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College Algebra: Solving Exponential Equations

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About this Lesson

  • Type: Video Tutorial
  • Length: 12:01
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 129 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra Review (30 lessons, $59.40)
Algebra: Exponential and Logarithmic Functions (36 lessons, $49.50)
College Algebra: Exponential & Log Equations (3 lessons, $5.94)

In this lesson, you will learn to solve equations in which the unknown is an exponent. To do so, you will use logarithms. In simple problems, the exponential equation can be solved by simply converting both sides of the equation in such a way that the bases of each side match (e.g. with (3/4)^x = 16/9). When equations are more complex (like 3^(x-2) = 4^(2x+1)), Professor Burger will show you how to manipulate the equations using logs and the rules of logarithmic equations, which will allow you to change the exponents to coefficients of logarithmic expressions.

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 http://www.thinkwell.com/student/product/collegealgebra. 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

Thinkwell
Thinkwell
2174 lessons
Joined:
11/13/2008

Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

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Recent Reviews

Nachan_homepage
great lesson!
01/15/2009
~ nachan

This is a great introductory lesson for solving exponential equations. He shows you how to flip fractions to combine exponents, he also shows you how to find the value of x in more difficult problems using logarithms. Very useful when learning logs and exponents.

Nachan_homepage
great lesson!
01/15/2009
~ nachan

This is a great introductory lesson for solving exponential equations. He shows you how to flip fractions to combine exponents, he also shows you how to find the value of x in more difficult problems using logarithms. Very useful when learning logs and exponents.

So, let’s ask ourselves how can we solve equations where the exponents actually are the unknowns. So, let’s start figuring out how to solve exponential equations. We saw some simple cases of this a while back, where the idea was if you could get the same base then you can just set the exponents equal to each other. But how do you do that when things get a little bit ugly? Well, then the logarithm actually comes to our rescue.
So, let’s start off with a very simple example, just to get us sort of in the right mood. This would b (¾)x and suppose that equals 16/9. Now, what I want to do is figure out what value of x would make this thing actually hold. Well, there’s a variety of ways of thinking about it. Let me actually share a couple with you. One is to say, “Okay, I want to try to get this thing to be the same base as this.” You could see it’s looking actually not bad. I can sort of make this thing without too much effort into a 4/3 by just squaring. That was pretty easy. It looks so close to this, I actually want a ¾. Well, how do I make that a ¾? I just take 4/3 and flip it twice. 4/3, ¾, 4/3. So, flipping it twice means flip it here and then put a negative sign up there.
I flipped it here, but then a negative flips it again. So, it just comes right back to 4/3, it doesn’t change the value. Now, I’m in great shape, because since the bases are the same, that means the exponents must be the same. So, x must equal -2. Pretty easy. So, when I can actually massage one side to match up perfectly with the base of another, I’m in hog heaven. But consider the following more exotic example.
Suppose I take 3x-2 and I want that to equal 42x+1 and I want to find out the value for x, which actually makes this true. Well, there’s no way for me to take the 3 and make it into a 4 to a power or a 4 and make it into a 3 power in some sort of easy straightforward way. So, what would I do? Well, here’s the key idea. The key idea is to go back to these formulas and in particular this one. Let’s think about this formula. This says, if I have log of something to a power that power can be rewritten as just a coefficient. I can bring a power out in front.
Now, if you notice, with an exponential equation like this the unfortunate thing, the thing that I hate is that the unknowns are in the exponent. I want unknowns to be on ground level, so I can do all the arithmetic that I already know about arithmetic to solve, but the problem is I don’t know much arithmetic about how to deal with things when they’re in the exponent. So, I want to bring them down. I want to get down. So, how do you get down? Well, if I took the logarithm of both sides then I could use that property to pull the things out in front. Then exponents become coefficients. So, whenever you want an exponent to become a coefficient, whenever you want an unknown that’s in the exponent to be able to be usable, just take logs of both sides.
If two things are equal, their logs will be equal. Now, you can take any log you want. You could take log7, you could take log3, log4, loge, or just log10. I’ll just take log10 because maybe at the end of the day, I might just want to use my calculator to actually figure out what the values are. So, let’s log both sides. If you log both sides, we have log(3x - 2) and that will equal log(42x + 1). Okay. Well, now can use this fantastic property of logarithms, which is crucial, that exponents can be written out in front as coefficients. That saves us, because now I can write this as (x - 2)log3, log3 is just some number, equals (2 + 1)log4. Notice these parentheses, by the way. It’s because this log has to hit both terms. This log has to hit both terms. I have to distribute. Don’t make the cardinal, standard, classic mistake of not distributing here.
If I do distribute, and remember log3 and log4, they’re just some numbers; I don’t know exactly what they are numerically. I could use a calculator and figure them out, not a problem, but they are numbers. So, if I distribute, I see (log3)x - 2log3 = (2log4)x + log4. Notice I put the little parenthesis around here. You know why? If I didn’t put the parenthesis around there, you may have thought that maybe--imagine this without parenthesis. In fact, I’ll just show it to you without parenthesis. If I write it without parenthesis, it would look like this and you may have thought maybe it was log of that whole thing. It’s not log of this whole thing. It’s just log of 4, but I multiply the whole thing by x. So, the x is way on the outside. The x is not part of the log. You see the x is way over here, times log4.
Okay. Well, now what can I do here? I want to solve for x. So, I’ll just bring all the x’s over to this side and if I do that, I see (log3)x and then I have a minus, I subtract this. So, I have (-2log4)x and what does that equal? Well, I’ll bring the constant stuff to this side. So, I’ll see a 2log3 + log4. So I want to solve this. I can actually simplify this quite a bit. Let’s take a look at how we can actually simplify this if we wanted to. In fact, you know what, this is going to be sort of an intense problem. I’m just going to take myself out of the picture, quite literally. Watch this. I’m going to it by my own hand. So, you how strong I am. Ready? I’m just going to squeeze myself out of the picture. Here go.
Okay, now that I’m out of the picture we can get to work, finally get something done. Here we go. So, the first thing I do is notice I can factor out the x here. So, if I do that, I’ve got a log3, but actually I’m going to do a little simplification at the same time. Because notice this 2 can be brought upstairs to this exponent on the 4 and I could write that as minus log of 4 squared, which is 16x. So, I factored out the x and also did that step at the same time. Now watch this big step I'm going to do right here. The big step I’m going to do right here is first of all take this 2 and pull it up on top, so it becomes the log32 which is log9 and then I have a plus log4.
Now, I can actually combine a lot of these guys. I see the difference of logs. Remember that’s the log of the quotient. So, I could write this as (log(3/16))x and that will equal log of--and if I’m adding logs, I can multiply them. So, this is just a 9 times 4, which is 36. If I divide through by this thing, that’s a number, I get what x equals. So, x it turns out equals’ log36 divided by log(3/16). So, that’s the answer. What is the x that satisfies this? It turns out the answer is some exotic x. It’s that. What does that equal numerically? Well, you can just pop on a calculator and take a look at it if you want. Log36 divided by log(3/16) equals’ -2.1407 stuff. So, basically, the x that works here and satisfies this is equal to -2.1407 stuff. You can see it’s sort of a weird answer, but that’s the exact answer to make these all equal. Wow, that took a long time didn’t it? Lot of work, but not that hard. Just a lot of work.
Let’s try one last one here. Suppose I want to solve 22x - 3 = 5-x - 1. I want to find the value of x which makes this thing hold. So, what do I do? Well, what I do is, I see that there’s no way of making a 5 a 2 or 2 a 5 an easy way so I’ll just take logs of both sides. If I take logs of both sides--so, log(22x - 3) = log(5-x - 1). Then I can use laws of exponents to pull this out in front, pull that out in front and using the log stuff I see the quantity (2x - 3)log2 equals the quantity (-x - 1)log5. Well, now all the x’s are down below. They’re not in the exponents anymore. Of course there’s a price you pay for that. I now have logs in the picture, but that’s okay. They’re just logs of number, so what do I care about it.
I can distribute these logs and I’d see, (2log2)x - (3log2) equals, here I’d see a (-log5)x and here I’d see a -log5. What can I do now? Well, I can bring this stuff with x’s over to this side, so it becomes a plus. This 2 out in front, I can pull that up and make it a log22, so I’ll just write log4, because that’s 2². I bring this over, it becomes a plus log5 and that’s all times x. I factored out the x. If I bring this minus 3log2 over to this side, what does it look like? It becomes a plus 3log2, which is just a fancy way of saying log8. So, I take that 3 and put it up as an exponent, it’s 2³ and 2³ is 8. But I still have that -log5. Well, log4 + log5, remember the laws of logs, that’s just log of the products. So, that’s just log of 20x and that equals log of--and the difference of logs is just the log of the quotient. So, log8 over 5.
Great. So, now I can actually solve for x exactly. I’ll do that right here and I see that x equals log8 over 5 all divided by log20. Okay, you see all the little steps there. There are a whole bunch of little teeny steps, like bringing the 2 up here as an exponent, becomes a log4. Bringing the 3 up here as an exponent becomes a log8 and so on. But now there’s the answer and you can actually compute that away if you wanted to. Let’s see if we can do that really fast. Log8 divided by 5, and I take that answer and divide it by log20. So, numerically this equals, x = .15689 stuff. So, the x value that makes this whole turns out to be sort of an exotic number. It’s .15689 stuff.
The method was just to take logs and the logs have the feature that it turns these exponents into coefficients. Really powerful with a small price to pay, just some logs of numbers, which calculators can easily spit out to you. Try these. Have some fun with them.

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