WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about solving exponential and logarithmic equations.
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At this point, we have a good understanding of how both exponentiation and logarithms work.
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However, we haven't seen much about how to solve equations involving them.
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For example, how do we solve something like e^2x - 3 = e^5x - 12? Or log(x - 2) = 1?
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We haven't really talked about how to do that.
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We briefly touched on it in the final example in the last lesson, but now we are going to really explore it.
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We will go over two different ways of approaching such equations.
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First, we will discuss how we can use the 1:1 property, and then we will see how we can apply the inverse property for more complicated equations.
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Make sure you already understand how exponents and logarithms work.
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The previous few lessons explain this stuff in detail--explain how all of these things work and their properties.
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So, make sure you get an understanding of that before you try to get into these equations.
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While you might be able to understand it, it will make a lot more sense if you already have a grasp
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of how exponents and logarithms work and what their properties are.
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Then, you will really be able to throw yourself into the equations.
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If you don't have a good grasp of that, first make sure that you have watched those lessons previously,
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because they will really help out for understanding this lesson.
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All right, notice that the exponential and logarithmic equations are both one-to-one: different inputs imply different outputs.
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If we put a different value into the function, a different output always will come out.
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We can see this in the graphs, because both function types pass the horizontal line test.
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If we cut an exponential function's graph with a horizontal line, it is only going to intersect at a maximum of one point.
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This means that it is a one-to-one function.
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If you want a better understanding of one-to-one functions, you want to check out the Inverse Functions lesson.
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The same with logarithmic functions: if we cut at any place with a horizontal line, it will only intersect one time.
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Thus, they are one-to-one functions: for any input, there is only one output; for every output that can come out of it, there is a unique input that creates it.
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As an example, let's consider some specific numbers: if we wanted to see 3⁴ = 3 to the something,
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what could go into that box to give us 3⁴?
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Well, if we think about it for a while, we will probably think, "Well, there is only one choice to go into that box."
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It doesn't make sense for there to be anything else; the only other thing that could be on the other side of the equation is also 3⁴.
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The only thing that is going to make 3⁴ with a base of 3 is 3⁴.
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So, the only thing that can go inside of this box is a 4.
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Similarly, over here with log₅(25) = log₅(something),
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the only thing that could possibly be in there...if we think about this for a while, we will see,
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"Well, the only thing that could make any sense to go inside of that box is a 25."
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It wouldn't work any other way; there is nothing else that we could take log base 5 of and get the same value as log₅(25).
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So, the only thing that could go in those boxes is the number on the other side of the equation.
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This is because they are one-to-one--this one-to-one property.
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Since we have an input of 4 over here, and it gives us an output of 3⁴, 81,
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we know that the only other input that is going to be able to do that is no other input.
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This 4 is the only input that can give us 81 as our output; so the same thing must be here--it is one-to-one.
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If an input gives an output, the only input that can give that output is that original input.
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There is no other input; it is unique to the output.
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We call this the one-to-one property; it says that, if the base of a is raised to x, and we have that equal to the same base, a,
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raised to y, then it must be the case that x equals y.
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Similarly, if we have log<font size="-6">b</font> acting on x, and we know that it is equal to log<font size="-6">b</font> acting on y, then we know that x = y.
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And notice that, in both of these cases, the base has to be the same on each one.
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We have a^x and a^y; we have log<font size="-6">b</font> and log<font size="-6">b</font>--
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log<font size="-6">b</font>(x) and log<font size="-6">b</font>(y)--we have the same bases in both of these cases.
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And that is why we have this one-to-one property working: we know that x has to equal y in both of these situations.
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All right, with this property in mind, we can now solve equations where we have an exponent or a logarithm of a single base on both sides of the equation.
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For example, if we have e^2x - 3 and e^5x - 12, well, since we have the same base on both sides,
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we know that 2x - 3 has to be the same thing as 5x - 12.
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Otherwise, we wouldn't have equality.
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So, we know that 2x - 3 = 5x - 12; and at this point, we go about it just like we are solving a normal algebra equation.
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Add 12 to both sides; we get 2x + 9 = 5x; subtract 2x from both sides; we get 9 = 3x; divide by 3 on both sides; we get 3 = x.
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If we want to check it, we can plug in our 3 = x: e^2(3) - 3 = e^5(3) - 12.
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So, we have e^6 - 3 = e^15 - 12, so e³ = e³; and that checks out just fine.
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The same basic idea over here: log₇(x + 5)...and let's move the other log₇(2x - 3).
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We will add log₇(2x - 3) to both sides; so it will now appear on the right side.
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At this point, we know that what is inside of both logs has to be the same thing.
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They are both log₇, so we know that they must be taking logarithms of the same object.
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x + 5 and 2x - 3 must be the same thing; otherwise, we couldn't have equality.
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We have x + 5 = 2x - 3 by this one-to-one property.
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We can add 3 on both sides; we will get 8; so x + 8 = 2x.
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We can subtract by x on both sides; we will get 8 = x; and there is our answer.
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And if we wanted to, we could check that one the same way.
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The inverse property: previously, we have talked about exponentiation and logarithm, if they have the same base, are inverse processes.
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If they are applied on after another, they cancel each other out.
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So, if we have natural log on e², well, since natural log is just the same thing as log base e,
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and it is operating on something that is base e, to the 2, then the log base e here and the e cancel out, and we are left with just 2.
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So, sure enough, 2 = 2; that is the idea of what is going on there.
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The same thing over here: 5 as our base, raised to the log₅(125)--they cancel out, and we get 125.
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That is how this stuff is coming out.
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If you want a more in-depth exploration of this, check out the previous lesson, Properties of Logarithms,
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where we will actually prove this stuff and see why it has to be the case.
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We call this the inverse property: so log<font size="-6">a</font>(a^x) is equal to x.
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And b = x; so we have this cancellation.
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If we end up having the same base like this in log base, and then exponent base, or in exponent base and then log base,
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we will cancel out, and we will just get the thing that is at the end.
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So, in this case, that is x here or x here that we end up getting out of it.
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Solving by inverses: with an equation or inequality, we can do algebra.
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Now, algebra is just applying the same thing to both sides; we are doing the same operation--whatever it is.
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When we first learned to do algebra, we just used simple arithmetic (things like addition, subtraction, multiplication, or division).
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what we just saw in the previous thing when we were solving after we used these special properties.
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But as we learned more in our work in algebra, we realized that we could do more than just apply simple operations on both sides.
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We could even do more complicated things, like squaring both sides, or taking the square root.
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We realized that, as long as we are doing the same action to the left side and the right side--
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we are doing the same thing to two things that are equal--
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we know that it is going to remain being equal, even after the action goes through.
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So, by that same idea, we can also make both sides of an equation exponents, or take the logarithm of both sides,
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because we are doing the same action to both sides--the equality is still based in it.
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If we have wavy stuff equals loopy stuff, then we know that a, as a base for wavy stuff, is going to be equal to a as a base for loopy stuff,
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because a to the something-on-the-left equals a to the something-on-the-right, and we were told that left side and right side are the same.
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So, it must be the same still, even when they are working as bases for something else,
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or when something is a base underneath them and they are now exponents.
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A similar idea: if we were told that wavy equals loopy, then we know that log<font size="-6">b</font>(wavy) must be the same thing as log<font size="-6">b</font>(loopy).
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We are applying the same action, whether it is turning them into exponents on some base,
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or we are taking the logarithm of some base of both sides.
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We are doing the same action to both sides, so we have this equality still holding.
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Combining this idea with the inverse property allows us to get rid of exponent bases or logarithms that are in the way of solving an equation.
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For example, if we have log(x - 2) = 1 (remember, if it is just log, then that is the common log, so it is log base 10 of x - 2 equals 1);
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what we can do is come along and raise both sides with a 10 underneath them.
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So, we are not raising to a power, like squaring them; we are actually causing this exponent base to erupt underneath them.
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We have 10 to the log<font size="-6">10</font>(x - 2); and that is going to be equal to...10 has erupted underneath the 1.
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All right, at this point, we have the inverse property; we are solving by inverses, so we have 10 and log<font size="-6">10</font>.
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So, these cancel out, and the x - 2 just drops down; and we get x - 2 =...10 to the 1 is just 10.
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So now, we just solve it normally: x = 12--there is our answer.
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A similar idea is going on over here: 3^2x = 7--well, let's get rid of that base of 3; that is getting in our way.
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So, we will take log₃(3^2x), and that is going to be equal to log₃ of what is on the right side, so log₃(7).
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log₃(3^2x)...those will end up canceling out, and the 2x will just drop down;
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so we will have plain 2x = log₃(7).
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Now, if we want to figure this out with a calculator...
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log₃(7) is still correct, but if we want figure it out with a calculator--if we actually want a decimal version--
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we will have to turn it through change of base; let's take natural log--I like natural log.
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ln(7)/ln(3): remember the change of base formula that we talked about in the previous lesson.
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x =...now we are dividing by 2 on both sides, so it will show up in the denominator: 2 times the natural log of 3,
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which will end up working out to approximately 0.8856.
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Now, notice: 0.8856 is approximate; it is not the exact answer.
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This is actually the exact answer; once we have calculated through with ln(7) and ln(3), we end up getting something
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that is very, very close, but it is no longer precise, because we are having to cut off some of the decimal places.
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And if we wanted to, we also could have used any other base.
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We could have used log<font size="-6">10</font>(7)/log<font size="-6">10</font>(3)--a common log there.
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We would have had log(7)/2log(3), which would end up coming out to be the same thing when we used a calculator on it.
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And this would also be just equally as correct an answer.
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A useful property: one particularly useful property of logarithms is this ability to bring down things.
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If we have log<font size="-6">a</font>(x^n), then that is the same thing as if the n had been in the front, if it was n times log<font size="-6">a</font>(x).
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So, we can bring down exponents with any logarithm.
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This means that we can use logarithm bases that we have on our calculators.
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That might be convenient for us sometimes.
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So remember: e is the same thing as natural log; 10 is the same thing as log without a number on it.
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So, if we want, we could just start by taking the natural log of 3^2x = the natural log of 7.
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At this point, we can bring down the 2x; it comes down in front by this property up here.
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So, we have 2xln(3) = ln(7), so 2x = ln(7)/ln(3), or x = ln(7)/2ln(3), which is the exact same thing that we just had on the previous slide.
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The thing to notice here is that we have two different ways of doing this.
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We could go about it by using the change of base, or we could go about it by just using this property where we can bring down exponents.
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Sometimes it will be more useful to use the bringing down the exponent property.
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Sometimes it will be more useful to do the change of base.
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They will both end up working out; it is just that sometimes one will be a little bit more work than the other.
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And you will get a feel for which one you want to use as you work on these things.
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Many of the properties we have discussed about exponents and logarithms can be useful in solving exponential or logarithmic equations.
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If the problem is complicated, try to figure out if you can first simplify it with some of the various properties we have learned.
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We have learned a lot of properties by this point, about how exponentiation works and about how logarithms work.
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And sometimes, by combining things or breaking things apart, you will make the problem easier to do.
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And we will see some of that in the examples later on.
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Extraneous solutions: while solving these equations, it is important to watch out for extraneous solutions.
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An extraneous solution is a value that appears over the course of solving, but isn't actually a solution--
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that, if we were to try to use it, would just fail or cause our equation to break apart or not work or not be defined for some reason.
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The easiest way to see how this works is to just see an example.
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So, let's look at this: we have the natural log of x² - 2 equals natural log of x.
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By the one-to-one property, we see that x² - 2 has to be equal to x.
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Alternatively, if we wanted, we could put e's underneath it, and just cancel out both of them.
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Inverse or one-to-one property--both end up working as ways to look at this.
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x² - 2 = x: at this point, it looks like the polynomials we are used to solving from that section.
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We move it over: x² - x - 2 = 0; so we can factor that.
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We get (x - 2)(x + 1) = 0; we solve both of those, so x - 2 = 0; x + 1 = 0; we get x = 2 and x = -1.
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Now, if we go back and try to work this out, if we plug in x = 2, things are pretty reasonable.
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We get ln(2² - 2) = ln(2); so this is ln(4 - 2) = ln(2); and then, ln(2) = ln(2).
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That is perfectly reasonable; but if we try x = -1, we will see some problems very quickly.
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ln((-1)² - 2) = ln(-1); and as soon as we see this right here, we get suspicious,
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because what is the problem here? You can never take the logarithm of a negative number.
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So, as soon as we see ln of a negative number inside, we know that this is not possible.
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We can't take the natural log of -1; we can't have logs of negative numbers at any point showing up.
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This is an extraneous solution--it is something that appeared over the course of solving, because we turned it into this quadratic form.
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And in the quadratic form, it was a solution; but up here, in the original form that we have, it fails to be a solution.
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It can't be a solution, because we end up having this logarithm of a negative number; so we knock it out--it is an extraneous solution.
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And our only answer is x = 2.
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So, it seems at first as though we have two answers, because we are solving a quadratic.
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But as we work our way through the quadratic, we realize that if we were to actually plug this in
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and try it out to see if it works as a solution, it would cause the whole thing to blow apart.
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So, we end up seeing that it can't actually be used as a solution; so it is called an extraneous solution--
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something that appears over the course of solving, but can't actually be used as a solution.
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So, we have only one out of this, even though at first it seemed as though there would be two.
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All right, we are ready for some examples now.
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The first one: Solve for x exactly if we have 3/7^x + 2 and (49/9)^x - 2.
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So, we look at this, and we think, "Well, we could take logarithms of both sides; we could bring down our exponents."
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But things are going to get pretty messy; we will have to actually figure out what the logs are of 49/9 and 3/7,
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and we will have to work out a bunch of numbers.
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It is going to get really, really ugly: we could work it out that way, but we would end up having approximations,
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because we would be taking the logs of these numbers, and they would come out to be decimals.
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So, that won't end up working in the end.
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But if we look at it, we might realize that 49/9...there is a connection to 3/7.
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Well, if we want, we could rewrite this as 9/49 to the -1.
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And then, we might realize that 9 is just 3²; 49 is just 7²; it is still all to the -1.
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We can pull out the 2's, and we have (3/7)^-2.
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That is what we started with on the left side; so we can use that one-to-one property.
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So, we take this fact here; we can swap them out; so the same thing is still on the left side, (3/7)^x + 2,
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is equal to...we swap out 49/9 for (3/7)^-2, so we have ((3/7)^-2)^x - 2.
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Well, that is going to be equal to (3/7)...we can bring that -2 out by the property of exponents.
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It will multiply everything that is already out there, so we have to have that and the quantity as well.
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So, we have (3/7)^x + 2; at this point, we can use the one-to-one property, because we have the same base here and here.
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We have (x + 2) = -2(x - 2); x + 2 = -2x + 4; add 2x to both sides; we get 3x; subtract by 2 on both sides, and we get 2,
00:18:36.600 --> 00:18:42.300
and we get 3...sorry, we now divide by 3; we don't divide by 2--that would be going the wrong way.
00:18:42.300 --> 00:18:48.000
Divide by 3 on both sides; we get x = 2/3; and there is our answer for x.
00:18:48.000 --> 00:18:51.600
And if we wanted to, we could plug that in and check--use our calculator and end up working it out.
00:18:51.600 --> 00:18:55.600
We would get decimal answers that would end up being the same thing; we would see that that ended up working.
00:18:55.600 --> 00:19:10.800
So, you can check this by using a calculator if you want to; you could do a check, and you would have (3/7)^2/3 + 2 = (49/9)^2/3 - 2.
00:19:10.800 --> 00:19:17.200
You would have to use a calculator to work this out, but if you did, you would get decimal answers that were very, very, very close--
00:19:17.200 --> 00:19:20.600
actually, they should be exactly the same, because we solved for x exactly.
00:19:20.600 --> 00:19:23.800
The only problem might be if your calculator has just a little bit of sloppiness in it.
00:19:23.800 --> 00:19:29.300
But they should get decimal answers that are well within 5 or 10 decimal digits of each other.
00:19:29.300 --> 00:19:32.900
And so, you will end up seeing that it checks out when you use your calculator.
00:19:32.900 --> 00:19:36.700
Or you could also just work through each of these, and then use the properties of exponents.
00:19:36.700 --> 00:19:41.000
And you could see that it is exactly the same thing--there are two ways to do it.
00:19:41.000 --> 00:19:46.700
Solve for a exactly if we have log(a³) - log(a²) = 2 - log(a).
00:19:46.700 --> 00:19:51.000
All right, let's use the properties of logarithms to bring some things together and simplify things a bit.
00:19:51.000 --> 00:19:58.100
Remember: we have subtraction here--subtraction of logarithms is the same thing as division inside of the logarithm.
00:19:58.100 --> 00:20:05.800
So, we have a³/a² = 2 - log(a) (not base a).
00:20:05.800 --> 00:20:10.100
Now, the only issue we would have is...what if a was equal to 0?
00:20:10.100 --> 00:20:14.500
Well, if a was equal to 0, we would already have problems, because we would be taking the log of 0.
00:20:14.500 --> 00:20:19.400
So, we don't have to worry about that; so we know that a is not equal to 0, so we are good there.
00:20:19.400 --> 00:20:22.800
We know that a is not equal to 0, so we can do this cancellation.
00:20:22.800 --> 00:20:24.100
We don't have to worry about that..
00:20:24.100 --> 00:20:31.300
log(a³/a²)...well, we will get just log(a), because a³/a² will cancel two of the a's on top.
00:20:31.300 --> 00:20:32.700
We will be left with just one.
00:20:32.700 --> 00:20:38.500
2 - log(a): at this point, we can add log(a) to both sides, so we will get log(a).
00:20:38.500 --> 00:20:46.600
And now, there are 2 of them, because we added, and they are of the same type; so log(a) + log(a) is 2log(a) = 2.
00:20:46.600 --> 00:20:51.400
At this point, we can divide by 2 on both sides, and we get log(a) = 1.
00:20:51.400 --> 00:20:55.900
We want to know what that is: well, remember, log is just common logarithm, so it is base 10.
00:20:55.900 --> 00:21:05.800
So, we can raise both sides to the 10; so that cancels this out, and we have a = 10¹, which means a = 10.
00:21:05.800 --> 00:21:18.200
Great; if we wanted to, we can check this; so as a check, we have log(10³) - log(10² = 2 - log(10).
00:21:18.200 --> 00:21:25.400
log(10³) comes out to be 3, because it is base 10, so what do you have to raise 10 to, to get 10³?
00:21:25.400 --> 00:21:32.200
Well, you have to raise it to 3; the same idea is over here--log(10²) is just 2, equals 2 - log(10).
00:21:32.200 --> 00:21:37.500
log base 10 of 10 is just going to be 1; 1 = 1; great--that checks out.
00:21:37.500 --> 00:21:44.000
All right, the third example: Solve for x to four decimal places: 5^ x + 4 = 11^2x.
00:21:44.000 --> 00:21:58.200
So, we could write this log₅ acting on 5^x + 4, and then log₅ on the other side, as well, acting on 11^2x.
00:21:58.200 --> 00:22:07.100
So, since we have log₅ and exponential base 5, they cancel out, and we have x + 4 on the left side.
00:22:07.100 --> 00:22:12.800
On the right side, we see that we have 2^x raised to an exponent; so if we want, we can bring that out to the front.
00:22:12.800 --> 00:22:20.800
We have equals 2x times, and then our remaining log₅(11).
00:22:20.800 --> 00:22:27.800
All right, so at this point, we can divide by 2x on both sides, because we want to try to get our x's together.
00:22:27.800 --> 00:22:49.100
Or actually, better yet, we can subtract x on this side, and we will get 4 = 2x(log₅(11)) - x.
00:22:49.100 --> 00:22:53.600
Now, at this point, we can see that there is an x here, and there is an x here; we can pull out the x's.
00:22:53.600 --> 00:23:05.000
And we will get 4; pull out the x's; x times 2 times log₅(11) - 1.
00:23:05.000 --> 00:23:19.100
We can divide this over, so we have x =...dividing over, we have 4 divided by what we are dividing over, 2log₅(11) - 1.
00:23:19.100 --> 00:23:26.300
We can use the change of base formula, x = 4 over (because we probably wouldn't be able to use a calculator,
00:23:26.300 --> 00:23:33.600
and lots of calculators can't do log base 5) 2 times log₅(11)...let's go with natural log,
00:23:33.600 --> 00:23:40.700
just because I like natural log; so ln(11)/ln(5) - 1.
00:23:40.700 --> 00:23:44.500
Now, that one looks kind of ugly, and it is; it is going to take some work through a calculator.
00:23:44.500 --> 00:23:55.400
But you work it through with a calculator, and you will get that that is approximately equal to 2.0204 once you round it down.
00:23:55.400 --> 00:24:00.400
So, there you are: another way to have done this would have been to take the natural log of both sides.
00:24:00.400 --> 00:24:09.000
We could have taken ln(5^x + 4) = ln(11^2x), and this would be true with any base.
00:24:09.000 --> 00:24:11.500
We could be doing this with the common log, as well, if we wanted to.
00:24:11.500 --> 00:24:16.700
So, we can bring down our exponents; we will get (x + 4), remember, as a quantity, because it is the whole exponent,
00:24:16.700 --> 00:24:21.700
times the natural log of 5, equals 2x times the natural log of 11.
00:24:21.700 --> 00:24:36.000
At this point, we could move the natural log of 5 over, and we would have (x + 4) = 2x[ln(11)/ln(5)].
00:24:36.000 --> 00:24:39.000
It is over the whole thing, but we can also just compact it into that one thing.
00:24:39.000 --> 00:24:50.500
And then, if we want, we could move the x over as well, and we would have 4 = 2x[ln(11)/ln(5)] - x.
00:24:50.500 --> 00:25:00.900
Now, we do the same trick and pull out our x's: we get 4 = x times 2 times ln(11)/ln(5) - 1.
00:25:00.900 --> 00:25:15.200
We divide that over, and we get 4 over 2 times ln(11)/ln(5) - 1 = x, which is the exact same thing that we had when we did it by using log base 5.
00:25:15.200 --> 00:25:23.000
So really, it is just a question of if we are prolonging the change of base or causing the change of base to happen as we take a log in a different base.
00:25:23.000 --> 00:25:25.000
So, there are two different ways to do it.
00:25:25.000 --> 00:25:29.800
If you don't end up realizing this x trick, where the fact that we have x here and we have x here
00:25:29.800 --> 00:25:35.300
means that we can pull them both out to the front and get x at the front, you can also just work this out
00:25:35.300 --> 00:25:40.100
through a bit more arithmetic and working at it, moving things around.
00:25:40.100 --> 00:25:46.200
You can eventually get it; you could also just do it by evaluating what is log₅(11).
00:25:46.200 --> 00:25:52.400
You could figure out what log₅(11) is, get that down to 8 decimal places, and then multiply that be 2x.
00:25:52.400 --> 00:25:56.800
And you could work it out--just work with a whole bunch of decimal places for a while and solve for what x is,
00:25:56.800 --> 00:26:00.600
and then just cut it down to four decimal places--that is another way to do it.
00:26:00.600 --> 00:26:03.100
There are a bunch of different ways that you can approach problems like this.
00:26:03.100 --> 00:26:07.900
Just remember all of the properties that we have talked about, and just work through it and pay attention to what you are doing.
00:26:07.900 --> 00:26:11.200
And then, at the very end, check your work--do a quick check.
00:26:11.200 --> 00:26:23.400
You can do a check that 5^2.0204 + 4 = 11^2(2.0204).
00:26:23.400 --> 00:26:27.000
And in fact, you will end up finding out that they come out to be very, very close.
00:26:27.000 --> 00:26:31.900
They will end up being different after the fifth or sixth digit, because there was a little bit of rounding error.
00:26:31.900 --> 00:26:36.100
We did, after all, only round to four decimal places, and it just keeps going forever.
00:26:36.100 --> 00:26:43.000
But beyond that little bit of rounding error, after the fifth or sixth digit, you will end up being really, really close to being exactly the same.
00:26:43.000 --> 00:26:46.500
So, you will see that it checks out--that you do have the right answer.
00:26:46.500 --> 00:26:51.000
The fourth example: Solve for t exactly--we are back to solving for t exactly,
00:26:51.000 --> 00:26:55.300
so we can't really use these numerical methods that we have been working out with a calculator.
00:26:55.300 --> 00:26:56.900
We need to figure out something clever here.
00:26:56.900 --> 00:27:02.900
The problem here is that we have 2t and t; it is not just the same thing happening up there.
00:27:02.900 --> 00:27:05.600
And then, we also have this confusion from the 12.
00:27:05.600 --> 00:27:09.400
If the 12 weren't there, we could just move them over, and we could use the one-to-one property.
00:27:09.400 --> 00:27:18.900
But we have this problem where we have this 4^2t - 4^t = 12...well, let's try moving things around and see if this looks like something we are used to.
00:27:18.900 --> 00:27:25.200
4^t - 12 = 0; at this point, we might have a moment of understanding.
00:27:25.200 --> 00:27:33.900
We might realize that this looks a lot like a polynomial--like a quadratic polynomial that we are trying to solve.
00:27:33.900 --> 00:27:39.300
We realize that this is squared; this is to the one; and this is to the nothing--this is the constant.
00:27:39.300 --> 00:27:45.100
So, we realize that 4^t is kind of like the x that we are used to; so let's say 4^t = u.
00:27:45.100 --> 00:27:50.800
We will do this as a u substitution, where we will replace something complicated with something simple, this nice u.
00:27:50.800 --> 00:27:53.900
If it is 4^2t, it is now going to be u²,
00:27:53.900 --> 00:27:59.700
because u² would be (4^t)², which is the same thing as 4^2t.
00:27:59.700 --> 00:28:05.500
Minus u minus 12 equals 0--at this point, it is really easy to solve.
00:28:05.500 --> 00:28:14.100
We see (u - 4) (u + 3) = 0; great--that makes perfect sense.
00:28:14.100 --> 00:28:21.200
u times u is u², plus 3u minus 4u--that is our -u; -4 times +3 is -12; it checks out.
00:28:21.200 --> 00:28:26.500
So, we solve each one of these independently: u - 4 = 0, so we get u = 4.
00:28:26.500 --> 00:28:30.900
Now, what is it?--because we are actually solving for the t; we are not solving for u.
00:28:30.900 --> 00:28:38.000
So, we replace it with 4^t = u; so we have 4^t = 4, and there is only one thing that is going to end up giving us that.
00:28:38.000 --> 00:28:41.900
It must be 4¹ there, so we have t = 1.
00:28:41.900 --> 00:28:54.000
t = 1 is one of our answers; u + 3 = 0, so we have u = -3; we swap out for 4^t...equals -3...
00:28:54.000 --> 00:29:00.600
and at this point, we realize that that is madness; there is no possible way that 4 to the t can equal a negative number.
00:29:00.600 --> 00:29:05.900
There is no real number that we can raise 4 to that is going to give us a negative number.
00:29:05.900 --> 00:29:10.600
So, this way lies madness, so we knock it out; there are no possible answers there.
00:29:10.600 --> 00:29:17.700
So, our only answer is t = 1; it seemed like we were going to get this answer, but then we realized that that is not possible.
00:29:17.700 --> 00:29:34.600
If you want to check it out, we plug it in as a check: 4^2(1) - 4¹ = 12; 4² - 4 = 12; 16 - 4 = 12.
00:29:34.600 --> 00:29:37.500
Indeed, that is true; so it checks out.
00:29:37.500 --> 00:29:42.200
The final example: Solve for x exactly; then give an approximation.
00:29:42.200 --> 00:29:45.300
We have the x that we are looking for; it is in the denominator.
00:29:45.300 --> 00:29:52.100
We don't like things stuck in the denominator, so let's multiply it on both sides; we will multiply it by (ln(2x) - 3) on both sides.
00:29:52.100 --> 00:29:59.600
So, we get 4 = 2 times (ln(2x) - 3) on both sides.
00:29:59.600 --> 00:30:10.100
We can distribute the 2: 4 = 2 times (ln(2x) - 3); the thing we want to figure out is...it is like we are solving for ln(2x).
00:30:10.100 --> 00:30:17.000
And then later, we will crack the log; but for now, let's just figure out solving for ln(2x), and then we can solve the log.
00:30:17.000 --> 00:30:26.300
So, we see the -3 here; we add 3 to both sides, so we get 7 = 2 times...oops, I made a mistake!
00:30:26.300 --> 00:30:36.000
An important thing to notice: it is 2 times (ln(2x) - 3), so it is 2 times -3; it is not minus 3; it is minus 6.
00:30:36.000 --> 00:30:42.200
All right, so I have to always be careful with distribution, because it can catch anyone, including me and including your teachers.
00:30:42.200 --> 00:30:44.600
You always have to watch out for distribution.
00:30:44.600 --> 00:30:55.300
Add 6 to both sides; we get 10 = 2(ln(2x)); we divide by 2 on both sides; we get 5 = ln(2x).
00:30:55.300 --> 00:31:00.800
We now raise both sides to the e; so e⁵, e^ln(2x).
00:31:00.800 --> 00:31:08.900
Since these are the same base, natural log and e, they cancel out, and we have e⁵ = 2x.
00:31:08.900 --> 00:31:17.100
So, we have that e⁵/2 = x; that is our exact answer, e⁵/2.
00:31:17.100 --> 00:31:24.300
Now, e is this complicated irrational number; we can't really turn it into...it is not exactly a decimal number.
00:31:24.300 --> 00:31:29.400
But sometimes you want to have a decimal approximation, because that makes it easier for us to work with things.
00:31:29.400 --> 00:31:33.300
So, e⁵/2 is the exact answer; that is what it is precisely.
00:31:33.300 --> 00:31:45.800
But if we want an approximate answer, e⁵/2 ends up being approximately 74.207; great.
00:31:45.800 --> 00:31:51.600
Now, if we want to do a quick check, we can do a numerical approximation, where we try plugging it in using a calculator.
00:31:51.600 --> 00:32:03.100
So, we would have 4 over the natural log of 2, times...replace our x, which we know is approximately 74.207, minus 3.
00:32:03.100 --> 00:32:12.100
You work that through with a calculator, and you end up getting approximately 1.99999, and then it ends up changing after that.
00:32:12.100 --> 00:32:15.300
But that is really, really close; so we know that this checks out numerically.
00:32:15.300 --> 00:32:20.500
We have gotten that close to being exactly 2; so we see that numerically (because remember:
00:32:20.500 --> 00:32:27.200
there is some rounding error when we get an approximation), with that little bit of rounding error, we are still extremely close.
00:32:27.200 --> 00:32:28.800
So, we know that that is a good answer.
00:32:28.800 --> 00:32:34.500
If we wanted to, we could also work it out and show that it ends up being a precise answer, as well--
00:32:34.500 --> 00:32:38.500
that e⁵/2 is going to be equal to precisely what x has to be.
00:32:38.500 --> 00:32:50.100
We can check by plugging that in: 4 over the natural log of 2, times e⁵ divided by 2, minus 3.
00:32:50.100 --> 00:33:01.100
So, 2 times...dividing by 2...they cancel out; so we have 4 over the natural log of e⁵, minus 3.
00:33:01.100 --> 00:33:11.400
e⁵ is going to end up being...4/ln(e⁵) has to be 5, because what do we raise e to
00:33:11.400 --> 00:33:17.600
(since the natural log is base e) to get e⁵? Of course, we raise it to a 5 exponent.
00:33:17.600 --> 00:33:26.000
5 - 3...we could also think about the fact that natural log is log base e, and we have an exponent base e, so they cancel each other out.
00:33:26.000 --> 00:33:31.600
4 over 5 minus 3 becomes 2, which equals 2; that checks out.
00:33:31.600 --> 00:33:37.100
So, we can do it either numerically or precisely, and see that it worked in either case.
00:33:37.100 --> 00:33:41.000
All right, great; now I have a really good understanding of even probably the most complex kind
00:33:41.000 --> 00:33:45.500
of logarithmic and exponential equations that will be thrown at you at this point in math.
00:33:45.500 --> 00:33:50.400
So, with this sort of knowledge, you can go and do all kinds of problems, from the easy to the hard ones.
00:33:50.400 --> 00:33:52.300
And you will be able to solve them if you follow these things.
00:33:52.300 --> 00:33:55.800
Remember: be careful--mistakes happen; you even saw one happen to me.
00:33:55.800 --> 00:33:59.500
So, they happen to everybody; it is really useful to do these checks.
00:33:59.500 --> 00:34:02.600
By checking your work, you can make sure that you didn't make a mistake.
00:34:02.600 --> 00:34:08.000
And if you see that something went wrong in the check, you can go through and carefully analyze your work and figure out where things went wrong.
00:34:08.000 --> 00:34:10.000
All right, we will see you at Educator.com later--goodbye!