerical answer 125 to go on the inside this is sometimes referred to as the argument so 5 to the third power is equal to 125. Let's try doing this backwards. Suppose you were given a logarithmic expression like the log base b of a is equal to c and I'm using letters here to try to disguise what their meaning might be and so the question is how could I write this in an exponential form? Anybody have an idea how you'd write that? You could write it b to the c equals b to the c power equals a. Okay Tony do you agree with that looks like you were going to give an answer to. Okay so b to the c power is equal to a. Yeah because here's base b so that's going to be the base over here c is the exponent because it's equal to the log so c goes as a superscript and then this is equal to a. So now we're converting from a logarithmic expression back to an exponential expression. Okay back to our episode 17 list. You should be familiar with common and natural logarithmic expressions and how they're abbreviated that is that base 10 or common logs are abbreviated as L-O-G and natural logarithms are abbreviated as L-N for natural logarithms it means base e and remember that e is about 2.718 approximately and then finally you should be able to compute logarithmic values with a calculator so here's a place where you want to use your calculator as well as those compound interest formulas I think you'd want to use your calculators that item given at the very top there is a place where you'd want to use your calculator. So you should be able to find the natural log and the common log of numbers using your calculator. Okay let's go to episode 18 and here we started getting into some of the more fundamental properties of logarithms such as the four laws of logarithms. Now you know in a lot of textbooks you'll find that they list more than four laws of logarithms but I've tried to keep it abbreviated to only four logarithms to make them easier to apply and many of the other laws that you see sometimes in textbooks are merely consequences of these four so let's come to the green screen and see what those four laws are. Let's see the first law of logarithms says that if you take the log base let's say base b of a number m plus the log base b of a number n then you get what when you add those together? Log base b of m times n. Right log base b of m times n. What do you think is the most common wrong answer that students put right here? m plus n. Yeah see when you see the addition of two logarithms then it seems natural that you might want to be adding the m and n here but it's actually a product and the reason is a logarithm is an exponent. So what I'm doing is I'm adding two exponents. The exponent that you put on base b to get m plus the exponent you put on base b to get m is the exponent you put on base b to get mn. When you add two exponents you get the exponent that belongs on the product. In the same way the log base b of m minus the log base b of n. Tony what do you think this one would be? It's probably the log base b of m over n. m over n that's exactly right and this time you see we're subtracting two exponents that is we're subtracting two logarithms and so what you get is the exponent that goes on the quotient. Now if necessary you can put parentheses around that to separate it but sometimes putting in those parentheses makes a little bit harder to read I think so you can either put parentheses on that quotient or not. Okay the third law of exponents says that if you have the log base b of m to the c power and I think I'll put parentheses on this one this time then this c comes out the exponent inside a logarithm comes out but let's see what will it be when I bring it out. Yeah Tony. It's c log base b log base b of m exactly right so this is c times that logarithm if any parentheses are needed I think maybe it would be right here it's c times that logarithm so what's surprising is if you have something in the exponent inside it's not an exponent outside it's a coefficient and as a matter of fact one of the common misinterpretations of the rule is to take a c on the outside and when you bring it in it's a coefficient rather than an exponent but it should be an exponent of course so these laws are two-way streets you can go either way with this if there's an exponent inside you can bring it out as a coefficient if there's a coefficient outside you can bring it in as an exponent and the last the last rule it looks a little bit different or the last law is a little bit different from the others and it but it says if you have base b and you raise it to a logarithmic power such as the log base b of m power then you get m for the answer and this is merely saying that a logarithm is an exponent so what I've done is I place the logarithm in the exponential position where you'd normally expect to see an exponent this is the exponent you'd put on base b if you wanted to get m for the answer so look what we've done we placed it on base b and so not surprising we get m for the answer let me just ask you a couple questions right below here about these laws and we'll keep these laws written up above what would be the log base 5 of 3 plus the log base 5 of 7 now you know I don't know what the log base 5 of 3 is or the log base 5 of 7 is but there is a way that I can reduce this Tony be the log base 5 of 21 log base 5 of 21 exactly now you notice these two bases have to be alike if this had been base 5 and this had been base 3 then I wouldn't be able to combine them they have to be the same base and I get that base in the answer but I get the product okay very good another problem another question might be what is one half the log base 9 of 81 well let's see there are several ways I could work this but I'm thinking let's use law number 3 that says if there's a coefficient you can bring it in but it'll be an exponent so if I bring in the one half how would I rewrite this as a logarithm with the one half inside log base 9 of 81 raised to the one half 81 to the one half power yeah exactly now what does the one half power mean square root take a square root so what I'll do is write this as a square root that is the one half power now the square root of 81 is 9 so this is the log base 9 of 9 and what is the exponent I'd put on base 9 to get 9 for the answer one one so this answer into this answer ends up being one in this case this one I could actually reduce okay one more one more problem that goes with these laws of logarithms suppose I had suppose I had 6 raised to the 2 times log base 6 of 10 power 6 raised to the power 2 log base 6 of 10 well now there's a base 6 here and there's a base 6 here but what are we going to do with that 2 out in front Tony exactly see even up here in the in the exponent I can move the 2 into the logarithm and it will be an exponent so this is 6 to the log base 6 of 10 squared or 100 right so in other words we just used law number 3 there and now let's use law number 4 to reduce this ultimately so what's the final answer going to be be a hundred right be a hundred because B to the log base B of Amazon 6 to the log base 6 of 100 is 100 so sometimes we're able to we're able to simplify these logarithms eventually and sometimes we can merely we can simplify them but but not give them a numerical value okay let's go back to our list for episode 17 know the four laws of logarithms be able to apply these laws and we just saw some examples of that know that logarithmic functions and exponential functions are inverses okay now this is a fundamental property that's certainly important and you'll get some sort of a question about this apps for sure just like you will for those four laws we just went over when you're taking your test so be prepared to know this next fact let's come to the green screen and discuss what that what that actually means suppose I have a function f of x equals 5 to the x power so this is an exponential function and then there's a corresponding logarithmic function that is log base 5 of x so I would refer to this as a logarithm function and this exponential function is base 5 and this logarithmic function is base 5 now what's significant about the two functions is these are inverses of each other you remember to say that a function is an inverse if this is the domain of a function f and this is the range of a function f we saw this not too long ago then if you take a number x in the domain of f and you send it over to a number y which we might call f of x by way of f then this is sort of the the basic relationship here in a function that is that every x is assigned to unique y now if this is a one-to-one function I can reverse the process and I can send this function I can send the number y back to the x using the inverse function and this becomes the domain of f inverse the set on the right and the set on the left becomes the range of f inverse and so what happens is if you start off with x and if you send x away by f and then afterwards you apply f inverse you come right back and we would summarize this by saying that f inverse of f of x is x because you start with x f sends it away f inverse brings it back and you're right back where you started now this is sometimes written as a composition of functions in this form now the same thing holds true if I had chosen my x over here in the domain of f inverse f inverse could send it away and then f could send it back afterwards and I'd come right back where I started you notice this x is actually different than that x because this x is chosen in the in in the set on the right not the set on the left and in that case I would say that f composed with f inverse of x is equal to x now this is only going to be the case for functions and their inverses and let's see if that's the case with the two functions we have at the top of our of our screen here so the way I'm going to find out if these two functions are really inverses of each other is I'm going to take their composition let's try taking the composition f inverse composed with f of x oh except I better call it g here I guess not f inverse we don't know it's the inverse function yet so I'm going to take the composition of g with f now the way we interpret this is this is g of f of x and f of x is five to the x so this is five to the x power and now what is g due to that g takes the logarithm of any number that you put inside of g so if I put in an x I get the log base five of x if I have five to the x what I get is the log base five of the quantity five to the x now you might say well Dennis this doesn't look like x I thought you said the composition was going to give you x well let's see what happens if I reduce it by bring the exponent out in front then I get the log base five of five and the log base five of five is one so this is x times one is x yes and so we're all excited here because we've started off with x and I got x when I finished now if I work this in the reverse order I don't think I'll work this one out but I'll leave this one for you to check if I work this in the reverse order I will eventually get x again so what that tells me is these are inverse functions of one another and that's going to help me draw graphs of these functions coming up now this also explains I think why exponential and logarithm logarithmic functions are discussed in this chapter and are together for this exam is because they related to each other they're inverses of one another okay let's go back to our to our episode 18 list be able to graph logarithm functions well now that's a consequence of what we've just been over right here let's see how we'd go about graphing the log base five of x for example let's see we just had g of x equals the log base five of x and I want to be able to graph that so what I'm going to do is keep in mind how I would graph its inverse function five to the x so let's get your graph right here okay so what I want to do is to graph the function g but I'm going to be keeping in mind with the graph of the function f looks like you remember to graph this exponential function at the origin we started off it would went up one well for the inverse function I'm going to be flipping this over the 45 degree line I'm going to be flipping that over and instead of going up one I'm going to go to the right one and for the exponential function if I went to the right one I'd go up five so now when I flip it over if I go up one I'll go over five and for the original exponential function if I went to the left negative one I went up the reciprocal of the base so now when I flip it if I go down one I go over the reciprocal of the base or one fifth and I get a graph that looks like this and it approaches a vertical asymptote at the y-axis so this is the graph of g of x equals the log base 5 of x you know one of the characteristics of a fundamental logarithmic functions is they all cross the x-axis at one they have an x-intercept but they have no y-intercepts and a fundamental characteristic of the exponential functions is they all cross the y-axis at one and they have no x-intercept so you see everything's been reversed because of this inverse relationship. Let's try sketching a transformation of a logarithm function while we're at it. For example suppose I wanted to graph the function f of t equals the log base 2 of x plus 3 plus 1 well basically we're graphing well I should put a t in there shouldn't I because I said f of t so basically we're graphing the function f of t equals the log base 2 of t but there are two changes being made in this graph the plus 1 here causes the graph to do what? Yeah it's going to move up one thank you right it's going to move up one and the t plus 3 is going to shift the graph in which way? To the left. It's going to move it to the left 3 okay and I think those are the only changes we're going to be making in this fundamental function we there's no negative in front so we're not going to flip it over there's no other coefficient in front that's going to cause a stretch or a compression so if we set up our coordinate system let's see we're going to we said we're going to raise the graph one unit so when I go up one that's going to make this point at one my new origin except I still have to shift it over 3 and my vertical asymptote is going to move over 3 units I'll put a dotted line in right here this would be at t equals negative 3 because this is the t axis and the y axis so that says that right here that's at negative 3 plus 1 this is my new origin and from my new origin I should go to the right one and if I go up one I should go over two and if I go down one I should go over one half so here are my three target points and on this basis I can sketch the graph yep and so this is the graph of capital F now when you graph these functions you should plot only the target points and not make a table of values you might say well Dennis what's what's wrong with making a table of values well you see what we're trying to do is to graph these by speed and not necessarily accuracy if I plotted more points I could get a more accurate graph but I'll never get a truly accurate graph because I just can't plot all the points that the graph goes through so we try to have a minimal number of points that we can plot and so we plot only these three target points and this is what we'll be looking for when we grade your exam so don't plot more points don't make a table of values but just know how to plot these three you should be familiar enough with with those points based on our previous discussions in the homework that you've been doing okay so those were transformations of logarithm graphs and there are only three more graphs three new graphs that we talked about in this course those are the conic sections that will come up later on okay let's go to episode 18 oh excuse me I'm sorry we want to go to episode 19 okay we want to be able to solve exponential and logarithmic equations and there are quite a variety of these let's just take a few examples to kind of remind you what they might look like now first of all let's take a problem that you can solve that you could have solved actually before you took you took this course that's an exponential equation so if you come to the green screen suppose we had seven raised to the x minus three power equals 49 now this is an equation because I have two quantities equal with a variable in it and because there's an exponential expression in it I'll call this an exponential equation but that doesn't make it difficult to solve this one's actually fairly straightforward I think what we have to recognize is that on each side I can find an expression with a base seven x to the seven to the x minus three and over here seven squared so seven to this power equals seven to this power so what's the natural conclusion that we're going to draw here well okay that that is going to be the answer but what I'm thinking is I know that this exponent should equal two so x minus three equals two and you see what significant here is this is now a linear equation it's like the most fundamental equation that we could solve so I no longer have an exponential equation I have a linear equation and if I solve it tummy says x will equal five that's exactly right so here's our solution and we found it by actually changing the problem into a simpler form and then solving that so x is equal to five here let's just check that if I substitute in a five right there then I'm going to have seven to the five minus three power that seven squared seven squared is forty nine that's right okay now let's take a problem that's maybe not quite that easy to solve but it's still an exponential equation and for this next one we'll be using a calculator suppose we had ten to the two x plus one power is equal to let's say seventy four is equal to seventy four well you see the problem now is I have a base ten here but I don't see any way I could make that a base ten I don't see any way I could make these have a common base so now I'm going to have to solve this exponential equation but what I'll be doing is applying logarithms so I'm going to take advantage of the fact that exponential functions and logarithmic functions are inverses of each other I should take a log on both sides now you know technically I could use any base I wish but what do you think would be the most natural base to you maybe natural the wrong word use here what do you think would be the most appropriate base to use maybe to well not to but since there's a base ten here and I have based in on my calculator I think I'll use base ten or a common logarithm base two might seem appropriate because these both have a factor of two in them but I think since I have based in on my calculator I'll use that so log base ten of ten to the two x plus one equals the log base ten of seventy four now why am I not writing a subscript on these logarithms if you don't write a log or if you don't write a subscript it's just assumed that it's ten yeah if you don't write a subscript it's assumed to be ten now you know I should give you a bit of warning here if you're going to take more mathematics and science courses there are some later textbooks that will write L.O.G. to mean a natural logarithm as opposed to L.N. you remember L.N. means a natural logarithm in our course so you have to be kind of careful in some in some more advanced math courses the natural log is used so commonly that when they write L.O.G. they mean base E but we'll be assuming that L.O.G. means base ten this is the way it's it would be written on your calculator now I'm going to use a property of logarithms to bring the exponent out in front two x plus one times log ten equals the log of seventy four and what is the log of ten one is one yeah this number is just a one because remember a logarithms and exponent what's the exponent you put on this base to get this answer well you'd put exponent one on base ten to get a ten for the answer so that's one so this tells me that two x plus one is equal to the log of seventy four now I could calculate this on my calculator but rather than doing that yet I want to get the exact answer I'm going to solve for x two x is equal to the log of seventy four minus one and then I'm going to divide by two and I have the log of seventy four minus one all over two now I would call this the exact answer and you see what I did to get to arrive at that is I solve this linear equation that I had in the middle this is this is a linear equation even though it may not look linear because of this expression but this is merely a number so two x plus one is equal to this other number and I isolated the x now of course most people would want to approximate that to find out roughly how much that is so if I go to my calculator I'll find out how much this expression will be if we can zoom in on my calculator right here and here's one of those case one of those places where on the exam you might want to have a calculator available to you so I'm going to take the log of seventy four let's see the log of seventy four minus one oops excuse me minus one and I'm going to total that up in the divide by two so that's the log of seventy four minus one now divide that by two and I get point four three let's round that off four three five I'm gonna round that off to three decimal places point four three five so I'll write that as zero point four three five okay now I was using a TI 82 to work that problem now let's go to a different calculator and just see how the difference in the keys will be so okay now one of the one of the differences here is that the order of the of the key usage is going to change just a little bit so this is a TI 30 and let's see how the problem be solved here now to find the log of seventy four I'm first going to enter seventy four and then push the log button and then I'm going to subtract one equals okay and then afterwards I divide by two and you see when I round this off to three decimal places it'll be zero point four three five so that's exactly the same number that we calculated on the other calculator by the way while we're at it I might point out that occasionally on two different calculators this last digit will be slightly different like on another calculator you could see a 158 instead of a 159 that's because all of these logarithmic values are approximated using formulas that are embedded within the calculator and some of the formulas are abbreviated depending on the the power of the calculator and it will sometimes cause that last digit to change so you never know if you see two different calculators with a different digit on the end which one is correct if either one of them is correct but those digits have all been rounded off the last digit according to the program that they have stored in them the same thing holds for square roots and cube roots and for exponential values okay well you can see well the answer that we just calculated point four three five is the same on both on both calculators okay now those are two examples of exponential of exponential equations now there is a formula I don't think I've mentioned yet today that you need to be aware of and that's the change of base formula so let's go to the green screen and let me just remind you how that goes for example suppose you wanted to find the log base two of seven but you go to your calculator you don't have a log base two button so what you would do is use the change of base formula that says you can write log seven using any other base say base ten log seven but then you have to divide by that same logarithmic base of the old base two so the log of seven divided by the log of two another way to write this would be to say take the natural log of seven that's log base e 2.7 approximately but then you have to divide by the natural log of the old base two and on our calculator we have an l o g button and we have a natural log l n button those are the only two they've given us but we don't have a log base two button so if you wanted to find this logarithm with this base two you'd have to convert it using the change of base formula to either common logs or natural logs okay I don't think I mentioned that in a earlier today but you certainly want to know that formula as well okay now let's try solving a logarithmic equation and suppose the equation were something like this let's say we have the log base three of x minus the log base three of x plus one is equal to two is equal to two now here I have an equation but I have logarithms involved in it rather than exponentials so I would refer to this as a logarithmic equation and when you see a problem something like this on the exam what you should do is combine the logarithms together so I'm going to write this as a single logarithm log base three of now let's see I've subtracted two logarithms so that'll be the log of the what do I put here x over x plus one x over x plus one right that was using our second law of second law of logarithms that we saw earlier so the log base three of this expression is equal to two now I'm going to change this to exponential form that by removing the logarithm and saying three raised to the second power is equal to the quantity inside three to the second power equals x over x plus one okay well now what I have is a rational equation because I have a rational expression and I have this ratio so this is a rational equation and I'd like to solve that well this is nine is equal to x over x plus one and let's multiply both sides of x plus one and that'll be nine times x plus one is equal to x well you know I no longer have a ratio this is a linear equation so I'm slowly getting this down to a form that I'd like to solve and I think this one we can work with this says nine x plus nine is equal to x or that says that eight x is equal to negative nine and that says that x is equal to negative nine over eight negative nine over eight so therefore do you think that this is the solution to that problem well you would think so but you know what we have to throw this one out and we have to say that there's no solution now why is that you can't take the log of a negative number yeah you see back here there was there was a there was an assumption that we were making about x you can't take the log of a negative number you remember logarithm graphs have this general shape we said that they cross at one and therefore when you take a logarithm you can't take a log of a negative number because there's no graph over here so we were assuming all along that x was bigger than zero and I was also assuming that x plus one was bigger than zero well it turns out if x is bigger than zero x plus one has to be bigger than zero so this really wasn't of any consequence in this problem so if I come up with any number that with with any number down here that is negative I have to throw it away it's a solution of the of the rational equation and it's a solution of all the linear equations all the way down but it wasn't a solution of the original logarithmic equation they might say Dennis would have been a solution of this one turns out it would have been because if you had substituted in negative nine over eight you would have had a negative in the numerator and a negative in the denominator and that ratio would have been positive and I could have proceeded from here so it's actually a solution of this equation but it's not a solution of the original equation where the two logs had been separated so this problem has no answer okay let's take another example okay let's try solving this problem it says two log x equals log two plus log of let's say three x minus four okay so this time I have log rhythms in three different expressions so as a general rule I would say get all the logarithms on one side now in this case you could actually leave the log of two separated from the other logs because it has no variable in it but just to kind of keep things in the same order I want to get all the logs on one side so I'm going to move these two guys over here to the left two log x minus log two minus log three x minus four equals zero now when I get all the logs on one side then the idea is to combine them well let's see I've got a coefficient on this one and I didn't on either of the others what can I do with the coefficient change it to log x squared log x squared okay so let's write this one more time as the log of x squared minus the log of two minus the log of three x minus four equals zero okay so I have a logarithm I'm subtracting this one and I'm subtracting this one that says I should write this as a as a quotient inside a logarithm and both of these will be in the denominator because they've both been subtracted so I'll write this as the log of x squared divided by now let's see I'll have to put a two in the denominator and I'll have to put the three x minus four in the denominator and this is equal to zero okay so this is a base ten log and ten to the zero power which is one ten to the zero powers one is equal to this expression so this tells me that x squared over two times three x minus four is equal to ten to the zero power and we said that was equal to one so if I multiply both sides by two times three x minus four this says that x squared is equal to two times three x minus four times one but we don't need to bother showing the times one what kind of an equation do I have here this isn't a linear equation what is it's a quadratic equation yeah so I had logarithmic equations I had a rational equation and now I have a quadratic equation okay so you just never know what's going to come up as you reduce these but I think we're in better shape than when we had logs in there so to solve this I'm going to multiply that out let me just draw a line here that's going to say x squared is equal to 6x minus 8 how would you go about solving that move everything over to one side move everything on one side let's keep the x squared positive so move everything to the left x squared minus 6x plus 8 equals zero now we could use the quadratic formula to solve this but there's a faster way and what is that Tony what would you say Jeff what would you say I'm not sure complete the square we could complete the square but there's even a better way yet how about factoring let's see if this will factor if we can factor it that's usually the quickest way out of these things there's an x squared so I know I have to put an x in front of each of these there's a plus on the 8 that tells me the signs are alike and there's a negative so something's negative if the signs are alike they must both be negative so now I'm looking for two numbers to fill in here that will give me that quadratic what should they be two and two and four right very good two and four so either x minus two is zero or x minus four is zero now I tell you what if x minus two is zero I'm running out of room what would x have to be two two okay and the other one if x minus four was zero x would be four now do all these numbers make sense because you see we may have to throw one of these roots out like we did before let's take the two if I substitute the two back here log two that's okay log two well that one's constant there and this would if I plug in a two that will be the log of two as a matter of fact I think you can see this says the two times the log of two is the log of two plus the log of two yeah that would be two times the log of two this one certainly checks and it's it's just fine how about the four two times the log of four equals the log of two plus the log of how much will that number be when you plug in a four twelve minus four is eight yeah twelve minus four is eight but I don't see any negatives coming up inside the logarithm so this number should check that problem if you if you work that out those two sides should be equal so both of these numbers are should we say keepers if I had to throw one out the term they use for that is an extraneous root but this time we don't have an extraneous root we did have an extraneous root in that last problem that we work though okay let's go to episode 20 and look at applications of exponentials and logarithms and first of all let me point out that in our preview in our the episodes that we've just finished recently in the last few days we talked about the pH of a solution we talked about the decibel level of sound we talked about the Richter scale I'm not going to ask you about any of those on the exam so you don't need to know those formulas and I won't be asking about those primarily because we didn't spend a lot of time on that so what I will ask you about are things like what you see here you should be able to solve problems that deal with population growth radioactive decay and Newton's law of cooling now in the few minutes that we have left let's take a problem like let's say radioactive decay suppose I have some material like a blob of material here and let's say this is radioactive don't be scared this is only an example okay let's say this is radioactive and what if we have four grams of radioactive material this could be uranium or some other radioactive element now suppose this material has a half-life of oh let's see let's don't make the half-life too long what what should we pick half-life of let's say in days how many days seven seven days okay seven days one week okay that's the half-life seven days so in other words if at this moment I have four grams of the radioactive material seven days later how much of this will still be radioactive two grams two grams half of it has become a stable element the other half is still radioactive and after another week how much of this will be radioactive one gram one gram because every time half of the remaining material becomes stable and half is still radioactive so we might ask this question how much of this material that is these four grams is radioactive after let's say 23 days 23 days because I want to pick a number that's not just a multiple of seven otherwise I think we could work it out without much difficulty okay so our formula is this the amount of radioactive material at time t is equal to the initial amount of radioactive material times e to the r t power where r is the rate of radioactive decay okay well the initial amount is four so I know that a of t is equal to four e to the rt and I know that the half-life is seven days so I know that after seven days a at seven will be two oh you're right put a negative on there thank you very much that should put a negative on there because this is radioactive decay so this is decreasing and that'll be four e to the negative r times seven you see I'm plugging in a seven for t so I have to put a seven right here so I want to solve for r and that says that one half is equal to e to the negative seven r a seven r and then I'll take a natural log of each side so I get ln of one half equals negative seven r what I've done is I brought I have brought the negative seven r out in front and the natural log of e is one so this says that r is equal to negative one seventh ln of one half so if I want to find out how much is left after 23 days I'll have to compute a at 23 and that's going to be four times e to the negative r power let's see now I already have a negative on that so the negative r power is the one seventh ln one half times t which is four times one half to the t over oops that should be 23 for t we'll put a 23 right there 23 over three see if I put 23 here I put 23 in over seven okay now I think I'm just about out of time but if I if you multiply this out on your calculator you'll have the number of grams of material that's radioactive after 23 days well I will see you next time for episode 28