 Welcome back to our lecture series, Math 1050, College Outsford for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misaline. In lecture 42, we're going to talk some more about solving exponential and logarithmic equations. We did a few examples like this in our lecture 41 in this series. We're going to do some more. And so take a look at the example we see on the screen right here. We want to solve the equation, the log base 5 of x plus 6 plus the log base 5 of x plus 2 is equal to 1. The key to solving this one is to use the first law of logarithms, which we saw in lecture 41 here, which told us that if we take the log base 5 of some quantity A and add it to the log base 5 of some quantity B, then you can actually condense the logarithms together, log base 5 of A times B. And it honestly doesn't have to be base 5. It can be whatever base you want in this situation. So we can combine together these logarithms. The sum of two logs becomes a log of a product. And so we get log base 5 of x plus 6 times x plus 2. So at some point, we're going to have to foil out that product there of x plus 6 and x plus 2. But we'll do that later. And so this is equal to 1. The next thing we want to do is we want to switch this thing to logarithmic form. We want to switch it from logarithmic form to exponential form. We don't want the logarithm around because that's where the variables are. We've got to free them. So we want to move the log base 5 to the other side of the equation. It doesn't just magically disappear because we don't want it to happen. We need to apply the same operation to both sides of the equation, in which case to move the log base 5 to the side, we have to switch it to its inverse operation, which should be the exponential base 5. So we're going to take 5 to the first, a.k.a. 5. And now we have this quadratic equation. So now it's a good idea to foil out the left-hand side. So we'll get x squared plus 2x plus 6x plus 12 equals 5. Combine some length terms here. Nothing to do with the x squared. We get 2x plus 6x, which is 8x. If we subtract 5 from both sides, we'll get 12 minus 5, which is 7. And so we're looking for factors of 7 that add to the 8th. That's a pretty simple one. We get x plus 7 and x plus 1 equals 0. And that's going to give us two potential solutions. x equals negative 7 and negative 1. And I say potential solutions because whenever you're dealing with an equation with logarithms, logarithms have restricted domains. There's only certain numbers that can be plugged inside of these logarithms to produce the correct solution because there's some domain restrictions. So what numbers are acceptable? You'll notice if you plug in negative 7 for x, you're going to get negative 7 plus 6, which is negative 1. The log of negative 1, that's outside the domain. So that's not going to work. And you see the other side as well. Negative 7 plus 2 is going to give us a negative 5. And this is the issue. You had a negative 1, negative 5. When you bring them together, you're going to get negative 1 times negative 5. That actually gives you a positive 5. And that's what kind of tricked our algebra right here. When we went from this to this, the double negative became a positive. And so we picked it up as a solution. So negative 7 doesn't work. So we have to throw it out for consideration right here. No negative 7. Now, just because negative 7 didn't work, doesn't mean negative 1 will work, or 1 doesn't tell us anything. These type of logarithmic equations, they could have two solutions because you are solving a quadratic at one point. You could have one solution. You have no solutions. We have to check both of them. Now, you're going to see what happens. This time, we take negative 1. You're going to get negative 1 plus 6, which is a 5. And then you're going to take negative 1 plus a 2, which is a 1. And sure enough, if you take log base 5 of 5, you're going to get a 1. And if you take log base 5 of 1, you're going to get 0, 1 plus 0 is equal to 1. So it turns out our only legitimate solution is going to be x equals negative 1. So it's important when you're working with logarithmic equations that you always check your solutions. Because even if you did everything right, if you don't check your solutions, you might have numbers that don't belong because they actually fall outside the domain of our logarithm here. Let's look at another example. This time, let's take the natural log of x and set it equal to the natural log of x plus 6 minus the natural log of x minus 4. And we want to solve this equation. Well, to solve this one, we have to again use one of our laws of logarithms. We're going to use the second law this time that tells us if you take the natural log of some function and subtract from it the natural log of another value there, you can bring these together as a quotient. You get the natural log of a divided by b. And again, we don't have to work base e. This log of logarithms works for any base whatsoever. So let's combine together the right-hand side. The left-hand side will just be the natural log of x. Nothing to do there yet. On the right-hand side, we get the natural log of x plus 6 divided by x minus 4, like so. And so now, because the left-hand side and the right-hand side are logarithms, we can use the fact that logarithms are one-to-one. That is to say, we're going to use the principle that if we have the natural log of a is equal to the natural log of b, this implies that a equals b. We can essentially cancel out the logarithms. Because since the function is one-to-one, the only way that the y-coordinates can be the same is if the x-coordinates are the same. So canceling out the logarithms on the left-hand side and the right-hand side, we end up with x is equal to x plus 6 over x minus 4. We have a now a rational equation. I'm going to multiply both sides by the denominator x minus 4 so that I can clear the denominators on the right-hand side. We then get x times x minus 4, which if you distribute the x, you're going to end up with an x squared minus 4x. This is then equal to x plus 6, okay? So we're looking at this equation right here. This is a quadratic equation. I'm going to move everything to the left-hand side so we get x squared minus 4x minus x minus 6 equals 0. Combining like terms, you get x squared minus 5x minus 6 equals 0. So we need to find factors of negative 6 that add up to be negative 5. Let's be really careful about this one because notice like if you take 6 minus 1, that gives you 5, but you have 2 plus 3, that equals 5, right? Make sure you pay attention to the signs. We need factors of negative 6. So to get two numbers to multiply to be a negative, one has to be positive, one has to be negative. And then to add them together to be negative 5, the bigger factor needs to be the negative 1 in terms of absolute value and the smaller one needs to be positive. So that actually gives us that the factorization is going to be x minus 6 times x plus 1. Sometimes we trick ourselves and say x minus 2 and x minus 3. That's a very common mistake in this situation, but you can always foil it out again to double check there, right? 6, negative 6 times 1 is a negative 6, and then you're going to get negative 6 plus 1, which is going to give you the negative 5 right there. Okay, so with those solutions in hand then, well excuse me, with the factorization hand, we then see that the solutions are potentially 6 and negative 1. But like we've seen before, we have to double check the domains, right? If I plug in 6, what happens here? If you plug in 6, you get natural log of 6, which gives me no concern. You get the natural log of 6 plus 6, which is 12, and then you get the natural log of 6 minus 4, which would be natural log of 2. There's no domain issues right there. 6 seems to be kosher. If we stick in negative 1 though, you see on the left-hand side, we already have a problem. Natural log of negative 1 is, it's not a real number. We'd get negative 1 plus 6, that's a 5, that's okay, but then you get negative 1 minus 5, which is again, excuse me, negative 1 minus 4, which is negative 5. So there's some domain issues. So we're gonna have to throw out negative 1 from consideration, and we see that the solution to this logarithmic equation is going to be 6. And these examples then demonstrate a basic strategy for solving equations, evolving logarithms. Use those properties and laws of logarithms to combine things as much as possible. You wanna condense all the logarithms. You have a log on the left-hand side or on the right-hand side so that you can cancel like we saw here. On the previous example though, if you don't have a log on the right-hand side, you just have it on the left-hand side, that's fine, just switch it to exponential form, and that'll do the same thing for you.