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M16: Derivation of the Base-Changing Property

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Hi, my name is Rebecca Muller. During this session, we're going to look at properties of logarithms. Now, you may recall that when you first learned about exponents, you learned a bunch of properties of exponents. So it makes sense that we're now going to look at that in terms of logarithms as the fact that logarithms are really exponents. So here are some specific topics we want to concern ourselves with. We want to look at log base b of 1, log base b of b to the nth power, log of a product, log of a quotient, log of a power, expanding a logarithm, and condensing a sum or difference of logarithms, the one-to-one properties, and then the base-changing formula. So let's begin now. We're going to start with looking at a couple of properties that you should be familiar with already from the fact that we started already on logarithmic functions in an earlier section. So if we look at log base b of 1, we'd like to be able to determine what that value is going to equal. Well, this says I'm going to take the base b, I'm going to raise it to an exponent power, and I'm going to equal the value of 1. So if you think about the exponent property, we have b raised to a power equals 1. What is the power that I'm going to have to raise be to in order to end up with 1? And our base to the 0 power equals 1. So what that tells us is log base b of 1 equals 0. Now, I find that the best way to remember these properties is to do so in words. So what I'm going to recall this as is the following. The log of 1-- that's a G-- is 0. Remember that when we're working these problems with logarithms, we're still going to have the same restrictions on our base that we had when we were talking about logarithmic functions. That is, our base has to be positive and not equal to 1. Let's now look at the second property of logarithms that, again, you might be familiar with already. That is log base b of b. Now, what would that have to equal? I'm looking for the exponent that when I raise b to that power gives me back b. In other words, I'm having b raised to some power equals b. What would that power have to be? Well, we recognize that b to the first power equals b, so log base b of b equals 1. And again, in words, log base b of b is 1. Now, I'm going to generalize this one a little bit. And that is, if we look at log base b of b to any power n, what can we say about that? It just says we're looking for the exponent that we raise the base b to in order to end up with b raised to the nth power. It's almost redundant to say it, but exponentially what it looks like is this. b raised to some power equals b raised to the nth power. Because of the one-to-one property of exponential functions, we know that the power has to equal n. So again, in words, log base b of b to the nth power is n. Now we're going to review some of the exponent rules with which you should be familiar already, and see how they translate into logarithmic properties. So I know you're familiar with this one. If I have a base raised to the nth power and I multiply it times the same base raised to the nth power, what do we do to the exponent powers? a to the nth times a to the nth is going to equal a raised of the m plus n power. So the reason for showing you this is because when I get to logarithms, I want you to make the connection between on the left-hand side, notice we are multiplying something. On the right-hand side, we end up adding the exponents. The property of logarithms that follows this rule for exponents is as follows. It's log base b of the product m times n equals log base b of m plus log base b of n. And so what we notice, again, in words, is the log of a product-- that's what happens over here-- is equal to the sum the logs. Let's give a second one now before we go on to examples, and that is the one that's comparable to this, if instead of multiplying, you're dividing. So with exponents, you may recall that if I have a to the nth power divided by a to the nth power, to combine those, that is, we're going to rewrite this as a to the m minus nth power. And so with logarithms, that rule looks as follows. Log base b of m divided by n is equal to log base b of m minus log base b of n. And if we want to put this into words, we can say that the log of a quotient-- remember that the result of division is called a quotient-- is equal to-- and remember, subtraction, the word is the difference. So we're going to have the difference of the logs. Now, before we move off of these properties, let's go ahead and look at a couple of examples using those. Actually more than a couple. I'm going to do about four. So the first one looks like this. What if I have the log of 25 plus the log of 4? Now, recall that when the log base is not written, that is the common logarithm, which means we're looking at base 10. So individually, I don't know what those values are going to equal. Like, what is the exponent I raise 10 to in order to end up with 25? I don't know. Or what is the exponent I raise 10 to in order to end up with 4? I don't know that offhand. However, if we use that first property-- only I'm going to use it kind of going this way-- that is, I have the sum of logs here, I can rewrite this as a log of a product. This is going to be the log, base 10, of course, of the product of 25 times 4. And 25 times 4 is 100. Now, what is the exponent power that I need to put on the base 10 in order to end up with 100? And we know that 10 squared is equal to 100, so that result is 2. As a second example, let's look at the natural log of 1 divided by the square root of e. Again, recall that the natural log means log base e. Well, now I have a log of a quotient. So I can rewrite this as the natural log of 1 minus the natural log of the square root of e. And we know that the log of 1 is equal to 0. So we can rewrite it in this way. And what's going to happen over here? I'm looking for the exponent power that I need to put on base e in order to end up with e to the-- that is going to be a square root. And e to the 1/2 power equals the square root. So we're going to have natural log of square root of e is the same as the natural log of e to the 1/2 power. And now we're looking at that property that we saw a few moments ago, where we have the base. The log of a base where the base is going to be to the same power is going to equal the power. So we're going to end up with 0 minus 1/2, and that's going to give us a negative 1/2. Let's continue with our examples. We have log base b of 1 divided by n. And so again, this is a log of a quotient. We can rewrite this as log base b of 1 minus log base b of n. Log base b of 1 is log of 1, which is equal to 0. So 0 minus log base b of n. And so this is going to give us negative log base b of n. Now, I'm actually going to come back to this particular example in a few minutes when I show you another property. So keep it in mind. The log base b of 1 over n is the same as negative log base b of n. And then finally, let's look at log base 2 of 4 divided by x. And we want to add that to log base 2 of x. So we have a sum of two logs. We're going to end up combining those into the log of a product. So this is going to give us log base 2 of 4 over x multiplied times x. And that's going to simply equal log base 2 of 4. And what is the exponent that we need to raise 2 to in order to end up with 4? Well, we know that 2 squared is 4, so that is going to be our result here. So it simplifies quite nicely. Now here's another rule for exponents that you may recall. Has to do with taking a base, raising it to a power, and then raising that to another power. Do you recall what you need to do to the exponents here? Well, you might remember that what we do is multiply. So we end up with a to the m then raised to the nth power, equals a raised to the m times nth power. Now, what does that look like when we're dealing with logarithms? Here's an example. We're going to have log base b of M raised to the y power. And what happens is-- notice that we have this part we can kind of think of as an exponent. And then we have this part that's an exponent. And what is going to occur is I'm going to take that exponent that's up here, and I'm going pull it right in front and multiply it times log base b of M. Now, that may seem like a strange property at first, but if you recall that we're thinking of logarithms as exponents, what you're really doing over to the right is multiplying the two exponents together, like what we did in the property of exponents that you're familiar with. So again, let's think about this in words. This is the log of a power is equal to the product of the power and the log. That one perhaps is not going to come as trippingly off the tongue as the other two were, but still, that's a way to think about it in words. So I told you I was going to revisit a problem, and I'm ready to do that now. Let's look at log base b of 1 divided by n. We used the property for the log of a quotient a few minutes ago. But now I'd like to use this property to demonstrate that we get the same result. One thing we can do is recall that if you have 1 over N, that is the same thing as N raised to the negative 1 power. And so now, using this new property, we can say, oh, I'm going to take this exponent power of negative 1. I'm going to pull it into the front of the logarithm and multiply it times the log, as it looks here. And therefore, negative 1 times-- this is going to give us negative log base b of n. So of course, we end up with the same result, which is part of the beauty of mathematics. Now, there are occasions when you're working with a very complicated logarithm that it's nice to be able to expand it into a sum or difference of multiples of logarithms. And I'm going to demonstrate that now with the next two examples. So we're going to look at an expansion of our natural log. And in the numerator, I'm going to have 3x squared. In the denominator, I'm going to have the square root of 5y. So here, what you're trying to do is think about properties you might be able to use to where this is now not going to look as complicated as it starts off looking. And the first thing that strikes me is the fact that you notice you have a fraction here, which means you're looking at the log of a quotient. The property we can use is that the log of a quotient is equal to the difference of the logarithms. So I'm going to start there. I'm just going to go ahead and write down here. This is going to be equal to. We're going to have the natural log of the numerator, which is going to be 3x squared. And then we're going to have a subtraction here, because we need to go the difference, of the natural log of the square root of 5y. Now, what can we do further? Well, I'm going to look at each term individually next. So in the first one, I notice that I have 3 multiplied times x squared. So I can notice that this is a log of a product. The log of a product is equal to the sum of the logs. So I will go ahead and take this first term and rewrite it as the natural log of 3 plus the natural log of x squared. What can we do with the second term? Well, you have to recall that the square root of 5y is the same as 5y, both parts, 5 and y, raised to the 1/2 power. So be a little careful about that part. Don't just think it's the y to the 1/2 power. Now, continuing on, what I notice is in the first term, I can't really do anything. That's just going to be the natural log of 3. In the second term, I can use the idea of the log of a power. So this exponent power, too, can be pulled in front as a factor, and I end up with 2 natural log x. Well, the same thing can occur to the last one. I can take the 1/2 from the exponent, and pull it in front, and end up with 1/2 times the natural log of 5y. So are you done yet? Almost. I actually can do one more step. And that is the fact that I notice that I have the natural log of a product. So that's going to be the sum of the logs. Be a little cautious here. You see how you have that 1/2 multiplied times that? It's going to have to be multiplied times everything that comes afterwards. So I'm going to write it down in two different steps so you can see it in all its glory. We're going to have the natural log of 3 plus 2 natural log x minus-- the 1/2 is going to have to be multiplied times everything that follows. And that's going to be a sum. It's going to be the natural log of 5 plus the natural log of y. And then we can distribute the multiplication by 1/2 to give us our final result, natural log of 3 plus 2 natural log x minus 1/2 natural log 5 minus 1/2 natural log y. Now, none of these terms can be combined. And of course, I don't want to combine them. I just now spent time expanding it. When you look at this fact that there are four terms here, of course that makes it more complicated in that there are four terms, but it makes it easier. Each term individually is easier to work with. So there are going to be times when you want to be able to do an expansion of a logarithm into this sum or difference of multiples of logarithms. Let's look at a second example where we do an expansion now. So this one, we're going to look at expanding. And we're going to have log base 5. And then we're going to have 25 multiplied times x squared minus 4. So the first thing that I notice about it is that we definitely have a product inside the bracket. We have 25 multiplied times x squared minus 4. That means we can rewrite the log of a product as the sum of the logs. So I'll do that first. We have log base 5 of 25 plus log base 5 of x squared minus 4. Continuing on, we notice that log base 5 of 25 is actually something we can evaluate pretty easily. What is the exponent I'm going to put on base 5 to end up with 25? That result's going to be 2. In the second one, it looks like I can't do anything with it, because there is no rule for taking the log of a difference. But it turns out that that expression is factorable. So I can go a little bit further with this one. This is going to be plus. We can rewrite this as log base 5 of. And I'm going to go ahead and factor the difference of two squares as x minus 2 times x plus 2. And now I'm looking at the log of a product. So I can use that rule once more to give us 2 plus log base 5 of x minus 2 plus log base 5 of x plus 2. Now, sometimes when you're using these expansions, it's as important to know when you can't go further as it is to know when you can. So notice that at this point, this was a difference. I could not go further just in the fact that it was a difference, but I could because it was factorable. Right here, you cannot go any further here. The log base 5 of a difference, x minus 2, is going to have to stay right there. And so this would be the expansion that we come up with for this expression. It's time for a quick quiz. Use properties of logarithms to simplify the expression natural log open parenthesis x times e to the x close parenthesis. Is the correct result A. Natural log x times natural log e to the x. B. Natural log x plus x. Or C. natural log x plus e. Choose A, B or C. Your correct, the simplification is natural log of x plus x. Sorry the correct simplification is B. Natural log of x plus x. We should begin by using the property that says we can rewrite the log of a product which is what's going on inside the parentheses x times e to the x as the sum of logarithms. So the first step is to rewrite natural log of x e to the x as the natural log of x plus the natural log of e to the x. Now the next step is to rewrite the second term as simply x because the first term is as simple as it gets said natural log of x plus in the second term we're looking for the exponent that we need to put on base e in order to end up with e to the x power. That's what that really is saying to you. So what is the exponent we need to put on e is going to be the value x. So our answer is natural log x plus x. Now, it turns out that sometimes in equations or expressions, you may be given a string of logarithms, where putting them together will actually make it an easier process for you to do whatever's next, or putting them together just makes it look simpler. So this is, in essence, the opposite of what we just did. We had taken expressions that were together as a logarithm and we expanded them out. Now we're going to do something called condensing them. So here's the first example. We want to condense the following string of logarithms. So this is going to have the natural log of x squared minus the natural log of x plus 1 plus 1/2 times the natural log of x. Again, the idea is to combine these into a single logarithm. So one thing you might notice is, well, let's just look at the first two to start with. We see that we have a difference of two logarithms. So we're going to go ahead and use the rule that the difference of logs is equal to the log of a quotient. That means I can rewrite these first two terms combined as the natural log of, and we're going to take the x squared and divide by x plus 1. And then we have our third term. Now, before I'm able to combine these terms, you see the 1/2 that's right here? I cannot combine them right now. I need to use the log of a power rule in order to help us to where we're going to end up only seeing natural log of something plus natural log of something. In other words, I want to take that 1/2 and pull it into the exponent power. So I'm going to rewrite my first term. And in the second term, I'll write this as the natural log of x raised to the 1/2 power. Now that I have a sum of two logs, where I don't have any kind of values in front, no multiples, we're going to be able to use our rule once more for the sum of two logs equals the log of the product. So I'm going to multiply this expression times x to the 1/2. And as I'm doing it, just note that x to the 1/2 is the same thing as the square root of x. I'm going to go ahead and write it like that. So this is going to equal the natural log of. When I multiply square root of x times this fraction, it ends up in the numerator. So we'll end up with x squared multiplied times the square root of x divided by the x plus 1. And we've been able now to combine these three logarithms into a single logarithm. Now let's look at a second example where we're going to end up condensing logarithms. Here's the problem we want to consider. Let's condense 1/3 log of 8 minus 2 log of x. And what I'm going do is pause for a minute, because what I'd like you to do is to try to do this problem on your own. So let's stop the recording, and try it on your own, and then come back when you want to look at your work. So now let's see how you did. We notice that we have a difference of two logarithms, but be careful. You cannot write this as a log of a quotient yet. Notice you have that factor of 1/3 in front. That's going to have to be pulled up into the exponent before you're able to combine it with any other logarithm. Same thing here. The 2 that's in front will have to become an exponent. So I'm going to use the property for the log of a power in order to rewrite this as my first step. This is going to end up being log of 8 to the 1/3 power minus log of x squared. Now, this first one doesn't have any kind of variable in it, but we are raising 8 to the 1/3 power. And that can be evaluated. Recall that a 1/3 power means a cube root. So I'm going to go ahead and just write in that step. We've got the log of the cube root of 8. And I'll just rewrite the second one as is. And go ahead and evaluate it. What is going to be the cube root of 8? That's just going to end up being the value 2. So we can evaluate it any time that ends up being something that's going to help us simplify. So we have log of 2 minus log of x squared. Now that it's written as the log of m minus the log of n which mimics the property, we can rewrite this as a log of a quotient. So this is going to equal the log of the quotient, where 2 is in the numerator and x squared is in the denominator, and we have now condensed these two logarithms into a single logarithm. Continuing on with our discussion of properties of logarithms, we're going to look at the one-to-one properties. So for x greater than 0, b greater than 0, and b not equal to 1, if we have M equals N, then log base b of M equals log base b of N. Now, in words, what that says is if I have an equation where two sides are equal, of course, then I can take the logarithm to any base of both sides of the equation, and that will also be equal. Second part reads, if log base b of M equals log base b of N, then M equals N. And that second part is just the reverse of that, which says if I have two logarithms that are equal to each other and to the same base, of course, then I can equate the values that I'm taking the logarithm of. So let's see how we can use this in order to develop something called the base-changing formula. And here's the example I want to go with. Let's say we have 2 to the x power equals 3. Now, we're trying to solve for x in this equation. And we're going to do more on exponential equations in a future section, but right now, what can we do? Well, we know that we can change this into its logarithmic form. So an exponential equation has a comparable logarithmic form. And in this case, it's going to be log base 2 of 3 equals x. Well, that's great, but if I wanted a decimal approximation for that, I would be wanting to grab my calculator. Most calculators will do two bases for you. They'll do a base 10, the common log, and they'll do the base e, which is the natural log. And of course, here, I'm working with base 2. So I'm not going to be able to press a key on the calculator to figure out what this is going to equal, at least not in the current format. So let's just look at this a little bit. I'm going to take a generalized format. A to x power equals c. And now, we know that we can rewrite that in logarithmic form. So rewriting that would read log base a of c equals x. So I know that that's comparable, which is what we just did. But it turns out I can now use that one-to-one property that allows me to take an equation where two things are equal, and take the logarithm of both sides of the equation. Now, what I'm going to do is introduce a new base here. And just to be generalizable, I'm just going to say I'm going to introduce log base b. So this is now going to read log base b of a to x power equals log base b of c. Again, this is that one-to-one property. What can I do now? Well, we can use properties of logarithms. In particular, we have the log of a power. So we can take the exponent x, and pull it in front, and make it a factor as follows. x times log base b of a on the left-hand side is going to equal log base b of c on the right-hand side. And finally, we can solve for x now. Notice that we can divide both sides of this equation by the value that x is being multiplied by. So we'll end up with x equals. We'll have log base b of c divided by log base b of a. Now, let's go back a little bit and compare it to what I had over here. Notice the c ends up in the numerator, and you see the c is above the a here. So that's a key to remembering what's going on. Notice I'm introducing a new base. Well, what bases can you use with your calculator? Again, we can either use base 10 or we can use base e. So let's see what we can do with what we've just found out. Let's go back to the original equation. We had 2 to the x equalled 3. We changed it to its logarithmic form, log base 2 of 3 equals x. At this point, if we want to use a calculator, we need to have a different base than base 2. That is, we need to have base 10 or base e. I can introduce the new base using this. I'm going to go from this format to that format. And the base that I'm going to choose to use is base e. So I'm going to write this as a natural log. Notice that the value in the numerator is above the base. So that's log base b of c over here. The value in the denominator is the base itself, so it's log base b of a. So rewriting this, I can have the natural log 3 divided by the natural log of 2. Now, both of these are exact answers. If you want to use a calculator now and you have one, go out and follow along with me. You'll substitute into your calculator the natural log of 3, and then divide that by the natural log of 2. You should come up with the answer that is approximately equal to 1.585. And just to check ourselves, let's understand what we found out. We know now that 2 raised to the 1.585 power should approximately equal the value of 3. Of course, we rounded, so that's why it's not equal to 3. So we've done a number of problems dealing with properties of logarithms now. It's time for you to try some on your own. Good luck. It's time for another quick quiz. Solve the equation 10 to the x equals 5. Is the answer A. Log of 5 equals x. B. x equals natural log of 5 divided by natural log of 10. C. x equals log base 2 of 5 divided by log base 2 of 10. D. All the above are valid answers. Choose A, B, C, or D. Your correct all of the above are valid answers. Sorry but all of the above are valid answers. Let's begin by taking the exponential equation 10 to the x equals 5 and rewriting it in logarithmic form. So we can write log base 10 of 5 equals x. Now based 10 is the common logarithm which allows us to rewrite this expression as log 5 not writing down the subscript of 10, so A is the correct result. So why are B and C both correct results, well this stems from the fact that we have something called a base changing formula. Let's re-write... Let's write down what that base changing formula tells us that we're able to do. It reads that if I have log base a of C, I can say that's equivalent to log base b of c divided by log base b of a. In other words I can introduce a new base to the problem that didn't appear originally. So what that allows me to do is if we look at part B, I've introduced the new base of e. So I can take my answer which again was log base 10 of 5 putting the 10 back in there and re-writing it as log base e of 5 which is the natural log of 5 divided by log base e of 10 which is the natural log of 10. So that means that B is the correct answer. If I look at part C, we have the introduction of a new base 2 so again if I take log base 10 of 5 and introduced the new base of 2, I can rewrite this as log base 2 of 5 divided by log base 2 of 10. Which means that C is also correct. So all together D. Which allows me to have all of the above answers as valid is the correct choice.
Hi, my name is Rebecca Muller. During this session, we're going to look at properties of logarithms. Now, you may recall that when you first learned about exponents, you learned a bunch of properties of exponents. So it makes sense that we're now going to look at that in terms of logarithms as the fact that logarithms are really exponents. So here are some specific topics we want to concern ourselves with. We want to look at log base b of 1, log base b of b to the nth power, log of a product, log of a quotient, log of a power, expanding a logarithm, and condensing a sum or difference of logarithms, the one-to-one properties, and then the base-changing formula. So let's begin now. We're going to start with looking at a couple of properties that you should be familiar with already from the fact that we started already on logarithmic functions in an earlier section. So if we look at log base b of 1, we'd like to be able to determine what that value is going to equal. Well, this says I'm going to take the base b, I'm going to raise it to an exponent power, and I'm going to equal the value of 1. So if you think about the exponent property, we have b raised to a power equals 1. What is the power that I'm going to have to raise be to in order to end up with 1? And our base to the 0 power equals 1. So what that tells us is log base b of 1 equals 0. Now, I find that the best way to remember these properties is to do so in words. So what I'm going to recall this as is the following. The log of 1-- that's a G-- is 0. Remember that when we're working these problems with logarithms, we're still going to have the same restrictions on our base that we had when we were talking about logarithmic functions. That is, our base has to be positive and not equal to 1. Let's now look at the second property of logarithms that, again, you might be familiar with already. That is log base b of b. Now, what would that have to equal? I'm looking for the exponent that when I raise b to that power gives me back b. In other words, I'm having b raised to some power equals b. What would that power have to be? Well, we recognize that b to the first power equals b, so log base b of b equals 1. And again, in words, log base b of b is 1. Now, I'm going to generalize this one a little bit. And that is, if we look at log base b of b to any power n, what can we say about that? It just says we're looking for the exponent that we raise the base b to in order to end up with b raised to the nth power. It's almost redundant to say it, but exponentially what it looks like is this. b raised to some power equals b raised to the nth power. Because of the one-to-one property of exponential functions, we know that the power has to equal n. So again, in words, log base b of b to the nth power is n. Now we're going to review some of the exponent rules with which you should be familiar already, and see how they translate into logarithmic properties. So I know you're familiar with this one. If I have a base raised to the nth power and I multiply it times the same base raised to the nth power, what do we do to the exponent powers? a to the nth times a to the nth is going to equal a raised of the m plus n power. So the reason for showing you this is because when I get to logarithms, I want you to make the connection between on the left-hand side, notice we are multiplying something. On the right-hand side, we end up adding the exponents. The property of logarithms that follows this rule for exponents is as follows. It's log base b of the product m times n equals log base b of m plus log base b of n. And so what we notice, again, in words, is the log of a product-- that's what happens over here-- is equal to the sum the logs. Let's give a second one now before we go on to examples, and that is the one that's comparable to this, if instead of multiplying, you're dividing. So with exponents, you may recall that if I have a to the nth power divided by a to the nth power, to combine those, that is, we're going to rewrite this as a to the m minus nth power. And so with logarithms, that rule looks as follows. Log base b of m divided by n is equal to log base b of m minus log base b of n. And if we want to put this into words, we can say that the log of a quotient-- remember that the result of division is called a quotient-- is equal to-- and remember, subtraction, the word is the difference. So we're going to have the difference of the logs. Now, before we move off of these properties, let's go ahead and look at a couple of examples using those. Actually more than a couple. I'm going to do about four. So the first one looks like this. What if I have the log of 25 plus the log of 4? Now, recall that when the log base is not written, that is the common logarithm, which means we're looking at base 10. So individually, I don't know what those values are going to equal. Like, what is the exponent I raise 10 to in order to end up with 25? I don't know. Or what is the exponent I raise 10 to in order to end up with 4? I don't know that offhand. However, if we use that first property-- only I'm going to use it kind of going this way-- that is, I have the sum of logs here, I can rewrite this as a log of a product. This is going to be the log, base 10, of course, of the product of 25 times 4. And 25 times 4 is 100. Now, what is the exponent power that I need to put on the base 10 in order to end up with 100? And we know that 10 squared is equal to 100, so that result is 2. As a second example, let's look at the natural log of 1 divided by the square root of e. Again, recall that the natural log means log base e. Well, now I have a log of a quotient. So I can rewrite this as the natural log of 1 minus the natural log of the square root of e. And we know that the log of 1 is equal to 0. So we can rewrite it in this way. And what's going to happen over here? I'm looking for the exponent power that I need to put on base e in order to end up with e to the-- that is going to be a square root. And e to the 1/2 power equals the square root. So we're going to have natural log of square root of e is the same as the natural log of e to the 1/2 power. And now we're looking at that property that we saw a few moments ago, where we have the base. The log of a base where the base is going to be to the same power is going to equal the power. So we're going to end up with 0 minus 1/2, and that's going to give us a negative 1/2. Let's continue with our examples. We have log base b of 1 divided by n. And so again, this is a log of a quotient. We can rewrite this as log base b of 1 minus log base b of n. Log base b of 1 is log of 1, which is equal to 0. So 0 minus log base b of n. And so this is going to give us negative log base b of n. Now, I'm actually going to come back to this particular example in a few minutes when I show you another property. So keep it in mind. The log base b of 1 over n is the same as negative log base b of n. And then finally, let's look at log base 2 of 4 divided by x. And we want to add that to log base 2 of x. So we have a sum of two logs. We're going to end up combining those into the log of a product. So this is going to give us log base 2 of 4 over x multiplied times x. And that's going to simply equal log base 2 of 4. And what is the exponent that we need to raise 2 to in order to end up with 4? Well, we know that 2 squared is 4, so that is going to be our result here. So it simplifies quite nicely. Now here's another rule for exponents that you may recall. Has to do with taking a base, raising it to a power, and then raising that to another power. Do you recall what you need to do to the exponents here? Well, you might remember that what we do is multiply. So we end up with a to the m then raised to the nth power, equals a raised to the m times nth power. Now, what does that look like when we're dealing with logarithms? Here's an example. We're going to have log base b of M raised to the y power. And what happens is-- notice that we have this part we can kind of think of as an exponent. And then we have this part that's an exponent. And what is going to occur is I'm going to take that exponent that's up here, and I'm going pull it right in front and multiply it times log base b of M. Now, that may seem like a strange property at first, but if you recall that we're thinking of logarithms as exponents, what you're really doing over to the right is multiplying the two exponents together, like what we did in the property of exponents that you're familiar with. So again, let's think about this in words. This is the log of a power is equal to the product of the power and the log. That one perhaps is not going to come as trippingly off the tongue as the other two were, but still, that's a way to think about it in words. So I told you I was going to revisit a problem, and I'm ready to do that now. Let's look at log base b of 1 divided by n. We used the property for the log of a quotient a few minutes ago. But now I'd like to use this property to demonstrate that we get the same result. One thing we can do is recall that if you have 1 over N, that is the same thing as N raised to the negative 1 power. And so now, using this new property, we can say, oh, I'm going to take this exponent power of negative 1. I'm going to pull it into the front of the logarithm and multiply it times the log, as it looks here. And therefore, negative 1 times-- this is going to give us negative log base b of n. So of course, we end up with the same result, which is part of the beauty of mathematics. Now, there are occasions when you're working with a very complicated logarithm that it's nice to be able to expand it into a sum or difference of multiples of logarithms. And I'm going to demonstrate that now with the next two examples. So we're going to look at an expansion of our natural log. And in the numerator, I'm going to have 3x squared. In the denominator, I'm going to have the square root of 5y. So here, what you're trying to do is think about properties you might be able to use to where this is now not going to look as complicated as it starts off looking. And the first thing that strikes me is the fact that you notice you have a fraction here, which means you're looking at the log of a quotient. The property we can use is that the log of a quotient is equal to the difference of the logarithms. So I'm going to start there. I'm just going to go ahead and write down here. This is going to be equal to. We're going to have the natural log of the numerator, which is going to be 3x squared. And then we're going to have a subtraction here, because we need to go the difference, of the natural log of the square root of 5y. Now, what can we do further? Well, I'm going to look at each term individually next. So in the first one, I notice that I have 3 multiplied times x squared. So I can notice that this is a log of a product. The log of a product is equal to the sum of the logs. So I will go ahead and take this first term and rewrite it as the natural log of 3 plus the natural log of x squared. What can we do with the second term? Well, you have to recall that the square root of 5y is the same as 5y, both parts, 5 and y, raised to the 1/2 power. So be a little careful about that part. Don't just think it's the y to the 1/2 power. Now, continuing on, what I notice is in the first term, I can't really do anything. That's just going to be the natural log of 3. In the second term, I can use the idea of the log of a power. So this exponent power, too, can be pulled in front as a factor, and I end up with 2 natural log x. Well, the same thing can occur to the last one. I can take the 1/2 from the exponent, and pull it in front, and end up with 1/2 times the natural log of 5y. So are you done yet? Almost. I actually can do one more step. And that is the fact that I notice that I have the natural log of a product. So that's going to be the sum of the logs. Be a little cautious here. You see how you have that 1/2 multiplied times that? It's going to have to be multiplied times everything that comes afterwards. So I'm going to write it down in two different steps so you can see it in all its glory. We're going to have the natural log of 3 plus 2 natural log x minus-- the 1/2 is going to have to be multiplied times everything that follows. And that's going to be a sum. It's going to be the natural log of 5 plus the natural log of y. And then we can distribute the multiplication by 1/2 to give us our final result, natural log of 3 plus 2 natural log x minus 1/2 natural log 5 minus 1/2 natural log y. Now, none of these terms can be combined. And of course, I don't want to combine them. I just now spent time expanding it. When you look at this fact that there are four terms here, of course that makes it more complicated in that there are four terms, but it makes it easier. Each term individually is easier to work with. So there are going to be times when you want to be able to do an expansion of a logarithm into this sum or difference of multiples of logarithms. Let's look at a second example where we do an expansion now. So this one, we're going to look at expanding. And we're going to have log base 5. And then we're going to have 25 multiplied times x squared minus 4. So the first thing that I notice about it is that we definitely have a product inside the bracket. We have 25 multiplied times x squared minus 4. That means we can rewrite the log of a product as the sum of the logs. So I'll do that first. We have log base 5 of 25 plus log base 5 of x squared minus 4. Continuing on, we notice that log base 5 of 25 is actually something we can evaluate pretty easily. What is the exponent I'm going to put on base 5 to end up with 25? That result's going to be 2. In the second one, it looks like I can't do anything with it, because there is no rule for taking the log of a difference. But it turns out that that expression is factorable. So I can go a little bit further with this one. This is going to be plus. We can rewrite this as log base 5 of. And I'm going to go ahead and factor the difference of two squares as x minus 2 times x plus 2. And now I'm looking at the log of a product. So I can use that rule once more to give us 2 plus log base 5 of x minus 2 plus log base 5 of x plus 2. Now, sometimes when you're using these expansions, it's as important to know when you can't go further as it is to know when you can. So notice that at this point, this was a difference. I could not go further just in the fact that it was a difference, but I could because it was factorable. Right here, you cannot go any further here. The log base 5 of a difference, x minus 2, is going to have to stay right there. And so this would be the expansion that we come up with for this expression. It's time for a quick quiz. Use properties of logarithms to simplify the expression natural log open parenthesis x times e to the x close parenthesis. Is the correct result A. Natural log x times natural log e to the x. B. Natural log x plus x. Or C. natural log x plus e. Choose A, B or C. Your correct, the simplification is natural log of x plus x. Sorry the correct simplification is B. Natural log of x plus x. We should begin by using the property that says we can rewrite the log of a product which is what's going on inside the parentheses x times e to the x as the sum of logarithms. So the first step is to rewrite natural log of x e to the x as the natural log of x plus the natural log of e to the x. Now the next step is to rewrite the second term as simply x because the first term is as simple as it gets said natural log of x plus in the second term we're looking for the exponent that we need to put on base e in order to end up with e to the x power. That's what that really is saying to you. So what is the exponent we need to put on e is going to be the value x. So our answer is natural log x plus x. Now, it turns out that sometimes in equations or expressions, you may be given a string of logarithms, where putting them together will actually make it an easier process for you to do whatever's next, or putting them together just makes it look simpler. So this is, in essence, the opposite of what we just did. We had taken expressions that were together as a logarithm and we expanded them out. Now we're going to do something called condensing them. So here's the first example. We want to condense the following string of logarithms. So this is going to have the natural log of x squared minus the natural log of x plus 1 plus 1/2 times the natural log of x. Again, the idea is to combine these into a single logarithm. So one thing you might notice is, well, let's just look at the first two to start with. We see that we have a difference of two logarithms. So we're going to go ahead and use the rule that the difference of logs is equal to the log of a quotient. That means I can rewrite these first two terms combined as the natural log of, and we're going to take the x squared and divide by x plus 1. And then we have our third term. Now, before I'm able to combine these terms, you see the 1/2 that's right here? I cannot combine them right now. I need to use the log of a power rule in order to help us to where we're going to end up only seeing natural log of something plus natural log of something. In other words, I want to take that 1/2 and pull it into the exponent power. So I'm going to rewrite my first term. And in the second term, I'll write this as the natural log of x raised to the 1/2 power. Now that I have a sum of two logs, where I don't have any kind of values in front, no multiples, we're going to be able to use our rule once more for the sum of two logs equals the log of the product. So I'm going to multiply this expression times x to the 1/2. And as I'm doing it, just note that x to the 1/2 is the same thing as the square root of x. I'm going to go ahead and write it like that. So this is going to equal the natural log of. When I multiply square root of x times this fraction, it ends up in the numerator. So we'll end up with x squared multiplied times the square root of x divided by the x plus 1. And we've been able now to combine these three logarithms into a single logarithm. Now let's look at a second example where we're going to end up condensing logarithms. Here's the problem we want to consider. Let's condense 1/3 log of 8 minus 2 log of x. And what I'm going do is pause for a minute, because what I'd like you to do is to try to do this problem on your own. So let's stop the recording, and try it on your own, and then come back when you want to look at your work. So now let's see how you did. We notice that we have a difference of two logarithms, but be careful. You cannot write this as a log of a quotient yet. Notice you have that factor of 1/3 in front. That's going to have to be pulled up into the exponent before you're able to combine it with any other logarithm. Same thing here. The 2 that's in front will have to become an exponent. So I'm going to use the property for the log of a power in order to rewrite this as my first step. This is going to end up being log of 8 to the 1/3 power minus log of x squared. Now, this first one doesn't have any kind of variable in it, but we are raising 8 to the 1/3 power. And that can be evaluated. Recall that a 1/3 power means a cube root. So I'm going to go ahead and just write in that step. We've got the log of the cube root of 8. And I'll just rewrite the second one as is. And go ahead and evaluate it. What is going to be the cube root of 8? That's just going to end up being the value 2. So we can evaluate it any time that ends up being something that's going to help us simplify. So we have log of 2 minus log of x squared. Now that it's written as the log of m minus the log of n which mimics the property, we can rewrite this as a log of a quotient. So this is going to equal the log of the quotient, where 2 is in the numerator and x squared is in the denominator, and we have now condensed these two logarithms into a single logarithm. Continuing on with our discussion of properties of logarithms, we're going to look at the one-to-one properties. So for x greater than 0, b greater than 0, and b not equal to 1, if we have M equals N, then log base b of M equals log base b of N. Now, in words, what that says is if I have an equation where two sides are equal, of course, then I can take the logarithm to any base of both sides of the equation, and that will also be equal. Second part reads, if log base b of M equals log base b of N, then M equals N. And that second part is just the reverse of that, which says if I have two logarithms that are equal to each other and to the same base, of course, then I can equate the values that I'm taking the logarithm of. So let's see how we can use this in order to develop something called the base-changing formula. And here's the example I want to go with. Let's say we have 2 to the x power equals 3. Now, we're trying to solve for x in this equation. And we're going to do more on exponential equations in a future section, but right now, what can we do? Well, we know that we can change this into its logarithmic form. So an exponential equation has a comparable logarithmic form. And in this case, it's going to be log base 2 of 3 equals x. Well, that's great, but if I wanted a decimal approximation for that, I would be wanting to grab my calculator. Most calculators will do two bases for you. They'll do a base 10, the common log, and they'll do the base e, which is the natural log. And of course, here, I'm working with base 2. So I'm not going to be able to press a key on the calculator to figure out what this is going to equal, at least not in the current format. So let's just look at this a little bit. I'm going to take a generalized format. A to x power equals c. And now, we know that we can rewrite that in logarithmic form. So rewriting that would read log base a of c equals x. So I know that that's comparable, which is what we just did. But it turns out I can now use that one-to-one property that allows me to take an equation where two things are equal, and take the logarithm of both sides of the equation. Now, what I'm going to do is introduce a new base here. And just to be generalizable, I'm just going to say I'm going to introduce log base b. So this is now going to read log base b of a to x power equals log base b of c. Again, this is that one-to-one property. What can I do now? Well, we can use properties of logarithms. In particular, we have the log of a power. So we can take the exponent x, and pull it in front, and make it a factor as follows. x times log base b of a on the left-hand side is going to equal log base b of c on the right-hand side. And finally, we can solve for x now. Notice that we can divide both sides of this equation by the value that x is being multiplied by. So we'll end up with x equals. We'll have log base b of c divided by log base b of a. Now, let's go back a little bit and compare it to what I had over here. Notice the c ends up in the numerator, and you see the c is above the a here. So that's a key to remembering what's going on. Notice I'm introducing a new base. Well, what bases can you use with your calculator? Again, we can either use base 10 or we can use base e. So let's see what we can do with what we've just found out. Let's go back to the original equation. We had 2 to the x equalled 3. We changed it to its logarithmic form, log base 2 of 3 equals x. At this point, if we want to use a calculator, we need to have a different base than base 2. That is, we need to have base 10 or base e. I can introduce the new base using this. I'm going to go from this format to that format. And the base that I'm going to choose to use is base e. So I'm going to write this as a natural log. Notice that the value in the numerator is above the base. So that's log base b of c over here. The value in the denominator is the base itself, so it's log base b of a. So rewriting this, I can have the natural log 3 divided by the natural log of 2. Now, both of these are exact answers. If you want to use a calculator now and you have one, go out and follow along with me. You'll substitute into your calculator the natural log of 3, and then divide that by the natural log of 2. You should come up with the answer that is approximately equal to 1.585. And just to check ourselves, let's understand what we found out. We know now that 2 raised to the 1.585 power should approximately equal the value of 3. Of course, we rounded, so that's why it's not equal to 3. So we've done a number of problems dealing with properties of logarithms now. It's time for you to try some on your own. Good luck. It's time for another quick quiz. Solve the equation 10 to the x equals 5. Is the answer A. Log of 5 equals x. B. x equals natural log of 5 divided by natural log of 10. C. x equals log base 2 of 5 divided by log base 2 of 10. D. All the above are valid answers. Choose A, B, C, or D. Your correct all of the above are valid answers. Sorry but all of the above are valid answers. Let's begin by taking the exponential equation 10 to the x equals 5 and rewriting it in logarithmic form. So we can write log base 10 of 5 equals x. Now based 10 is the common logarithm which allows us to rewrite this expression as log 5 not writing down the subscript of 10, so A is the correct result. So why are B and C both correct results, well this stems from the fact that we have something called a base changing formula. Let's re-write... Let's write down what that base changing formula tells us that we're able to do. It reads that if I have log base a of C, I can say that's equivalent to log base b of c divided by log base b of a. In other words I can introduce a new base to the problem that didn't appear originally. So what that allows me to do is if we look at part B, I've introduced the new base of e. So I can take my answer which again was log base 10 of 5 putting the 10 back in there and re-writing it as log base e of 5 which is the natural log of 5 divided by log base e of 10 which is the natural log of 10. So that means that B is the correct answer. If I look at part C, we have the introduction of a new base 2 so again if I take log base 10 of 5 and introduced the new base of 2, I can rewrite this as log base 2 of 5 divided by log base 2 of 10. Which means that C is also correct. So all together D. Which allows me to have all of the above answers as valid is the correct choice.