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College Algebra: Match Log Function to its Graph


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About this Lesson

  • Type: Video Tutorial
  • Length: 8:49
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 95 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
Algebra: Exponential and Logarithmic Functions (36 lessons, $49.50)
College Algebra: Logarithmic Functions (4 lessons, $5.94)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Let's start looking at the graphs of some slightly more exotic log functions. Now here it's always sometimes a good idea to plot some points and in particular see if you can find the x intercepts. Those are the values for x which makes the function equal to 0. That will sort of help you solidify where exactly these log functions go. For the most part, nothing too surprising here except for the big finish at the end. We'll see if we can watch out for that one.
Okay. So, let's graph the first thing. We'll start off, just to get us on solid footing, f(x) = log[3]x. So, what does that look like? Well, we've got some axes here. Okay. So, what this looks like is, roughly speaking, this. Here's 1. At 1 we're going to have 0, because here's the x intercept. If I put in a 1 here, 3 to what power will give me 1? Well, that's 0, so that's the intercept. It's asymptotic to the y-axis, it gets closer and closer to the y-axis. I'll try to color code these things. So this reddish thing is actually this function. That's just to put us into place. For example, that means that at 3, this height must be the number 1. Right? Why? Because if I put in 3 here, 3 to what power makes 3? Well, 3^1. So at 3 we have 1 and then you can put these other points in here, 9 would be 2 and so forth.
Well armed with that, let's try another one. How about g(x), will be a green one, will be log[3], but not of x. Now, how about ^x/[3]? So what happens if I were to multiply x by ^1/[3]? That's going to somehow stretch this thing in some direction. Let's figure out how. First let's find the x intercept. So, what plays the role of the 1 here? Well, what value of x will make this thing equal to 1? Well, the answer is three. When I put a three in here, I'll see ^3/[3] which is 1 and 3 to what power will give me 1? 0. So, now over at 3, this point is going to be over here now. What happens to the other points? Well, let's think about this for a second. What about 9? ^9/[3] is going to give me a 3 here. What power do I have to raise 3 to in order to get 3? 1.
So, now I've got to go all the way over to 9 just to get to a height of 1. So, this height of 1 is going to happen all the way over at 9. It's going to be actually off the screen here. This is a very slow growing function, a very slow growing function and it looks like this. Very slow growing and also will be asymptotic to this. So, this will want to kiss this, but it won't be able to kiss it because the red is always going to be the chaperone in between. They're all going to squeeze together at the end. So anyway, that's what happens here. So when you multiply the x by a number smaller than 1, in this case ^1/[3], what you see is, the thing sort of moves a little bit and then sort of tones down a little bit, moves and tones down a little bit
What happens if we do sort of the reverse kind of process? What if I multiply by big number like h(x), which would equal log[3](3x). What's the effect there? Well, first let's find the x intercept. Where does this thing cross the x-axis? Well, wherever this number is 1. This whole number is 1. So, what value of x makes that 1? Well, ^1/[3]. So now, instead of being at 1, I've got to shift over to a ^1/[3], it's way over here. You can argue that what happens here is that this is going to be sort off a movement in this direction. So, I'm going to now over cut and so the brown function is actually going to look like this. It's going to come a little bit this way. Again, that little sharp kink there, really shouldn't be there. It's very smooth and gradual. Now, when the red wants to kiss the y-axis, we've got the brown as a chaperone. Everyone's going to head toward kissing, as close as you want to get, but all these people will come into the race in that order.
Now let's try to really stir things up a little bit and take a look at this one. How about, I'll call this s, for stir things up, s(x) that will equal log[3](x - 2). Now what's that going to look like? Well, actually, let me try to show you that on a different set of axes. Just not to get everything messed up with what we previously had. Let me just draw these in. I'm also going to put in the very first one, because I just want to remind you--compare everything to the standard thing. This is going to be at 1 and it looks like this. That's just the standard one, that's the original one again. That's just for reference.
Now, what would this one look like? Well, notice that's just exactly this, but the x has changed to x - 2. So that's going to be a shift. I'm just going to take this thing and physically move it, either right or left. Which way? Remember, add to x go west. So, if you add, you go this way. If you subtract you're going to go this way by 2 units. So I take this whole picture and originally move it. Quite literally. Watch. One, two, and that's the graph. So, what it would like is this point at 1 would now be at now 3 and the asymptote that use to be at x =0 will now be at x = 2. So, I have to move the asymptote over. Now this is the new asymptote and I literally just copy that picture exactly as it is, by shifting it over. So, that is the orange function and it's literally a shift. Each point was just gone over 2. Each point went one, two. This point one, two and so forth. So this kind of thing is just a shift.
Okay? So that stirs things up a little bit. Let's now close by stirring things up a lot and look at this one. So let's call this l for a lot, l(x), purple, is log[3](-x). Now what in the world is that going look like? Well, again, let me just start with a clean slate, just so that you can really see how this is going to fit in. So, here are my axes. This is the big finish. So, first I'll put back this person, our favorite. So, our favorite is just in red, asymptotic over here, goes up there. That's the log[3]x. Now what I want to do is graph log[3](-x). What does that mean? It means that where I'd plug in this value for x, to get that value I'd have to put a negative that. Right? For example, if I put a 1 into here, log[3](-1) doesn't exist. Because there's no power of 3 that makes it -1.
However, notice how I put in -1 for x, that would be -(-1), which would make a positive 1. So, if I replace x by -x, I just see this reflection over the y-axis. This point will now be the analogous point to this one and this point here, for example, at 3 I use to be at one. Right? If I put in 3, 3^1 = 3. But now, I'd have to put in -3 because -(-3) gives me three. So, in fact, I get the exact same picture, but mirror imaged right over and that is the graph of log[3](-x). So, if you look at the graph of log[3] of a negative x, it's just the same thing as the log based three of x, but flipped over that way. So, it's sort of a surprising thing, which you just can figure out by potting some points and seeing how it goes.
So a dramatic difference. Doesn't look at all like what it usually looks like. That's because I've modified it with a -x, that's a flip over the y-axis. We talked about flipping over x-axis and y-axis before and this is just a special case with the log function. Anyway, I hope you enjoyed these pictures, those pictures and all the other pictures of log functions and hope you give the following a shot and practice. The more you practice, by the way, the better you'll feel about this. Okay. See you soon.
Exponential and Logarithmic Functions
Solving Logarithmic Functions
Matching Logarithmic Functions with Their Graphs Page [2 of 2]

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