 In a previous video in this lecture series, we talked about how exponential functions are one to one and therefore we can use the injectivity of the function that is the fact is one to one to solve equations. If ever you had an equation of the form AB equals AC, you could cancel out the basis and get BC because the function's one to one. This is also true for logarithms. Logarithms likewise are one to one functions. When you look at a graph of a logarithm like this, it passes the horizontal line test. It's one to one. So whenever you have an equation of the form, log of B equals log of C having the same base A, you can cancel out the logs and just get B equals C. All right, so that's gonna be a very useful thing in solving logarithmic equations. That's one of the things you can use a lot. Another thing you have to do when you're solving logarithmic equations, if you have an equation in a logarithmic form, you can switch it over to its exponential form and solve it, or you can go the other way around if that's at all useful to you. So look at this first example, example A right here. I have log base three of four X minus seven equals two. If I wanted to solve for X, what I could first do is get rid of the log and I have to apply the inverse function. So that means moving the log base three to the other side. But whenever you move the number to the other side, you have to switch the operation. So if you have log base three on the left, it has to switch to its inverse function, which is going to be the exponential base three. So you get three squared, which is equal to nine. And then after that, moving things to the other side is what we're quite used to, right? We wanna move the seven to the other side. How do we move seven to the other side? Well, you're gonna add seven to both sides. So you get nine plus seven, which is equal to 16. Now you wanna move the four to the other side. Well, how do you do that? You stick it with a four, but you have to switch the inverse operation. You get X equals 16 over four, which is equal to, of course, four. Now, whenever you're working with logarithmic equations, you have to be cautious. Their domains are restricted. Not all real numbers can be plugged into a logarithm. So you have to make sure this number fits inside the domain. And if you stick it in right here, you get four times four, which is 16 minus seven, which is nine, that's inside the domain. And then log base three of nine is in fact equal to two. The second power of three gives you nine. So that gives us, that we checked our solution, we're good to go there. Let's look at another equation here. So unlike the first example where we had a logarithmic equation where the variable was in the operand here, you'll notice here that now the variable is in the base. We don't know what the base is right here. That's okay. We still, it's kind of like if you had the equation, you know, like two over X is equal to five or something like that. Well, the first thing to do is just times both sides by X, right? Why are we times both sides by X? We're trying to move the X to the other side of the equation by clearing the denominators. We're gonna do the same thing here that we move the X to the side. You know, how do we clear the proverbial denominator in this situation? Well, you still switch it from logarithmic form to exponential form, in which case we get that 64 is equal to X squared. The base is X here. Now we have a polynomial equation, 64 equals X squared. We wanna move the two to the other side, right? But as this is the power of two, we move to the other side by taking the square root. We get the square root of 64 plus or minus, for which we then get plus or minus eight as our solutions right here. Now I have to warn you that kind of like the previous question here, there are some domain issues, right? What are acceptable values of X? As this is the base of a logarithm, X must be positive and not equal to one. Those are our stipulations. So if you actually forgot the negative eight as a solution, good at you, you gotta write for the wrong reason, which is better than getting the wrong, right? But still we wanna make sure we don't make this a mistake in the future. So eight is the only acceptable base for this problem right here. So notice that what power of eight gives you 64, the second power. So this is in fact a solution to that equation. All right, let's look at another one here. Let's take the logarithmic equation two times log base five of X is equal to log base five of nine. Notice that both sides of the equation are log base five, that's good. This is now an example where I wanna use the one-to-one property of logarithms here. Notice I have log base five of a and this equals log base five of b. I can cancel out the logs and we actually get that a equals b. So that's what we wanna do here. The right hand side's already a logarithm, that's great. But what about the left hand side? It's two times a logarithm, that doesn't quite apply. But we can actually move the two inside of the logarithm by the third law of logarithms. This then gives us the log base five of X squared is equal to log base five of nine. For which then we can use the one-to-one property to cancel out the logs. We get X squared equals nine, for which then if we take the square root again, we'll get X equals plus or minus three. Again, domain issues though, because logarithms have some restrictions on the domains, you need to put them back into the original equation and see what happens. If I take X equals three, you'll get two times the log base five of three, which is, is that equal to log base five of nine? In effect is, that one's perfectly fine. When you put the negative three in though, you see a problem, you get two times the log base five of negative three. Does that equal log base five of nine? The issue over here is there's a domain issue, right? You can't stick a negative number inside of the logarithm, log of negative three for any basis undefined. So this is actually, because of domain restrictions, that's undefined. So that's not a solution. And so therefore there's only one solution here, X equals three. Make sure you check your solutions so that everything's good at the end. Now let's look at another example. I really like this one here. We have two logarithms on the left hand side. So we have the log base four of X plus three plus log base four of two minus X, that's equal to one. How can we solve this equation? Like we did on the last example, if we condensed the logarithms into a single log, that'll actually make it work for us. Notice we have a log plus a log. So by the first law of logarithms, we get that this is equal to log base four of X plus three times two minus X, like so, and this is equal to one. And so once we condensed the logarithm, we can switch it to exponential form. We can move the log base four to the right hand side and we get X plus three times two minus X is equal to four to the first, which is equal to four. Sometimes people just magically make the log disappear, which you don't get X plus three times two minus X is equal to one. That's actually the mistake. And you see four to the first, which is equal to four. Like I said, right there. So we can solve this by switching to exponential form. I also wanna mention that you could also write the right hand side as a logarithm because if you have a one right there, that can be written as log base four of four, right? Notice four to the, what power of four if you're four of the first power, you can then cancel the logs and get the exact same thing we had here. X plus three times X minus two is equal to four. Either of those approaches is perfectly fine. And so now we have a quadratic equation. We're gonna foil up the left hand side. We get two X minus X squared plus six minus three X is equal to four. Combining like terms, I'm just gonna move everything, everything, everything to the right hand side. That's gonna give us zero is equal to X squared. Notice you have a two X minus three X that's a negative X when I move to the other side it becomes a positive X. And then when you move the six to the other side you get four minus six which is a negative two. We didn't have to solve this equation. It's a quadratic, so I'll probably factor or use the quadratic formula, which factoring works out really nice here. You get X plus two times X minus one. And so that suggests to me that X should equal negative two or one. But we should check our solutions. Do they make sense? Now just because we have a negative number does not mean it's actually ruled out as a possibility. If you plug it into the original equation, right? If we consider negative two for a moment this would actually give you negative two plus three which is a positive one. And this will give you two minus negative two which is equal to four. There's no domain issues there. Log of zero, log base four of zero is gonna give you, excuse me, log base four of one is gonna give you zero and log base four of four is gonna give you one. One plus zero is one. That actually checks out. So negative two passed the test. Just because one works doesn't mean the other one won't work. And just because one's negative doesn't mean it won't work either. So if we try one, you're going to get one plus three which is four. And you're gonna get two minus one which is one. So you get log base four of four which is one plus log base four of one which is zero, one plus zero is one again. So actually both of these solutions work out in this example here. So we see that when we check the domains, both of them since they worked out X could equal negative two or X could equal one.