College Algebra: Solving Logarithmic Equations

in a tight spot

05/01/2009

~ thomas1

I had a university instructor who was great at algebra, but was a poor teacher. These videos really helped me out.

Logs aren't that difficult!

01/15/2009

~ nachan

Professor Burger teaches you an easy way to solve a problem when dealing with logarithmic equation. He teaches the importance of checking your answers and illustrates how to easily combine all logs to solve the equation. He also explains the meaning of the exponent in logs. Very clear and informative. Really great for simple log problems.

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When you want to solve an exponential equation, that is an equation where you have the unknowns in the exponents, standard theme is going to be to take the log of both sides and then use the fact that if you take log of something that has an exponent that exponent becomes a coefficient. Really a powerful idea that allows you to solve these things. What about if you’ve got an equation with logs in it? How do you solve an equation that has logs? Well, the way you do that is try to get all the logs together and then use properties of logarithms to write it as one big momma log. Once you’ve got that, untangle what the log says by remembering log is an exponent.
So, the theme here is first get a one big momma log on side, numbers on the other and then convert a log back to the exponential equivalent form and solve that. Let me try to explain this to you and illustrate it with an example. But, before I do the first example, I need to tell you a caution. This is important. I really want you to listen to me on this one. That is whenever you have a log equation you must always check your answers, because sometimes you may have--just like with square roots equations--you may have extraneous roots. So you’ve really got to be careful with the log, that you’re going to check your answers. I’ll illustrate that the first time and the other examples I’ll let you check the answers for yourself.
So, first up, let’s solve the following equation, log2x + log2(x - 3) = 2. Now, I want to solve that for x. You notice that’s an equation that has logs in it. So, what do I do? I’m going to try to make one big log. I notice that these are two logs with the same base, happily, and they’re being added. So, I can use the properties of logs that says that if you have the sum of logs, that’s the same thing as the log of the product. So, this is identical to log2(x(x - 3)) = 2, because this side is identical to this. I just used one of the fundamental properties of logarithms. Log of a product, sum of the logs. Okay, well now I’ve got a big log here, so I just untangle by remembering my little mantra, log is the exponent. So, 2 is the exponent I have to raise 2 to in order to get this. So, what that says is that x(x - 3) = 2 raised to the exponent power. Well that’s just 4.
So, I’ve taken this log equation that looks really pretty daunting and converted it to this little teeny--gosh, look at that equation. Look how easy that is and it came from there. Piece of cake. All I do here is bring everything over, it’s quadratic, and I solve. Can it factor? Let’s see, (x - 4) (x + 1). Sure can. Great. So, either x = 4 or x = -1. So, I have my two answers. Neat. You see how I took that log, combined it into one big log and then just untangled the log stuff, got the equation solved. But, you must always remember to check your answers. Let’s see what happens. So, you go back to the original thing. Not this, not this, not this, not this. The original thing. Let’s see what happens when we plug in.
If I put in a 4 in here--so let’s do a little check, checks and balances, right here. If I put a 4 in here, I see log24 and then I add to it log2(4 - 3). Well, 4 - 3 is 1. Now what does that equal? I hope it equals 2. Well, log24, this is the exponent I have to raise 2 to in order to get 4. Well, 2² is 4. So, this number equals 2. What is log2 1? That’s a standard one. What power do I have to raise 2 to in order to get 1? Well, 20, so it’s 0 and that equals 2 and that equals 2 and it checks. Great.
What about this equation of -1? Well, here I have log2(-1) + log2(-4). What’s log2(-1)? Let’s see, it’s the exponent I have to raise 2 to in order to get -1. But wait a minute, 2 to any power at all will always be positive. We can’t take logs of negatives. So, in fact, this is impossible. Not only does it not equal 2, it doesn’t equal anything. So, this is an extraneous root. You see how important it is to check your answers? This only has one answer, namely x = 4. Just like a square root, you can’t take logs of negatives. So, it’s real important to check your answers.
Okay. Let’s try some other questions. Let’s solve this equation, log(5x - 1) = 2 + log(x - 2). Remember, if there’s nothing underneath the log, if it’s a naked log, that means an invisible base 10. Well, what do you do here? Well, again, you always try to bring all the logs together and write it as one big momma log. So, I’m going to take this term and subtract it. I’m going to subtract log(x - 2) to this side. So, I’ll see still log(5x - 1) - log(x - 2) and that equals 10. I just took that term, subtracted it over to here. But now I’ve got the logs together, I’ll see if I can use properties of logs to combine it into one big log.
Well, notice that if I have the difference of logs, if I have the difference of logs from one of the properties, we see, we know that that is the log of a quotient. So, this is just log(5x - 1) divided by x - 2 and that equals 2.
So, now I’ve got one big log equals 2. So what can I do? I can actually now untangle the log thing by writing in terms of exponents. Now, remember there’s an invisible 10 there. Of course now it’s visible. So, you have to remember what that base is though. That means log is the exponent, 2 is the exponent I raise 10 to in order to get this. That’s saying 5x - 1 over x - 2 is equal to 10², which is 100. So, I took this really complicated equation with logs and I converted it to this little thing. Well, what I’d do? Well, I’ve multiplied both sides by x- 2. If I multiply this side by x - 2, I can cancel the bottom and then the other side by x - 2 and I’ve got this.
Now I’ve got a linear equation. This is really easy to solve. I see 5x - 1 = 100x - 200. I can solve by bringing this 5x over to this side. That becomes a 95x when I subtract. I could take this -200, bring it to this side and that becomes a positive 200 minus that 1, gives me a 199. Then I see x = 199/95. Neat. Now, that’s the answer we got, but it’s not going to be a checked answer until we plug it back in to here and make sure that this is really satisfied. Now, I’m not going to check this answer, but you are responsible for checking that. Plugging back in and making sure that these two things are equal. You can use a calculator if you want or whatever. I turns out this will check, but I’m going to let you do that. This problem is not done. This is incomplete until you check it.
Okay. Let’s try one last one. ln x = ½ ln(2x + 5/2) + ½ ln 2. This is a big one. So, let’s think about how exactly we sort of pull this of. The first thing I’ll do is try to combine this side into one logarithm. So how would I do that? Well, let’s see, logx--by the way, there’s not like only one way of tackling this problem. If you would tackle it some other way, that’s fine too. Just make sure you’re tackling correctly. I could take these ½’s and pull them on top as exponents. Coefficients can become exponents if there’s a log between them. So, I see, ln(2x + 5/2)½ + ln2½. Well, now I see the sum of logs, so I can know that’s actually the log of the product. So, that’s actually the log of the product. So that’s actually ln((2x + 5/2)(2))½ and that equals this.
Well, now I could actually do what? I could--there’s a variety things I guess I could do. I guess I could subtract this log to this side, if I wanted to. I'm going to do this in one step right now--minus ln x, but then it becomes 0 on this side. So, if I subtract both sides, I just get a 0 here. Then what do I notice? This is now the difference of two logs, so I can write it as the log of the quotient. So, what I see here is 0 equals the log of a big quotient. Where on the denominator I have the x, on the numerator I have this, but by the way, what’s ½ exponent mean? It means square root. So, I have square root of and if you distribute this I would see 4x + 5. So, I’m moving pretty fast here. We’ve got ln x = ½ln of that stuff plus½ln2. I combine these things into one log and then I brought this lnx over to this. It became now a log of quotient because the difference of natural logs is the natural of a quotient and this thing to the ½ power is the same thing as taking a square root. I just distributed the 2 everywhere and I got 4x + 5.
Well, there’s still that pesky log in there, but remember that's loge, so this is actually loge here and I can actually uncover this by figuring out what this says. Log is the exponent. So, 0 is the exponent I have to raise e to in order to get this. What’s e0? What’s anything to the 0? It’s 1. So, in fact, this thing in here, equals 1. So, I can actually take this awful mess that we have and convert it to the amazingly simple 4x + 5 divided by x equals 1. So, I went from this horrendous thing with three logs into one little thing like this. Multiply through by the x, I see square root 4x + 5 = x. I square both sides to solve--remember when given radicals, a good technique is to square both sides, and I’ve got a quadratic to solve.
So, that’s going to be x² - 4x -5 = 0. Let’s see if that can be factored. I have x and x, the opposite signs. How about 5 and 1? Looks great to me. So, I see x = 5 and x = -1. So, I have two answers again, x = 5 and x= -1. But, you’re not done until you check and you haven’t checked until you’re done. It goes both ways. Amazing. So, here’s the original question again. The question is--well, is this a solution. Let’s try it. Well, you could put I an x = 5 in here and plug away and make sure that in fact these two sides are going to be equal. I’ll let you try that. What about when you let x be -1? Well, I run into a major problem right here, because I can’t take the natural log of a negative. So, in fact, this is an extraneous root, right there. So, that one’s not going to be good. I’ll let you try and verify that x = 5 is a good answer. But remember, with a log, it’s just like a square root, you ain’t finished until you’ve checked. See you up next.