 So what we've done so far is, again, gone through some basic techniques of how to solve equations, how to deal with things being added, subtracted together, being multiplied or divided, or exponents or radicals, right? So basically we've done, you know, we start off with simple things, like 2x plus 1 is equal to 0. Solve for this, you grab the 1, bring it over, divide by 2. So you're going to have 2x is equal to negative 1. Sine changes, divide by 2, divide by 2, so x is equal to a negative 1 over 2. So over here we have 1x on this side, 1x on that side. We're going to try to bring all the x's to one side. So what we're going to do here is cross multiply this up, right? So you're going to have 2x plus 1 is equal to 15x, right? And then bring all the x's to one side. So I'm going to grab this one and bring it over on this side. So it's minus 2x, because I'd like to keep my x's positive. 15 is bigger, so I'm bringing in the smaller one over, making it negative, right? Now this side is just 1, line up your equal sign. I'm going to put the 1 here because I'd like all my variables to be on the left side, and that's, I think, the norm of the way it goes. So 15x minus 2x is going to be 13x, and then divide by 13, divide by 13. So x is equal to 1 over 13, right? No restrictions in this equation, because there are no variables in the denominator, right? If there are no variables in the denominator, you don't have any restrictions. So over here, your solution for this would be x is equal to 1 over 13. If they ask you to check this solution, what you would do is, let's erase this part, so x is equal to 1 over 13, that's our solution, and if they ask you to check this equation, you would go back to the original, or after you've done the work, so the work would be here, and below here, you would go check, right? You would go left-hand side of the equation, is this, the right-hand side of the equation, is this, and we're checking for x equals 1 over 13. So wherever you see x, you put in 1 over 13. So over here would be 2 times 1 over 13 plus 1 over 5. Over here would be 3 times 1 over 13 over 1. And you would do this and see if this side of the equation equals this side of the equation. Obviously, you don't need the 1 over here, right? Now, I'm not going to bother doing this, hopefully it works out, hopefully I did the question properly. If it doesn't work out, send me a message, please. Go to question with exponents and radicals, and just kick it out one more level higher and see where that goes, and it's going to be the same thing. You're going to use the same rules to solve for the equation, right? So let's say you had 2 x squared minus 1. So what you're going to do, again, you're trying to get to the x, right? So bring the 1 over, so this side is going to be 2, the fifth root of x squared is equal to 3. Over here, you're going to divide by 2, divide by 2, so you're going to have the fifth root of x squared is equal to 3 over 2. And now you want to get the x by itself. And if you remember, this is just, you take the 5 and you write it as an exponent here. So this becomes x to the power of 2 over 5 is equal to 3 over 2, right? So because this goes in the denominator of the exponent. And to do the opposite of this, all you do, you take this to the power of 5 over 2. So you've got to take this side to the power of 5 over 2. So what you have here is going to be, let me have a room here. So these guys come up here. I'm just going to rewrite this so it looks clear. So this is going to be x. 2 over 5 to the power of 5 over 2 is equal to 3 over 2 to the power of 5 over 2. So this, this kills this, that kills that. This side is just x is equal to whatever that is. What is this? 3 to the power of 5, 3 times 3 times 3 times 3 times 3 times 3. 3, 3 is 9, 3, 3 is 9. That's 81 times 3. I should know this. I think it's 243 or something like this. 5 to the power of 2, or 2 to the power of 5, which is 32, all squared. So whatever it is, it becomes a square root of 3 to the power of 5 over 2 to the power of 5. And you would do that without your calculator. You wouldn't write it. For me, this is about this good enough. That to me shows that you know what you're doing. But in general, you would have to take it down to this level. Here you had this function where you got the cube root of x squared times 2x to the power of 3 over 4 plus 1 is equal to 0. Now this thing is fairly straightforward because what we have here is a multiplication. And we're able to combine these two x's together. If we had a plus here, then we would have a problem. We wouldn't be able to solve this equation without bringing other techniques to do it. Because the simple moving things around and taking the opposite of what's being done to this thing is not going to work for us. The plus side not being able to combine the two x's together would have created a problem for us. But right now these two things are being multiplied together so they're not creating a problem for us. So what we're going to do is, the way we're going to combine, multiply this x and this x, we're going to have to write this thing as an exponent. Take away the radical, right? And as we talked about before, this guy goes into the denominator in the exponent, right? Now what I'm going to do is do this at the same time as I'm moving the 1 over there. So over here, we can just grab this guy, break it over, it becomes negative 1, right? So in here we have negative 1. Over here we're going to have x to the power of 2 over 3 times 2x to the power of 3 over 4. And you know, series 2, you must know how to do this. If you don't, go back to series 2. But the way you multiply these things is you add the exponents. So the 2 here is just a 1 here. When there's no number here it just means 1. So it's just a 2. And this is going to be x to the power of 2 over 3 plus 3 over 4 is equal to 1. And the way we solve for this is we're going to find common denominator here, right? To add these guys. What I'm going to do at the same time as I'm finding common denominator here, I'm going to divide by 2 here, divide by 2 here. So get rid of the 2 at the same time as well. So divide by 2. Common denominator here is going to be 12. So let's do this thing on the side here. And what we can do is write the solution somewhere else, right, afterwards. So just adding 2 over 3 plus 3 over 4, common denominator is 12. And you multiply this by 4 so that becomes 8. Plus multiply that by 3, 9. So you add those 2 guys together. You got 17 over 12, right? Are we on the board? We're still on the board. So the power here becomes 17 over 12. We divide it this side by 2. This kills this. And that's 1 over 2. What I'm going to do, let's see, do we have room? Let's just go up here. Isn't that fantastic? So what we have here now is x to the power of 17 over 12 is equal to 1 over 2, right? And the way you get the x by itself now is you take this side to the power of 12 over 17. That means you have to take this side to the power of 12 over 17. So the way it works is 12 kills 12. 17 kills 17. And on this side we have x is equal to a half to the power of 12 over 17, whatever that is. I don't know. If you want to punch it out, you can punch it out. But this is sufficient enough. What you can do is actually another way you could express this, let's make it a, you know, test what we learned in series 2. 1 over 2 you can write down as 2 to the power of negative 1, right? Because the negative 1 kicks the 2 down. So this could be 2 to the power of negative 1. And when you have an exponent to an exponent, they just multiply each other. So this would be negative 1 times this. So it would be negative 12 over 7. And that would be the solution, 2 to the power of negative 12 over 17. That would be the solution. This guy here. It's as straightforward as this. These kinds of, you know, radicals and exponents scare you off of, you know, try to attempt to solve an equation. It's really straightforward. As long as you're able to get 1x on one side by itself, you're in the clear. Everything we talked about before works. All you have to do is do the opposite to what is being done to the x to solve for it.