Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Properties of logarithms, the product rule log base B of MN, equal log base B of M plus log base B of N Let's do the proof. We're going to start with letting X equal log base B of M&Y equaling log base B of N We're going to put it into exponential form. So B to the X equal M&B to the Y equal N. Now we're going to take and we're going to multiply the left sides together and we're going to multiply the right sides together. Now if we're multiplying and we have the same bases, we add the exponents, then we're going to change that exponential form back into logarithmic form. So log base B of MN is going to equal X + y. But in the original we said X was really just log base B of M&Y was really just log base B of N. So there we've proven the first of our properties of logarithms. The next one is the quotient rule, and it says that the log base B of m / n is equal to log base B of M minus log base B of N. Let's start this proof the same way. So we're going to put them both into exponential form, B to the X equal M&B to the Y equal N This time, instead of multiplying the two sides, we're going to divide the left side and divide the right side. If we're dividing and the bases are the same, we subtract the exponents. We're going to take this exponential form and put it back into logarithmic form. And then we're going to do substitution because in the original it said that X was just really log base B of M&Y was log base B of N. Another property is called the power rule, and the power rule says that log base B of the quantity M to the P is really just P times log base B of M We're going to start it the same way. We're going to take and let X equal log base B of M and put it into exponential form. This time we're going to take each side to the P power and powers to powers. We multiply, and we can multiply in any order we want. So I could think of it as P * X. Now we're going to put it back into logarithmic form using the same base as originally, which was that P. So we get log base B of MPM to the P equaling PX. Well, X was given originally to be log base B of M The last of the traditional logarithm properties is called change of base, and it says if we have log base B of M, it equals log base A of M divided by log base A of B. Starting with X equal log base B of M, putting it into exponential form. This time we're going to take the logarithm of each side in terms of a new base. So let's call it base A. The log base A of B to the X is going to equal log base A of M by our power property, we're going to be able to bring that X down in front. To get the X by itself, we're going to divide each side by log base A of B. And finally we're going to do substitution. We knew that originally that X was log base B of M so now we have that equaling log base A of M over log base A of B. Some examples of the properties of logarithms. If we have log base B of 7 * B, that really is log base B of seven plus log base B of B. But log base B of BB to what power gives us out B? That's the power of 1. So log base B of B is really just one. So that first example would simplify down to log base B of 7 + 1. Now it's not 7 + 1 that log base B of seven is altogether as a number. The next example, if I have log base 5 sqrt X / 25, the first thing we're going to do is we're going to split it up into subtraction. Then we're going to realize the square root of X is really just the same thing as the 1/2 power. So if it's to the one half power, we're going to bring that 1/2 down in front. So 1/2 log base 5 of X. And then the second part 5 to what power gives us out 25. Well, 5 ^2 gives us out 25 S log base 5 of 25 is really just the number 2. The next example is got four different things that are being multiplied. Two of them are in the numerator. So two of my terms are going to be positive when I add them. Two of my terms are in the denominator. So if it's in the denominator, it's going to be a subtraction. Now if it's got an exponent in front or if it's got an exponent, we're going to bring it down in front. So 3 log X + 1/2 log that X ^2 + 1 is still a quantity -4 log X + 1 minus log X -, 1. Now, if we want to go the other direction, they're added and subtracted and we want to rewrite them. We're going to start by getting rid of the coefficient by taking it up as a power. That one third's going to go up into the X + 9 to the one third power. That 4 is going to go up into X to the fourth. Now it's a log plus a log, so we're going to rewrite it as a single log. And because it was addition, we're going to multiply those two things together. And we can multiply in any order we want here. If we have two logs that are positive and two logs that are negative, the positive ones are going to go up on top, so X * X ^2 -, 1. The negative ones are going to go down on bottom 7 * X + 2. Technically, we could factor out that X ^2 -, 1 and if it could reduce, we would want to. In this case it can't. The last example is going to be a change of base formula. Most calculators don't have base 5 on them. Some of the newer ones do, but most the calculators that are out and available only have base 10 and base E So we're going to rewrite log base 5 of 13 as log understood 10. If there's not a number there of 13 over log understood 4:50 or log 13 over log 5, we could also rewrite that as the natural log of 13 over the natural log of five. It's a ratio. And when we put it in our calculator, it will come out the same whether we use base 10 or base E. Thank you and have a wonderful day.