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College Algebra: Change of Base Formula


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About this Lesson

  • Type: Video Tutorial
  • Length: 9:28
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 101 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: Evaluating Logarithms (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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So we know how to find the numerical value of logarithms if they're base 10, in which case on the computer or on the calculator you would just push the "log" key, that's all there is to it. Or, with a natural log, which appears a lot, you just push the natural log key, the ln key, and you've got that set up. So no problem there. But what if you want to actually numerically evaluate something like this? Like maybe log[7] 12? Well, there's no log[7]^ key on your computer--I guarantee it; or on your calculator. How about if you want log[b] ? No log[3] key on your calculator. What if you want log[3/2] 2? Well, there's no log[3/2] button on your calculator or computer as well. So, in fact, if you wanted to find these values, what would you do? Well, it turns out there's a really cool way of actually taking different bases like this and converting them, changing them, to a common base, for example, a log[10] or a natural log, log[e]. Now, I want to tell you the formula for that and then show you where that comes from. You can actually use the calculator to compute all these things once you know how to make the change of base.
So let me show you the change of base formula. I'm going to write it up first, and then I'm going to show you why it's actually true. So here's the formula. If you have log[b] x, and you want to change the base here to something else, let's say to base a, all you do is take log[a] x and divide it by log[a] b. That is the formula. So if you have a logarithm that's in terms of b as a base, but you want a logarithm that's in terms of a as a base, well, then this is a conversion. Because log[b] x is the exact same thing as log[a] x--that's the right base you want, and then you have to divide by log[a] b. Now, where does this come from? I hope you don't memorize this, I hope you just think about it. I want to show you how to think about it.
Let me take this thing right here, this side, the left hand side, and I'm going to call that y. So let's let y just represent that thing. My fantasy is to show you that y is actually also equal to this side. That would mean these two things are equal. Okay, well now how would you arrive at that fantasy? Well, all I'm going to do is figure out what this means by using a little mantra that I've been chanting for quite a while now--log is an exponent. So this is the exact same thing as saying that log is the exponent, so y is the exponent I have to raise b to in order to get x. So what that means is b^y = x. These two statements are identical. Okay, great. Now what do I do?
Now what I'll do is take the logarithm with respect to base a of both sides. If two numbers are equal, their logarithms, logging both sides is completely okay. So let's log both sides. So if I take the logarithm of both sides, that would mean that log[a] b^y would equal log[a] x. See, if two numbers are equal, their logarithms will be equal as well. So, in fact, I just took the log[a] of both sides. But now I use that great property of logarithms, that if I have a log of something to a power there, that can be pulled out in front. So if I do that, what do I see? I see ylog[a] b = log[a ]x, and if I solve this for y, I just divide both sides by log[a] b. It cancels here, and what do I see? I see that y equals log[a] x divided by log[a] b. But wait a minute. Remember that y equals log[b] x. So if y equals that and y equals that, these things must be equal. I just proved that formula to you. It wasn't that big of a deal at all. All I did was call this thing y, untangled it, took the log of both sides, and I got the formula. So this change of bases formula actually works, and now you can see why. So, coming back to this thing, not a problem.
How would we handle this? Well, let's try these and see. On the calculator you have an option. You can either use the log key or the natural log key; it doesn't make a difference. I'll do natural log for the first one just to show you that it doesn't really make a difference, because if it's natural log that I want, then all I know is that this is the exact same thing as natural log of 12 divided by natural log of 7. It's that easy. That's log[7 ]12, and so what does that equal? Well, you just open up the old calculator. Well, you don't open up the calculator. If you opened it up you'd see the backside. But if you turn it on and take the natural log of 17 [sic] and you divide that by the natural log of 7, and that equals 1.4559-stuff. And that makes sense, because what power do you have to raise 7 to to make it 12? Well, something a little bit bigger than 1, but certainly much less than 2, because 7^2 is 49. So this has got to be a little bit bigger than one.
Let's try this one. Log[3] . We could use natural log again. Natural log of divided by natural log of 3. It's real easy once you see the formula. Use the calculator. So we take natural log of and divide it by natural log of 3 and we get minus--oh-oh, is that okay--.6309-stuff. Is minus okay? Let's think about it for a second. Well, I'm asking what power do I have to raise 3 to in order to make it ? Well, to get it to flip I actually do need a negative sign, so that checks. And this has to be smaller than 1. I've got to be taking some sort of root of it to make a 3 a 2, so in fact, I take the .63 power and that does it. So that looks good.
Let's do one last one. And just for fun I'll do this one in logs. Log[2]over log[3/2]. You might want to check, by the way, and try doing some of these ones that I did before using the log key. You'll get the exact same answer. It doesn't make a difference. You'll get the exact same answer. So I'll use logs here now. Let's take the log of 2 and divide it by log 3 divided by 2, and that equals 1.7-something so it's 1.7095. And does that make sense? Well, let's think about that for a second. 3/2 have to be raised to some power in order to make it equal 2. So since 3/2 is less than 2, that power should be, in fact, bigger than 1. But if I were to square it that would be 9/4 and 9/4, if I were to square it--oh yes, that's a little bit bigger than 2. So, in fact, it has to be a little bit less than 2. Because 3/2 squared is 9/4 and 9/4, well, that's too big, because that's actually bigger than 2. So this has got to be a little bit less than 2.
Okay, now let's see. Everything okay here? Everything looks great except for one little typo I made. One little typo. I know it was flashing there all this time saying, "Can you find the typo, can you find the typo," and you thought, "gee, he didn't know there was a typo." I knew there was a typo. I knew. What was the typo? There was a typo right here. Instead of 17 this should have been 12. So there's a typo there. That 12 should have been 17. Now it does look like a 17, and it does look like a 12 if you do this. See, there's no typo. But, in fact, that was just a typo. That should be a 12 divided by 7. It probably changes the answer, by the way. Let me just quickly do that for you and report the actual news there. Actual retail value is--turns out that this is still bigger than 1; we've already argued that, but it turns out it's 1.2769 and it goes on. So it means 7 to that power gives me 12. Cool.
Anyway, if you try to remain typo-free and use this really neat formula, you can change any exotic base to either a natural log or a base 10 base without any sweat at all. Have some fun and try these on your own.
Exponential and Logarithmic Functions
Evaluating Logarithmic Functions
Using the Change of Base Formula Page [1 of 2]

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