Congratulations. You've decided to start studying for your Algebra 2 final exam. But what are you supposed to study? Everything. Well, in this video, I'm going to go over 10 problems with some of the most important concepts that you need to know for your final exam. So, here's the first problem that we'll be doing in this video. We need to solve for x in this inequality. So, there are multiple ways that we can do this. One way I'm seeing that you might have thought of is we can just take this minus2 and divide it out. But this pattern here also means that we can use the distributive property. So the distributive property says that when you have a number like this in front of parenthesis, then you need to multiply the number out to all the parts of the parenthesis. In this case, we only have two, but you can have more parts than that in the parenthesis. So we're going to multiply the minus2 out to the 4x first and then the minus 3. So -2 * 4x. Remember that the minus signs always have to stay. So 2 * 4 is 8. And then we have one minus sign. So the minus sign stays. And then we have an x. So the x stays. And then here we have - 2 * -3. So 2 * 3 is 6. And then we have two minus signs. And two negatives make a positive. So that's going to become pos 6. And then on the other side, the less than -2 is just going to stay the same. So now our goal is to just get x by itself. And we can do that by first getting rid of this minus 8 or this plus 6. Uh I think it's better to get rid of the plus 6 first because the8 is more attached to the x. So we'll get rid of that last. So let's go ahead and subtract six from both sides. And the reason that we're doing that is because the opposite of addition is subtraction. So these would cancel out. On this side we just have -8x and on the other side we have less than -2 - 6. Think of this as dollars. You lost $2 and lost another six. So, you've lost $8. And now we just need to get x by itself. And now we just need to get rid of this minus 8 to get x by itself. But what's the opposite of this? There's no sign at all. Well, when there's no sign, that means it's multiplication. So, we're going to imagine there's a multiplication sign here. And the opposite of multiplication is division. So, we'll divide both sides by8. Thegative8s cancel out. And on this side, we just have x. And on the other side here, we have8 /g8. Well, dividing anything by itself makes one. And you'll notice that I've been avoiding writing the sign here. And there's a reason for that. Because we divided by a negative number. We divided by8. And you also have to do this if you multiply by a negative number. Whenever you do that, you have to flip the inequality sign. So it was less than. Now it's greater than. So, our final answer is x is greater than -1. And if you wanted to check your work, you can always plug in a number that's greater than one into the equation and see if it's true. And then maybe try plugging in a number less than one and make sure that the equation is not true. Now, before we get into this next problem, I just wanted to let you know that for most of the problems we do in this video, I have a full video about that type of problem that you can watch in the description. So, our second problem here is we need to solve for x and y. We have two equations. And remember, you can only solve for a certain amount of letters if you have that many equations or more. So if we had like three letters and only two equations, then we couldn't solve it. So there's two main methods to solve a system of equations. There's substitution and there's elimination. Elimination would be if we added these two equations together and it canceled out a variable. So that would be really convenient if this number was a two, for example, and then we added these together and the plus 2 y and minus 2 y would cancel out. Now, we could use elimination with this by multiplying this first equation by two and then solving it, but I think substitution will probably be easier in this case. For substitution, we need to take one of these equations and get a letter by itself. And y here is already pretty much by itself. So, I think we're going to use this equation. So, how can we get y by itself here? Well, the only thing in its way is this - 3x. So, to get rid of it, we can do the opposite of minus 3x, which is + 3x. So, we'll add plus 3x to both sides. These cancel out. And on the left side, we just have y. And on the other side, we have 2 + 3x. Well, the two doesn't have a letter and the 3x does have a letter. So, we can't actually add these. So, we're just going to leave it as 2 + 3x. So now that we know that y is equal to 2 + 3x, we can take this second equation, and it's important that you do the other equation, the one we that we didn't just use, and we replace y with 2 + 3x. So we're going to rewrite the equation, but instead of y, we're going to put in parenthesis 2 + 3x. And then everything else is going to stay the same. So now we can see that we don't have any more y's in this equation. we only have one letter which is x. I mean x shows up in two different places but they're both x. So all we need to do is simplify this down and solve for x. So immediately I'm noticing that here we have the same pattern that I was showing you last problem which is the distributive property. So we can multiply out this two to both parts. But remember that this isn't just a two. This is a minus2. So we also have to distribute the minus sign. So here we have -2 * 2. Well, 2 * 2 is 4. And then we have 1 minus sign. So the minus sign is already here. I already wrote it. Then here we have -2 * 3x. 2 * 3 is 6. And then the x stays. And again here we have one minus sign. So the minus sign stays as well. And then the minus 6 is going to stay the same over here. So what else can we simplify here? Well, we have two numbers with an x in them, which means that we can combine them. That's called combining like terms. So we have 4x - 6x. We had four and then lost six. So you're down two. So that's - 2x. And then everything else is going to stay the same. So the minus4 we didn't touch. So that stays the same. And then this - 6 is still the same. And now we can just get x by itself with a two-step problem. Again, we'll get the minus4 out first because it's less attached to the x. So we have minus4. The opposite of minus is plus. So we'll add four to both sides. The fours cancel out and on the left side we have min - 2x and on the right we have -6 + 4. So you lost $6. You gained four of them back but you're still down two. And now we have multiplication here. So we can do the opposite of multiplication which is division. We'll divide both sides by minus2. The minus2's cancel out. And I'll write this over here because we're running out of room. we have x = -2 / -2. Well, dividing anything by itself makes 1. So, x is equal to 1. So, we've successfully solved for x. But the problem is asking that we need to solve for x and y. So, how would we solve for y in this case? Well, all we need to do is plug in one for x in either of these equations and see what we get for y. I'm going to use this first equation because y is closer to being by itself. So, it's going to be easier to use this one. but you can use the bottom one as well. So, we're going to rewrite the first equation, but instead of x, we're going to write 1. So, we're going to have -3 * 1 + y = 2. -3 * 1. Anything * 1 is itself, so that's - 3. And then everything else is going to stay the same. And now we can just get y by itself by getting rid of this minus 3. The opposite of minus is plus. So, we'll add three to both sides. These cancel out and we just have y = 2 + 3 which is 5. So y is equal to 5. And our final answer is x = 1 and y = 5. Another way to write your answer for a system of equations is to write them as coordinates of a graph. So if you have parentheses and then you put the two numbers, x always goes first and then y. So you could just write it as 1, 5. And this is another way to write your answer. All right, here's our third problem. And this is going to be a graphing problem. So, we need to graph y equals well, what do these lines mean, right? Well, these lines mean absolute value. So, this is saying y equals the absolute value of x + 3. So, how do we graph this? Well, you could plug in numbers for x and see what you get for y and then, you know, you'll get a bunch of points and then you'll just connect them. But, we can actually do this a faster way. For this problem, we can use something called transformations. So we need to first find what the parent graph is and then find out what this plus three is doing to it and then just adjust it. So our parent graph is just what the equation would be but without all the extra things added on. So the three main types of parent graphs are y = x^2, there's y = the absolute value of x and then there's just y= x. So in this case, we're using y equals the absolute value of x. And this plus three is what's being added on afterward. So we need to first graph the parent graph. So the y equals the absolute value of x is in the shape of a v. And the v starts in the middle here and then just goes diagonally like this. So you'll want to make sure that it hits all these little corner points in the boxes here. And then it's going to go the same the other way. And if you weren't sure if it was a Vshape, well, you can just think about it logically. Uh, absolute value just means that you take whatever the number is and make it positive if it's not already positive. So y is the same as x here, right? If you go to the right one, you go up one. If you go to the right two, you go up two. So it's always the same x and y. Except on the left side where x is negative, it's just y but positive. So now we need to figure out what this plus three is doing. So basically the graph is going to go up, down, left or right. Uh there's also other types of transformations like for example if you put a two here then you're going to squish the graph together or if you put a minus sign here then it's going to flip the graph upside down. I have a full video of the types of transformations if you want to watch that. But here we need to figure out what this plus three is doing. Now, if the plus three was on the outside of the absolute value signs or parenthesis, if we were doing like x^ squ, for example, then it would move the graph up. Or if it was minus3, it would move the graph down. But when the number is inside of the parenthesis or the absolute value signs, then it moves it left or right. And when we're moving left or right, it's the opposite of what you probably think. Normally, positive means numbers to the right, and negative means numbers to the left. But when we're moving graphs left and right, it's actually the opposite. Plus three is going to move it to the left. So if the problem asks to describe the transformation, then you would just say it moves to the left three. But now we need to graph that. Well, we're just going to take this little point, this little anchor point, I guess, the point where the tip of the V is, and we're going to move it to the left three squares. And this will be the new tip of our V. And now we just need to regraft our V, but three to the left. So, we're going to go ahead and draw a V just like we did before. And now this parent graph, because we've already graphed our final graph, we don't need this anymore. So, we can just erase it. And this here is the final graph of y equals the absolute value of x + 3. All right, let's get into some imaginary numbers. Remember those. So, we just need to simplify this expression. And in this case, that means we're multiplying because we have two sets of parentheses and they're just right next to each other without a sign. So, that still means multiplication. And in this case, we would actually call these complex numbers. So this would be a complex number because we have both a real part. 3 is a real number and an imaginary part. 2 I is imaginary. And it's important to remember what I even means. So remember that I is just a short form for writing the square root of -1. Remember that square roots of negative numbers don't exist. They're impossible. So that's why they're imaginary. They don't exist in the real world. So with both imaginary numbers and with real numbers as well, whenever you have two sets of parentheses here with a plus or a minus sign in each of them, these are called binomials. And when we're multiplying two of them, we have to use foil. So to multiply these, we need to follow the steps of foil. So the first one is F, which stands for first. So we need to multiply the first parts of each of the sets of parenthesis. So 3 * 5 is 15. The second step of foil is O which stands for outer. So we multiply the outer parts of the problem. So 3 * not 4 I -4 I remember that the minus signs always stay. So 3 * 4 is 12. And then we have one negative sign so the negative sign stays. And we also have one I. So the I stays as well. Then we need to do I which stands for inner. So we multiply the inner parts here. So 2 I * 5. 2 * 5 is 10 and then the i stays. And then finally we need to do l which stands for last. So we multiply the last parts. So 2 * 4 is 8 and then we have 2 i's. So that makes i^ 2. And then we have one minus sign. So the minus sign stays. And now we just need to simplify this down. So just like if i was a variable or like x or something, we can combine these two like terms. We have two numbers with an i in them. So -12 I + 10 I. You lost 12. You gained 10 back, but you're still down two. So that's going to be - 2 I. And the rest of the equation is going to stay the same. So there's no more like terms that we can combine. So you might think that we're done. But remember that I is the square root of -1. And we don't need to change all the i's into square roots of1. But we do have to do that for i^2 because it might turn into something else. So we're going to rewrite this equation except instead of i^2 we're going to replace i with the square<unk> of -1. And when we square it, you'll notice that we have contradicting things here. We have a square root and then it's being squared. So those actually cancel out because those are opposites of each other. So the square root and the squar cancel. And when we rewrite this equation, then we're only going to have -8 * -1. That's all that's left here, minus one. So you'll see that when we squared an imaginary number, it turned into a real number again. So we can simplify this as well. So we'll have 15 - 2 I and then8 * -1. 8 * 1 is 8 and two negatives make a positive. And then we can actually simplify this as well because we have 15 + 8. Those are both real numbers. 15 + 8 is 23. And then the minus 2 I stays the same. And there is now nothing that we can do. So this is our final simplification. All right, here's the fifth problem in this video. We need to solve for x in this equation. But in this case, we can't just get x by itself like normal because we have an x squ. And that makes this a quadratic equation. Now, there are tons of ways to solve quadratic equations. There's uh you can graph the equation, you can factor it, there's completing the square, there's the quadratic formula, but some of those methods don't work all the time. Like factoring, you can't always factor. Graphing is just really dumb. So although we could probably factor this, there's one method that works all the time and that's the quadratic formula. So I think that'll just be easier to do here. So the quadratic formula is x =b plus or minus the square<unk> of b^2 - 4 a c. And that's all going to be over 2 a. And if you didn't already have that memorized, just go look up a quadratic formula music video and you'll never forget it. So all we need to do is to use the formula. We just need to plug in a, b, and c and see what we get for x. But it would be helpful to know what a, b, and c are in the first place, right? So a is going to always be the number that's in front of x squ. So in this case, a is going to be three. b is going to be the number that's next to x. So that's going to be minus2. The minuses stay as well. The minuses basically always stay in algebra. And then our c is going to be the number by itself. In this case, that's minus 8. And a, b, and c were in order this time. A was on the left and then b and then c. But that's only because the equation was already in order. We had it x^2 x and then the number by itself. So if the equation's out of order, you'll probably have to put it in order before you can just go a b c. So now let's just go ahead and use this formula. So all we have to do is plug in the numbers. It's just the hard part is being careful about all the signs and making sure that we're not making a mistake. So let's go ahead and start plugging in our numbers. we have minus b and b is -2. So this is going to be minus min - 2 plus or minus the square t of b^ 2. b is - 2. So that's going to be - 2^ 2 - 4 and then a. a is 3 and then c. c is8. And then this is all going to be over 2 * a. And a is 3. So it's going to be 2 * 3. And now we just need to simplify this down, which may take a while, but we'll just take it one step at a time. So, first we have -2. Two negatives make it positive. So, that's just going to become pos2 plus or minus the square<unk> of -2^ 2. Remember, when you're squaring, that just means multiplying by itself. So, -2 * -2 becomes pos4 because two negatives make a positive. And then here we have -4 * 3 *8. Well, you could put this in your calculator if you wanted. 4 * 3 * 8 would be 4 * 3 is 12 * 8 is 96. But in this case, we actually have two minus signs. We have a minus sign on the 4 and a minus sign on the 8. Two minuses make a positive. So it's going to be + 96. And then that's going to be over 2 * 3, which is 6. And now we can simplify this even more. So 2 plus or minus the<unk> of 4 + 96, which is 100 / 6. And now I'm going to go ahead and move this up. So we can simplify this even more. We have 2 plus or minus the<unk> of 100, which is 10 because 10 * 10 is 100. And now we don't have anything else to simplify except for the fact that we have a plus or minus here. And you might have been wondering what this even means. Well, this means that we are actually going to have two answers. There's an answer if this was a plus and there's an answer if this was a minus. So we need to split this up into two problems. We have 2 + 10 / 6 and we have 2 - 10 / 6. So let's simplify the first one. 2 + 10 is 12 / 6 and 12 / 6 is 2. So 2 is one of our answers. And now if we look at the other one, we have 2 - 10 / 6. 2 - 10 is -8 over 6. And this is just going to stay as a fraction. But we can actually simplify this fraction because we can divide both the top and the bottom by two. So8 / 2 is -4 and 6 / 2 is 3. And you can't simplify 4 over 3 anymore. So that's going to be our second answer. -4 over 3. So our final answer is that x is equal to -4 over3. and positive2. And if you wanted to check your answer, you could plug in both of these numbers into the equation and see if it gets you zero. So, let's go back to the graph and talk about domain and range. If you've done any sort of graphing problem in algebra, you've probably been told that you need to find the domain and range of the graph. But what what do those even mean? Well, here's the easiest way to think about it. In my opinion, domain is just all of the numbers that you can put in for x. So any numbers that you can put in possibly are going to be in the domain. And then the range is the same thing. It's all the numbers that you can possibly put in but for y. And remember that x is left and right on a graph and y is up and down on a graph. That's important to know. So first of all, let's look at the domain for x. What numbers can we put in for x? Well, it doesn't look like there's anything stopping us from putting in any number for x. And by the way, when I say put in a number, I mean like when when you have put in X as your input and then you get something out for Y. So like for example, we could put in one for X because one means going one to the right. And if we put in one, then we get 1 2 for Y. So when we put in one for X, we get two out for Y. So it doesn't look like there's anything stopping us. And you might think, well, wait, wait a minute. We can't put in, for example, four. Four is four squares to the right and we don't hit anything for y, right? But these lines actually go on infinitely. And for a parabola, I know it looks like it's going to stop at some point, but these actually are going to keep going infinitely to the right and this is going to go infinitely infinitely to the left. So we can put in whatever we want for x. So that means our domain is going to be all real numbers. Or another way to write this is to say that you can go from negative infinity to positive infinity. So the way that we do that is we put it in parenthesis and we say that we have negative infinity as the lowest we can go and then positive infinity as the highest we can go. So now what about our range? What can we put in for y? Well, for example, we could put in two for y because if we have two here, that's two upwards. Then we hit the graph, right? We get two numbers for x actually. But unlike for the domain, unlike for x, there are actually numbers that we can't put in. You'll notice that if we put in anything in this area of the graph, we're never going to hit the graph because the graph only exists from here and above. So, what is the lowest number that we can put in for y? Well, the lowest that this graph goes is right here. And to get here, we have to start in the middle and go up one square. So, we can't put in anything less than one. It has to be one or higher. So we're going to write this the same way we did for the domain. We're going to put one which is the lowest we can go and then comma infinity which is the highest we can go. We can go up any infinite number of squares. We're actually not going to only put parenthesis. We have to first ask ourselves is one exactly is exactly one part of the numbers we can put in. And in this case yes it is. If we put in exactly one then we will hit the graph. So whenever we can put in the exact number, then instead of a parenthesy, we're going to put in a bracket. But we're still going to put parentheses around the infinity because it's not a number. You can't actually put in infinity, right? So our range is 1 to infinity. And there are other ways to write domain and range, too. Like for example, you could just put that y is greater than or equal to 1. That's a pretty easy way to write the range. So it it just depends on what your teacher or professor wants you to do. So, make sure you check to see what format your teacher prefers. And with that, you are now over halfway done with the problems in this video. There's only four more, but this one is polomial division, which can seem pretty tricky, but just follow it step by step, and you'll understand. So, just like tons of the things that we're doing in this video, there are multiple methods to dividing polomials. The first one is long division, which works every time, but it's long. And then there's synthetic division, which is a lot faster, but only works if this is x plus a number or x minus a number. Now, technically, we could use synthetic division here if we divided everything in this problem by two. But if if this isn't already x plus something or x minus something, we can always just use long division. So, even though long division gets kind of confusing and there's a lot to keep track of, it's actually pretty much the same as the long division you did in elementary school with regular numbers. So the first part of our division problem, the big the big part is what we're going to put inside of our little box. So we're going to write 2x 3r + 5x^2 - 11 x + 6. And we're going to put this inside our little division box or whatever you would call this. And then the thing that we're dividing by 2x - 1, we're going to put on the outside. And now we just need to follow the same steps as elementary school long division which are divide, multiply, subtract, and then bring down. And in the wise words of my fourth grade teacher, you can remember this as does McDonald's sell burgers or she actually said cheeseburgers because there was the step of C, which was to check your answer. But I guess you don't have to do that. So the first step is to divide. We're going to take the first parts of both parts and we're going to divide them. So, we're going to divide 2x the 3r by 2x. So, we'll divide the numbers first, then the letters. So, 2 / 2 is 1. We don't have to write the one cuz that's just not going to be necessary. And then we have x the 3r / x. When you have x the 3r and you divide out one of the letters, you're basically just subtracting it. So we had three, we're dividing out one. So we're only going to have two left. So x^2. And now you can see why we didn't have to write the one because putting one in front of a letter doesn't do anything. So now that we've divided, we can go to the next step, which is multiply. So we need to multiply the entire thing that we're dividing by by the answer that we just got. So 2x * x^2. Well, the 2 * the invisible 1. 2 * 1 is 2. and then x * x^2 you're when you multiply two letters you're basically just adding them together. So you're adding another x to the x^2 that makes x the 3r and then we multiply the minus1 by the x^2 so anything * 1 is itself so that's just going to be x^2 but we do have a minus sign so the minus sign stays and you'll see that I've been lining these up with the numbers that are above it and that's because we're now going to do the third step which is to subtract. So we need to subtract these two parts here. But if we were to just add these together, that would be adding. So to make it subtraction, we just have to change all the signs of the bottom. So this becomes a minus and this becomes a plus. And now we just add these together. So 2x 3r - 2x 3r. Those are just canceling each other out. So those don't do anything. And then 5 x^2 + another x^2. This would be just like if you were combining like terms. 5 x^2 + 1 x^2 is going to make 6x^2. And now finally we can just do b which is bring down. So we bring down the next part which is - 11x just like if you were bringing down a digit in long division. And we're going to go ahead and repeat the process. So we need to do divide divide 6x^2 by 2x the first parts only. So 6 / 2 is 3. and x^2 / x, you had two x's. And then you remove one, you just have one. And then we're going to put a plus sign in between all of these. Then we have multiply. So we're going to multiply the 3x that we just got to this whole thing here. 2x - 1. So 3x * 2x. 3 * 2 is 6. And x * x is x^2. And then we have 3x * -1. Anything * 1 is itself. So that's 3x. And we have a minus sign. So the minus sign stays. And now we can subtract. So we change all the signs of the bottom and we add them together. So 6x^2 - 6x^2. Those cancel. Those don't do anything. And then - 11x + 3x. So we're down 11. We got three back, but we're still down eight. So - 8x. And now we can bring down this final part, which is the plus 6. And do it one more time. So divide - 8x / 2x. 8 / 2 is 4. and we have x / x. Those cancel out and we do have one minus sign here this time. So the minus sign is going to stay. Then we multiply. We multiply the -4. Remember the minus sign stays by 2x - 1. So 4 * 2 is 8 and the x stays. And we have one minus sign. So the minus so the minus sign stays as well. And then -4 * -1. 4 * 1 is 4. And we have two minus signs. Two minuses make a positive. So we have pos4. Then we subtract. So we change all the signs at the bottom. So this becomes a plus. And then this is going to become a minus. And we add them together. So - 8x + 8x. Those cancel. And then 6 - 4 is 2. So what we have at the top here is our final answer. If the bottom part here that we end with is zero. But this isn't zero. It's two. So that means we're going to have a remainder. And to do a remainder, all you have to do is attach the remainder to the end and you have to make it a fraction. You have to put it over what we divided by, which is 2x - 1. And that's all you have to do with the remainder. So this is our final answer here to this division problem. x^2 + 3x - 4 plus our remainder here 2 over 2x - 1. All right, let's get into some functions or f ofx. And we have an f ofx and a g ofx here. And we need to find fog of x. Well, this is a composite function and this reads f of g of x and we can actually rewrite this as f of with g of x inside. If this was g composite f or goth, then it would be g of x with f ofx inside. So remember that when you're solving a function, like for example, if you had f of 3, then you just replace x with 3 with whatever's inside the parenthesis. So this would become 3 * 3 - 2. So we're just going to be doing the same thing except with an entire function which is g of x. So we need to take f ofx which is 3x - 2 but replace x with g of x which is x^2 + 1. So we're going to rewrite this but with g of x as x. So we're going to have three and instead of x we're going to write g which is x^2 + 1. So we are going to put that in parenthesis and then minus 2. that stays the same. And now we just need to simplify this down. So we can use our favorite pattern in algebra, the distributive property. So this 3 needs to be multiplied to both parts of the parenthesis. So 3 * x^2. You can't multiply a number in a letter, so it's just going to stay as 3x^2. And then 3 * 1 is 3. And then this -2 is still hanging out here. So that's going to stay the same. And now we can simplify this even further. We have a 3 and a minus2. Those are both just numbers. So we can combine those. So 3 - 2 is a 1. And there's nothing else we can simplify. So there we go. We've already found f composite g of x. It's 3x^2 + 1. All right. Two more problems. For this one, we need to solve for x again, but this one we have a cube root in it. Now a cube root is basically the same thing as a square root, except that a square root does the opposite of squaring. But a cube root does the opposite of cubing, which is putting something to the power of three. And you can put any number you want here. You could put a four here, and then it would be the opposite of putting something to the power of four or whatever number you want here. So this cube root is kind of just this big clump of math that we can't do anything with yet. But we can get rid of this plus three first. So we can get rid of this plus three by doing the opposite of plus, which is minus. So we'll subtract three from both sides. And remember, you have to do it to both sides of the equation. So these cancel and on this left side we have our big clump of math. And on the other side we're going to have 5 minus 3 which is two. And now that the plus three is gone we can do stuff with this big clump of math. So we have a cube root here. The opposite of cube rooting is cubing to the power of three. So we're going to put everything to the power of three. And we have to do the same thing to the other side as well. So the cube root and the power of three cancel. And on the left side we have x^2 + 1. And on the other side we have 2 the 3r which is 2 * 2 * 2. So 2 * 2 is 4 * 2 is 8. So 2 3r is 8. And now we just have a two-step algebra problem where we can get x by itself. So I'm going to get rid of this plus one first because it's a less attached to the x. So the opposite of plus is minus. We'll subtract one from both sides. These cancel out and we just have x / 2 is equal to 8 - 1 which is 7. And now we just have one step. So remember that a fraction just means division. So the opposite of division is multiplication. So we'll multiply both sides by two. The twos cancel out. And on the left side we just have x and on the other side we have 7 * 2 which is 14. So our final answer is x is equal to 14. And as always, you can plug in 14 back into x and see if it makes the equation true. That's how you can check to make sure your answer's correct. All right, here's the final problem of this video, and it's a logarithm problem. So, I know that logarithms can seem a little scary, but there's actually two different ways that we can solve this. I'll show you the regular just pencil and paper way, and then how you can put any log into a calculator. So, what is a log? Well, a log is basically like the opposite of an exponent, sort of. The answer to a log is some sort of exponent here. And the exponent is going to be what we need to put on the small number to get the big number. So in other words, to find this log, we have the small number three and we need to figure out what exponent we need to put to make it the big number, which is 243. So what exponent do we need to put here? Well, we can we can basically just do trial and error for this, right? So remember that exponents just mean multiplying over and over until we get a number. So let's just start multiplying by threes. 3 * 3 is 9 * another 3 is 27 times another 3 is 81. And you can put this in your calculator to make sure. And then times another three. If you put this in your calculator, you'll get 243, which is the number we're looking for. But to get to 243, we had to multiply three five times, which means the exponent here is going to be five. So our final answer is that the log base 3, that's how you say it's the base, the log base 3 of 243 is 5. And that's our final answer. Now, what if you didn't know that? What if you weren't sure how to find the log? Or what if the log was decimals or fractions? Then there's no way you would have just trial and narrowed it. Right now, your calculator probably does have a log button if it's a scientific calculator. And you'll notice that it asks for you to find the log of some number. So, like the log of 50. But what does that even mean, right? We don't have a base to go off of. But whenever you don't have a base for a log, that means that it's base 10. But that's all your calculator lets you do. It only lets you do base 10. But there's actually a way that you can put in any log into a calculator. All you have to do is find the log of the big number. Ignore the small number. So in our calculator, we would put log of 243, but then we have to divide it by the log of our small number. So we divide it by the log of three. And if you were to put in log of 243 / log of 3 in your calculator, you should get exactly five, which is what we found before. So, this is how you can find any log on a calculator. So, there you go. Those are 10 of the most important things that you need to know for your algebra 2 final exam. And remember that if you were stumped on any of those problems and you think you need more practice, you can go in the description where I have full videos on most of the topics covered in this video. But if you want even more test practice just like this, then I have a midterm video with 10 more problems that you'll need to know. Some of them are pretty similar, but some of them are very different and cover stuff I haven't talked about in this video, such as the equation of a straight line. So, if you want to watch that video, you can watch it right here. And as always, if you have any questions about this video, go ahead and leave it in the comments and I'll be sure to answer your question. Thanks for watching and good luck on finals.