College Algebra - Lecture 29 - Logarithmic Functions

[Music] now as you see from the title on the screen we're going to be talking about logarithmic functions and then eventually their relationship to exponential functions but before we discuss logarithmic functions we want to discuss what an inverse function is because a logarithmic function will be an example of such an animal so let's look at inverse functions and in order to set the scene here let's observe something about a familiar function that you know the familiar function that I will look at is f of x equals x squared and this will repeat some of what we saw back in the one-to-one function segment but it's worthwhile to relook at this this is an example as you call of what I named a many-to-one actually I think I said more than 1 X to 1 f of X this is a many to one situation so if you're looking at a function in the form of my little function machines here then what we've got is a situation where we have a function that might take in two or minus two for example and produce on the other end the same number for because 2 or minus 2 when squared is 4 so this is a function that is our f of X function here and it's a squaring function now notice this is not reversible as I said there if I start with 4 and I go backwards and I ask where did this come from there's no way to tell there are two possibilities so going backwards is not a function not a function in Reverse ok and another way to look at this of course is to look at the graph so let me draw that here because we want to have both images in our mind the graph of f of x equals x squared of course is this parabola which we know well and here say is 2 and here say is minus 2 and we know that they both go to the same height on the graph the height 4 which is their f of X value for the function f of x equals x squared the problem of course seems to be that although we can go forward take the VEX values up to the y-value and have no contradiction we have a legitimate function going backwards is the problem so it seems to be if I try to analyze this that the problem seems to be if I have x1 not equal to x2 then I've got them both going to the same Y that seems to be the problem I'm facing here that's what makes this not reversible well of course that is exactly the kind of problem that is overcome by a one-to-one function a one-to-one function doesn't have that so we see that a one-to-one function which doesn't have any of the problems of non reversibility we saw before a one-to-one function can be what we might say is undone it can be undone let me give you an example of what I mean by that here's a nice example G of x equals 2 times X and I'll say that that is undone by H of X which is 1/2 X now that makes sense when you think about what these functions mean this is the doubling function this is the having function so if you first double something and then have it you're back where you started and in Reverse if you cut something in half and then double it you're back where you started so the kind of picture you might have here if I'm allowed to draw a little picture here imagine we start with X we apply G to it so G of box is 2 times box we end up with that 2x at the end and then if we continue the process we take 2x and apply H of box which is 1/2 of box to 2x what do we end up with we end up with 1 hand off of 2x which of course is equal to the original X so we've gone forward and this was one to one going forward and the reverse is also one to one going backwards so maybe we have a little bit of an aha moment here and we have a pretty good idea what sorts of functions are going to be reversible so I think it's time to write down a definition of what we'll call an inverse function the word inverse is used and not reverse but that's just a piece of terminology we can get used to definition the inverse of a function say F so the inverse of a function f and what it will have its own notation it'll be denoted F inverse little minus 1 up there we'll talk about that in a minute is a function such that and here's the definition such that this is the key point if you take F inverse of f of X so what's happening here f of X is being applied to F and then F inverse is following that application you end up back with X where you start it and if you start with F and if you start with F inverse rather the middle one and then go outward and do F you end up back with X now this is the key idea here so I'll even put key here up in the corner that's the key idea for an inverse function now we have to be a bit more specific since F is the function that's starting this composition here the X's that come out this is going to be for all X in the domain of F the domain of F where F begins but if we look at this previous example we begin to notice something this X over here on the left is a member of the domain of G isn't it because it goes into G also realize that on the return it becomes a member of the range of H and that is in fact always true so this is true for X is in the domain of F and the domain of F will be the same thing as the range of F inverse and likewise here if you start with F inverse that means this is true for all X in the range or rather the domain let's start with that one first the domain of F inverse and that will be the range of the original function f so with that definition in mind let me go ahead and draw a little bit more elaborate picture down here imagine that we have two sets here this will be the domain of F and here's a point X and X will be acted upon by F and result over here in a little point f of X that's what functions do now this will then of course in that case be the range of F but now if I want to act upon f of X via a function inverse F inverse this now becomes the domain of that new function f inverse and this becomes the range of that new function f inverse and what do I have here I have f inverse of f of X which of course is equal to X that's exactly this statement right there so the range of F and the domain of F inverse are the same the domain of F and the range of F inverse are the same they're just reversing that's all not a mystery so this is the definition of an inverse function what sorts of functions have inverses here is the question the answer to that question fact if f is a one-to-one function that's based on all the work we just did if F is a one-to-one function then F inverse exists for F so f has an inverse function that undoes it if it's one-to-one that's automatically true now I said I'd give you a little bit of warning about the notation let's do that right now so we can get that out of your system notation warning first of all in verse is read as I just spoke it F inverse there is no other way you should be reading that and let's be clear this minus one up here although it means inverse it also has another meeting you may recall but in this case F inverse is not the same thing as one over F not true okay do not use that when an F inverse when the inverse is used the minus one is used in the context of a function it means the inverse function it does not mean 1 over F so be a little careful with that [Music] okay well let's look at a couple of functions that are inverses of each other and let's just verify just to get into practice here verify to be inverses here we go suppose we take the function G of X is X cubed and take the function G inverse of X of course I'm claiming it's G inverse of X I think you'll believe me though it's the cube root of x okay and to verify that these are inverses what must I do I must check that if I apply G and then G inverse I end up with X if I apply G inverse and then G I end up with X in either case so let's start with the first one G inverse of G of X now I'm doing this to remind you of all of those composition exercises we did long ago back in unit 2 G inverse then what is G of X just copy down G of X is X cubed and now as you remember this is where you have to think about what this means so it was wise to rewrite G inverse here as G inverse a box is the cube root of box and now I want to do G inverse of X cubed so I'll place the X cubed into the boxes and I'll have the cube root of x cubed and as we know that's X so it does work one way and this helped us here going the other way G of G inverse of X well G of G inverse of X we just copy is the cube root of x and then I want to apply G again it's probably wise to think of this as G of box equals box cubed and then I'll take that idea down here the box is going to be filled by cube root of x so this will be cube root of x cubed and of course that also is X so I now have verified that no matter which way I apply these functions they undo each other and therefore they are inverses of one another well I think you should do an example like this just to get your algebra and Comp addition juices flowing here is the function verified to be inverses these two functions f of X is X minus 5 over 2x plus 3 and G of X is 3x plus 5 over 1 minus 2x and if you remember what I just did on the problem I showed you I use the boxes to help me tease apart what I needed to do at each stage I would recommend that you do that here I'll come back and show you what I did alright let's see what I did on this problem now you see what I already did I I knew that the boxes ideas would help me so I wrote reroute both of the functions with boxes in them now I'll go ahead and see if they are inverses of one another first I'll try F of G of X in that order F of G of X of course is this with X's in the place of the boxes so that's F of 3x plus 5 over 1 minus 2x now that's the easy step as usual the harder step is the second step where I need to put this object here in for the boxes from F a box there'll be a box there in a box there so I have to be a little careful about that 3x plus 5 over 1 minus 2x minus 5 and that's all over the much larger bottom which is 2 times 3x plus 5 over 1 minus 2x minus 3 okay and actually let me see two at box plus 3 I'm sorry that's got the right function there the original function I just checked is a plus so let's put a plus in there make sure that's right okay this is a very large fraction I have as I see fractions on the top and the bottom with the same denominator one slick way to get that denominator out of there is to multiply top and bottom by 1 minus 2 X so that will knock out this denominator and this denominator of course they will multiply by 5 and 3 but at least I won't have any more double fractions so what do I get out of this 3x plus 5 minus 5 times 1 minus 2x all over 2 times 3x plus 5 plus 3 times 1 minus 2x just multiplying everything out and with a little bit of work here we see that we've got 3x plus 5 minus 5 plus 10 X multiplying out everything on the top and 6x plus 10 plus 3 minus 6 X on the bottom well what happens here on the top 5 and minus 5 - 0 3 X + 10 X ad - 13 X on the bottom the 6x and the minus 6x add to 0 10 + 3 add to 13 13 X / 13 is X and so in one direction at any rate F of G of x equals x so it looks like these might be inverses but you need to check both directions you cannot just check one so let's check the other direction G of f of X now recalling the f of X from the previous page that'll be G of X minus 5 over 2 X plus 3 okay don't want to lose my plus this time equals and now I remember what G looked like G look like this so I want to put in my expression into three box plus 5 over 1 minus 2 box it's easier if you have this all in one page 3 times X minus 5 over 2 X plus 3 plus 5 all over 1 minus 2 times X minus 5 over 2x plus 3 and again I use the same idea to simplify I'll find out the common denominator here is 2x plus 3 multiply top and bottom by that that will now yield if I multiply everything out on the top 3 times X minus 5 plus 5 times 2x plus 3 and on the bottom I'll have remember the one has to multiply by the 2x plus 3 so I get 2x plus 3 minus 2 times X minus 5 while multiplying everything top and bottom I have 3x minus 15 plus 10 X plus 15 on the bottom 2x plus 3 minus 2x plus 10 and once again things simplify nicely the minus 15 and the 15 add to 0 the 3x and 10x give me 13 X once again 2x and minus 2x add to 0 3 and 10 add to 13 and lo and behold I have X again since I got X in both directions G f of X and the previous F of G of X the two functions are in fact inverses of one another now we won't be spending a lot of time doing this I wanted to do this so that you would get a reminder of some of the algebra we did earlier in the course which will be helpful and how to deal with compositions which will also be helpful okay there's one last thing we need to talk about before we leave the topic of inverses and that is graphing F and F inverse okay we need to learn how to graph these two and here is the basic idea if you start with a and you apply F to a we've seen this picture before you get F of a then if you apply F inverse to the F of a and go backwards you end up with F inverse of F of a of course and that is a if these are inverses so another way to say that is if a F of a that point is on the graph of F so that'd be the point a F of a would be on the graph of F then the reverse F of a a is on the graph of F inverse because if I go backwards F of a becomes the domain element and a becomes the range element so that makes it very easy to produce a graph let me show you how this looks in general with a nice picture I hope this works out well here is my y-axis and here is my x-axis and let me mark in here and you'll see why in a minute this y equals x 45-degree line okay I've got that marked on there now let's mark a few other points let me imagine that a is here on the x-axis and then when it goes up to the curve wherever the curve is this is the height F of a okay so that a goes up there and this is a point on the curve okay now that is then the point let me label it here this is the point a F of a alright from what I just said on the previous page if a f of a is on the curve y equals f of X then the inverse curve has the point F of a a so let me come down here and Mark a point let's see right about there that would be F of a and when we go up to the curve it will go to a height of a so it will have this kind of picture and this will be the point F of a a which lies on the inverse function curve so there are the two points and this is the y equals x line in between and if I continue the dots out here you'll see that they intersect this point here has height F of a it also has x value f of a so it is on the y equals x line because remember the points on this line are all of the form XX same thing here this is the point a a so we've got this little square here and the opposite ends of the square are the points that lie opposite each other from the original curve to the to the inverse curve in fact they lie on this diagonal and they are exactly at right angles to the y equals x line so this is an observation that will allow us to draw the inverse curve if we know the original curve the original curve looks like that we're going to know that the opposite curve the inverse curve looks like that because it's a reflection there's a reflection here across instead of the X or Y axis across the y equals x line now that's very different from what we've done before we're usually reflected across the x or y axis now we reflect across this line to get the difference between an original functions graph and the inverse functions graph so let me put that down in the form of a theorem and clean this picture up a bit now that I've seen the analysis I could just write down the graphs here's the theorem the graph of or the graphs of F and it's inverse F inverse R and here is the fact about them they are symmetric across the ye quux line so that is the connection between the graph of the original function and the graph of its inverse which means if you know the original function and you know its graph you can immediately draw the graph of the inverse without plotting any points so here's the idea once again that if you have your x and y axes and this is the y equals x line there's the x and y axes and this is your original curve say something like this if this is y equals f of X then I immediately know that the graph of the inverse function looks like that y equals F inverse of X and I did that by recently reflecting across this line so that a point a B here becomes directly across at right angles to this 45-degree line becomes the point B a on the inverse functions graph so this is it a reflection across the Y equals X line is how you get the graph of the inverse function very nice very straightforward well with that we finish our information about inverse functions in general and now we're going to go on ahead and look at a particular one the logarithmic function now it's time to look at specific inverse functions and since the title of the table says logarithmic functions you might expect that that's we're going to look at and so it is logarithmic functions and their graphs so let's go ahead and get right to that recall and this will be the connection to exponential functions recall that since f of X equals a to the X remember the exponential function we looked at and of course I'll just remind you that a is greater than zero and a is not equal to one usual fact about the base since that function is one-to-one as we discovered earlier it has an inverse so an inverse exists for this function and we're going to give it a name and here we go definition the logarithmic function the real autorhythmic function and we'll put in parentheses here to the base a you'll see where the a comes in the logarithmic function to the base a in standard form is the following y equals log subscript base a of X I'll talk about the notation a little bit in a bit in in a moment but log it's a subscript down here a and that's where the base goes and the object you're operating on the domain element goes here the parentheses are not always necessary as you'll see if there's a single object there it's often left out but I think you should use it most of the time and what is I mean that's just the name of it what is it defined to be it is defined to be the inverse of f of X equals a to the X the exponential function with that base so this a down here is what we're calling the base here just as we called the a for a to the X the base for the exponential function and so we've given the name log base a of X to a to the X's inverse function we're not using that inverse notation we solve for inverse functions we have this special name log base a so we'll have to get used to how to play with this and manipulate it so to put things down we have the following statement that y equals log base a of x that holds if and only if x equals a to the Y in other words they undo each other and another way to put that is the following that a to the log base a of X is simply X now see this is unusual this is not the F and the F inverse sitting next to each other what we seem to have here is a risen to the power log base a of X so we have an exponentiation going up because of the peculiar nature of the exponential function that it writes things upward but when you apply these two functions in order you get X and if you do it the other way if you do log base a of a to the X you also get X now I had boxed this in on the page with inverses as the key point here we need to specify more about what these X's are allowed to be the X's here start out in log X and then get hit by the a so these are 4 X's that are in the domain of the log function log base a of X but the domain of the log function is the range of the a to the X function why is that because the log in the a to the X are inverses the domain of 1 is the range of the other but we know what the range of the eight of the X function is because we've examined it and that's zero to infinity a to the X is never zero and always positive log base a of a to the x equals x in this case the X can be anything that is allowed in a to the X so this is for X in the domain of a to the X but the domain of a to the X is everything all real numbers since the domain of a function equals the range of its inverse that means this is now the range of log base a of X so by putting together the facts we knew about inverses together with the two functions here a to the X and log base a of X we've got all the basic information but I know that this is a little bit odd sometimes to get used to so let me draw you a couple more pictures that I hope will clarify a little bit of this these are pictures similar to what I did before suppose I start with a value X and I apply the function log base a of box to that value X and I end up of course with log base a of X and see there's a place where I might not write the parentheses around X because there's only a single thing there then if I go backwards and I apply the function a to the X or a to the box the exponential function I end up with back here a to the log base a of X follow this around X becomes log base a of X and then is exponentiated and of course since they're inverses those are equal so that's the process that's happening here in one direction now we can do this in another direction suppose we start with X over here and first we apply a to the box and end up with a to the X so that's our first direction and now we continue apply backwards log base a box and end up with log base a of a to the X which of course is equal to X because these are inverses so there you have it written in both directions the inverse functions undo each other and this is sort of the picture you want to have in mind of how they do that X first let's look at finding domains of the log functions that's worth a little bit of effort here find the domain or domains I'll do a couple of problems here the first ones straightforward suppose f of X is equal to log base 2 of 1 minus X okay I would like to know what X's are allowed in there well I know that we want 1 minus X to be greater than 0 because if you recall from the previous page the domain of the log function is always greater than 0 and so of course that means that 1 is greater than X and so that can be written as X is in the interval from minus 1 to 1 this is X less than 1 after all and if you like the little graphical picture here's one and we're going everything to the left not including 1 okay that one was fairly straightforward these can be a little bit trickier say for example with G of X the algebra just can be Messier the idea is still simple log base 5 of say 1 plus X over 1 minus X we're just to make things easy on ourselves we'll assume that X is not equal to 1 so there's no division by zero here so what do I want here again the log only operates on things that are positive so I want X 1 plus X over 1 minus X to be greater than 0 that's what I desire well if you remember this is an inequality with a rational expression in it and we did these when we did inequalities by looking for split points and here the split points are easy to find remember how we got split points they were where the top or the bottom would go to 0 here that's easy that's X equal minus 1 or 1 and even though the bottom point is not allowed it might split the real number line in two places where this is positive and negative so we need to consider it and then we can do our little chart if you recall I'll make this short and sweet minus 1 to 1 and we'll pick an easy number on each side maybe minus two zero here and two and I'm going to put these numbers directly in the entire expression so if I put minus two in here 1 plus x over 1 minus X with a minus 2 in it that's a negative on the top and a positive on the bottom so it'll be negative over all the expression here will be I'll just write box to save me writing will be equal to 0 here at minus 1 it's at 1 however it's undefined so this is undefined because that's where we would have division by 0 and in here we'll do the same kind of operation what is 1 plus x over 1 minus X here putting 0 in for X I get 1 so that makes this positive and 1 plus X over 1 minus X put a 2 in I get a 3 over a negative so this is negative so it looks like the region I want is the region inside here not including minus 1 or 1 because that would be equality and inside where this is greater than 0 so putting this all together that means X is in the interval from minus 1 to 1 and that is the domain of this function again pictorially that's minus 1 and this is 1 it's open they're open there and everything in between so there's just a little bit of practice with domains and here now let me do a little practice with the use of the log and the exponential as inverses of one another this is kind of stuff you want to practice on your own suppose you have y equals log base 3 of x and suppose you want to solve for X now we're going to spend time later looking at how to solve logarithmic equations but the basic idea here is not anything fancy it is simply you want to undo the log well to undo the log you use its inverse function the exponent to the same base this is base 3 so I will now we might say exponentiate where I'm going to raise both sides up so three up to the y equals three up to the log base three of X so I've raised both sides up by exponentiating with the same base now by design the right hand side I have exponentiated and log to the same base so I get back X so I have 3 to the y equals x and I'm done because that's all I wanted to do here okay let's try another one that's the opposite of this suppose I have one point two to the 3 power equals M and I would like to somehow rewrite this so that I don't have an exponent involved well I can bring this exponent down this is an exponential function by changing the base or rather by using the log function which has the same base as this so what I like to say is hit it with a log hit it with a log so it will be log base 1.2 of 1.2 to the 3 power equals log base 1.2 of M and then this left-hand side by design I'm logging and exponentiating will give me 3 and on the right I just have what I had before log of 1.2 and I'll just write M because there's only a single object there and there you go now that was just a little bit of practice these are the kind of operations you want to do in in larger numbers so that you feel confident when these things occur in equations later here's another thing that will occur later when we do equations it's the result of this fact if you have two Exponential's a to the U equal a to the V same base different powers and remember of course I'm talking about a greater than zero a not equal to one as usual then the conclusion is that if they have the same base then the powers must also be the same I won't prove this but it's a fact that I think you believe and then I can apply it to an example here where I can ask the question find exactly the value of log base 3 of 1/3 now there are several ways to do this I want to show you one way just to get this pattern down if you know enough about logs and powers you can write this down pretty quickly without any difficulty but let me just show you how you might do this in general what you first want to do is set y equal to the object you're looking for say y equal log 3 of 1/3 and then what I want to do again is get rid of the log so I'm going to because the log is in the way of me seeing what this is exactly so I'll get rid of the log and hope that I'll be better off I want to exponentiate then by what exponentiate by the exponential function with base 3 so this will now be 3 to the Y again and 3 to the log base 3 of 1/3 well by design the right hand side exponentiation and log with the same base leaves me with whatever the X is which is 1/3 of course 1/3 can be written as 3 to the minus 1 so what I have now is 3 to the y equals 3 to the minus 1 and by the fact above that means that Y is equal to minus 1 and if you remember Y was the original expression so I now know the original expression is equal to minus 1 and I've done what I set out to do okay good there's a little bit of practice with logs now let's look at say the graph of the function graph of f of X equals log base a of X well you know I shouldn't have to work at all to do this because as I saw in the previous segment the graph of the inverse of a given function is simply the given functions graph reflected across the Y equals x line the 45-degree line so since I know the graph of a to the X I ought to be able to produce this graph without much effort so let me write that down all I have to do is reflect the graph of a to the X across the y equals x line and that's exactly what I've done here in this drawing that I want to show you here's the drawing and I had to do this in two different cases depending on whether a is less than greater than 0 and less than 1 or greater than 1 so let's look at this case first suppose a is in this interval from 0 to 1 well if you remember the graph of the exponential function looked like a graph that came up from below and came down like this if you can imagine that being flipped across this line the part that was low down here becomes high up here the part that was high up here has been flipped over down here so this is the graph of the log base a of X when the a is between 0 and 1 and you can see it has all the usual features 1 0 is on the graph a 1 is on the graph because a here goes up to 1 and it is a decreasing function it's going down to infinity it is going up to infinity in that direction and clearly the y-axis is a vertical asymptote the x equals 0 line and the point a 1 of course that's information we may not have had we now see that the log of 1 from this the log of 1 must be 0 we also see that the log of a must be 1 so we've gotten a couple of pieces of facts a couple of pieces of information from this that comes from simply looking at this as the reflection of a to the X's graph let's look at the other one the case where a is greater than one here's the graph I've drawn for this in this case a is greater than one so the original graph looked it started low down here passed through one and went on to infinity now reflecting it across the y equals x line we end up with a graph that looks something like this this is an increasing graph goes off to infinity here goes down to minus infinity here passes through one as the other one did so that the point 1 0 is on the graph and then the point a 1 is also on the graph and of course the y-axis once again has become a vertical asymptote because for the original exponential function the x axis was a horizontal asymptote and now they have switched places upon reflection around y equals x so this is the graph of the log base a where the graph where the a is greater than 1 ok well let's put all this together into a summary about logarithmic functions now here is the summary I have for you and this is very reminiscent of the one I had for exponential functions it should remind you of it logarithmic functions base a here's the way it looks log base a of X that a is a subscript the A's will come either from zero to one or above one the domain is zero to infinity the domain of any log function is always only positive numbers the range is everything that was the up and down range x intercept is 1 0 instead of the 0 1 we had for exponential there are no y intercepts the vertical asymptote x equals 0 is now is the vertical line which is the y axis afters 1 2 1 because the original function was 1 to 1 the original a to the X function was 1 to 1 and it passes through a 1 for a between 0 and 1 the graph looks like this without all of the extraneous notes it is a decreasing function and if a is greater than one it's an increasing function it looks like this and both of these you see have y as a vertical asymptote so with all of this information it's time for you to try a problem and here's one I will pose to you I want you to graph this problem graph the function f of X equals minus log base 2 of X plus 1 now I don't think you'll have any difficulty with it but I'll come back in a minute and show you what I did alright here's the function that I wanted you to graph should be fairly straightforward we'll start out and I'm going to do this by hand because it is easy to look at we'll start off with what might be called the core function here log base 2 of X ignoring the various the shift here and this reflection just take log base 2 of X now since the 2 is bigger than 1 I know what the picture of this has to look like the picture of this has to look like the following here is my axes now remember if this were the exponential the exponential would go up like this I'll just dot it in because I don't want that now that would be the exponential if this were log if this would 8 this were 2 to the X okay what we want is what happens when you reflect it around so the reflection gives me a picture something like this passing through 1 so that's a way to remember this in case you you've forgotten what it might look like ok let's go over to the right and the next thing we want to look at is log base 2 of X plus 1 now what is that that's a horizontal you remember leftward by 1 so this picture will shift leftward so I will have this and the 1 is struck here at 1 it will now strike at the origin so it will have the line minus 1 here as its asymptote and it will do something like this striking here at 0 now so that is the picture that I have for log base 2 of X plus 1 and what's left the negative which we know is going to be I'll put that down here this is minus log base 2 of X plus 1 and the minus will do what well it's taking the Y value and negating it this is the final function here so it's going to reflect this around the x axis so my picture will be something like this there's my Y and my x axis and this the reflects around the x axis will not change this asymptote so it'll still stay there at minus 1 but now instead of coming from above and going to the right like that it'll now have this shape more or less and that will be the graph that we were looking for in this function so I hope you ended up with that yourself that should be roughly the same kind of picture that you've got on your screen and this passes again through zero okay I think that's all I'm going to say about that one let me just for the record show you one more you can see another graph here suppose we look at G of x equals log one-third base of minus X plus one solution by hand will be the following the first picture will be log base one-third of X seems a natural thing to do and again here is how you might remember the picture in case you've forgotten it the picture the original exponential function would look something like that 1/3 to the X would look something like that we're now flipping that across the y equals x line so now we have a graph that looks like this asymptotic to the y-axis and coming down and going to infinity passing here at 1 okay let's see what we can do next we probably want to look at log base 1/3 of minus X next now since the minus is inside we're talking about a reflection across the y-axis this time so the picture will look pretty easy to draw just like that hitting here at minus 1 now there's Y and there's X and we've flipped it across the y-axis finally we have the plus 1 and we know what that will do that will simply shift everything up vertically so let me move that down here and we'll have log base 1 3 minus X plus 1 that's the function G of X that's what we were trying to get to and as I said it will be simply shifted upward so see if I can draw that this had passed - one so now it's going to pass here at this point which is right above minus one but at a height of one and it will still be asymptotic so it looks something like that passing through this point here which will be what minus one one will be that point so I've given you a little bit a couple of landmarks and that gives you a rough idea of what this graph will look like in the end so on that will stop and then we'll come back and talk about a special base the X the base e which gives us as you might expect the natural logarithmic function all right now we'll look at the analog to the natural exponential function the natural logarithmic function its inverse as you might expect so here we go definition get right to it the natural logarithmic function the natural logarithmic function is the function we denote by f of X equals log base e of X where the base is of course the number e that we examined now that would be what you expect me to write however no one writes the log base e function this way except to make the point that it's the inverse of e to the X however it is always denoted and this is the way I will use it from now on and so should you f of X equals log of X that way where I'm using an L and an N and where did the Nate what does this come from well n stands for log L stands for log and n stands for natural as you might expect so this is the notation I want you to use from now on for the natural logarithmic function do not use log base e of X that I just wrote that here to make the point that it is the inverse of the e to the X function but this is the notation you'll use from now on so as I said since these are inverses so we know that the natural log of e to the X will be equal to X and e to the natural log of X will be equal to X also this is the same pattern we've seen before except that now it's with these very important functions the natural log and the natural exponential functions now what do these hold for well the first one eetu the X the X can be anything so this holds for all X in the real numbers and that is the domain of e to the X and here it holds for all X's that are allowed in the log because the log is where the X's begin this is for X in the interval from 0 to infinity only positive numbers and that's the domain of course of the natural log of X so what we're seeing here are the facts about inverses that we've seen before so there's nothing new here I might make a couple of remarks one is use the Ln X whenever possible and really that goes for e to the X all so everything you do with logs and exponential functions can be turned into these two functions e to the X or natural log of X and that makes your work of memorizing things much easier if you know the trick to change everything into natural log or natural exponential the second remark I wanted to make here is about your calculator keys on your calculator there will be keys that look like this Ln key there will also be an L og key now be very careful that you consider that these are two different keys for two different reasons for for given reason natural log has base e l og is considered base ten the first is called natural the second is called common logarithms now we're going to spend some time talking about these in a little bit but I wanted to make sure that you're aware on your calculator to always use the Ln key when you want base E and L og will be base ten and for any other base we'll find out how to deal with that in a moment so I'll come back [Music] [Applause] [Music] [Applause] [Music]

Channel: UMKC

Views: 70,272

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Keywords: UMKCVSI, College, Algebra, Lecture29

Id: VuL890FP6iQ

Length: 54min 2sec (3242 seconds)

Published: Tue May 05 2009