 This video is part 2 of Properties of Logs. So we have some more Properties of Logs that we need to learn, and they really come from our exponent rule, so that's why we haven't both listed here. When we had exponents, we said if we had the same base, we could add the exponents, and if we had the same, when we were multiplying. If we had the same base and we're dividing, we said we could subtract the exponents, and if we had a power raised to a power, we said we could multiply the exponents. Well remember, let's remember up here that logs are exponents, so whatever I would be doing with my exponents, given the operation I'm doing, I should be able to do with my log. So this is log base B of A times C. It's A times C, and it's the same base, and I want to know, we're really looking for the exponent, this log exponent, log base B of A plus, because when you multiply, you add your exponents, log base B of C. There's my two exponents that I'm adding, and this tells me that I was multiplying the same base. Here it says I'm dividing the same base. When I divide the same base, I can subtract my exponents over here. So log base B of A minus log base B of C, and we always subtract top minus bottom, so it will always be the log of the top minus the log of the bottom. And now we have log base B of A raised to the C, so it's kind of like thinking this one's the easiest one to see, I think. This one is an exponent, log base B of A, but so is this. So what do we do with our exponents? We multiply them. We take C, and we multiply it times the exponent of log base B of A. Now you do need to be careful that you don't, the only ones we can use are multiplication in here, division, or an argument with an exponent in it. Alright, so let's use those properties to write it as a single log. So this, the addition here, means that I really have to multiply my arguments. Remember the arguments are what's in the parentheses. So I want one single log, and in here I'm going to have x plus 2 times 3x, and I can take that out and simplify it. So that would be ln of 3x squared when I distribute, plus 6x. That would be my final answer. When I'm subtracting, that means I have to divide my argument. So I'm just going to say arg for argument. So I take my log, and it has to be the same kind of log. So log base 6, and in here I'm going to divide the 30, the first argument, divided by the second one, which would be 10. And we can always simplify. So log base 6 of 3. Make this a single log. Again I'm subtracting, so we take our single log, and then in here I say my top is the first argument, divided by the bottom, which is my second argument. And then we have one more. So we're going to multiply, and we'll have log base 9 of 2 times 15, or log base 9 of 30. That just takes some practice. So now we want to go the other way. We want to take the properties and use the logs. So we have a couple properties going on here. Remember the first one was multiplication. So we have a times c. But we also have a raised to the c going on right here with the n squared. A times c said I get to add my logs. So I'm going to have a log plus another log. And I take each factor is a different argument. So m squared plus the log of n. You don't have to have these parentheses. I just like to differentiate. And then we have, so we've taken care of this property, but now we need to take care of this property. And if you refer back or remember, it says that my exponent can actually come out front. Because I'm going to multiply my exponent of my argument times the exponent, which is my log. The log is the exponent. And then over here we didn't have anything more to do. So 2 log m plus log n. What do we have going on here? Again, we have the a times the c. And we really have the a raised to the c. I want to remind you that a cube root of a is the same thing as a to the one third. Let's rewrite this then as lnpq to the one third. But it's on the outside. Like if I had the cube root of x, I would have to cube that before I could start working with my x. So the order of operation says you have to take the exponent before the multiplying. So we'll take the one third in front of the log first. And it's actually going to be times the log of this pq. And then we have to work inside to get that log separated because of the multiplication. So one third ln of something plus since it was multiplying. ln of something else. All in parentheses. And inside the little parentheses we put in our lnp and our lnq. And we're done. So we have a division. So we have a over b. And we also have that a to the c. So this means to track back to my logs. And this means that we have to multiply, bring it out front. So I'm going to take it one step at a time. I'm going to separate this first. That's my subtraction. So ln of m squared minus the natural log of n cubed. Now let's take each one separately. lnm squared. I've got that second property going on which says bring that exponent out front. And then multiply it times the log which is another exponent. So 2 times lnm minus. And then again I have to bring this one out front. So 3 ln of n. We've got one more property that we like to use with logs. And that is the change of base formula. So we have these positive real numbers where m is our argument and we've got a and we've got b depending on. Those are going to be bases in this particular problem. And a and b are neither one going to be one. Because we're letting them be bases. So here's my original problem. Log base b of m. And the formula says that that can be the log of the m over the log of my base. And the way I always remember it is the base is the subscript and the base is the bottom or the base of my fraction. So log base b goes on the bottom and the log of the argument goes on top. And the only difference between this one and the second one is to show you that you can go into any base you want to. So we can say alright I think we could go into a log because I can use my calculator with just log base 10 or common log. But what if I wanted to use a natural log? That would be acceptable too. That's the log we're choosing or base that we're choosing to go to. So we just take the natural log of m divided by the natural log of b. Remember you've got to take the same kind of log and this m and b are always going to be in that order. The argument over the base. And then the third one here is just saying okay what if I want to go to some totally different base. Maybe I'm in base 2 and I want to go to base 5. Same thing. You choose the base that you want for your log. And then you take the log with that base to the argument or the log with that base to the base. We're going to say that y is equal to that log base b of m. And now we have b to the y is equal to m in exponential form. So when we write it as a log and we want to go to base a. Then we can take the base a log of both sides. So log base a of b to the y is equal to log base b of m. I just took the log of both sides, same base. Now I'm going to rewrite this, use my properties so I can bring the y out front. So y times the log of a base a of b is equal to the log of base a of m. Because there was no exponent over here. I want to solve for y because I know something about y over here. And you won't have to do this. I'm just proving it for you. So y will be equal to log base a of m divided by log base a of b. And then by substitution we can really say that log base b of m instead of y is equal to log base a whatever I choose of m over log base a of my base. So that just proves it. You won't ever have to go through this. I just wanted to show you that it's not just something that they threw out of the air. This really is a property that can be proved. So we want to use that change of base for either base 10 or base e. And we want an exact form and an approximate. And I'm not sure if we talked about exact or not before, but let's talk about it now. If we change the base on this, I know I'm going to go to a log first. And we have log the arguments on top. So log of 92 over the log of my base, which is 8. This is what we call an exact answer. There's no rounding involved here at all. Bring in my calculator then. I can find the, not the exact, I can find the approximate. And it says the 5 decimal places. So I go to my calculator and I use log because I said I went to a log, not a natural log of 92 and then close your parentheses, divided by literally what you see. Log, it opens the parentheses of 8, close the parentheses. And we end up with 2.17452. 2.17452, I think is what it was. What if I had wanted to go to a natural log? Could I do that? Sure. We could say it's equal to the natural log of 92. Instead of the log, we're taking base E, so natural log, and ln of 8. Again, this is an exact answer. And if I call it my calculator, I can try to see what that one is. So natural log of 92, divided by the natural log of 8. And look, what do you notice? It's exactly the same answer. So what does that tell you about choosing the log to go to? It should tell you that it doesn't matter which type of log that we want. Do we want common log? Do we want natural log? Any kind of log I want, I'm always going to get the same answer. We're going to do logs and natural logs. No matter which one we use, we're going to get the same answer.