 In this video, we provide the solution to question number four for practice exam number four for math 1050, we have the logarithmic expression log base five of the fraction, the cube root of x squared plus one over x squared minus one. And we want to expand this log and that is we want to break it apart into smaller simpler pieces as far as we can go. And so the three laws of logarithms are going to come into play here. So first of all, we're taking the logarithm of a quotient. And so therefore using the second law of logarithms, we can break this up into a difference. This will become the log base five of the cube root of x squared plus one minus the log base five. The base doesn't change when we do these calculations log base five of x squared minus one like so. So that's the first thing we want to keep on going here. So with the first expression, notice that if you have the cube root of x squared plus one, the cube root is really just the one third power. And that one third power can be factored out of the logarithm via the third law of logarithms that we've learned about. So you end up with a one third log base five of x squared plus one. Now x squared plus one, it's a sum of squares. It's not going to factor anymore. But on the other hand, we have a law for the second law, we have x squared minus one that does factor as a difference of squares. So that would factor as the log base five of x minus one plus the log base five of x plus one. And utilizing the fact here that of course that x squared minus one, it factors as x minus one and x plus one. That's what we're using here. Now, of course, we broke apart the product, but there was this negative sign in front of them. We have to distribute this onto both pieces. So you're going to end up with one third log base five of x squared plus one minus log base five of x minus one and then also a minus log base five of x plus one. Don't forget to distribute the negative sign right there. And so then we see, let me get this out of the way because this is now covering up the correct answer. We see the correct answer is going to be D one third log base five of x squared plus one. That's exactly what we had right here. Then we're going to have negative the log base five of x minus one just like we saw and then a negative because again, this second one will be negative as well. We distribute the negative sign onto the log base five of x plus one.