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 have an introduction to logarithms.
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At this point, if we want to find the value of a number raised to an exponent, it is easy.
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We use our exponentiation rules and evaluate; if it is something simple, like 2³,
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then we know that is 2 times 2 times 2, 2 times itself 3 times, which we figure out is 8.
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And we also figured out how to do numbers that weren't just simple integer exponents.
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But we have all of these nice rules from our previous few lessons.
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But what if the question was inverted, and what if we knew the base and the end result, but we don't know the exponent that we need to get there?
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If we knew that we had 2 as the base, and we wanted to have an end result of 8,
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but we had no idea what exponent we had to use to get to 8, how could we figure out that exponent?
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This is the question that leads us to explore the idea of the logarithm, which we will be looking at over the next few lessons.
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We define the logarithm as a way to talk about this unknown exponent.
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The logarithm base 2 of x, denoted log₂(x), with this little 2 as a subscript right here,
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is defined to be the number y such that 2^y = x.
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So, when we see log₂(x) = y, then we know that that would be 2^y = x.
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In other words, we are looking for the number that...when the logarithm goes onto some number,
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we are trying to figure out what value, if it raised to this space here, would become the number that we took the logarithm of.
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So, for the example of log₂(8) = 3, the reason why that is the case is because
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we are looking for what number here we have to raise to to get 8.
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So, the answer there is 3; if we raise 2³, we get 8, so we know that log₂(8) = 3.
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It can be a little bit confusing to remember how this works at first--what the notation means.
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So, when you see log₂(x) = y, you can think of this as...if the 2 were under the y,
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if we had 2^y, it would make the x that we are taking the logarithm of.
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So, we take the logarithm of some number in regards to some base.
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And that tells us what number we would raise the base to, to get the original number we are taking the logarithm of.
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It will make more sense as we see more examples.
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Here are some other examples: log₂(32) = 5, because if we raise 2 to the 5th, we get 32.
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2 times 2 times 2 times 2 times 2 is equal to 32; 4, 8, 16, 32.
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log₂(1) = 0 because, if we raised 2 to the 0, from our rules about exponents, we know that that is just the same thing as 1.
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log₂(1/4) = -2, because we know that, if we raise 2 to the -2, then that is the same thing as 1/2 raised to the 2,
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where you flip to the reciprocal; and 1/2 squared would become 1/4, which is what we initially started with.
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So, a logarithm is a way of taking a logarithm of a number, so that you figure out
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what you would have to raise some base to, to get the thing you took the logarithm of.
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We can expand this idea to something beyond just base 2, to a general idea.
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The logarithm base a of x, log<font size="-6">a</font>(x) (that little a, right down here, is a subscript), where a > 0,
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and a is not equal to 1 (our base has to be greater than 0, and our base has to not be equal to 1),
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is defined to be the number y such that a^y = x.
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If we take log<font size="-6">a</font>(x), then we know that that gives us some y, and that a^y = x.
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So, once again, it is the same idea, where, if we take this base, and we put it under the y, we would get a^y.
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And then, we would have the thing that we were originally taking the logarithm of, which is what we have there.
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That is what is occurring right here.
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The idea of the logarithm is that you take the log, and it tells you something that you can raise a number to, to be able to get this other value.
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It is a little bit complex the first time you get it; but as you do it more and more, it will start to make more sense.
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And we are going to see a whole bunch of examples to really get this cleared up.
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Now, notice: we have these restrictions on what our base can be.
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We know that the base has to be greater than 0, and the base is not equal to 1.
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The base a of a logarithm has the same restrictions as the base of an exponential function.
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This is because exponentiation and logarithms are inverse processes; they do the opposite thing.
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And we will see more about how they are inverses in the future.
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But they do reverse things, so they have to have the same restrictions, because they are basically working with the same idea of a base.
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They are being seen through different lenses, and it will make more sense as we work on it more and more;
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but we have to have the same restrictions on it--otherwise the idea of a logarithm will just not make sense, or not be very interesting.
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So, we have these restrictions: that we have to have our base greater than 0, and we have to have our base not equal to 1.
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Let's look at some examples to help clear this idea up.
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log₇(49) = 2, because, if we move this base over,
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then 7² is just 7 times 7, which gets us 49, which is exactly what we started with.
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So, this is the case, because 7² = 49; or maybe let's write it in the way that we had it originally here: 49 =...
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moving the base over, moving our base underneath the right side...we have 7², like this.
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The same thing over here: we know that log<font size="-6">10</font>(10000) = 4,
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because, if we move our base under, we know that 10⁴ is equal to 10000.
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The question is: if we want to know what number we have to raise 10 to, to get 10000--that is what log<font size="-6">10</font>(10000) is effectively asking.
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It is saying, "What number do we have to raise 10 to--raise our base to--to get 10000 as the end result?"
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10 to the what equals 10000? The answer to that is 4.
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So, we take 10⁴; we get 10000; and sure enough, we see that that is 10 times 10 (100) times 10 (1000) times 10 (10000).
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The same thing is going on over here: if we have log₅(1/125), then we move that under, and we get 5^-3 = 1/125.
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Let's check that out: 5^-3 is the same thing as 1/5 to the positive 3; we flip to the reciprocal.
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And then, 5 times 5 times 5 is 25...125; so we get 1/125; sure enough, it checks out.
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Finally, if we take log₄(2), then we see that 4^1/2 is equal to 2.
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What is 1/2? 1/2 is an exponent that means square root; 4^1/2 is the same thing as √4, which is 2; so once again, that checks out.
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It is the question of what exponent I am looking for to be able to get this base to become the number that I am taking the logarithm of.
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We can even do this with more unusual bases on our logarithms.
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For example, if we have log<font size="-6">1/2</font>(1/16), then we can see that that will become 4, because (1/2)⁴ is equal to 1/16,
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because 2, 4, 8, 16...1/2, 1/4, 1/8, 1/16; so we see that that is the same thing.
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If we take log<font size="-6">π</font>(π), then we know that it has to come out as 1,
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because π¹--of course, that is no surprise that it is going to equal what it already started with.
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So, if we are taking log<font size="-6">π</font>(π), then the thing that π has to be raised to, to get π, is just 1.
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So, we have 1 as the thing that comes out of that.
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If we take log<font size="-6">√2</font>(4√2), then we get that that has to be 5, because (√2)⁵--sure enough, that is equal to 4√2.
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We can check that out: √2 times √2 times √2 times √2 times √2.
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Well, √2 times √2 becomes 2; √2 times √2 becomes 2; and we have this one √2 left.
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So, 2 times 2 is 4, and we are left with 4√2; it checks out.
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The final one: log<font size="-6">e</font>(1) becomes 0, because if we move this over, e⁰, just like anything raised to the 0, becomes 1.
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So, it checks out; so we have some idea of how it works.
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A logarithm is, "For this base, what number do I have to raise it to, to get the number that I am originally taking the logarithm of?"
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When I take the log<font size="-6">10</font>(10000), it is a question of what number I have to raise 10 to, to get 10000.
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I have to raise 10 to 4 to get 10000.
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When I take log₇(49), it is a question of what number I need to raise 7 to, to get 49; the answer to that is 2.
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That is the idea of a logarithm.
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The two most common logarithmic bases to come up are the numbers e (remember, e is the natural number;
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we talked about it previously, when we talked about exponential functions--a very important idea) and 10.
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As such, they have special notation, because we have to write them so often.
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The base of e is expressed as ln; so when we want to talk about base of e, the shorthand for that is ln.
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It is called the natural logarithm; remember, e is called the natural base.
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So, when we are taking a log<font size="-6">e</font>, we call it a natural logarithm, and we use ln, because originally,
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the French were the ones who came up with this; so it was *logarithme naturel* (excuse my French--I am not very good at speaking French).
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So, the natural log of x is equivalent to log<font size="-6">e</font>(x); ln(x) is just a shorthand way of saying log with a base of the number e.
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Base 10 is expressed with just log on its own; notice, it has no subscript--there is no little number down there.
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If no base is given, it is assumed to be base 10; since base 10 comes up a lot, it is just an easy way to write it; this is normally what it means.
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It is called the common logarithm, because it is a commonly-used logarithm.
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So, if you see log(x), notice that it has no little subscript--no little number down there.
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Then, we know that that is going to mean log<font size="-6">10</font>(x).
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Well, we can find the value of expressions like log₂(8); we know that that came out to be 3,
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because the number that we raised 2 to, to get 8, is 3.
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How do we figure out the value of more complicated expressions?
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Like if we wanted to figure out the natural log of 12.19--and as we just saw, that would be the same thing as asking, "What is log<font size="-6">e</font> or 12.19?"
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Well, e is a very complicated number; it goes on forever--it is irrational.
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12.19 is not a very friendly--looking decimal number; so how are these two things going to interact?
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We can guess that it is probably not going to come out very cleanly, in a nice way.
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Sure enough, it doesn't: it comes out to be approximately 2.500616; and precisely, it would keep going forever, as well.
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So, for calculating logarithms, just like exponentiation, we can find the expressions, or a very good approximation, by using a calculator.
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Any scientific or graphing calculator will have natural log and log<font size="-6">10</font> buttons to take logarithms base e and 10, respectively.
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However, many calculators will not have a way to take logarithms of arbitrary bases.
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So, if we had log₃, most calculators won't have an easy way for us to just get what log₃(some number) is.
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But there is a way around this, and it is called change of base.
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So, if you do need to take the log base 3 of some number, check out the next lesson, Properties of Logarithms,
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where we will explore how you can change from one base to another.
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So, the way that you calculate complicated logarithms like this is: you generally just use a calculator.
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The calculator has a way, a method, to be able to figure out what that comes out to be.
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Now, just like with exponentiation that we talked about before, we should note that there are ways to calculate these values by hand.
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We could do this by hand and figure it out; and you will learn about this in more advanced math classes.
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But we won't learn about it in this course right here.
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Doing this takes a lot of arithmetic, though; and so we designed calculators to speed up the process.
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It is something that we could do; it is not like we are completely reliant on calculators for figuring this idea out.
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Logarithms weren't something that we only got once we had calculators created.
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We have been able to have this idea for a very long time--since the 1600s, in fact.
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But being able to calculate what these numbers come out to be--that takes a long time; it is a slow process.
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So, we have calculators to be able to figure this out for us very quickly and very easily.
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So, it speeds things up, but it is not that we are dependent on calculators.
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It is just that they are a useful tool that we can apply in this situation.
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Graphs of logarithms: so now, since we can evaluate logarithms however we want,
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because we have these nice calculators as tools, we can plot graphs of them.
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So, let's look at some graphs: f(x) = log₂(x) (this is in red); g(x) = log₅(x)
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(that one is in blue), and finally h(x) = log<font size="-6">10</font>(x) (that one is in green).
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Notice how short the y-axis is; it only goes from -3 up until positive 5.
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But we go all the way out to positive 100 on the x-axis.
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We can see that here--right here; it is hard to see--that is a 1 value on the x-axis.
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And that is going to end up corresponding to 0, because log of anything--log of any base of 1--will come out to be 0,
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because the number that you raise anything to, a⁰, = 1.
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So, if we would take log base anything of 1, it is going to always come out to be 0,
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because that is the number that we raise anything to, to get 1 in the first place.
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So, that is why we see a common height of 0 there.
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And notice how slowly they grow: at 16, log₂ is only going to be at a meager 4.
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But for log<font size="-6">10</font>, when we look at log<font size="-6">10</font>, it takes getting all the way up to 100 to even get to 2.
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If we go out here to the 2, it takes all the way to 100 to be able to get that from log<font size="-6">10</font>, because 10² = 100.
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We are seeing a similar thing for log₅: it has to get all the way up to 25 before it hits this height of 2, as well, because it is 5².
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And we aren't even going to see it hit height 3, because it is not going to hit a height of 3 until it manages to get to 125
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as an input value, because 5³ becomes 125.
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So, notice how slowly these graphs grow.
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These graphs grow really, really slowly, because for logs, it takes a really big number to be able to get even slight increases in our verticals.
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And the farther out they go, the even slower they are going to grow.
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Now, notice that they approach the y-axis asymptotically.
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So, as they get smaller and smaller, they get really, really close to this y-axis right here.
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They never touch or pass it, although that might be hard to see in this picture, since it looks like it is right on top of it.
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But they get very close; they won't actually touch it, but they get very close to the y-axis.
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We will talk about this behavior of how it gets really close to the y-axis,
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and why it can't actually touch the y-axis or go past it, soon, when we talk about the domain of a logarithm.
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The logarithm is the inverse process of exponentiation.
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For example, let's consider base 2: if we have log₂(x) = y, then we can see its flip of 2^y = x.
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We just change the x and the y location.
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So, if we take log₂(8), that becomes 3, because remember: 2³ = 8.
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So, when we take log₂(8), we get 3.
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But then, if we take that 3, and we plug it into the other one--we take the 3, and we plug it in up here--
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we look at 2³--look: we are right back where we started.
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We have the same thing on both sides.
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We take this log, and we do something to it, and then we do the reverse process with the same base as the exponentiation.
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We get back to the original input that we put in.
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The same thing: we did it the other way, where we did exponentiation first.
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If we take 2^-2, then that is going to flip to (1/2)²; so we would get 1/4.
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And then, if we take log₂(1/4), we are going to get -2.
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So, exponentiation and logarithms are doing inverses: one goes one way, and one goes the other way.
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Together, they cancel out; we will be discussing this idea a lot more in the coming lessons.
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It is a very important thing; we will also be proving it in general.
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We can see this as one in a general form for any logarithm.
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The exponential function of base a is the inverse of the logarithmic function of base a.
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It is critical, though, that they do have the same bases; our exponential function is base a, and our logarithmic function must be base a.
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If they are not the same base, it won't work.
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Let's see why this is the case: if we have f(x) = log<font size="-6">a</font>(x), f^-1(x) = a^x.
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Then, we can take f^-1(f(x)) and see what happens.
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Now, remember: we are talking about stuff from our lesson on inverse functions.
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If you need more background on inverse functions, make sure you go and check out that lesson.
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It will help you understand what is going on here.
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So, f^-1(f(x)) =...well, we could do this as...since this is a^x, then it is going to be...
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well, we will apply f^-1 next; first, f(x) = log<font size="-6">a</font>(x).
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We have log<font size="-6">a</font>(x); then we apply the f^-1, and we have a.
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Now, what does that end up coming out to be?
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Well, remember: log<font size="-6">a</font>(x) = y is the same thing as saying a^y = x.
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So, that is what log<font size="-6">a</font>(x) is: it is this y, some y such that if we were to put it as an exponent on a, we would get x.
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So, log<font size="-6">a</font>(x) = y: we can just say, "Whatever the number log<font size="-6">a</font>(x) is, let's call it y."
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So, we can swap that out, and we can say, "a^y," just because we are saying we will call log<font size="-6">a</font>(x) y.
00:18:46.100 --> 00:18:53.500
That is what we have over here; but remember, we defined this idea of what log<font size="-6">a</font>(x) is based on a^y = x.
00:18:53.500 --> 00:18:58.700
Well, we now have a^y = x; so if a^y = x, then that equals x,
00:18:58.700 --> 00:19:04.400
which means that f^-1(f(x)) = x.
00:19:04.400 --> 00:19:09.400
Whatever we put in as our input comes out as our output if we do these two functions, one on top of the other.
00:19:09.400 --> 00:19:14.000
We have inverse functions, because one function cancels out the effects of the other function.
00:19:14.000 --> 00:19:17.500
We will talk about this more in future lessons.
00:19:17.500 --> 00:19:21.500
We can also see this in the graphs of exponential and logarithmic functions.
00:19:21.500 --> 00:19:28.400
So, if we take two graphs of, say, 2^x (that one is in red) and log₂(x), we see them like this.
00:19:28.400 --> 00:19:34.400
And then finally, we also have y = x in yellow here, coming through the middle.
00:19:34.400 --> 00:19:39.800
Now, remember from our lesson about inverse functions: when we learned about inverse functions,
00:19:39.800 --> 00:19:47.900
we know that if two functions are inverses, they mirror over the line y = x.
00:19:47.900 --> 00:19:52.800
They are swapping x and y coordinates; this shows us that they have to be inverses.
00:19:52.800 --> 00:20:01.400
For example, if we look at what log₂ at 2 is, it comes out to be a height of 1; here is a height of 1.
00:20:01.400 --> 00:20:12.400
And then, if we look at what our 2^x at 1 is, at 1 it is a height of 2.
00:20:12.400 --> 00:20:22.700
So, for this one, we have (I'll color-code it back to what it had been) (1,2).
00:20:22.700 --> 00:20:26.100
But for the blue one, we have (2,1).
00:20:26.100 --> 00:20:35.800
They flip x and y locations, and that is going to be true wherever we go on this, because we see that they do this thing with y = x,
00:20:35.800 --> 00:20:42.700
where they mirror across it; their x and y locations swap, showing us that they are inverses.
00:20:42.700 --> 00:20:48.000
Notice all the graphs that we have seen of logarithms: they never pass, or even touch, the y-axis.
00:20:48.000 --> 00:20:52.600
They never pass the y-axis; they never even manage to touch the y-axis.
00:20:52.600 --> 00:20:56.700
This is because the domain of a logarithm is 0 to infinity.
00:20:56.700 --> 00:21:00.200
And notice: there is a parenthesis on that 0; so it says it is not inclusive--
00:21:00.200 --> 00:21:06.600
so, not including 0, everywhere up from 0 (but not including 0), all the way up to positive infinity.
00:21:06.600 --> 00:21:08.600
We can see this for a couple of reasons.
00:21:08.600 --> 00:21:15.000
First, since exponentiation and logarithms are inverses, that means that the range of an exponential function is the domain of a logarithm.
00:21:15.000 --> 00:21:20.000
The range of f(x) = a^x is going to be 0 to infinity.
00:21:20.000 --> 00:21:27.800
a^x...if we put in any base a that is greater than 0 and not 1, it is going to go anywhere from 0 up until infinity.
00:21:27.800 --> 00:21:37.100
If we look at 2^x, by varying what we plug in for x, we are going to be able to get anything between 0 and positive infinity.
00:21:37.100 --> 00:21:46.200
Now, let's talk briefly about this idea: if we had a pool of numbers that we called a, the set of things that we are allowed to use,
00:21:46.200 --> 00:21:55.400
and then we had another pool of numbers that was b, the set of things that it is possible to get to through some function f...
00:21:55.400 --> 00:22:02.000
we have some function f that takes numbers from a, and it goes to b; then we call the numbers over here domain.
00:22:02.000 --> 00:22:05.800
We talked about this when we first talked about the ideas of functions.
00:22:05.800 --> 00:22:11.500
So, here is the domain of f; and over here is the range of f.
00:22:11.500 --> 00:22:17.900
The domain of f is everything that f is able to take in; the range of f is everything that f is able to put out.
00:22:17.900 --> 00:22:23.200
So, for the example a^x or the example 2^x, as a specific example,
00:22:23.200 --> 00:22:31.900
the domain is anything; it can take in any number at all--negative infinity to positive infinity--any real number whatsoever.
00:22:31.900 --> 00:22:35.400
But it is only going to be able to give out numbers from 0 to infinity.
00:22:35.400 --> 00:22:39.000
So, in this case, we see that it is going to have its range as 0 to infinity.
00:22:39.000 --> 00:22:47.700
Now, notice: if we do the reverse of this, if we want to see the reverse of this, a function that does the opposite of what f does,
00:22:47.700 --> 00:22:53.100
f^-1, then it is going to have to go, not from a, but from b.
00:22:53.100 --> 00:23:00.500
So, its domain, the domain of f^-1, is going to be going in the other direction.
00:23:00.500 --> 00:23:05.600
Since it is taking what f did and reversing it, it has to be able to take the things that f does as outputs.
00:23:05.600 --> 00:23:11.600
Whatever f makes as outputs--whatever f puts out--is what f^-1 will take in.
00:23:11.600 --> 00:23:22.400
So, the domain of f^-1 is the range of f, which means that the range of f^-1 is also the domain of our original function, f.
00:23:22.400 --> 00:23:26.700
f goes from a to b; f^-1 goes from b to a.
00:23:26.700 --> 00:23:32.200
Now, we saw that, for any exponential function, its range is 0 to infinity.
00:23:32.200 --> 00:23:38.800
So, that means that the domain of f^-1, the domain of a logarithmic function,
00:23:38.800 --> 00:23:44.700
since it is the inverse of exponentiation, must also be from 0 to infinity.
00:23:44.700 --> 00:23:50.000
So, that is why we have this domain here; the domain of any log has to be from 0 to infinity,
00:23:50.000 --> 00:23:54.400
because the range of any exponential function is from 0 to infinity.
00:23:54.400 --> 00:24:01.600
So, they are going to be done as opposites; the range of an exponential function is the domain of a logarithmic function.
00:24:01.600 --> 00:24:05.700
So, that is a fancy way to be able to understand why this has to be the case,
00:24:05.700 --> 00:24:10.400
because we can say what we learned about inverse functions applies here, because we have an inverse function.
00:24:10.400 --> 00:24:16.400
But alternatively, we can just see that it would not make sense--it just is nonsense if we look at it otherwise.
00:24:16.400 --> 00:24:23.200
Consider if we tried to take log₂(0); then we know that that has to be equal to some number b for it to be a possible thing.
00:24:23.200 --> 00:24:30.600
Then that means that 2^b has to somehow be equal to 0.
00:24:30.600 --> 00:24:34.600
But that doesn't make any sense: no such number b exists.
00:24:34.600 --> 00:24:40.100
No possible number could exist that would be able to take 2 and turn it into 0.
00:24:40.100 --> 00:24:46.200
2^b can't ever become 0; if we plug in any number, we can make very small numbers; but we can't actually get all the way to 0.
00:24:46.200 --> 00:24:49.100
We can't touch 0; the same is going to go for negative numbers.
00:24:49.100 --> 00:24:54.800
If we wanted to say 2^b = -4, there is no number that does that.
00:24:54.800 --> 00:25:00.800
We can't raise 2 to some number and make it negative--it started out positive, so we can't possibly make it negative.
00:25:00.800 --> 00:25:08.900
So, this is impossible; this is impossible; this is impossible; so it means that log₂(0) is an impossible idea.
00:25:08.900 --> 00:25:13.400
We can't take the logarithm of 0 or anything that is going to be negative,
00:25:13.400 --> 00:25:19.400
because it just won't be possible for it to work over here, where we are trying to figure out what would be the exponential version of it.
00:25:19.400 --> 00:25:23.000
So, since it just doesn't make sense to take the logarithm of a number that is 0,
00:25:23.000 --> 00:25:28.200
or to take the logarithm of a number that is negative, it must be that the domain is always positive.
00:25:28.200 --> 00:25:33.400
We have to go from 0, but not including 0, all the way up to positive infinity.
00:25:33.400 --> 00:25:37.200
We can take in any of those things, but we can't take in 0; we can't take in negative numbers.
00:25:37.200 --> 00:25:40.100
That explains why our domain has to be this.
00:25:40.100 --> 00:25:46.400
We can think about it that way, or we can think about it as this flipped idea of the fact that exponentiation and logarithms are inverses.
00:25:46.400 --> 00:25:52.300
So, we can have this more complex idea of the domain and range of what those things have to be.
00:25:52.300 --> 00:25:59.100
But we can also just go to the fact that it does not make sense--it would be nonsense; that is a reasonable idea, too.
00:25:59.100 --> 00:26:01.300
All right, we are ready for some examples.
00:26:01.300 --> 00:26:03.700
Let's evaluate these numbers without a calculator.
00:26:03.700 --> 00:26:08.600
If we are looking at log₆(216), then that is going to be equal to some number,
00:26:08.600 --> 00:26:12.500
such that, when we raise 6 to that number, we get 216.
00:26:12.500 --> 00:26:17.400
216 = 6^?: we want to figure out what this is, so let's see.
00:26:17.400 --> 00:26:18.700
What are some numbers that we could get out of this?
00:26:18.700 --> 00:26:28.600
6¹--that is just 6; 6²--well, that would be 36; 6³ is 180 + 36 = 216.
00:26:28.600 --> 00:26:40.300
That is what we are looking for; so it must be the case that it is 3: log₆(216) = 3; that is our answer.
00:26:40.300 --> 00:26:49.000
If we have log(1/10000), the first thing we want to do is remember: if we have just log, then that is a way of saying it is log base 10.
00:26:49.000 --> 00:26:53.000
So, log<font size="-6">10</font> (1/10000)--once again, we are asking what that is going to be.
00:26:53.000 --> 00:27:04.500
Well, that is going to be the number such that 10 to whatever that number is is going to be equal to 1/10000.
00:27:04.500 --> 00:27:09.600
So, let's look at possible numbers for 10; if we go positive, we have 10¹ = 10.
00:27:09.600 --> 00:27:11.700
Well, that is not going to work, because we are going to need a fraction.
00:27:11.700 --> 00:27:17.700
So, we notice that 10^-1 is 1/10; and then, if we think about that, 10^-1 would be 1/10;
00:27:17.700 --> 00:27:25.500
10^-2 would be 1/100; 10^-3 would be 1/1000; 10^-4 would be 1/10000.
00:27:25.500 --> 00:27:35.100
So, 10^-4 = 1/10000; we can also see this, because we can count the number of 0's: 1, 2, 3, 4.
00:27:35.100 --> 00:27:41.500
So, that is 10⁴, and since it is 1/10⁴, then that must be 10^-4.
00:27:41.500 --> 00:27:48.100
We have that -4 is what we have to raise 10 to, to get 1/10000.
00:27:48.100 --> 00:27:53.600
The natural log of e^17: well, remember: natural log is just another way of saying log base e.
00:27:53.600 --> 00:28:02.700
So, log<font size="-6">e</font>(e^17): what number do we have to raise e to?
00:28:02.700 --> 00:28:11.500
e^? =...well, the thing we are working with is e^17, so e^17 would be e^?.
00:28:11.500 --> 00:28:15.000
Well, that is pretty clear: the thing that the question mark has to be is the 17.
00:28:15.000 --> 00:28:21.800
Otherwise they will never match up; so it must be e^17 that we want here, so 17 is our answer,
00:28:21.800 --> 00:28:26.800
because if we raise e to the 17, it is no surprise that we get e^17.
00:28:26.800 --> 00:28:33.700
Finally, log₄(32): once again, we are saying, "What is the number that we have to raise 4 to, to get 32?"
00:28:33.700 --> 00:28:40.000
So, we move that over; we can think of this as 4^? = 32.
00:28:40.000 --> 00:28:45.100
So, 32 = 4^?: well, let's start looking at some possible numbers for 4.
00:28:45.100 --> 00:28:49.700
We could have 4¹; that would just be 4--not big enough.
00:28:49.700 --> 00:28:53.100
4² would be 16; we are starting to get close.
00:28:53.100 --> 00:28:57.900
4³ would be 64--it looks like we overshot.
00:28:57.900 --> 00:29:06.500
Well, we might notice that 16 times 2 equals 32; so if we could somehow get 2 to show up, we would be good.
00:29:06.500 --> 00:29:12.700
Notice: how is 4 connected to 2--what is the connection between these numbers?
00:29:12.700 --> 00:29:21.900
Well, the square root of 4 is equal to 2; but we also had another way of saying that: 4^1/2 is the same thing as saying square root.
00:29:21.900 --> 00:29:30.600
4^1/2 = 2; so we see that 4² times 4^1/2 equals 32.
00:29:30.600 --> 00:29:38.500
4 squared times 4 to the 1/2 (4 squared is 16; 4 to the 1/2 is 2)--16 times 2--gets us 32.
00:29:38.500 --> 00:29:50.300
So now, we just need to combine those: 4² times 4^1/2 is just another way of saying 4^4/2 times 4^1/2.
00:29:50.300 --> 00:29:55.000
We can add them--they are on a common base; so, 4^5/2...
00:29:55.000 --> 00:30:02.800
So, the answer for this--the number that we have to raise 4 to, to get 32, is 5/2.
00:30:02.800 --> 00:30:09.400
All right, what if we were doing the other direction--if we wanted to write an exponential equation in logarithmic form?
00:30:09.400 --> 00:30:13.900
We have these exponential equations: 3⁴ = 81--and now we want to do it in the logarithm form.
00:30:13.900 --> 00:30:23.800
Remember, we have that log<font size="-6">a</font>(x) = y is the same thing as saying a^y = x.
00:30:23.800 --> 00:30:30.000
Remember, our base here--we can think of it as popping up under what is on the other side of the equation.
00:30:30.000 --> 00:30:38.400
So, this over here is the exponential form; this here is the logarithmic form.
00:30:38.400 --> 00:30:42.700
Logarithmic form is this log stuff, and exponential is this a to the something stuff.
00:30:42.700 --> 00:30:47.400
So, what we have is exponential forms here; we want to flip it.
00:30:47.400 --> 00:30:58.100
3⁴ = 81: that is going to be log...what is our base? Our base here is a 3, so log₃...
00:30:58.100 --> 00:31:02.500
what is the number that we are raising to? That is the blue, so we are not going to use that.
00:31:02.500 --> 00:31:17.800
Finally, the number that we have is x...so, log₃(81) = the number that we have to raise to, 4, because 3⁴ = 81.
00:31:17.800 --> 00:31:27.300
If we ask what number we have to raise 3 to, to get 81, that is going to be 4; 3⁴ = 81.
00:31:27.300 --> 00:31:34.000
We can do this with any of this stuff: 10^2.4 = 251.18.
00:31:34.000 --> 00:31:42.000
Then, that is going to be...our base is 10, so we can write that as just log, because if we don't have a base, it just says log base 10.
00:31:42.000 --> 00:31:48.100
log of what number? Our number that we are going to get to is 251.18,
00:31:48.100 --> 00:31:54.700
and it actually keeps going, so we will leave those dots there to show that it keeps going.
00:31:54.700 --> 00:32:00.900
And that is going to end up equaling 2.4, because the number that we have to raise 10 to,
00:32:00.900 --> 00:32:09.400
to get 251.18, is 2.4, as was shown to us in our original exponential form.
00:32:09.400 --> 00:32:16.800
So, another one: our base here is e, so we can write that as natural log of this number.
00:32:16.800 --> 00:32:21.700
We were told that it comes out to be 4.
00:32:21.700 --> 00:32:35.200
Finally, our base is π: so log<font size="-6">π</font>(this number) = √(1/2), because we know that,
00:32:35.200 --> 00:32:41.900
if we raised π to the √(1/2), we would get 2.2466, and continuing on.
00:32:41.900 --> 00:32:50.800
So, that is how we were able to figure out that log<font size="-6">π</font>(2.2466...) must be the square root of 1/2.
00:32:50.800 --> 00:32:58.200
Graph f(x) = log₃(x): to do this, we want to start with a nice table to figure out the values.
00:32:58.200 --> 00:33:06.600
x; f(x); notice that we probably don't want to just toss in numbers right away.
00:33:06.600 --> 00:33:12.000
If we plug in 10, well, I don't know what number we have to raise 3 to, to get 10.
00:33:12.000 --> 00:33:14.100
That is going to be something complicated; we have to use a calculator.
00:33:14.100 --> 00:33:18.300
But we do know what it is going to be if we plug in numbers like, say, 3.
00:33:18.300 --> 00:33:24.000
If we plug in 3, what number do you have to raise 3 to--what is log₃(3)?
00:33:24.000 --> 00:33:26.400
What number do we have to raise 3 to, to get 3?
00:33:26.400 --> 00:33:29.400
That is easy: we just have to raise it to the 1--nothing at all.
00:33:29.400 --> 00:33:35.400
We don't have to raise it to anything, other than what is already there; so just something to the 1 is what it starts as.
00:33:35.400 --> 00:33:38.700
What about 9?--well, what number do we have to raise 3 to, to get 9?
00:33:38.700 --> 00:33:41.600
We have to square it, so we have to raise it to the 2.
00:33:41.600 --> 00:33:45.400
We can keep going in this pattern: what number do we have to raise 3 to, to get 27?
00:33:45.400 --> 00:33:46.900
We have to raise it to the 3.
00:33:46.900 --> 00:33:51.100
What number do we have to raise 3 to, to get 81? We have to raise it to the 4.
00:33:51.100 --> 00:33:52.500
And we can keep going if we want.
00:33:52.500 --> 00:33:56.800
What if we went in the other direction? Well, for 2, we don't know what number we would have to raise 3 to.
00:33:56.800 --> 00:33:59.500
But for 1, yes, we do know what number we would have to raise 3 to.
00:33:59.500 --> 00:34:07.600
3 to the what equals 1? Just like everything else, 3 to the 0 equals 1.
00:34:07.600 --> 00:34:12.700
We could go to 1/3: what number do we have to raise 3 to, to get 1/3? -1.
00:34:12.700 --> 00:34:17.300
What number do we have to raise 3 to, to get 1/9? -2.
00:34:17.300 --> 00:34:20.200
And it would get lower and lower and lower, the closer we got to 0.
00:34:20.200 --> 00:34:24.800
Once again, we will never actually be able to get to 0, because there is no number that we could raise 3 to, to get 0.
00:34:24.800 --> 00:34:27.000
But we can get really, really close to 0.
00:34:27.000 --> 00:34:29.400
So, at this point, we are ready to actually try plotting it.
00:34:29.400 --> 00:34:38.000
Notice: our x-values go pretty widely; so let's look at x-values going from -10 up to +100.
00:34:38.000 --> 00:34:42.900
And let's look at our y-values: our y-values, our f(x) values, don't really manage to change very much.
00:34:42.900 --> 00:34:52.600
So, we will look at y-values only going from -3...oh, let's make it -5...up to positive 5.
00:34:52.600 --> 00:35:02.900
OK, let's start drawing that in; we start here; here is our x-axis and y-axis.
00:35:02.900 --> 00:35:19.900
Make a scale; the scale for the x will be in chunks of 10, because we have to cover a lot of ground: -10, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.
00:35:19.900 --> 00:35:22.700
We can keep going if we want, but that is good enough for us.
00:35:22.700 --> 00:35:31.100
So, here is a 10; here is 100; I will mark 50 in the middle; 1, 2, 3, 4, 5....50, and -10.
00:35:31.100 --> 00:35:33.500
So, we can see the scale on it here.
00:35:33.500 --> 00:35:44.600
For our verticals, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.
00:35:44.600 --> 00:35:49.900
Here is -1 and -5, positive 1 and positive 5.
00:35:49.900 --> 00:35:52.000
Great; now we are ready to plot down some points.
00:35:52.000 --> 00:35:58.900
We see a 3; we are at 1; so we are very close here; we are a little under 1/3 of the way up to the 10.
00:35:58.900 --> 00:36:06.700
So, let's say it is about here; at 9, just a little bit before the 10, we are at 2.
00:36:06.700 --> 00:36:19.900
At 27 (10, 20, 30--a little bit before that, but a little under 1/3 of the way towards the other side), we are at 3.
00:36:19.900 --> 00:36:30.600
At 81 (100, 90, 80--just a hair in front of 80), we manage to be at 4.
00:36:30.600 --> 00:36:34.200
There we go; now we want to go the other way, as well.
00:36:34.200 --> 00:36:42.100
At 1/3...at 1, we are at 0, so we are really, really close to that y-axis already; at 1/3,
00:36:42.100 --> 00:36:46.600
now we are getting pretty close; we are at -1 at 1/9; we are practically on top of it,
00:36:46.600 --> 00:36:49.900
but we will never actually be on top of it; we will just get really, really close.
00:36:49.900 --> 00:36:56.500
And we can see that this pattern is going to continue: 1/27, -3; 1/81, -4.
00:36:56.500 --> 00:37:01.500
So, as it gets really close to the 0, it is going to just shoot down really quickly.
00:37:01.500 --> 00:37:08.200
So, let's draw this side in; it approaches this asymptotically.
00:37:08.200 --> 00:37:14.900
It gets really, really close, but it will never actually touch it; the part where it looks like it is touching it is just my human error at fault.
00:37:14.900 --> 00:37:19.000
But it is not going to ever quite touch it on a perfect graph.
00:37:19.000 --> 00:37:22.600
It might look like it, because of the thickness of the lines, but it will never actually do it.
00:37:22.600 --> 00:37:30.600
And as it grows more and more, it slows down, because it has to go even farther out to be able to get any growth.
00:37:30.600 --> 00:37:34.600
It slows down the farther out it gets; and we graph log₃(x).
00:37:34.600 --> 00:37:40.900
Cool; finally, what are the domains of these functions? f(x) = log₇(-x + 2).
00:37:40.900 --> 00:37:51.300
Remember: the idea we had was log<font size="-6">a</font>(stuff); then this stuff here must always be positive.
00:37:51.300 --> 00:37:56.500
So, it must be positive; otherwise, it just doesn't work.
00:37:56.500 --> 00:38:02.100
If we try to take the log of 0, it doesn't work; if we try to take the log of a negative number, it doesn't work.
00:38:02.100 --> 00:38:06.100
You always have to take the log of positive numbers, whatever the base is.
00:38:06.100 --> 00:38:10.600
For any base, this is going to be the case; so it doesn't matter if it is base 7 or base fifty billion.
00:38:10.600 --> 00:38:14.100
It is going to be the case that we have to have whatever is inside of the logarithm,
00:38:14.100 --> 00:38:20.300
whatever the logarithm is operating upon--it has to be greater than 0; it has to be a positive number.
00:38:20.300 --> 00:38:23.900
So, we know that the thing that log is operating on here is -x + 2.
00:38:23.900 --> 00:38:28.500
So, we know that -x + 2 must be positive; it must be greater than 0.
00:38:28.500 --> 00:38:34.200
We move the x over; we have that 2 has to be greater than x, so x has to be less than 2,
00:38:34.200 --> 00:38:38.600
and it can go all the way down to negative infinity, because the only restriction we have is that 2 is greater than x,
00:38:38.600 --> 00:38:45.300
which we could write out as...anywhere from negative infinity up until positive 2, but not including positive 2,
00:38:45.300 --> 00:38:49.000
which we show with a parenthesis to show that it is not inclusive.
00:38:49.000 --> 00:38:53.200
Over here, g(t) = 5t(log<font size="-6">π</font>(3t + 7)).
00:38:53.200 --> 00:38:59.400
Once again, the base doesn't really matter; it has to be positive, no matter what the base is.
00:38:59.400 --> 00:39:04.700
For any arbitrary base a, it has to be positive on what the logarithm is operating on.
00:39:04.700 --> 00:39:09.000
We look at the 5t part: we might get worried--"oh, is the 5t going to interact with it?"
00:39:09.000 --> 00:39:13.000
5t times log<font size="-6">π</font>...5t is really in its own world; it is doing its own thing.
00:39:13.000 --> 00:39:18.700
5 times t...we can do that for any number; we can multiply 5 times any number, so its domain is anything at all.
00:39:18.700 --> 00:39:21.900
It is not going to actually get in our way; once again, the only thing we are worried about is
00:39:21.900 --> 00:39:27.700
when the logarithm is going to try to take the log of a negative or 0 number.
00:39:27.700 --> 00:39:33.300
So, to avoid that, we have to have that 3t + 7 must be greater than 0;
00:39:33.300 --> 00:39:38.100
otherwise, we will be taking the log of something that we cannot take logs of, that would break our function.
00:39:38.100 --> 00:39:47.800
So, 3t + 7 is greater than 0; subtract 7; 3t > -7; divide by 3; t must be greater than -7/3.
00:39:47.800 --> 00:39:51.900
So, t starts at -7/3, but is not actually able to include -7/3.
00:39:51.900 --> 00:39:57.100
So it starts just above -7/3 and can go anywhere larger; so it can go all the way out to positive infinity.
00:39:57.100 --> 00:40:04.500
So, we have -7/3 shown with a parenthesis, because we can't actually include -7/3; we can just get arbitrarily close to it.
00:40:04.500 --> 00:40:07.500
It is going all the way out to positive infinity.
00:40:07.500 --> 00:40:08.900
And there are our two domains; all right, cool.
00:40:08.900 --> 00:40:12.700
We will talk a bunch more about logarithms in the next one, where we will explore the properties of logarithms.
00:40:12.700 --> 00:40:14.300
And then, we will see even more about how the two connect.
00:40:14.300 --> 00:40:18.000
We have a lot of really interesting ideas; it is new stuff, but once you start practicing it,
00:40:18.000 --> 00:40:20.500
as you do it a bunch of times, logarithms will really start to make sense.
00:40:20.500 --> 00:40:25.700
You will get this idea of "what am I trying to raise this number to, to get the thing I am taking the logarithm of?"
00:40:25.700 --> 00:40:29.800
What does this base have to be raised to, to get the number that I am taking the log of?
00:40:29.800 --> 00:40:31.000
All right, we will see you at Educator.com later--goodbye!