Common Core Algebra II.Unit 4.Lesson 11.Solving Exponential Equations Using Logarithms

The real key in everything we do today is going to be the third logarithm law. So recall that if we have log of log base B of a to the X, we can literally bring that X out. We can take that exponent and bring it out and form a product with it. And this is pretty critical because by doing that, we take the variable that we're trying to solve for, which used to be in the exponent, and now we just bring it on down, so it's just part of a product instead. And believe it or not, that'll make all the difference in the world in terms of solving equations. Let's just clear that out. So let's kind of see how this is done. All right, number one, we've got saw four to the X equals 8 using a, the method of common bases, and B, the logarithm law, shown above. Well, or at least on the previous slide. So pause the video now. See if you can remember how to solve this with the method of common basis. All right, let's go through it. I'm going to rewrite the problem down really quickly. Four to the X equals 8. Remember how the method of common basis works? I look at the four. I look at the 8. And I try to think of something that I can use as a common base. 

Well, four is equal to two squared. 8 is equal to two cubed. So I'm going to use two as my common base. So on the left hand side, I'll have two squared raised to the X and on the right hand side, I'll just have two to the third. Remember then I can use the exponent law that says that I will multiply those two exponents and get two to the two X equals two to the three. And as soon as we've got those common basis we can now set the exponent SQL. Two times X is equal to three. And X is equal to three halves. There we go. Right? The method of common basis. But let's take a look at how we can use logarithms. Now, of course, one of the golden rules of algebra. Is you can pretty much pretty much. Do anything you want to both sides of the equation as long as you do it to both sides of the equation. So what I'm going to do is I'm going to take the log of both sides of the equation. Now, what should I make my base? Well, due to the fact that the calculator has base ten, and it could have other basis, but it definitely has base ten. I'm going to use the common law. 

I'm going to use the common log, not the common law. That's something totally different. And what I can now do with this logarithm is I can use that log law that pops that X out. So I'll get X, and let me I'm going to put a little dot there for multiplication. Times the log of four is equal to the log of 8. Now, of course, I can solve for X by dividing both sides. By the log of four. So X is equal to the log of 8 divided by the log of four. Now it might be tempted to think that's the log of two, you know, because I do like 8 divided by four and I get two, but it's not. In fact, this is where I want to use the calculator. Now I'm not going to have the TI 84 plus open a lot today. But I do want to show you what it looks like when I enter that into my calculator. So let's just do it this one time. Hello TI 84 plus. All right, all I'm showing you is how to evaluate the log of 8 divided by the log of four. And it's easy enough. Let's do it. I'm going to hit the log. Put in 8. Divided by. The log of four. Got it all in there. Everything looks good. Nothing funny with parentheses or anything like that. And I hit enter. And look what my calculator tells me it tells me 1.5. Which is precisely what I had over here. All right? So it's really pretty impressive. 

You know, without using common bases or anything like that, that third logarithm law allows me to solve this exponential equation. Very, very fast. All right? Pause the video now and write down anything you need to. Letter B is very important to think hard about what we did there. Okay, let me get rid of it. Let's keep going. All right, exercise two solve each of the following equations for the value of X round your answers to the nearest hundredth. All right. Well, again, let me do the first one with you. Okay, and then I'll have you work on some of the other ones on your own. So we get 5 to the X equals 18. I'm going to do the log of both sides. Now don't skip this step. Some students want to go immediately to the next step. I guess that's okay. But I think it's really helpful to have this step first. The next step where that X comes out, right? X times the log of 5 equals the log of 18. So X is equal to log 18 divided by the log of 5. And if I put that into my calculator, just like I did before, what I'll find out is that's approximately 1.80. It's kind of cool, isn't it? That means 5, never forget to the 1.80. Is 18. It kind of makes sense 5 to the first is 5, 5 to the second is 25. So whatever 5 had to be raised to to get 18, had to lie between one and two. Then it did. It was one point, you know, roughly 8. 

All right, why don't you try letters B and C pause the video now and take a couple minutes. All right, let's take a look at them quickly. Take the log of both sides, log of four to the X equals log of 100. By the way, the log of 100 is two, but we don't even have to think about that. I can pop that X out now, but law X times log of four. Equals the log of a hundred. And X is the log of 100 divided by the log of four. Which, if we work that out on our calculator, we can't do it in my head, is about 3.32. Right, you can kind of see how this is all going, the log of two to the X equals the log of 1560. Pop the X out. X times log of two equals the log of 1560. And X equals the log of 1560. Divided by the log of two. Which ends up being what? About ten point 6 one. All right. Now, of course, if that was the entire deal, if the whole deal was that's it. Well, nonetheless, it would be over and we would call it a day. But we can do a little bit more with this. So let's keep going. Pause the video now, write down anything you need to, and then we'll take this up a notch. All right, here we go.

Okay, solve each of the following equations for X, round your answers to the nearest hundredth. All right. Now, I'd like you to take a look at both of these two equations for a moment. Okay. The big difference between these equations and the ones we were dealing with before is that they're more complicated. The exponents are more complicated. So I want to be more careful. I don't want to introduce any rounding errors or anything like that. And yet, I'm still going to do exactly what I did before. I'm simply going to take the log of both sides, let's work through the first one together. Then that X plus three is going to come out front now. This is critical. Keep that X plus three in parentheses. Okay? If all I do is write this. Right. As my left side. Then it's only the three. That's multiplying the log 6. And that's not the case. It's the entire X plus three. So we have to be very careful there. All right. Now, how do we get the three? Or how do we get the X all by itself? We'll think about this. I've taken X I've added three, and then I've multiplied by log 6 and gotten log 50. So the first thing I'm going to do is get rid of the multiplication by log 6 by dividing by log 6. 

Now you could evaluate that on your calculator, but I would. I would just leave it because then I will get rid of that plus three, right like that. And then I'll enter all of this on my calculator at once. Log 50 divided by log of 6 minus three. Now, if you haven't already tried this, try this really quick and then I'll give you the answer so you can check yourself. All right. Well, if you work through all of that, what do we end up getting? It's negative. We should get negative point 8, where are we? Yeah, negative .82 about. Again, can't do these in your head impossible. Negative .82. Let's try this one very, very similar, little more challenging. All right? Log of 1.03 to the X divided by two -5. Equals the log of two. Again, just like before we can bring this out, let's keep it in parentheses. X divided by two -5. Times the log of one O three equals the log of two. Now again, I really want you to understand. When we look at this equation, we think I'm solving for X let's talk about how to do it. We took X and we divided by two, then we subtracted 5. Then we multiplied by log of one O three, right? So we can easily undo this by first, dividing both sides by log of 1.03. You don't want to distribute that log of one O three. Just want to divide by it on both sides to get rid of it. 

Then we're going to add 5. And then we're going to multiply the entire quantity by two. Now that's your final answer, not decimal form, but it's your final answer. The issue now is just evaluating it. And you might want to evaluate it slow. You might want to do log two divided by log of one O three, enter. Plus 5 enter times two, enter. Right, so you can do it in phases. That's okay, but it's critical that you get the calculator work right. And if you do, you end up getting an answer for X it's about 56.90, much larger than our last one. All right. So pause the video now and write down anything you need to. All right, clear it out and moving on. Uh oh. We got to get back into that keynote. There we go. My apologies for that. You saw the inner workings of my Macintosh. All right, let's take a look at more of a modeling problem. I never understand why that happens. A biologist is modeling the population of bats on a tropical island, okay? Cool enough. When he first starts observing them, there are 104 bats. Okay. The biologists believe believes that the bat population is growing at a rate of 3% per year. It feels some exponential modeling coming on. But a race is right an equation for the number of bats, B of T as a function of the number of years T since the biologists started observing them. 

So this is something you should feel very, very comfortable with. Why don't you go ahead and try to write that function? All right. Hopefully. Because it wasn't that long ago. Hopefully this was a piece of cake. We know we start with a 104. That's our Y intercept. We know that every year we get to multiply it by 1.03, because it's growing at 3% per year. And then we raise it to the T all right, simple enough. Letter B says using your equation from a, algebraically, that's a very, very important word. We all know that, right? Algebraically, determine the number of years it will take for the bat population to reach 200. Round your answer to the nearest year. Well, before we would have had to have done this graphically. And it's kind of a pain to do it graphically. I mean, it's not terrible, but kind of a pain. We'd have to put this in. This in graph, get a good window, use the intersect command, et cetera. But now let's take a look at this. 

We want to solve for T now, given that what we've been doing, we keep taking the logarithm as our first step, it would be an understandable thing to just say, oh, I'm going to take the log of both sides, pop that T out. But the problem is we've got this product here first. So the best thing to do honestly is to begin to solve this problem by dividing both sides by a 104. Okay. So that we have 1.03 to the T equals 200 divided by one O four. Now, if this is a nice number in terms of a decimal, you know, go ahead. Even if it's not a nice number, you could use the decimal version. I'm going to leave it as the fraction. All right? And I'm now going to take the log of both sides. So I'm going to take the log of one O three to the T equals the log of 200 divided by 104. And of course that's going to pop the T out. So I'll have T times the log of one O 5 equals. The log of 200 divided by one O four. So now I can divide both sides by log of one O 5. Again, be careful. We got two divisions going on here. One that's inside of a log one that's sort of outside of the log. But ultimately, you just have to put this all into your calculator. Why don't you go ahead and try to do that, pause the video if you need to. Throw that into your calculator and make sure you can get the right answer. All right. Well, when I crank that all into my calculator, I get an answer that's about 22 years. 22 years, it'll take to reach a population of 200. 

All right, well, they're not growing very fast. Only 3% per year. I can buy 22 years. Pause the video now and write down anything you need to. Okay, let's clear out that text. All right. Exercise number 5, a stock has been declining in price. Oh, that's important. The fact that it's price is going down. Declining. At a steady pace of 5% per week. Okay. If the stocks started at a price of 22 50 per share, determine the number of weeks it will take for the price to reach $10. Round your answer to the nearest week. All right. Well, there's a couple pieces here, right? We have to model the problem. And then we have to actually solve some kind of an equation. So see if you can model this problem in terms of a function that models the price of the stock, and then a little bit of algebra. Okay? Take a minute. Or even more than a minute. All right, let's do it. Well, if we wanted to just a function, that predicted the price of this stock as a function of the number of weeks, right? We know it starts off at 22 50. We know it goes down 5% per week, but remember how we accomplish that is not by multiplying by .05 or 1.05 or negative 5. But by multiplying by .95, we have 95% of the price every week. 

Now, I'm going to set that equal to $10. And again, just like in the last problem, we could solve this graphically. But the problem specifies for us to do it algebraically. And it's a very, very similar solution. I'm going to divide both sides by 22 50. And just like before, if we want to deal with the decimal version of this, we can. We just wouldn't want to round it. I'm going to leave it as ten divided by 22 50. And then I'm going to take the log of both sides, and I can leave that fraction absolutely unevaluated inside of the parentheses. No need to figure out what ten divided by 22 50 is. The W would pop out, and then I would divide by the log of .95. So at the end of the day. We end up having this. And again, I would suggest to pause the video right now, make sure you're good in terms of getting that into your calculator, and then I'll give you the final answer. All right, well, let's do it. Our final answer is just about 16 weeks. It'll take for that price to get down there. Okay? So pause the video now right down anything you need to, and then we'll do one last problem that's a little bit different. All right, let's get rid of it. One of the things that the common core would like you to be able to do is solve exponential equations and leave them in terms of logarithms. 

Okay, so not get like a final answer like X equals 17. But get an answer like X equals the log base three of 5. Something like that. So exercise number 6 says find the solution to each of the following exponential equations find in terms of a logarithm. So when you see something like this, what it means is that there will be a logarithm in your final answer. It says in terms of a logarithm with the same base as the exponential equation. All right, well, let's see how it plays out. Let's solve this equation. Now take a look at what we have. We've got two rays to the X, then we're multiplying it by four, then we're subtracting three, and we're getting 17. So the last thing I did was subtract three, so the first thing I'm going to do is add three. So I'm going to get four times two to the X is equal to 20. Right then, I've got to get rid of that multiplication by four. So that's easy enough. I'll just divide by four. And I'll get two to the X equals 5. Now, what do they mean by solve this equation in terms of the same base? Well, what they want me to do is do the log base two, right? The same base. Of two to the X well, now I could bring the X out. I could. But think about this for a minute. The whole point of logarithms is that they're inverses of exponentials. They literally cancel. These cancel. Whoops. Oh, that was horrible. At the log base two of 5 on that side. I can't take the log on the left and not the log on the right. 

Anyway, so eventually I just get X equals the log base two. A 5. And that is what it is. You know, it's some decimal answer. But we're going to just leave it like that. Okay? Very similar letter B okay, look at this. We've got 5. Raised to the X divided by three. Times 17 equals four. So the first thing that was done to X is we divided by three, then we exponentiated it. Then we multiplied by 17. So the first thing I'm going to do is get rid of the 17. So I have 5 to the X divided by three. Equals four seventeenths. Now again, we want to rewrite the final answer in terms of a log with the same base. As what we have. So I'm going to do the log base 5 of 5 to the X divided by three. Equals the log base 5, a four 17. Now again, the idea is that these just cancel. And we're left with X divided by three. Is the log base 5. A four seventeenths. And then of course I can multiply both sides by three, we can get three log base 5. Of four seventeenths. That what I left that answer is you bet. 

All right. So a little practice in leaving final answers in terms of logarithms with a variety of basis in this case we had a base two, and we had a base 5 logarithm. Many of your calculators actually have a sort of a more generic logarithm base where you can say, hey, look, I'm going to use any base I want. All right? So it's not bad to have your answers like this. Okay, we'll pause the video. I'm going to clear out the text and we'll wrap up the lesson. Okay. Let's do it. Text is cleared. Let's finish up. So in my personal opinion, the third log law is the most important of them all. For one specific reason. It allows you to solve exponential equations algebraically, right? Now we can solve almost any equation graphically. So we always have that option. But when you want to have a quick answer to an exponential equation, three to the X equals 17. How do I, what do I do? What do I get? Well, being able to take the logarithm of both sides typically base ten. Allows us to solve for X using that third logarithm law. We had a lot of examples of that in this particular lesson. Very, very important one. For now, I'd like to thank you for joining me. For another common core algebra two lesson by E math instruction. My name is Kirk Weiler. And until next time, keep thinking and keep solving problems.