 So let's take a look at some of the basic properties of logarithms. So it helps to remember what a log is. Suppose log to base a of c is equal to b. Then our definition says that a to power b is equal to c. Now we can evaluate a to power b whether or not b is positive, negative, or zero. But since a must be greater than zero, then a to power b must be positive. And remember, equals means replaceable. If a to power b must be positive, then c must be positive. And so that means that the expression log to base a of c is only defined if c is positive. At least for now. Let's see what else we can determine. Suppose I want to find the log to base a of one. So again, we'll set this up as an equation. Let x be log to base a of one. This transforms into a new equation. a to power x equals one. And let's see if we can solve this. So the thing to remember is that since a to power zero is equal to one for any value of a, we know that x has to be zero. And so this suggests the following theorem. For any base, the log of one is equal to zero. Let's see how log plays with products. Suppose the log to base a of something is equal to c, and the log to base a of something else is equal to d. Can we say anything about the log to base a of the product p times q? Well, definitions are the whole of mathematics, all else is commentary. Think about what it means when we say that the log to base a of p is c, and the log to base a of q is equal to d. Both of these statements translate into exponential expressions. a to power c is equal to p, a to power d is equal to q. Now I'd like to say something about p times q, so let's multiply them together. Notice that I now have a product of exponents, so I can rewrite that product as a to power c plus d. Now I'll hit both sides with a log. And the important thing to remember about logs is that logs are exponents. So the log to base a of p, q is the exponent whatever I need to raise a to to get p, q. In other words, it's going to be c plus d. But wait, equals means replaceable. So I know that c is the same as log to base a of p, and I know that d is the same as log to base a of q. So I can replace them, and this suggests a general result. For any base, the log to base a of p, q is the log to base a of p plus the log to base a of q. Sometimes we say this as the log of a product is the sum of the logs. How about quotients? Again, suppose that the log to base a of p is something, and the log to base a of q is something else. What about the log of the quotient p over q? And once again, definitions are the whole of mathematics, all else is commentary. Because we know the log of base a of p is c, and the log to base a of q is d, we can rewrite these in exponential form. Since I want to know something about the quotient p over q, I can set that down. Since I know something about how to divide exponential expressions, I can simplify. And again, the log is the exponent. So the log to base a of p over q is the exponent c minus d, and equals means replaceable. Every time I see c, I can replace it with log to base a of p. Every time I see d, I can replace it with log to base a of q. And as before, this suggests a general result. Any base, the log to base a of p over q is the log to base a of p minus the log to base a of q. And again, we sometimes state this, the log of a quotient is the difference of the logs. How about powers? Suppose the log to base a of p is c. What about the log to base a of p to the n? So again, you know the definition. If log to base a of p is c, then we can rewrite that in exponential form. Since we want to know something about p to the n, we'll find p to the n. We'll apply our rules of exponents. We'll hit both sides with a log. And we know that c is the same as log to base a of p. And so this gives us a result. For any base, log to base a of p to the n is n times the log to base a of p. You could say that the power comes out front. So let's play around with these rules. Suppose that log of a is 5 and log of b is 3. Let's find the log of 1 over a and the log of square root of b. So again, how you speak influences how you think. And so you might want to speak of this 1 over a as 1 divided by a, which says that it's a quotient. So we might want to bring in our rule for the log of a quotient. So the log of 1 divided by a is log of 1 minus the log of a. But wait, there's more. We have this log of 1. And we know something about the log of 1. For any base, the log of 1 is equal to 0. So we can simplify our expression and we can do some arithmetic. How about the log of the square root of b? So remember a square root is the same as an exponent, 1 half. So this log square root of b is the same as log of b to the power 1 half. But wait, we know how to handle the log of a power, the exponent comes out front. And we know the log of b is equal to 3, so we'll replace that and do some arithmetic. How about the log of, well, frightening horrible mess. But that's okay, we'll take it one step at a time. And again, how you speak influences how you think. So here we have log of a cubed over, no wait, that's divided by b squared c to the fifth. So that's a divided by, so I have the log of a quotient is going to be the difference of the logs. So the first reduction, log numerator minus log denominator. Now let's take a look at these each separately. This is log of a to the third, that's a to power 3. So we use our power rule that exponent comes out front. This other term, this is log of b squared c to the fifth. Wait, no, that's b squared times c to the fifth. This is a product, so we can apply our log of a product rule. And importantly, because we're subtracting this log, we have to make sure that we use parentheses. This is log of b to the second power, so we can use our log of a power. And likewise for our log of c to the fifth. And we know our values of log a, log b, and log c. So we'll substitute those in and do a little bit of arithmetic.