WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today begins the first in a series of lectures on exponential and logarithmic relations, starting out with exponential functions,
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beginning with the definition: What is an exponential function?
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Well, an exponential function is a function of the form f(x) = a times b^x, where a is not 0,
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because if a were to equal 0, this would all just drop out, and you wouldn't have a function.
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b is greater than 0: we are restricting this definition to values of b that are positive; and b does not equal 1.
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If b were to equal 1, no matter what you made x, that would still remain 1.
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And then, you wouldn't really have a very interesting function.
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So, the base of the function here is b.
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Now, recall earlier on, when we worked with functions and equations, we have seen things like this: x² + 2x - 1...f(x) equals this.
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Here, the base was a variable, and here the exponent was a constant.
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Now, we are going the other way around: in this case, with exponential functions, we are going to be working with situations
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where the base is a number (a constant) and the exponent contains a variable.
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And that is what makes these functions fundamentally different from some of the other functions that we have seen so far.
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Examples would be something like f(x) = 6x: here the base is 6, the coefficient is just 1 (a = 1), and the exponent is x.
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Or you could have something a little bit more complicated: f(x) = (3 times 1/2)^4x - 2.
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I have an algebraic expression, not just a single variable, as the coefficient.
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So here, I have a base equal to 1/2; the coefficient is 3; and then the exponent is 4x - 2.
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Starting out by looking at the graphs of exponential functions: as we have done with other types of functions,
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we can use a table of values to graph an exponential function.
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And we are going to look at a few different permutations of these functions and see what we end up with.
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I am going to start out with letting f(x) equal 3^x; let's find some values for x and y.
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I am going to just rewrite f(x) as y.
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If x is -3, then what is y? Well, this is giving me 3 to the -3 power.
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Recall that a^-n equals 1/a^n; so what this is really saying is 1/3³, which equals 1/27.
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At this point, if you are not really comfortable with exponents and the rules and properties governing working with exponents,
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you should go back and review the earlier lecture on that, because we are going on to solve equations using these.
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So, you need to have the rules learned, as we will be applying them frequently.
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When x is -2, this is going to give me 3^-2, which is 1/3², or 1/9.
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And we continue on: when x is -1, that is 3^-1 = 1/3¹, or 1/3.
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Then, getting back to some more familiar territory: when x is 0, this gives me 3⁰.
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A number or variable, or anything, to the 0 power, is going to give me 1; when x is 1, f(x), or y, is 3.
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When x is 2, that is 3² is 9; when x is 3, that is 3³, to give me 27.
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Let's go ahead and plot these values out; let's do that so that this is -2, -4, -6, -8, so we can look at more values right on this graph.
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2, 4, 6, 8, 2, 4, 6, 8, 10; -2, 4, 6, 8, 10; so -10 is down here.
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When x is -3, the graph is just slightly above 1; it is 1/27; when x is -2, we get a little farther away from 1--it becomes 1/9.
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When x is -1, the graph rises up even a little more and becomes -1/3.
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When x is 0, y is 1; when x is 1, y is 3; when x is 2, y is right up here at 9; and then, when x is 3, y gets very large.
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This gives me a sense of the shape of the graph--that as x becomes positive, y rapidly becomes a very large number;
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as x is negative, what I can see happening is that f(x) is approaching 0, but it never quite gets there.
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Therefore, what I have is that, for this graph, the x-axis is a horizontal asymptote.
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So, this is the graph of f(x); let's look at a different case--let's look at the graph of...this is in the form f(x) = ab^x;
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let's look at the graph f(x) = a(1/b)^x.
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Well, a is just 1, so I am going to look at the situation g(x) = 1 over 3 to the x power.
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I am making a table of values, again, for x and y.
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Again, when x is -3, let's look at what y is; it is going to be 1/3 to the -3.
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Using this property, I am then going to get 3³, which is 27.
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For -2, I am going to get (1/3)^-2, or 3², which is 9.
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-1 is going to give me (1/3)^-1 = 3; 0...anything to the 0 power is simply going to be 1.
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(1/3)¹ is 1/3; (1/3)² is 1/9; and (1/3)³ is 127.
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Let's see what happens with this graph: here, when x is -3, y is very large--it is some value way up there, 27 (that is off my graph).
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Let's look at -2: when x is -2, y is 9 (that is right here).
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And I know that, when I get out to -3, this is increasing; so I know the shape up here.
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When x is -1, y is 3; when x is 0, y is 1; they have the same y-intercept, right here at (0,1).
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When x is 1, y is 1/3; when x is 2, we see that the graph is approaching the x-axis, because now, we are here at 1/9.
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And when x is 3, the value of the function is 127.
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So again, I see this x-axis, again, being the horizontal asymptote for both graphs.
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I know that y is getting very large as x is becoming negative.
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And I know that, as x is positive, the graph is approaching, but not reaching, the x-axis.
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Here, the y-axis also forms an axis of symmetry; so these two graphs are mirror images of each other.
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f(x) and g(x) are mirror images reflected across the y-axis.
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Let's look at one other case: let's look at the case...let's call it h(x) = 3^x, but let's take the opposite of that.
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Therefore, this will be simpler to figure out the values: let's just leave the x-values the same as they were for f(x),
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and let's just extend this graph out: these were my values for f(x), and now let's figure out what h(x) is going to give me.
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Well, if x is -3, and I put a -3 in here, again, I am going to get 1/27, but I want the opposite of that; so I am going to get -1/27.
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If x is -2, again, I am going to get 3^-2, which is 1/9, but I am going to take the opposite.
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So, all I have to do is change the signs on these to get my h(x) values.
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And then, we can see what this graph looks like.
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This is f(x); I have g(x) here; for h(x), when x is -3, h(x) is right here; it is very close to the x-axis, but not quite reaching it.
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And at -2, it is a little bit farther away at -1/9; at -1, we are down here at -1/3; at 0, here we have the y-intercept at (0,-1).
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When x is 1, y is -3, right about here; when x is 2, y is down here at -9; and when x is 3, we are going to be way down here at -27.
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So, for h(x), what I am going to have (I'll clean this up just a bit)...
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This is the graph of h(x), and again, I am seeing the x-axis here, acting as a horizontal asymptote.
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And I am seeing that, as x becomes positive, y becomes very large, but this time it is in the negative direction.
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It is giving me a mirror image here with f(x); but now, my values are going down in the negative direction.
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So, these are several graphs of exponential functions; let's go ahead and sum up the properties of these functions.
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What we saw is that f(x) is continuous and one-to-one.
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I am just very briefly sketching these three situations: I let f(x) equal 3x, g(x) equal (1/3)^x, and h(x) equal (-3)^x.
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So, for the graph of f(x), what I got is this...actually, rising faster; so let's make that rise much faster.
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And I saw that it is continuous; there weren't any gaps or discontinuities.
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When we worked with graphing some rational functions, we saw that there were actually discontinuities; there aren't any here.
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It is also one-to-one; and I could use the vertical line test.
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Recall that, if you draw a vertical line across the curve of a graph, if it only crosses the graph once,
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at one point, everywhere you possibly could try, you have a one-to-one relationship; you have a function.
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So, no matter where I drew a vertical line, I would only cross this curve one time.
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The domain of f(x) is all real numbers: you see that I could make x a negative value; I could make it a positive value.
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However, the range is either all positive real numbers or all negative real numbers.
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Here, the range is positive real numbers; for h(x), I graphed that, and that turned out like this.
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Here, the domain was all real numbers, but the range was the negative real numbers.
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We saw that the x-axis acts as an asymptote that the graphs of these functions approach, but never reach.
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The y-intercept is at (0,a); so for both g(x) and f(x), here a = 1, so the coefficient is 1; so the y-intercept is at (0,1).
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However, for h(x), a = -1; the coefficient is -1, so I have a y-intercept here at (0,-1).
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Finally, we saw that the graphs of f(x) = ab^x and f(x) = a/(1/b^x) are reflections across the y-axis.
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So, that was this graph, which shares the same y-intercept, but is reflected here across the y-axis.
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So, looking at these three different situations and their graphs can tell us a lot about what is going on with exponential functions.
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Introducing the concept of growth and decay: if you have a function in this form, f(x) = ab^x,
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and a is greater than 0 (it is a positive number), and b is greater than 1, then this represents exponential growth.
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And the concepts of exponential growth and decay are very frequently used in real-world applications.
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So, we are going to delve into this topic in greater depth in a separate lecture.
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But thinking about exponential growth: that could be something like working in finance, or thinking about your own savings and compound interest.
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That works in such a way that the growth is exponential.
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An example of exponential growth would be something like f(x) = 1/4(2^x).
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And that is because here, b is greater than 1; so this is growth.
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If I had another function, f(x) = 4((1/8)^x), here this is representing exponential decay,
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because b is greater than 0, but it is less than 1: it is a fraction between 0 and 1.
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And we sometimes talk about things such as radioactive decay and half-life in terms of an exponential function.
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Another way to think of this for a minute is: recall that a^-n = 1/a^n.
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So, I can actually rewrite this function as 4(8^-x).
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If I had a base here that was greater than 1, but the exponent was negative, then I also know that I have decay.
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If you write it in the standard form where the exponent is positive, then all you need to do is look and see the value of b.
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If it is greater than 1, you have growth; if it is between 0 and 1, you have decay.
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If it is a number greater than 1, and it is not in standard form, and you see I have a negative exponent,
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that also gives you a clue that you are looking at decay.
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But you can always put them in standard form, like this, and then just go ahead and look at the value of the base.
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Working with exponential equations: in exponential equations, variables occur as exponents.
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That is what I mentioned in the beginning, when I was talking about exponential expressions and exponential functions.
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But now, we are talking about equations: again, you are used to working with things where the base may be a variable, but the exponent is a number.
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And now, we are going to change that around and actually have situations where variables are exponents.
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There are some properties of this that we can use to solve these equations.
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If the bases are the same (if b^x equals b^y), then x must equal y.
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If the bases are the same, in order for the left half of the equation to equal the right half, the exponents have to be the same; they have to be equivalent.
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Look at this in a very simple case: 6^3x - 5 = 6⁷.
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6 is the same base as 6; so for this left half to equal the right half, if the bases are equal, these two must be equal.
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So, in order to solve these, I am just going to take 3x - 5 and put that equal to 7, which leaves me with a simple linear equation to solve.
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I am going to add 5 to both sides, which is going to give me 3x = 12.
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Then, I am going to divide both sides by 3 to get x = 4.
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Things get a little more complicated if the bases are not the same.
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If the bases are not the same, the first thing to do is try to make them the same.
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If that is not done reasonably simply, then we can use another technique that we are going to learn about in a later lecture.
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But for right now, we are going to stick to situations where either the bases are the same, or you can pretty easily make them the same.
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For example, I could be given an exponential equation 2^x - 3 = 4.
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These bases are not the same, so I can't use this technique.
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However, you can pretty easily see that you can make them the same base
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by saying, "OK, 2 squared is 4; so instead of writing this as 4, I am going to write it this way."
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Now, I am back to the situation where the bases are the same; I am going to set the exponents
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equal to each other (because they must be equal to each other) and solve for x: x = 5.
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You could have a little bit more complicated situation, where 3^x - 1 (a separate example) = 1/9.
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And I can see that I want these to be the same, and I know that 1/3 squared is 1/9.
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So, I am pretty close; but I need this to be 3.
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Well, recall that I could rewrite this as 3^-2: 3^-2 is the same as 1/3 squared.
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This also equals 1/9; and it is fine that this is a negative exponent--I can go ahead and use it up here, rewriting 1/9 as 3^-2.
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Now, -x - 1 = -2, so x = -1.
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OK, so basically, when you are working with exponential equations,
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if they are the same base, you simply set the exponents equal to each other,
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because this property tells us that that must be the case.
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If the bases are not the same, try to make them the same: that is going to be your first approach.
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And then, once you have written them as the same base, then you go ahead and solve by setting the exponents equal to each other.
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Now, let's look at exponential inequalities: exponential inequalities involve exponential functions.
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We just talked about exponential equations, and this is a similar situation, except we are working with an inequality, not an equation
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(greater than, less than, greater than or equal to, less than or equal to).
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And there are some properties that we can use to help us solve these.
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If b > 1, then b^x > b^y if and only if x > y.
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And this makes sense: if I have the same base (these are the same),
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the only way that this left half is going to be greater than the right half
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is if these exponents hold the same relationship, where x is greater than y.
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And this could be greater than or equal to, or less than, or less than or equal to.
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So, b^x < b^y if and only if x < y--the same idea.
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If the bases are the same, and this on the left is less than the one on the right,
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then that relationship, x < y, must be holding up.
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We are going to use this property to solve inequalities.
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For example, 4^x + 3 > 4²: I know that the bases are equal, so that x + 3 must be greater than 2.
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So, I just solve, subtracting 3 from both sides to get x > -1.
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Again, if you are trying to work with a situation where you are solving an exponential equation or inequality,
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and the bases aren't the same, see if you can make them the same.
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5^6x < 115 well, I know that 5² is 25; if I multiply 25 times 5, I am going to get 125.
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Therefore, 5² times 5...that is 5³...equals 125.
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I am going to rewrite this as 5³; now that the bases are the same, I can say, "OK, 6x is less than 3, so x is less than 1/2."
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Again, the idea is to get these in the form where the base is the same,
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and then use the property that the relationship between the exponents has to be maintained,
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according to the inequality, if the bases are the same.
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Looking at examples: let's go back to talking about graphing.
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We are asked to graph this function, f(x) = 3(2^x)--graphing an exponential function.
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x...and we need to find y, so that we can do some graphing.
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When x is -3, this is going to be 3 times 2 to the -3 power.
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Again, this is equal to 3 times 1 over 2³; this is going to be equal to 3 times...2 times 2 is 4, times 2 is 8; or 3/8.
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When x is -2, I get 3 times 2^-2; 3 times 1/2 squared...2 times 2 is 4, so this gives me 3 times 1/4, or 3/4.
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-1: 3 times 2^-1 equals 3 times 1/2¹...remember that this -1 tells us that we need to take 1/the first power.
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So, that is 3 times 1/2, which is 3/2.
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0 gives me 3 times 2⁰, or 3 times 1, which equals 3.
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Using 1 as the x-value gives me 3 times 2¹, or 3 times 2, which equals 6.
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Using 2 gives me 3 times 2 squared, which is 3 times 4, or 12.
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And then, one more: 3 times 2 cubed equals 3 times 8, or 24.
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Plotting these values: when x is -3, y is 3/8 (just a little bit greater than 0); when x is -2, y gets a little bit bigger; it is 3/4.
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When x is -1, then y becomes 3/2, or 1 and 1/2; 3/2 is going to be about here.
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When x is 0, y is 3; when x is 1, y is up here at 6, rising rapidly; when x is 2, y will be all the way up here at 12.
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I have enough points to form my curve; and I see the typical properties of exponential functions.
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Remember that the y-intercept is going to be at (0,a): in this case, a = 3, so my y-intercept is going to be at (0,3), and that is exactly what I see.
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And I can see that over here in the table of values.
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I also see that the x-axis is an asymptote, and that the graph is approaching, but never reaching, the x-axis.
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And I also can see that, as x becomes large in the positive direction, the value of y rapidly increases.
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This is a graph of a typical exponential function.
00:27:51.600 --> 00:27:57.300
Example 2: Does this function represent exponential growth or decay?
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Recall that, if a is greater than 0, then we can look at a function in the form f(x) = ab^x and evaluate b.
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If the base is greater than 1, then we have exponential growth.
00:28:17.900 --> 00:28:23.900
If b is greater than 0, but less than 1, we have exponential decay.
00:28:23.900 --> 00:28:31.000
But the caveat is that it has to be in the standard form; and I have an equation here that is not in standard form.
00:28:31.000 --> 00:28:36.100
But I can use my rule that a^-n equals 1/a^n.
00:28:36.100 --> 00:28:47.600
Therefore, I am going to rewrite this as f(x) = 4((1/5)^x).
00:28:47.600 --> 00:28:54.900
So, I just took the reciprocal of the base and rewrote it.
00:28:54.900 --> 00:29:02.200
And this is talking about a different 'a' than I am talking about here, with the coefficient: here, the coefficient is a.
00:29:02.200 --> 00:29:10.900
All right, now that I have it in this form, I can evaluate; and I see that my coefficient is greater than 0, and that b is actually between 0 and 1.
00:29:10.900 --> 00:29:16.000
Therefore, this is exponential decay.
00:29:16.000 --> 00:29:20.300
I also could have looked back at the original and recalled that, if b is greater than 1,
00:29:20.300 --> 00:29:25.000
and then you have a negative exponent, that would also give me a clue that this is exponential decay.
00:29:25.000 --> 00:29:33.000
But one way to go about it is just to put it in the standard form, and then evaluate the value of the base.
00:29:33.000 --> 00:29:39.300
All right, in Example 3, I have an exponential equation; and recall that, if the bases are the same,
00:29:39.300 --> 00:29:47.500
then the exponents must be equal in order for the equation to be true.
00:29:47.500 --> 00:29:53.300
The problem is that the bases aren't the same; so I need to get them to be the same.
00:29:53.300 --> 00:29:56.000
Looking at these, they are both even; so I am going to try 2.
00:29:56.000 --> 00:30:17.200
And we know that 2 cubed is 8; so let's start from there: 8 times 2 is 16; 16 times 2 is 32, so this gives me 16, 32...
00:30:17.200 --> 00:30:30.800
times 2 is 64; times 2 is 128; so 2 to the third, fourth, fifth, sixth, seventh, equals 128.
00:30:30.800 --> 00:30:38.900
If 2⁷ is 128, then 2^-7 = 1/128.
00:30:38.900 --> 00:30:44.600
So, I have this written as a power of 2; let's look at the right.
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I know that 2⁷ is 128; 128 times 2 is 256; 2⁷ times 2¹...
00:30:56.100 --> 00:31:02.900
I add the exponents, so this is actually 2⁸ = 256.
00:31:02.900 --> 00:31:07.300
Now, I can write this equation using the base of 2.
00:31:07.300 --> 00:31:15.400
On the left, I am going to have 2^-7 raised to the 2x power, equals...
00:31:15.400 --> 00:31:23.100
and then, on the left, 2⁸ times 3x - 1.
00:31:23.100 --> 00:31:37.100
So, this is going to give me 2 to the -7 times 2x (is going to be -14x), equals 2 to the 8, times 3x - 1.
00:31:37.100 --> 00:31:41.500
Now, I have it in this form, where the bases are the same, and I have an x and a y.
00:31:41.500 --> 00:31:50.800
Therefore, -14x = 8(3x - 1); and then I am just left with a simple linear equation.
00:31:50.800 --> 00:32:06.500
-14x = 24x - 8; I am going to add 14x to both sides to get 38x (I added 14 to both sides).
00:32:06.500 --> 00:32:10.600
At the same time, I am going to add 8 to both sides to get it over here.
00:32:10.600 --> 00:32:16.200
I could have done it the other way; then I would have had a negative and a negative, and then divided or multiplied by -1.
00:32:16.200 --> 00:32:28.800
Anyway, I am going to go ahead and divide both sides by 38; and this is going to give me x = 8/38.
00:32:28.800 --> 00:32:31.900
I can do a little simplification, because I have a common factor of 2.
00:32:31.900 --> 00:32:35.700
I can cancel that out to get 2/19.
00:32:35.700 --> 00:32:41.200
The hardest part for solving this was simply getting them into the same base.
00:32:41.200 --> 00:32:47.500
And I was able to do that because 2^-7 is 1/128, and 2⁸ is 256.
00:32:47.500 --> 00:32:55.700
Once I did that, it was simply a matter of putting the exponents equal and solving a linear equation.
00:32:55.700 --> 00:33:01.200
Here we have an exponential inequality: again, the first step is to get the bases the same,
00:33:01.200 --> 00:33:11.800
because I know that, if I have a base raised to a certain power, and it is less than that same base
00:33:11.800 --> 00:33:21.200
raised to another power, that the relationship between these two exponents has to hold up.
00:33:21.200 --> 00:33:27.800
Well, this time (last time I had some even numbers; I tried 2, hoping I could find a base there
00:33:27.800 --> 00:33:34.900
that I could easily make them into a common base, and I did), since I see a 27, I am going to work with 3's.
00:33:34.900 --> 00:33:48.200
So, I know that 3 cubed is 27; therefore, 3^-3 = 1/27.
00:33:48.200 --> 00:33:56.700
243...let's look at that: if 3³ is 27, I multiply that times 3; that is going to give me 81.
00:33:56.700 --> 00:34:03.800
If I multiply that by 3, fortunately, I am going to get 243.
00:34:03.800 --> 00:34:11.800
So, this is 3³ times 3; that is 3⁴, times 3, so this gives me 3⁵ = 243.
00:34:11.800 --> 00:34:22.900
Let's go ahead and work on this, then: rewrite this as, instead of 243...I am going to write it right here as 3⁵,
00:34:22.900 --> 00:34:34.100
raised to 4x - 3, is less than...instead of 1/27, I am going to write this as 3^-3, all to 6x - 4.
00:34:34.100 --> 00:34:47.000
So, this is going to give me 3 to the 5, times 4x - 3, is less than 3 to the -3 times 6x - 4.
00:34:47.000 --> 00:34:51.800
Now, I have the same bases: I can just look at the relationship between the exponents:
00:34:51.800 --> 00:35:01.100
that 5(4x - 3) is less than -3(6x - 4), and solve this linear equation.
00:35:01.100 --> 00:35:16.400
This gives me 20x - 15 < -18x + 12; that is 38x - 15 (I added 18 to both sides) < 12.
00:35:16.400 --> 00:35:23.500
So, 38x (I am going to add 15 to both sides to get this) < 27.
00:35:23.500 --> 00:35:29.200
And then, I am just going to divide both sides by 38 to get x < 27/38.
00:35:29.200 --> 00:35:35.100
Again, the most difficult step was just getting these to be written as the same base.
00:35:35.100 --> 00:35:39.900
And I was able to do that by using 3 as the base.
00:35:39.900 --> 00:35:50.600
And then, I had equivalent bases; I did a little simplifying, set them equal, and solved this linear inequality.
00:35:50.600 --> 00:35:58.000
That concludes this lesson of Educator.com on exponential equations and inequalities; thanks for visiting!