 In this video, we provide the solution to question number five for practice exam number four for math 1050. We have to find the domain of a logarithmic function f of x is given as the log base two of the ratio x minus one times x plus one on top and x plus two in the denominator. Now when it comes to finding the domain of a logarithmic function, that really just means that the operand that the expression inside of the logarithm has to be positive. It can't equal zero because the log of zero is undefined, that corresponds with the vertical asymptote of the logarithmic graph. This thing has to be greater than zero, so we have to solve this rational inequality. The fact that we're working base two really makes no difference whatsoever. As long as we have an acceptable logarithmic base, that is that the base is positive and not equal to one. Other than that, it makes no difference. You'll of course always have an acceptable logarithmic base on a problem like this. I'm going to solve this and I'm going to do it graphically, especially since the rational functions will be factored. I'm going to put my markers here on the x-axis. We have an x-intercept at x equals one. We have another x-intercept at negative one. Then we have a vertical asymptote at negative two. I'm going to mark that a little bit different here on the screen to indicate that the functions can behave a little bit different here. Then the next thing I want to do is consider, we've done, of course, the asymptotes and the x-intercepts. Let's think about their multiplicities, right? Each of them shows up once in the graph. That means we're going to cross the x-axis or we'll cross infinity at their respective locations. The next thing to do would be consider in behavior. Is there a horizontal asymptote here maybe? As x goes towards infinity, this function will be approximately the same thing as x squared over x, which is the same thing as x. What this tells us is there actually is an oblique asymptote, but I don't need that much detail for this one. In behavior of x is sufficient. That tells me I point up on the right and I point down on the left because that's what y equals x will do. Use the information, recognizing where the intercepts are, and the multiplicities at the asymptote and intercepts. We can actually finish this picture really nicely. Start on the left-hand side. We're coming from the bottom right. We don't have an x-intercept. We're going to have to go off towards our asymptote. We do that. We cross infinity. Come down. We cross the x-axis. We're going to have to turn around and cross the x-axis and go off towards infinity again, like so. That's the crude little picture we get, but that's okay. It doesn't have to be perfect. We just care about whether we're above or below the x-axis. Since our expression needs to be greater than zero, we're looking for things above the x-axis. We're above the x-axis to the right of one. We're above the x-axis to the left of negative one, but to the right of negative two. Therefore, we would say the domain is going to be negative two to negative one. We don't include negative two because that's in vertical asymptote. The function's undefined there. We don't include negative one because at negative one, x equals negative one, the expression here becomes zero and long with zero is undefined. Then we're going to jump towards the final port. We're going to get one. Again, x equals one is not included for the same reasons it wasn't for negative one. Then we go off towards infinity, like so. Our domain is going to be negative two to negative one, union one to infinity. While I drew over it a little bit, this then shows us that the correct answer is f.