WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we are going to talk about logarithms and logarithmic functions, beginning with the definition.
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What are logarithms? First, the restrictions: x cannot be 0, and b needs to be a positive number, but not equal to 1.
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So, the logarithm of x to the base b is written as follows: log < font size="-6" > b < /font > (x).
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Let's start out with log < font size="-6" > b < /font > (x) = y: in this case, this log is defined to be the exponent y, which would satisfy this equation.
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Over here, I am talking about the same base: the base b here is the base b right here.
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We worked with exponents already; and logarithms are actually just exponents.
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When you use logarithmic notation, you are just writing an exponential expression in a different way.
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The log < font size="-6" > b < /font > (x) = y is defined as the exponent y that, when you raise b to that power, will give you x back.
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So, you need to be able to get comfortable with going back and forth between the logarithmic notation and the exponential expression.
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These two statements are actually inverses of each other; and we will talk more about that relationship in a little while.
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But we are just starting out now, to get used to the idea of converting back and forth.
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And the reason that you need to be able to convert back and forth and understand the relationship between these two
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is: exponential expressions can help you solve logarithmic equations, and logarithmic expressions can help you solve exponential equations.
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Given log₄(16) = 2, you can rewrite that into the exponential expression.
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At first, you might need to just stop and analyze the various components.
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The base is 4; here, x is 16, and y is 2; so I am going to rewrite that: the base remains the same:
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4 to some power y (here it is 2) equals 16; and we know that that is true--that 4² is 16.
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You can also move from the exponential equation into the logarithmic equation.
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Looking at an example where you are starting out with 2³ = 8:
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the base here is 2; y = 3 (that is the power you are raising the base to); and x = 8.
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Now, I am going to write this as a logarithmic equation: the base is still 2; x is 8; and y is 3; so log₂(8) = 3.
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We can also use this for slightly more complicated situations, such as log₅(1/125) = -3.
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But it is the same idea, because I still have the base, 5, and I know that this is y, so 5^-3 equals 1/125.
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And I know that that is true, because 5^-3 would be the same as 1/5³, which is 1/125; so that holds up.
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I can use this to evaluate logarithmic expressions when I am trying to solve.
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Now, here I gave you all of the pieces; you already had all of the numbers, and it was just rewriting them a different way.
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Let's look at a situation where we are actually trying to find a value.
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You are given log₃(81) = y.
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I have the logarithmic expression; but I can solve this more easily if I rewrite it as an exponential equation.
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I have this base, 3; and what this is saying is that a logarithmic expression is defined to be the exponent y satisfying this relationship.
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So, if I take the base, and I raise it to the y power, I am going to get 81.
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I just need to figure out what I would need to raise 3 to, to get 81.
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3 squared is 9; 3 cubed is 27; 3 to the fourth is 81; therefore, y equals 4, because 3⁴ is 81.
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This shows you how being able to convert between the logarithmic equation and the exponential equation can help you to solve either one of them.
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OK, looking on at logarithmic functions: a logarithmic function is a function of the form f(x) = log < font size="-6" > b < /font > (x).
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We just introduced this idea of logarithms; now we are talking about logarithmic functions, where b is greater than 0 but not equal to 1.
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And you will call that these are the same restrictions that we had when talking about exponential equations.
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And this would be for the same reasons as discussed in that lecture, because again, logarithms are simply exponents.
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As I mentioned in a previous slide, these two are inverses; so f(x) = log < font size="-6" > b < /font > (x)
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is the inverse of the exponential equation, expression, or function g(x) = b(x).
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So, f(x) = log < font size="-6" > b < /font > (x) is the inverse of g(x) = b^x.
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And this is an important relationship, because it helps us to solve the equations that we will be working with shortly.
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Let's just take an example to make this more concrete.
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Let's let f(x) equal log₂(x): the inverse of that would then be g(x) = that same base, 2, raised to the x power.
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Let's look at some values for f(x): if x is 1, what is y?
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Well, think about what this is saying: this is saying log₂(x) equals some value of y.
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Rewriting that as a related exponential expression, this is telling me that, when I take 2 and I raise it to the y power, I am going to get x back.
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What I want to figure out here is: 2 to some power, y, equals 1; what would y have to be?
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y would have to be 0; therefore, 2⁰ = 1 satisfies this: x is 1 when y is 0.
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2 to some power y equals 2; what would y have to be? It would have to be 1.
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How about 4? 2 to some power y equals 4; 2² = 4; therefore y = 2.
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Let's let x be 8: 2 to some power y equals 8: well, 2³ is 8; therefore, y = 3.
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All right, so that is f(x); now let's look at g(x).
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If f(x)...this is f(x), but here I am calling it y; now let's look at g(x)...and g(x) are inverses,
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then what I am going to expect is that the domain of f(x) is going to be the range of g(x),
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and the range of f(x) is going to be the domain of g(x).
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So, I am going to go ahead and take these values right here that are the range of f(x);
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and I am going to use them as the domain of g(x), and see if I get these values back.
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So, when x is 0, y is 2 to the 0 power, or 1; when x is 1, y is 2; when x is 2, y is 2², is 4.
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When x is 3, y is 2³: it is 8; and I look, and the domain here is equal to the range; the range here is equal to the domain.
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Now, that doesn't prove anything: it just shows that this one example holds up.
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But our finding was, as expected, that if f(x) and g(x) are inverses, I do expect the domain of one to be the range of the other,
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and the range of this one to be the domain of that one.
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Looking at the graphs of these functions, we can use a table of values to graph a logarithmic function,
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just as we have used tables of values to evaluate functions earlier in the course, including exponential functions.
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So, we already started a table for f(x) = log₂(x), and for g(x) = 2^x, the inverse.
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So, let's keep going with those, but add on some values.
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Recall that I said, if f(x) = log₂(x), then if I take 2^y, I am going to get x.
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I am rewriting this in exponential form to make it easier for me to find y, because it is difficult to find f(x), or y, when it is in this form.
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Recall that I said that, when x is 1, then what this is saying is that 2^y = 1.
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And I said that y must be 0; when x is 2, 2^y = 2, so y must be 1.
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And I went on and did a couple of other values, and I did 8: I said 2^y = 8; therefore, y had to equal 3.
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Now, let's add some values: let's add some fractions to get a better idea of what this graph is doing as x becomes small, as it gets close to 0.
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When x is 1/2, that is telling me 2^y = 1/2.
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The way I would get that is if I took 2^-1; I would get 1/2; therefore, y is -1.
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1/4: 2^y = 1/4--if I took 2^-2, I would get 1/4, so y is -2.
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1/8: using that same logic, 2³ is 8, but 2^-3 would be 1/8, so I am going to make that -3.
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And then, I am going to graph these values.
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When x is 1, y is 0; when x is 2, y is 1; when x is 4, y is 2; and when x is way out here at 8, y is 3.
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So, the general shape is just going up like this.
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Small values: when x is getting smaller and smaller, what is going to happen with y as x is approaching 0?
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When x is 1/2, y is -1; when x is 1/4, y is -2; when x is 1/8, y is -3.
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And what I can see is happening here is that the y-axis is a vertical asymptote.
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And we talked earlier about the graphs of exponential functions: we saw that the x-axis formed a horizontal asymptote.
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Here, the y-axis is an asymptote.
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All right, let's go ahead and look at the graph of the inverse.
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And that is going to be very simple, because I know that, since this is the inverse, all I have to do
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is take the domain and make that the range, and then I take the range and make that the domain.
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And I am going to check and make sure I have all of these values correctly matched up: 1 and 2...yes, I do.
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Then, I am going to go ahead and graph them: when x is 0, g(x) is 1 (this is f(x)).
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When x is 1, g(x) is 2; when x is 3, g(x) is way up here at 8, about here.
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When x is -1, y is 1/2; -2, 1/4; at -3, it is 1/8.
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Now, I know that I have an exponential function here that I am graphing.
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So, as expected, the x-axis is an asymptote for g(x), written in this exponential form.
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So, looking at what this is saying: the vertical asymptote here is at x = 0, so x will never cross this axis.
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It will never become negative for f(x), and that makes sense.
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x can never be negative, because there is no value of y that I can take...2 to some value...and get a negative back.
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So, x cannot be negative, because there is no possible value of y that would turn 2 into a negative number.
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Therefore, the domain of f(x) is restricted to values greater than 0 (to positive values).
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Now, since this is the inverse, and the range of this g(x) is going to be the same as the domain of that,
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that means that the range of g(x) is going to just be all positive numbers.
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To sum up properties: the graph of a logarithmic function f(x) shows that f(x) is continuous and one-to-one.
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There are no gaps; there are no discontinuities; also, you can take that graph that we just did and try the vertical line test.
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No matter where you drew a vertical line, it will only cross the curve of f(x) once.
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Therefore, this is a function; there is a one-to-one relationship between the values of x and the values of y.
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We just discussed why the domain of f(x) must be all positive real numbers.
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The domain cannot include a negative number, because there would not be any value for y that would give you a negative number.
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You cannot end up with a negative value for x.
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The range, however, is all real numbers; so y can be a negative number--I can have -2 here, or something.
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We also saw that the y-axis is an asymptote, and that the graph is going to approach that axis, but it is never actually going to reach it.
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And finally, just sketching this back out again, the graph of f(x) looks like this.
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This is the graph of f(x), and I used log₂(x) for that.
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And the y-intercept was at (1,0), and illustrated here, the y-axis is an asymptote.
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The domain is only positive numbers; however, the range is all real numbers.
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Since f(x) equals b^x, and g(x) is log < font size="-6" > b < /font > (x), and they are inverses of each other,
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we can end up with the identity function when we use a composite function, or composition of functions.
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Recall that, when we talked about composition of functions, we ended up with something like this: f composed with g equals f(g(x).
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Applying that up here, that is going to give me f(log < font size="-6" > b < /font > (x)) = b...and this is going to be my x;
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so for x, I am going to insert this: log < font size="-6" > b < /font > (x).
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And since these are inverses, this two essentially cancel each other out; and I am just going to end up with my x back (the identity function).
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g composed with f should do the same thing: this is g(f(x)) = g of f(x), which is b^x.
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g here is log < font size="-6" > b < /font > (x), so I am going to take log < font size="-6" > b < /font > , and for x right here, I am going to substitute in b^x.
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And this will work as an identity function, giving me x back.
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And the inverse property is going to be very helpful to us, as we work on solving equations involving logarithms and exponents.
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Let's take a look at some methods for solving equations, starting with just very simple ones, and then advancing to more complex equations.
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Logarithmic equations are defined as those that contain one or more logarithms with a variable in them.
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The definition can be used to solve simple logarithmic equations.
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First, recall what the definition of the logarithm is: log < font size="-6" > b < /font > (x) = y if this base, b, when raised to the y power, generates x.
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This definition can be used to solve very simple logarithmic equations where there is a log only on one side of the equation.
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By log, I mean a log containing a variable.
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log < font size="-6" > b < /font > (x + 4) = 3: recall that I said that, if you have a logarithm, you can often use the exponential form to help you solve the equation.
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And later on, we will see that, if you have the exponential form, you can use the log to help you solve equations.
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So, thinking this out: I know that I have the base equal to 2 and that x = x + 4 and y = 3.
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So, I can rewrite this in this form: my base 2, to the third power, equals x, which is x + 4.
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Now, I have something I can solve: 2³ is 8; 8 = x + 4.
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All I have to do is subtract 4 from both sides, and I get x = 4.
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Now, you have to be careful when you are working with logs, because we can't take the log of a negative number.
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So, I am going to just check back in my original here and see...
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If I put this 4 in here, I am going to get log₂(4 + 4), which is log₂(8).
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And that is fine, because that is a positive number.
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I went ahead and solved this, and then I just double-checked that I ended up with a value that is allowed.
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Logarithmic inequalities are a similar idea: these are inequalities that involve logarithms.
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And if b is greater than 1, and x is a positive number, and I have log < font size="-6" > b < /font > (x) > y...
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(again, we are talking about just a very simple situation, where there is a log on only one side), then x is greater than b^y.
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Look at what we did here: we went from the log form to the exponential form.
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Recall that, with equations, it would look like this: we are doing this same thing, only we are doing it with inequalities.
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And this relationship still holds up: if the log < font size="-6" > b < /font > (x) > y, then x must be greater than b^y.
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It is a little more complicated with less than: so let's just start out talking about greater than, and illustrating it with an example.
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I am starting out with log₃(x + 4) > 2.
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I know that, if this is true, then I can convert it to this form to solve, because this relationship will hold up.
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What this is telling me is that, if I take my base equal to 3, and x is equal to (x + 4), and y is equal to 2, I can convert it to this form.
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So, x is right here: x + 4 is greater than my base b, 3, raised to the second power.
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Now I can go ahead and solve: x + 4 > 9, so x > 5.
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I always have to be careful with logs, and make sure that I don't end up taking the log of a negative number.
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So, log₃(x + 4): if I put 5 in here, or slightly more than 5 (x is greater than 5)...say 5.1, I am going to get, say, 9.1, which is positive.
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So, I am not worried about taking the log of a negative number, because in order to get a negative number,
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you would have to have something that would be less than 4, and that is not going to occur here.
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So, that is fine: it is more complicated with less than, and let's look at why.
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For less than, let's say I had log₄(x - 1) < 3.
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I can't just come up with some number, some value, "x is less than 2," because the problem is:
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then I could get into very, very small numbers--very negative numbers--
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where I could end up taking the log of a negative number, which we are not doing.
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I have to put a restriction on the other side; I have to make sure that x is greater than 0.
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So, when I convert to this form, instead of just x < b^y, I need to say, "but it is also greater than 0."
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So, the base here is 4; I have my 0 here; x in this case is x - 1; the base is 4; and y is 3.
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So, looking at this: the base is 4; x is x - 1; and y is equal to 3.
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Therefore, 0 is less than x - 1; 4 times 4 is 16, times 4 is 64.
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I am going to add a 1 to both sides; and this is going to give me that x is greater than 1, but less than 65.
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It has put a restriction, a lower limit, on this.
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OK, we just discussed solving equations with a logarithm on one side.
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In order to solve equations with logarithms on both sides, we use the following property.
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If the base is greater than 0, but not equal to 1, then if you have a logarithmic equation
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with a log on each side, and whose bases are the same, then x must equal y.
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So, the restriction here is that the bases need to be the same.
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And you might remember back to working with exponential equations: we said that, if you have an exponential equation,
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and the bases are the same, then the exponents must be equal.
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It is similar logic here, which is not surprising, since this is just a notation for working with exponents.
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For example, log₄(3x - 1) = log₄(x + 5).
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Since these bases are the same, then in order for this equation to be valid, 3x - 1 has to equal x + 5.
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This leaves me with a linear equation that I can easily solve.
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First, I am going to add a 1 to both sides to get 3x = x + 6; then I am going to subtract an x from each side to get 2x = 6.
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Then I will divide both sides by 2 to get x = 3.
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Remember, when working with logs, it is very important to check and make sure that you have a valid solution,
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because you can't take the log of something that is negative.
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So, I need to make sure that this expression is not going to end up negative, and this expression is not going to end up negative.
00:26:18.700 --> 00:26:24.000
So, I am going to plug in this value and see what happens.
00:26:24.000 --> 00:26:32.300
For this first one, I am going to get log₄(9 - 1), which is log₄(8); that is valid.
00:26:32.300 --> 00:26:37.000
Now, this should be the same in here, since these two are equivalent; but we will just double-check it anyway.
00:26:37.000 --> 00:26:44.500
log₄(x + 5) would be log₄(3 + 5), or log₄(8).
00:26:44.500 --> 00:26:52.600
And since that is taking a log of a positive number, that is allowed, and this is a valid solution.
00:26:52.600 --> 00:26:59.000
When working with inequalities with logarithms on both sides, we can solve these inequalities with the following formula.
00:26:59.000 --> 00:27:09.700
If b is greater than 1, then if you have a log with a certain base, log < font size="-6" > b < /font > (x) > log < font size="-6" > b < /font > (y),
00:27:09.700 --> 00:27:14.400
then this relationship is true if and only if x is greater than y.
00:27:14.400 --> 00:27:22.200
The relationship between these two is maintained as long as x is greater than y; so that is a given.
00:27:22.200 --> 00:27:28.500
And log < font size="-6" > b < /font > (x) is less than log < font size="-6" > b < /font > (y) if and only if x is less than y.
00:27:28.500 --> 00:27:35.200
This is similar to the logic that we use when solving logarithmic equations with one log on each side.
00:27:35.200 --> 00:27:37.700
We are doing the same idea, but with inequalities.
00:27:37.700 --> 00:27:46.000
You need to make sure that you exclude solutions that would require taking the log of a number less than or equal to 0 in the original inequality.
00:27:46.000 --> 00:27:53.800
So, we are going to look for excluded values at the end, and make sure that we remove those from the solution set.
00:27:53.800 --> 00:28:01.000
Solutions to the inequality need to actually make the inequality valid.
00:28:01.000 --> 00:28:05.300
And they need to be not part of the excluded values.
00:28:05.300 --> 00:28:18.400
I will illustrate that right now: log₅(3x + 2) ≥ log₅(x - 4).
00:28:18.400 --> 00:28:26.300
Since the bases are the same, I know that this needs to be greater than or equal to this expression.
00:28:26.300 --> 00:28:35.800
So, 3x + 2 ≥ x - 4.
00:28:35.800 --> 00:28:54.500
Therefore, just solving this inequality is going to give me 3x ≥ x - 6; 3x - x ≥ -6; 2x ≥ -6.
00:28:54.500 --> 00:28:58.100
Divide both sides by 2; I have x ≥ -3.
00:28:58.100 --> 00:29:04.900
And it might be tempting to stop there; but I need to go back and look at the original.
00:29:04.900 --> 00:29:12.400
I need to make sure that I am taking only values (in my solution set) that are not excluded.
00:29:12.400 --> 00:29:15.600
When we were working with equations, it was simple: we would get something like x = 2.
00:29:15.600 --> 00:29:20.300
And then, we just had to plug it in here and make sure that that was valid; we were fine.
00:29:20.300 --> 00:29:26.900
Here, we have a whole solution set; so I have to use inequalities to find excluded values,
00:29:26.900 --> 00:29:36.000
or to find what x must be to end up with a valid solution set.
00:29:36.000 --> 00:29:55.300
So, I am going to look at log₅(3x + 2): for this to be valid, I need for this in here to be greater than 0.
00:29:55.300 --> 00:30:08.200
All right, that would give me 3x > -2, or x > -2/3.
00:30:08.200 --> 00:30:18.900
So, this log right here will be valid, as long as x is greater than -2/3.
00:30:18.900 --> 00:30:24.800
If it is smaller than that, I will end up having an excluded value.
00:30:24.800 --> 00:30:38.000
Let's look at this log: log₅(x - 4): in order for this to be valid, I need to have x - 4 be greater than 0.
00:30:38.000 --> 00:30:40.600
That means that x would have to be greater than 4.
00:30:40.600 --> 00:30:49.500
If x is some value less than 4, I will end up taking the log of a negative number, or 0; that is not allowed, so that would not be a valid solution.
00:30:49.500 --> 00:30:54.700
Now, look at my solution set: my solution set says that x has to be greater than or equal to -3.
00:30:54.700 --> 00:31:05.700
But that would encompass values that are too small--excluded values--values that are not allowed, like...
00:31:05.700 --> 00:31:12.000
I could end up with 0, which would then make this -4, and then I would be taking the log of a negative number.
00:31:12.000 --> 00:31:20.500
So, I have to have a solution set that meets this criteria, this, and this--the most restrictive set.
00:31:20.500 --> 00:31:26.400
x needs to be greater than or equal to -3, greater than -2/3, and greater than 4.
00:31:26.400 --> 00:31:32.200
So, I need to go with this: x > 4--that is the solution set.
00:31:32.200 --> 00:31:38.300
So, when there is an inequality with a logarithm on both sides with the same base,
00:31:38.300 --> 00:31:46.800
you put this expression on the left side, keep the inequality the same, put this expression on the right, and solve.
00:31:46.800 --> 00:31:53.800
Then, find excluded values, and make sure that your solution set does not include excluded values.
00:31:53.800 --> 00:31:59.300
All right, in the first example, we have a logarithmic equation that only has a log on one side.
00:31:59.300 --> 00:32:03.000
And recall that we can use the definition of logarithms to solve this.
00:32:03.000 --> 00:32:10.500
log < font size="-6" > b < /font > (x) = y if the base, raised to the y power, equals x.
00:32:10.500 --> 00:32:24.800
Therefore, I can rewrite this in this form...so, just rewriting it as it is, here the base is equal to 2;
00:32:24.800 --> 00:32:29.800
x is equal to x³ + 3; and y is equal to 7.
00:32:29.800 --> 00:32:35.600
So, writing it in this form would give me 2⁷ = x³ + 3.
00:32:35.600 --> 00:32:51.600
Let's figure out what 2 to the seventh power is: 2³ is 8, times 2 is 16, times 2 is 32, times 2 is 64, and then times 2 is going to give me 128.
00:32:51.600 --> 00:32:59.000
So, this is 2 to the third, fourth, fifth, sixth, seventh: so 2⁷ is going to be 128.
00:32:59.000 --> 00:33:06.400
All right, I am rewriting this as 128 = x³ + 3.
00:33:06.400 --> 00:33:13.900
I am going to subtract 3 from both sides: x³ = 125.
00:33:13.900 --> 00:33:24.600
The cube root of x³ is x, and the cube root of 125 is 5, because 5 times 5 is 25, times 5 is 125.
00:33:24.600 --> 00:33:27.800
Now, I also have to make sure I don't end up with a negative value inside the log.
00:33:27.800 --> 00:33:30.800
So, I am going to check this solution to make sure it is valid.
00:33:30.800 --> 00:33:39.200
x base 2...I am going back to the original...x cubed plus 3...I need to make sure that this is not negative when I use the solution.
00:33:39.200 --> 00:33:54.300
log₂(5³ + 3)...well, I know that this is going to be positive; so that is fine--this is a valid solution.
00:33:54.300 --> 00:34:05.200
All right, log₂(2x - 2) < 4--now I am working with an inequality that has a log on only one side.
00:34:05.200 --> 00:34:17.000
And I am going to go back and think about my definition of logs--that log < font size="-6" > b < /font > (x) = y if b^y = x.
00:34:17.000 --> 00:34:26.900
And by rewriting this using the exponential form, I can much more easily solve it.
00:34:26.900 --> 00:34:50.300
Now, I am working with less than; so I need to recall that, if log < font size="-6" > b < /font > (x) < y, then x is greater than 0, but less than b^y.
00:34:50.300 --> 00:35:00.000
If I didn't put this restriction here, I could end up with a value that is too small, and that would be excluded.
00:35:00.000 --> 00:35:07.500
All right, so rewriting this in this form: I am going to have the 0 here, and x is greater than that.
00:35:07.500 --> 00:35:21.700
x is 2x - 2...is less than the base, which is 2, raised to 4.
00:35:21.700 --> 00:35:27.400
I am rewriting this in an exponential form, but making sure that I have this restriction, since we are working with less than.
00:35:27.400 --> 00:35:35.900
So, 0 is less than 2x - 2; 2 times 2 is 4, times 2 is 8, times 2 is 16.
00:35:35.900 --> 00:35:43.200
Now, I am going to add 2 to both sides to get 2x is greater than 2, but less than 18;
00:35:43.200 --> 00:35:50.500
and then divide both sides by 2 to give me x is greater than 1 and less than 9.
00:35:50.500 --> 00:35:59.900
So, I solved this by using this property of logarithmic inequalities, and making sure that I had the 0, since I am working with less than--
00:35:59.900 --> 00:36:04.700
that I had the restriction that this expression is greater than 0,
00:36:04.700 --> 00:36:13.000
so that this in here (what goes inside for the log) does not end up being negative.
00:36:13.000 --> 00:36:16.900
I don't need to worry about that with greater than.
00:36:16.900 --> 00:36:23.600
This time, we are going to be solving a logarithmic equation in which there is one log on each side of the equation.
00:36:23.600 --> 00:36:33.000
Recall that we talked about the property: if there are two logs with the same base, log < font size="-6" > b < /font > (x) = log < font size="-6" > b < /font > (y)--
00:36:33.000 --> 00:36:39.500
the same bases--then for this equation to be valid, x has to equal y.
00:36:39.500 --> 00:36:46.500
So, I have log₆(x² - 6) = log₆(x).
00:36:46.500 --> 00:36:54.400
Since these are both base 6, I can just say, "OK, x² - 6 = x."
00:36:54.400 --> 00:37:05.100
This is just a quadratic equation: I move the x to the left, and I am going to solve by factoring, just as I would another quadratic equation.
00:37:05.100 --> 00:37:12.300
This is x + a factor, and then x minus a factor of 6, equals 0.
00:37:12.300 --> 00:37:18.200
Factors of 6 are 1 and 6, and 2 and 3; and I need them to add up to -1.
00:37:18.200 --> 00:37:26.900
So, -3 + 2 = -1; I put the 2 here and the 3 here.
00:37:26.900 --> 00:37:35.400
Using the zero product property, I can solve this, because x + 2 = 0 and x - 3 could equal 0.
00:37:35.400 --> 00:37:38.800
And either way, this is going to become 0.
00:37:38.800 --> 00:37:43.300
x = -2 for this; and another solution is x = 3.
00:37:43.300 --> 00:37:53.100
Now, I have two possible solutions; I need to check these.
00:37:53.100 --> 00:37:57.000
Let's go up here and look at this, with the -2.
00:37:57.000 --> 00:38:12.400
I have log₆(x² - 6); if x = -2, then I am going to end up with log₆(4 - 6), or log₆(-2).
00:38:12.400 --> 00:38:19.300
That is not valid; I could have also just looked up here and said, "OK, if x equals -2, I would be taking log₆(-2)."
00:38:19.300 --> 00:38:28.300
So, that is not valid; the solution is not valid.
00:38:28.300 --> 00:38:34.300
Let's try x = 3: well, if x equals 3, and I take the log base 6 of 3, that is OK.
00:38:34.300 --> 00:38:49.600
Let's check this one out: log₆(x² - 6): log base 6...and we are letting x equal 3 here.
00:38:49.600 --> 00:38:59.000
of 3 squared minus 6; so that is log₆(9 - 6), or log₆(3), which is positive.
00:38:59.000 --> 00:39:04.200
That is allowable; log₆(3) is allowable; so these are both allowable.
00:39:04.200 --> 00:39:10.700
Therefore, the solution is simply x = 3.
00:39:10.700 --> 00:39:20.100
We came up with two solutions; one was extraneous; we checked and found that we have one valid solution, which is x = 3.
00:39:20.100 --> 00:39:26.800
Example 4: Solve this inequality that involves a logarithm on each side of the inequality.
00:39:26.800 --> 00:39:41.000
And I am going to recall that, if the bases are the same (which they are), then log < font size="-6" > b < /font > (x) > log < font size="-6" > b < /font > (y) only if x > y.
00:39:41.000 --> 00:39:44.700
x must be greater than y; that relationship has to hold up.
00:39:44.700 --> 00:39:53.700
I am just going to go ahead and look at what I have in here, which is x² - 9 > x + 3.
00:39:53.700 --> 00:40:00.700
And I am going to move the 3 to the right by subtracting a 3 from both sides.
00:40:00.700 --> 00:40:10.600
So, x² - 9 - 3 > x; so x² - 12 > x.
00:40:10.600 --> 00:40:21.900
I am going to subtract an x from both sides; it is going to give me x² - x - 12 > 0 (the 0 will be left behind).
00:40:21.900 --> 00:40:28.500
So, this gets into material that we learned earlier on in the course; and it is a little bit conceptually complex.
00:40:28.500 --> 00:40:32.700
But if you just think it out, you can solve this.
00:40:32.700 --> 00:40:39.300
Let's factor this out; that is always a good first step.
00:40:39.300 --> 00:40:56.500
This gives me (x - 4) (x + 3), because the outer terms are -3x - 4x; the inner terms will give me -x; and -4 times 3 is -12.
00:40:56.500 --> 00:41:05.500
If I wanted to graph this out, I could say, "OK, (x - 4)(x + 3)...let's turn it into the corresponding equation and find the roots."
00:41:05.500 --> 00:41:17.600
x - 4 = 0, so x = 4; I am finding the roots of this corresponding quadratic equation, just as we did earlier in the course when we were solving quadratic inequalities.
00:41:17.600 --> 00:41:27.200
Also, x - 3 = 0 would satisfy this equation if x = 3.
00:41:27.200 --> 00:41:33.400
Actually, that is x + 3 = 0, so x = -3.
00:41:33.400 --> 00:41:45.500
What this is telling me is that the corresponding quadratic equation has roots at -3 and 4; the zeroes are there.
00:41:45.500 --> 00:41:54.900
Something else I know is that the leading coefficient here is positive; so this is going to be a parabola that faces upward.
00:41:54.900 --> 00:41:58.500
And this is really all I need to solve this.
00:41:58.500 --> 00:42:03.300
And then, I need to go back to my inequality and say, "OK, if this graph looks like this,
00:42:03.300 --> 00:42:13.000
and what I want is this function, the y-values, to be greater than 0, where are those going to be?"
00:42:13.000 --> 00:42:20.000
The graph crosses the x-axis right here, at -3: when x is -3, y is 0.
00:42:20.000 --> 00:42:25.200
For all values of x that are more negative than -3, y is positive.
00:42:25.200 --> 00:42:40.400
So, for this portion of the graph, when x is less than -3, y is greater than 0, which is what I want.
00:42:40.400 --> 00:42:54.100
I look over here, and the graph crosses the x-axis at x = 4; so for all values of x that are greater than 4, y is greater than 0.
00:42:54.100 --> 00:43:07.000
Therefore, in order to satisfy this inequality, x can be less than -3, or x can be greater than 4.
00:43:07.000 --> 00:43:11.400
Now, I can't stop there and say that I have solve this inequality, because the problem is
00:43:11.400 --> 00:43:14.700
that I need to make sure that I am not dealing with excluded values.
00:43:14.700 --> 00:43:17.100
So, let's look at what the excluded values are going to be.
00:43:17.100 --> 00:43:28.800
If I have log₇(x² - 9), I need for this x² - 9 to be greater than 0.
00:43:28.800 --> 00:43:39.900
So, if I factor that, I am going to get (x + 3) (x - 3) > 0.
00:43:39.900 --> 00:43:53.100
In order for (let's rewrite this a little bit better)...(x + 3) (x - 3) needs to be greater than 0.
00:43:53.100 --> 00:43:59.200
Therefore, x + 3 needs to be greater than 0, so x needs to be greater than -3.
00:43:59.200 --> 00:44:07.300
And x - 3 needs to be greater than 0; so x needs to be greater than 3.
00:44:07.300 --> 00:44:18.600
This is a restriction, and this is a restriction, that if I don't meet these restrictions in my solution set,
00:44:18.600 --> 00:44:22.100
I am going to end up with excluded values, and it is going to be invalid.
00:44:22.100 --> 00:44:30.200
So, I am going to look at my solution set: x > 4 meets these criteria, that x > -3 and x > 3.
00:44:30.200 --> 00:44:35.600
And I don't need to worry about this one, because I already have that factor covered right here: x + 3 > 0.
00:44:35.600 --> 00:44:39.400
This meets the criteria: it doesn't include any excluded values.
00:44:39.400 --> 00:44:49.100
However, when I look at x < -3, it doesn't meet all of the criteria, because here, this says x has to be greater than -3.
00:44:49.100 --> 00:44:57.700
If I take a value, -4, it is not going to meet this criteria; it is not going to meet this criteria, either (that x needs to be greater than 3).
00:44:57.700 --> 00:45:04.900
Therefore, this is not valid; and my solution is just going to be x > 4,
00:45:04.900 --> 00:45:13.400
because it solves the inequality--it satisfies the inequality--and it doesn't include excluded values.
00:45:13.400 --> 00:45:21.300
That one was pretty difficult: first you had to realize that, with the same bases, then you could just take this expression
00:45:21.300 --> 00:45:25.700
and say that it is greater than this expression and go about solving the inequality.
00:45:25.700 --> 00:45:29.800
That was a little bit tricky, because then you had to think about what this meant,
00:45:29.800 --> 00:45:35.800
solve this quadratic inequality, and then realize that x < 3, or values of x that are greater than 4,
00:45:35.800 --> 00:45:42.400
would satisfy the inequality, but there were some excluded values as part of this solution set.
00:45:42.400 --> 00:45:46.100
So, we had to just go with x > 4.
00:45:46.100 --> 00:45:54.000
That concludes this lesson of Educator.com on logarithmic equations and inequalities; and thanks for visiting!