WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about the properties of logarithms.
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In the previous lesson, we introduced the idea of a logarithm, which was defined as log<font size="-6">a</font>(x) = y for a^y = x.
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So, the logarithm of a number is what we would have to raise the base to, to get the number.
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So, log<font size="-6">a</font>(x) = y means a^y = x.
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It is a little bit of a complex idea the first time we talk about it, so the previous lesson is really useful.
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If you haven't already watched the previous lesson, Introduction to Logarithms, I really recommend that you watch it,
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because that will get you a grounding in how these things work--it will really explain things, if it is confusing you.
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Previously, when we investigated exponentiation, we found all sorts of interesting properties,
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such as x^a times x^b = x^a + b--that we could add exponents--
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or that x^-a = 1/x^a--that we flip when we have negative exponents.
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Since logarithms and exponents are so deeply connected (we have this idea that log of something
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equals this other version in exponent world), we might expect logarithms to also have some interesting properties.
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Indeed, they do: this lesson will be all about looking at the properties of logarithms.
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Let's just start with some really basic properties--remember, this is the definition that we will be working with the whole time.
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It is a really good idea to understand this.
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From this, we can immediately see two basic properties: log<font size="-6">a</font>(1) = 0, because a⁰ = 1 for anything;
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that was one of the things that we figured out when we were working with exponents;
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and also, log<font size="-6">a</font>(a) is going to equal 1, because a¹ is just equal to a, because it is just 1a.
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Those are two basic properties that we get, just from the simple definition.
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Next, we can talk about inverses: logarithms and exponentiation are inverse processes.
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If they have the same base, they cancel each other out.
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This is clear when a logarithm acts on exponentiation; if we have log<font size="-6">a</font>(a^x),
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then we are going to get x out of it, because by the definition of a logarithm,
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the number that we need to raise x to--what do we have to raise a to, to get a^x?
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Well, that is going to be a^x; so a is our base for the right side; a^x = a^x, so log<font size="-6">a</font>(a^x) = x.
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It cancels out; a logarithm on exponentiation...they cancel each other out, because we have an a here and an a here as bases.
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So, they cancel out, and we are left with just the x that we originally had.
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So, logarithms cancel out exponentiation, if they are of the same base.
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We just get left with the exponent at the end.
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To see the reverse process of exponentiation, canceling a logarithm, we first need to realize a useful idea.
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Notice that, for any exponent base where a is greater than 0, our exponent base is positive, and it is not equal to 1.
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And for any number x such that x is a positive number, there exists some real number b such that a^b = x.
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In other words, through exponentiation, this a to the something, we can get to any positive number x.
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For any positive number...you say 58 as your x...then with my a, I can raise it to something that will get it to 58.
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And then, if we want, we can name the exponent that will do this; and we called it b here.
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There is some number b such that a^b = x.
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So, for example, we could say that 10 to the something has to be equal to 100.
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Our question is what number it is as our something: 10 to the what equals 100?
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Well, it is going to be 2; we raise 10 to the 2--10²--we get 100.
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So, in this case, our b, if we want to name the number in the box, was 2.
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We could do this for something else: we could say, "3 to the box equals 47."
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Now, this is considerably more complex than 10 to box equals 100.
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3 isn't going to have some nice integer, some nice even power, that we can raise it to, that will put out 47.
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But we can note that there has to be something that fits in that box.
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We don't know what the number is right now, but we are confident--we are sure that there has to be something that we could raise 3 to, to get 47.
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So, we can just call it b; in fact, if you were to work it out through other stuff that we will talk about as we go on in this class,
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if you were to work it out, you would find out that b is approximately equal to 3.504555.
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If you plug that into a calculator, you will see that it is very, very close to being precisely 47.
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So, there is some number out there; we can't get it exactly, but we can get a very good approximation through things that we work out in this course.
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We can be sure that this number does exist--it is out there somewhere, even if we don't know it.
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But that doesn't mean we can't talk about it; we can not know what something is precisely, and still talk about it as a general idea.
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So, we know that this has to be true, because every exponential function, a to the something, has a range of 0 to infinity.
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We have talked about this before, and you can see it in the graphs of it.
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Since it has a range of 0 to infinity--since any number can come out of it, as long as it is a positive number--
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then there has to be some b that will make that number from between 0 and infinity.
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For any x contained in 0 to infinity, for any positive number x, there has to be some b that,
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when we plug it in here, we will end up getting a^b = x.
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We will end up getting that x when we plug it in.
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This idea that there is always an exponent for creating any positive number (it won't work if it is negative;
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but for creating any number) is really important; we will use this fact to prove a variety of properties about logarithms.
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So, we want to keep this idea that, if we have some x, and we know we have some base a,
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then there has to exist some real number b to make a^b = x.
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So, you start with a base a; you start with some sort of destination x; you are guaranteed that there is a b that will get you to that destination through exponentiation.
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With this idea in mind, we can now show inverses in the other direction.
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Consider a: now, by our new idea, we know that there has to be some b such that a^b = x.
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Here is our x; and we just choose the base a--it is going to be connected to the fact that we have the base a here, showing up here.
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So, we know that a to the something equals x; and we are guaranteed that there has to be some b there.
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So, we are talking about it here: we know that there is some b such that a^b = x.
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So, we can say a to the log<font size="-6">a</font> of...and now we replace it with a^b = x,
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so we replace right here, and it becomes a^b, so a.
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Now, we just showed that log<font size="-6">a</font>(a^b) = b, because logs on exponents cancel out.
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So, log<font size="-6">a</font>(a^b) = b, so we have the same thing here.
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log<font size="-6">a</font>(a) will drop us down to just having the b, so that b will move over to being the exponent, and we will have a^b.
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But how did we define a^b in the first place? It was a^b = x.
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So, we know that a^b = x; since we created b based on the fact that a^b = x,
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we have now shown that what we started with ends up being equivalent to x.
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So, we have that a raised to the log<font size="-6">a</font>(x) is equal to x.
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So now, at this point, we have shown inverses in both directions.
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If you take log of an exponent, and they both have the same base, they cancel out, and we are just left with whatever was our power.
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If we take some base and raise it, through exponentiation, to a log, and that base there and the base on the log are the same thing,
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we end up just having what the log was taking as its quantity.
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So, we have inverses shown in both ways; we can cancel out in both directions.
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All right, we will move on to something else: the logarithm of a power.
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Consider if we had some positive number x, and any real number n; then what if we raised to the n and took its log?
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So, we have log<font size="-6">a</font>(x^n); by our key idea,
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we know that there exists some b such that we can write a^b = x--the same thing.
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So, if we want, we can swap out our x here for a^b.
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So, we have a^b to the n now; from our rules about exponents, we know that a^b to the n,
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is the same thing as just a to the b times n; so now, we have b times n, or we could write it as n times b.
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log<font size="-6">a</font>(a) to the n times b becomes canceled out, because we have inverses there,
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since they are the same base, and we are left with n times b.
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Now, do we have another way to express b?
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Well, from the beginning, we know that a^b = x, so we could write that in its logarithmic form.
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And we would show that log<font size="-6">a</font>(x) = b, because a^b = x.
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If we move this over here, a raised to the b becomes x from how we had this originally.
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So, we see that log<font size="-6">a</font>(x) = b, which means that we can swap the b out here for the log<font size="-6">a</font>(x).
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And we have that log<font size="-6">a</font>(x) here; we started with log<font size="-6">a</font>(x^n),
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and now we see that we can take this n here and move it out front, and we have n times log<font size="-6">a</font>(x).
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So, if we have an exponent, we can move it out front.
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Let's look at an example: if we look at log₃(3²), by this property,
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we see that we could take this 2 and move it out front, and we would have 2 times log₃(3).
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Well, log₃(3)...log base something of something, if they are the same something...is just 1.
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What do you have to raise 3 to, to get 3? Just 1.
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So, we have 2 times 1, which is equal to 2.
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What if we look at it the other way?
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Well, log₃(9)...what do we have to raise 3 to, to get 9?
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We just have to raise it to 2; so we have 2 here and 2 here; either way we go at it, we end up getting the same thing.
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We see this property in action.
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OK, now let's consider two positive numbers, m and n (that should be "numbers, m and n"): log<font size="-6">a</font>(mn).
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log base a of the whole quantity, m times n: now, from our key idea, we know that there exist m and n,
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such that we can raise a to the m, and will get M; and we raise a to the n, and we get N.
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So, we can swap these things out: we can swap out here, and we can swap out here.
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And we end up getting log<font size="-6">a</font>(a^ma^n).
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Now, from our work about exponents, we know that a^m times a^n is the same thing as a^m + n.
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The same base means that we can add the exponents, so we have log<font size="-6">a</font>(a^m + n).
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And then, since it is log<font size="-6">a</font> on an exponent base of a, we get cancellation once again through inverses;
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log<font size="-6">a</font> cancels out with that, and we are left with just m + n.
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Now, we can ask ourselves, "Do we have another way to express m and n?"
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Well, from the beginning, we know that a^m = M; that was how we set this up.
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So, log<font size="-6">a</font>(M) = m, because we know that if we raise a to the m, we get M; so we have that.
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We can express it in its exponential form or its logarithmic form.
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So we now can say, "What about looking at it through its logarithmic form?"
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The same thing goes over here with a^n = N; we can express it, instead, as log<font size="-6">a</font>(N) = n.
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So, at this point, we have two different new ways to be able to describe m + n.
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We can swap that out; and so, log<font size="-6">a</font>(M) here becomes here, and log<font size="-6">a</font>(N) = n goes here.
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So, you see this here; you see this here; we can swap that out.
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So, what we started with, log<font size="-6">a</font>(MN), is going to be the same thing as log<font size="-6">a</font>(M) + log<font size="-6">a</font>(N).
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So, if we have a logarithm of a product, we can split it into the sum of the logarithms of the numbers that made up each part of that product.
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Let's look at an example to help clarify this.
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We could look at log<font size="-6">10</font>; we will write it as just log, since the common log can be expressed as just log.
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So, log(1000): we could write this, if we felt like it, as log(100 times 10).
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And then, by this rule that we have here, we could split it; we have products 100 and 10, so we can split it into log(100) + log(10).
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Now, what number do we have to raise 10 to, to get 100?
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We have to raise it to 2; what number do we have to raise 10 to, to get 10? We only have to raise it to 1.
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So, 2 + 1...we end up getting 3.
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Alternatively, we could have done this as log(1000): what number do we have to raise 10 to, to get 1000?
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10 is 1; 100 is 2; 1000 is at 3; so we could also see that log<font size="-6">10</font>(1000) = 3 over here.
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So, it works out either way that we want to approach it.
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So, we see now that we can break up products into two different things being added together.
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What if we took the logarithm of a quotient, log<font size="-6">a</font>(M/N)?
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Well, we could rewrite that as M times N^-1, since we could rewrite M/N as M times 1/N.
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And then, we can rewrite 1/N as just its flip, so we would have N^-1.
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We can swap it around like that.
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From the rule that we just saw, the splitting of products, we can write this as...
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here is the M, so we have log<font size="-6">a</font>(M); and here is the N^-1, so we have log<font size="-6">a</font>(N^-1).
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So, log<font size="-6">a</font>(M)...and then, we also have our rule that we can bring down exponents.
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log(x^n) becomes nlog(x); so we bring this down in the front.
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Since we are bringing down a -1, it just becomes a minus sign here; so we have log<font size="-6">a</font>(M) - log<font size="-6">a</font>(N).
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Thus, log<font size="-6">a</font>(M/N) is equal to log<font size="-6">a</font>(M) - log<font size="-6">a</font>(N).
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So, the logarithm of a quotient becomes the difference of the two logarithms.
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Let's look at an example to help clarify this one, as well.
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So, if we look at log₂(32/2), we could write that, by our new rules, as log₂(32)
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(is the part on the top), and then the part on the bottom is 2, so minus log₂(2).
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log₂(32): what number do we have to raise 2 to, to get 32?
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Well, 2 is 1; 4 is 2; 8 is 3; 16 is 4; 32 is at 5--so we raise 2 to the 5, and we get 32; so log₂(32) is 5.
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What is log₂(2)? -1: 2 to the 1 equals 2, so just 1; so 5 - 1...we end up getting 4.
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What if we had looked at it, not through breaking it apart, but just simplified it first--32/2?
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Well, we could write that as log₂(16); what number do we need to raise 2 to, to get 16?
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2 to the 1 becomes 2; 2 squared becomes 4; 2 to the 3 becomes 8; 2 to the 4 becomes 16; so this would end up being 4.
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So, we end up getting the same thing, either way we look at it.
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Great; I want to caution you that there is no rule for log(M + N).
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Notice: none of these properties were ever of the form log(M + N) inside of there.
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That is because there is just no nice formula to break apart log(M + N).
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So, if you have a log of a quantity, and inside of that quantity it is something plus something else, there are no special rules.
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Sorry--there is just no easy way around it; you are going to have to work things out there in a complicated way.
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There is no way to be able to just break things apart or put things together anymore.
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Lots of people make the mistake of thinking that log<font size="-6">a</font>(M + N) is equal to log<font size="-6">a</font>(M) + log<font size="-6">a</font>(N).
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or that log<font size="-6">a</font>(M - N) is equal to log<font size="-6">a</font>(M) - log<font size="-6">a</font>(N).
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Those are not true--not true at all; you can't write them and split them apart like this.
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These things do not work; it is the same thing as, if you have √(2 + 2), saying, "Oh, I will just split that into √2 + √2."
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That does not work; you can't just split on square roots; you can't just split on logarithms, either.
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So, this idea of just splitting because you see an addition sign--you can't do that.
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You have to work out what is the logarithm of everything inside of there; there are no clean rules to do that.
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It is an easy mistake to end up making; but don't let it happen to you.
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Be vigilant; watch out for this; don't let it happen to you; don't do the same mistake.
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Remember, it only works with M times N and M divided by N.
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If it is plus or minus, there are no special rules; you just have to work it out by figuring things out,
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simplifying, hopefully, if you can, like they are actually numbers; but there is really no easy way around it.
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At this point, we have seen a lot of properties for logarithms; so let's review them.
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Our base ones right at the beginning were the log base any a of 1 equals 0,
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because any a raised to the 0 becomes 1; and also, log base a of a equals 1, because any a raised to the 1 is just a.
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Then, we also have our inverse properties, log<font size="-6">a</font>(a^x) = x,
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that it cancels out when you have logarithm on exponentiation if they are the same base;
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and then a raised to the log<font size="-6">a</font>(x) = x when we have exponentiation acting on logarithms.
00:18:12.600 --> 00:18:14.900
It cancels out if they are the same base.
00:18:14.900 --> 00:18:18.200
Then, we have the fact that we can bring down powers.
00:18:18.200 --> 00:18:25.400
If we have log(x^n), then we can bring the n in front: we have n times log(x).
00:18:25.400 --> 00:18:30.800
If we have log(MN), then we can split that into log(M) + log(N).
00:18:30.800 --> 00:18:36.500
We have the two different pieces, M and N, so it splits into log(M) and log(N).
00:18:36.500 --> 00:18:41.600
Multiplication inside of a logarithm becomes addition outside of the logarithm.
00:18:41.600 --> 00:18:53.000
log<font size="-6">a</font>(M/N) is log(M) - log(N); if we have M here and N here, then we have this minus sign right here.
00:18:53.000 --> 00:19:01.200
So, division inside of the logarithm becomes subtraction outside of the logarithm, once we split it into two logs.
00:19:01.200 --> 00:19:05.400
So, it seems like a lot of rules, and there are a fair bit of new things that you have to get used to here.
00:19:05.400 --> 00:19:13.400
But they are all based off of our original definition of what it means to be a log--the idea that log<font size="-6">a</font>(x) = y means that a^y = x.
00:19:13.400 --> 00:19:16.700
This is really what it is: you can either write it in exponential form or logarithmic form.
00:19:16.700 --> 00:19:18.300
It is just a way of denoting things.
00:19:18.300 --> 00:19:25.000
So, a underneath that y, a^y, becomes what we were originally taking the log of.
00:19:25.000 --> 00:19:30.900
And so, for any base a, we also figured out this key idea that for any base a, and any positive number x,
00:19:30.900 --> 00:19:37.500
there was a b that allowed us to get to that x: a^b = x, for any x that we wanted to get to.
00:19:37.500 --> 00:19:46.100
So, these two ideas--you can put them together, and you can figure out pretty much any one of these things right here with those ideas.
00:19:46.100 --> 00:19:51.800
And then, these ones are all just coming off of the basic definition, right from the beginning.
00:19:51.800 --> 00:19:54.100
But if you take that key idea, as well, you can figure these out.
00:19:54.100 --> 00:19:57.600
So, if you ever forget them on a test, in a situation where you can't just look them up,
00:19:57.600 --> 00:20:00.500
you now have a way of hopefully being able to figure them out on your own.
00:20:00.500 --> 00:20:04.200
They are not quite as easy as being able to figure out all of the things that made up our exponential rules,
00:20:04.200 --> 00:20:08.700
our rules for exponentiation, but we are able to figure these things out on our own.
00:20:08.700 --> 00:20:12.200
And as you get more used to working with them, and get some practice in them,
00:20:12.200 --> 00:20:14.300
it will be even easier for you to work them out on your own.
00:20:14.300 --> 00:20:17.200
And they will also just stick in your head that much easier.
00:20:17.200 --> 00:20:19.700
All right, now we are going to switch to a new idea.
00:20:19.700 --> 00:20:25.400
In the last lesson, we mentioned that most calculators only have buttons to evaluate natural log of x and log(x),
00:20:25.400 --> 00:20:34.500
ln(x) and log(x), that is log base e (that is what natural log means) and log without a number (means base 10).
00:20:34.500 --> 00:20:37.500
So, how could we evaluate something like log₇(42)?
00:20:37.500 --> 00:20:48.300
Well, 7 to the 1 equals 7, and 7 squared equals 49; so we see that there is no easy integer number that we raise 7 to, to get 42.
00:20:48.300 --> 00:20:54.300
It is not going to be an easy thing, so we need to use a calculator, because 42 is not an integer power of 7.
00:20:54.300 --> 00:20:57.500
But since it is base 7, we don't have a button on our calculators.
00:20:57.500 --> 00:20:59.700
What we want is some way to transform the base of the logarithm.
00:20:59.700 --> 00:21:06.000
If we could transform from log base 7 into log base 10 or log base e, we would be able to use a calculator,
00:21:06.000 --> 00:21:11.600
because then we have our natural log and common log buttons on our calculator; we can just punch it in.
00:21:11.600 --> 00:21:15.600
Now, you might have a calculator that lets you just put in an expression like this.
00:21:15.600 --> 00:21:19.000
But even if you have that sort of calculator, this is still sort of a useful thing to learn.
00:21:19.000 --> 00:21:24.900
As we will see in some of the examples, there are ways to apply changes of base beyond just using them to get what these numbers are.
00:21:24.900 --> 00:21:33.100
And also, lots of times, you won't have a calculator that is able to do this, and you will only have natural log or plain common log, log base 10.
00:21:33.100 --> 00:21:40.600
And you will need to be able to have this change of base, so that you can change when you need to take a base that isn't e or 10.
00:21:40.600 --> 00:21:44.200
To help motivate the coming formula and its derivation, let's look at a specific example.
00:21:44.200 --> 00:21:48.600
Consider the expressions log₃(81) and log₉(81).
00:21:48.600 --> 00:21:55.500
log₃(81): well, we could rewrite 81 as 3⁴, so we see that that is just 4 over here.
00:21:55.500 --> 00:22:02.900
Now, log₉(81) we could rewrite, also, as 9²; so with a base 9, that would come out as 2.
00:22:02.900 --> 00:22:12.700
So, we see that log₃(81) = 2log₉(81), because 4 is equal to 2 times 2.
00:22:12.700 --> 00:22:21.600
So, we see that log₃(81) is equal to 2 times log₉(81).
00:22:21.600 --> 00:22:25.600
And this is somehow related to the fact that 3² is equal to 9.
00:22:25.600 --> 00:22:28.100
You square 3, and you manage to get a 9 out of it.
00:22:28.100 --> 00:22:30.500
Now, we can probably intuitively know that that holds, generally.
00:22:30.500 --> 00:22:37.700
We can figure out that normally (or in fact, always), we are going to have log₃(x) = 2log₉(x).
00:22:37.700 --> 00:22:45.100
But let's prove it, instead of just assuming--instead of getting a feel that that makes sense, let's actually prove definitely that that is the case.
00:22:45.100 --> 00:22:48.800
We start by noting that x is equal to 9 to the log₉(x).
00:22:48.800 --> 00:22:58.000
Remember: if we wanted to, we could cancel those things out, because we have a base of 9 and log base 0.
00:22:58.000 --> 00:23:02.300
So, they would cancel out, and we would end up having just x; so this here makes sense.
00:23:02.300 --> 00:23:06.600
But having it, 9 to the log₉(x), written in this funny way, is a complicated idea.
00:23:06.600 --> 00:23:09.800
It is hard to see where we pulled it; it is kind of like just pulling a rabbit out of a hat.
00:23:09.800 --> 00:23:15.900
But with this idea, if we leave it in this form, we will be able to do some cool tricks that will let us show what we want to get to.
00:23:15.900 --> 00:23:18.100
If we want, we can take log₃ of both sides.
00:23:18.100 --> 00:23:30.300
We know that x is equal to 9^log₉(x), so it must be the case that log₃(x) is the same thing as 9^log₉(x).
00:23:30.300 --> 00:23:32.800
So, log₃(9^log₉(x)).
00:23:32.800 --> 00:23:39.500
We can take log base 3 of both of the things on either side, because that equals sign means that it has to be equal for whatever happens to it.
00:23:39.500 --> 00:23:43.600
So, log₃(x) is equal to log₃(9^log₉(x)).
00:23:43.600 --> 00:23:47.200
OK; with that idea in mind, we can start applying our rules that we have.
00:23:47.200 --> 00:23:53.200
We know that we can bring down exponents; so in this case, we have effectively an exponent of log₉(x).
00:23:53.200 --> 00:23:57.300
We actually have it exactly as an exponent, so if we want, we can bring that down in front.
00:23:57.300 --> 00:24:02.600
We see that log₃(x) is equal to log₉(x) times log₃(9).
00:24:02.600 --> 00:24:11.700
Now, what is log₃(9)? That comes out to be 2, so we can simplify that as log₃(x) = 2log₉(x).
00:24:11.700 --> 00:24:15.800
And we have proven what we originally wanted to show.
00:24:15.800 --> 00:24:20.400
We follow a similar structure to create a formula to change between any two bases, u and v.
00:24:20.400 --> 00:24:24.900
If we start as x = v, which we know is true, because of inverses,
00:24:24.900 --> 00:24:30.800
then we can once again take a log on both sides; and we will take log<font size="-6">u</font>, because we want to get v and u.
00:24:30.800 --> 00:24:35.000
We want to get both of those logs into action, so we take log<font size="-6">u</font> on both sides.
00:24:35.000 --> 00:24:43.600
And then, we can bring this down in front, because it is an exponent; and we have log<font size="-6">u</font>(x) = log<font size="-6">v</font>(x) times log<font size="-6">u</font>(v).
00:24:43.600 --> 00:24:48.300
At this point, we can create a formula that will have log<font size="-6">v</font> on one side and log<font size="-6">u</font> on the other side.
00:24:48.300 --> 00:24:53.500
We divide log<font size="-6">u</font>(v) over, and we rearrange; we swap the order of the equation.
00:24:53.500 --> 00:25:03.500
We have log<font size="-6">v</font>(x) = log<font size="-6">u</font>(x)/log<font size="-6">u</font>(v).
00:25:03.500 --> 00:25:09.900
Notice that this allows us to change from an expression log<font size="-6">v</font>(x) into an expression that only uses log base u.
00:25:09.900 --> 00:25:15.200
Now, if we choose our u to be either e or 10 or whatever is convenient for the problem we are working on,
00:25:15.200 --> 00:25:18.100
we will be able to evaluate it with any calculator at all.
00:25:18.100 --> 00:25:23.600
Since every calculator we will be using has natural log and common log (base 10 log) buttons,
00:25:23.600 --> 00:25:28.100
we will be able to evaluate with any calculator, because we will be able to change the u's over here.
00:25:28.100 --> 00:25:38.400
So, whatever we end up having--if it is log₇(42), then we can change it into log<font size="-6">10</font>(42)/log<font size="-6">10</font>(42).
00:25:38.400 --> 00:25:44.400
Or alternatively, we could have changed it into ln(42)/ln(42).
00:25:44.400 --> 00:25:48.700
Both would end up giving the same thing; and we will see what that is in the examples.
00:25:48.700 --> 00:25:51.900
All right, the first one: Write as a sum and/or difference of logarithms.
00:25:51.900 --> 00:25:57.900
Our first example here: we are working with base 5, but that doesn't affect how any of our properties work.
00:25:57.900 --> 00:26:10.000
So, remember: if we have log(M/N), for any base a, then that is equal to log<font size="-6">a</font>(M) - log<font size="-6">a</font>(N).
00:26:10.000 --> 00:26:13.400
The same log base is on both, and it splits into subtraction.
00:26:13.400 --> 00:26:20.500
So, in this case, we have, on the top, x⁵; so we will have log₅, the same thing,
00:26:20.500 --> 00:26:36.400
minus what is on the bottom, y√z, so - log₅(y√z).
00:26:36.400 --> 00:26:49.600
Great; now, we also have the rule that log<font size="-6">a</font>, for any a of M times N, equals log<font size="-6">a</font>(M) + log<font size="-6">a</font>(N).
00:26:49.600 --> 00:26:53.600
So, we can split with addition, as well; so we have multiplication here.
00:26:53.600 --> 00:26:57.500
y times √z is what is really there.
00:26:57.500 --> 00:27:05.400
log₅(x⁵) -...now, notice: we are splitting all of this here.
00:27:05.400 --> 00:27:12.400
We are still subtracting by all of it, so we want to put parentheses around it, because it is substitution that we are doing here.
00:27:12.400 --> 00:27:17.000
So, we now work on this thing here, log<font size="-6">a</font>(M) - log<font size="-6">a</font>(N).
00:27:17.000 --> 00:27:28.500
So, our M is y; our N is √z; and we have log...still the same base...of y + log, still the same base (5), of √z.
00:27:28.500 --> 00:27:41.800
Simplify this out a bit: we have log₅(x⁵) - our subtraction distributes...minus here, as well... log₅(√z).
00:27:41.800 --> 00:27:45.500
Now, that is technically enough, because we have a sum and/or difference of logarithms.
00:27:45.500 --> 00:27:48.600
But we can also take it one step further, and we can get rid of these exponents.
00:27:48.600 --> 00:27:54.600
We can get rid of "to the fifth"; we can get rid of √z, because we also have the rule that log<font size="-6">a</font>,
00:27:54.600 --> 00:28:02.400
for any base, of x^n, is equal to n times log<font size="-6">a</font>(x).
00:28:02.400 --> 00:28:06.100
So, in this case, we have to the fifth; and how can we rewrite √z?
00:28:06.100 --> 00:28:14.900
Well, remember: any square root is just like raising to the half, so we can see this:
00:28:14.900 --> 00:28:20.700
bring the 5 to the front, so we have 5log₅(x).
00:28:20.700 --> 00:28:31.100
We continue to just bring down our log₅(y) -...we will rewrite the z...log₅(z^1/2).
00:28:31.100 --> 00:28:38.000
And now, we can bring down this, as well...not to there, but we have to move it all the way to the front of the log.
00:28:38.000 --> 00:28:52.200
There we go: we have 5log₅(x) - log₅(y) - 1/2log₅(z).
00:28:52.200 --> 00:28:59.300
And there we go: we have managed to write this entirely using very simple things inside of our log: just x, y, and z.
00:28:59.300 --> 00:29:05.200
We have managed to break up this fairly complicated expression inside of the log into a fairly simple expression
00:29:05.200 --> 00:29:09.200
inside of the log by just breaking it up into more arithmetic.
00:29:09.200 --> 00:29:13.300
We can do the reverse, and we can also compact things: write the expression as a single logarithm.
00:29:13.300 --> 00:29:20.700
We will compact all of these log expressions into one tiny log with a more complicated structure on the inside.
00:29:20.700 --> 00:29:29.400
So, first, we have, once again, that subtraction becomes division: so 1/3ln(a) + 2...
00:29:29.400 --> 00:29:34.400
actually, the first thing: we have these coefficients out front.
00:29:34.400 --> 00:29:40.400
We can also bring the coefficients in, so 1/3 can hop up onto an exponent on that a.
00:29:40.400 --> 00:29:54.300
The 2 here can hop up to an exponent on that b, so we have ln(a³) + 2(ln(b²)) - ln(c).
00:29:54.300 --> 00:30:01.600
If you forgot, remember: natural log, ln, is just a way of saying log base e, where e is a special number, the natural base.
00:30:01.600 --> 00:30:05.600
ln(a^1/3) + 2(ln(b²)) - ln(c).
00:30:05.600 --> 00:30:12.000
Now, we can compact what is inside of those parentheses, because we see we have subtraction.
00:30:12.000 --> 00:30:21.400
So, that is natural log of b²/c; subtraction of logs is the same thing as division inside of the logarithm.
00:30:21.400 --> 00:30:28.800
And now, natural log a^1/3 plus...well, now we have this 2, so we can take this 2,
00:30:28.800 --> 00:30:34.500
and it is hitting a log; it is times a log, so it can go up and also become an exponent.
00:30:34.500 --> 00:30:44.500
So, ln(b²/c)...make sure we remember that we are doing the whole logarithm of that whole thing,
00:30:44.500 --> 00:30:53.400
so ln(a^1/3) + ln...we distribute that...(b⁴/c²).
00:30:53.400 --> 00:31:00.400
And we can also bring this in now: natural log of...addition of logarithms becomes multiplication inside of the logarithms.
00:31:00.400 --> 00:31:08.100
So, a^1/3 times the rest of it...b⁴...it will show up in the numerator, divided by c².
00:31:08.100 --> 00:31:13.000
And we have the whole thing compacted into a single logarithm; great.
00:31:13.000 --> 00:31:20.000
All right, the next one: Evaluate each of the following; use a calculator and the change of base formula.
00:31:20.000 --> 00:31:28.600
Remember our change of base formula: if we have log₇(42), then we can change to any base...
00:31:28.600 --> 00:31:39.300
I'll put it as a square right now...of 42; the thing that we are taking our log of initially, divided by...
00:31:39.300 --> 00:31:44.900
the base has to be the same between here and here; these have to be the same base.
00:31:44.900 --> 00:31:54.700
And it is going to be of our original base; so since it was log₇(42), we now have log of something (42), divided by log of something (7).
00:31:54.700 --> 00:32:06.800
All right, that is how it works: that is what it meant when we saw log<font size="-6">v</font>(x) = log<font size="-6">u</font>(x)/log<font size="-6">u</font>(v).
00:32:06.800 --> 00:32:15.000
In this case, for this one, our v is 7; our u is whatever we are about to choose.
00:32:15.000 --> 00:32:17.800
So, we can make it any u we want; we can make it 50, and it would work.
00:32:17.800 --> 00:32:22.000
We could make it .1, and it would work; but let's do something that shows up on our calculators.
00:32:22.000 --> 00:32:28.100
So, let's choose e: we can put in an e here and an e here.
00:32:28.100 --> 00:32:37.000
So, we will rewrite that as ln(42)/ln(7); we punch that into our calculator, and that will end up coming out to be...
00:32:37.000 --> 00:32:45.400
it will go on with lots of decimals, so let's cut it off, and it will end up being approximately 1.9208.
00:32:45.400 --> 00:32:48.400
Now, if you wanted to, you could have also done this as something else.
00:32:48.400 --> 00:32:56.600
It would have also been the same as log<font size="-6">10</font>(42)/log<font size="-6">10</font>(7).
00:32:56.600 --> 00:33:00.800
That would be the same if you used common log, something that also shows up on a lot of calculators.
00:33:00.800 --> 00:33:06.300
And you would end up getting the same thing; it would come out to be approximately 1.9208.
00:33:06.300 --> 00:33:10.100
Now, if we want to check our work--we want to make sure that that did work out--
00:33:10.100 --> 00:33:17.900
we would check it by saying that 7^1.9208 does come out to being 42.
00:33:17.900 --> 00:33:24.200
And it does come out to being approximately 42, so it ends up checking out if we punch that into a calculator.
00:33:24.200 --> 00:33:27.600
Let's do log<font size="-6">π</font>(√17); it is the same basic idea here.
00:33:27.600 --> 00:33:41.800
We can do it as ln(√17)/ln(π); we punch that out in a calculator, and we get approximately equal to 1.2375.
00:33:41.800 --> 00:33:49.300
And if we wanted to, we also could have done that as log<font size="-6">10</font>(√17)/log<font size="-6">10</font>(π).
00:33:49.300 --> 00:33:52.000
And we would have ended up getting the exact same thing.
00:33:52.000 --> 00:33:54.600
It would have come out to be approximately 1.2375.
00:33:54.600 --> 00:33:58.500
Either one you work with--they are both going to end up working out to give you the same answer.
00:33:58.500 --> 00:34:00.800
Great; and if you wanted to, you could also check this, as well.
00:34:00.800 --> 00:34:09.400
You could check and make sure that π raised to approximately 1.2375 does come out to be approximately √17.
00:34:09.400 --> 00:34:12.700
And indeed it does, if you want to check that this does work.
00:34:12.700 --> 00:34:21.400
Great; the fourth example: Given that log₅(a) = 6, and log₅(b) = 1.2, evaluate each of the following.
00:34:21.400 --> 00:34:24.300
At this point, we want to use the rules that we have to split things apart.
00:34:24.300 --> 00:34:27.900
Splitting it apart will allow us to use the pieces of information we have.
00:34:27.900 --> 00:34:35.200
We have to see a log₅(a) before we can swap it out for 6; so we have to get that sort of thing to show up--the same for log₅(b).
00:34:35.200 --> 00:34:43.800
So, we split things up; we have multiplication between each one of these--that is what it means when they are just stacked on top of each other.
00:34:43.800 --> 00:34:54.700
So, log₅(5) + log₅(a) + log₅(b³):
00:34:54.700 --> 00:35:00.200
log₅(5) is just 1, because it is the same thing as its base, and what it is operating on,
00:35:00.200 --> 00:35:04.200
plus log₅(a); we were told that was 6, so we get 6;
00:35:04.200 --> 00:35:07.000
plus...now, log₅(b³)...we can't do that yet.
00:35:07.000 --> 00:35:11.900
We need to get it as just a b inside; but we see that there is an exponent.
00:35:11.900 --> 00:35:17.500
We can bring that out front; so we have 3log₅(b).
00:35:17.500 --> 00:35:38.700
So now, we can use the fact that it is 1.2: 1 + 6 is 7, plus 3 times 1.2 (is 3.6); 7 + 3.6 becomes 10.6; and there is our answer.
00:35:38.700 --> 00:35:44.300
Work on the other one: log<font size="-6">25</font>(√a); this one is a kind of a problem.
00:35:44.300 --> 00:35:49.600
We have 25 here, but we were told our base for working with the stuff--the information we were given was log base 5.
00:35:49.600 --> 00:35:53.400
So, we have to use change of base; now we see a time when you have to use change of base,
00:35:53.400 --> 00:35:58.100
not just for calculating numbers (if you have a calculator that can do change of base on its own,
00:35:58.100 --> 00:36:01.900
that doesn't need you to do it, then there is still a use for it for problems like this).
00:36:01.900 --> 00:36:08.600
Change of base: we have that this log<font size="-6">25</font>(√a) is the same thing
00:36:08.600 --> 00:36:18.900
as log of same base of √a, divided by log of some base of what our original base was, 25.
00:36:18.900 --> 00:36:24.100
So, what do we want to use there? We probably want to use 5, because that is the thing we have all of our information on.
00:36:24.100 --> 00:36:33.200
So, 5 and 5 here; log₅(√a)...now, √a is not what we have--we have a.
00:36:33.200 --> 00:36:36.500
Is there another way to write √a that would involve a?
00:36:36.500 --> 00:36:43.300
Yes, √a is the same thing as a^1/2: so we can write that as log₅(a^1/2).
00:36:43.300 --> 00:36:49.300
divided by log₅(25)...well, we see that that is log₅(5²),
00:36:49.300 --> 00:36:53.400
because what number do you have to raise 5 to, to get 25? You have to raise it to 2.
00:36:53.400 --> 00:37:06.400
So at this point, 1/2log⁵(a)/2log₅(5)...log₅(5) is just going to be 1.
00:37:06.400 --> 00:37:19.200
You swap out; we know that we have 6 up here, so it is 1/2 times 6, divided by 2; 1/2 times 6 is equal to 3, still divided by 2.
00:37:19.200 --> 00:37:21.400
There we go--cool.
00:37:21.400 --> 00:37:24.500
All right, the final example: Solve the following equations.
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We didn't talk about how to solve equations like this very much in detail, because hopefully this idea will make sense.
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But don't worry; we are going to talk about this in great detail in the next lesson, where we will really get into this.
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But we are just going to start with some simple ones, just in case you end up having any problems like this already that you are working on.
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So, solve the following equations: log₃(x + 5) = 2.
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Now, remember: we have that inverse property--we know that a to the log<font size="-6">a</font> of "something"
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ends up being equal to something, because these cancel out.
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Now, notice: we have an equals sign here, so we know that log₃(x + 5) is the same thing as 2.
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So, we can use either one either way we want to.
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So, that means that a to the something equals a to the something, if it is the same something.
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So, why don't we choose 3 as our a? 3 to the stuff is equal to 3 to the stuff, as long as it is the same stuff.
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Let's put 2 over here and log₃(x + 5) over here.
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3² and 3^log₃(x + 5): 3 and log₃ end up canceling out,
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and x + 5 just drops down; that equals...there is nothing to cancel on the right side; it is 3²;
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but we know what 3² is: 3 times 3 is 9.
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x + 5 = 9: we subtract 5 on both sides, and we get x = 4.
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Great; and if we wanted to, we could check that that does end up working out.
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Check: we plug in our x = 4, so log₃(4 + 5) = 2; does it?
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log₃(4 + 5)--let's see if that ends up coming out to be 2.
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log₃(9): what number do we have to raise 3 to, to get 9?
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We have to raise it to 2; so it checks out.
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Great; e^x - 8 = 47: to do this one, we remember that log<font size="-6">a</font>(a^x) = x.
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A log on an exponent, as long as they are both the same base, also cancels out.
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So, what is the base for e? it is natural log, ln; we could also do log<font size="-6">e</font>, but normally it is done as ln.
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We can take the natural log of both sides; just as we had 3 to the something equals 3 to the something,
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the natural log of something equals the natural log of something, as long as it is the same something.
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So, ln(stuff) is equal to ln(stuff), as long as it is the same stuff.
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Well, we have this right here; so we know that we can plug in 47 over here, and e^x - 8 we can plug in over here,
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because we are guaranteed by that equals sign that it is the same stuff on either side, that they end up being the same thing.
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So, ln(e^x - 8)--well, natural log is just log base e, so these cancel out; and the x - 8 drops down.
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And that equals ln(47)...well, that is going to end up coming out to be a pretty not-simple number.
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It is going to have a lot of decimals; so let's just leave it as ln(47) for right now.
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And we end up getting, by adding 8 to both sides, ln(47) + 8.
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Now, alternatively, we could also figure out what this is as a decimal approximation.
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So, if we punch ln(47) into our calculator, and then add 8 to it, we get approximately 11.85.
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That would also be approximately 11.85.
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It is precisely ln(47) + 8, but if we want to take ln(47) and have a number that we can work with, it ends up coming out to something with a lot of decimals.
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It is just like when you have 2 times π you can leave the answer precisely as 2π.
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But if you want to, you can also approximate that into 6.28...and there is more stuff to it.
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So, 2 times π is the exact answer; but you also might want a decimal answer to work with, so you can approximate it by multiplying it out.
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The natural log of 47 is the same sort of thing as in the π example.
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It is something that is a complicated number, so we might want to leave it precisely, or we might want to get it approximately.
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We can check with this number; and we have that e to the...let's use our decimal approximation,
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so we can actually put it into a calculator...11.85, minus 8...what do we put that in?
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That will become e^3.85; e^3.85 ends up being approximately 46.99.
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So, that ends up checking out, because ultimately, remember, we just said it was an approximation, not perfectly the answer.
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The thing that is perfectly the answer is this one right here.
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This is pretty great stuff that allows us to get a whole bunch of applications worked out with this,
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as we will see two lessons from now, when we talk about applications of the stuff.
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And in the next lesson, we will really dive into how we solve equations like this.
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So, if you want more information on that, check out the next lesson,
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where we will really see some really complicated examples, and get a really great idea of how these things work.
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And we will really understand how to solve all of these sorts of equations.
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All right, we will see you at Educator.com later--goodbye!