Ignore the bad grammar and stuff :D
What is 039 * 8? If you said 039039039039039039039039, you'd be correct. But we have a problem. What is 8 * 039? How many 8s should there be? Well, we can either define 8*039 as 8*39, but then we lose the ability to divide both sides by 8. I don't really see another option here, but I'll get back to this in a future part probably.
Solve the equation: 3x+4=334. Of course, 3*2+4=33+4=334, so x=2. But what if we had a more complex equation? Let's say we have 44x+2=4444442. First, we subtract 2 from both sides, and get 44x=444444. Now we can divide both sides by 44, and get x=3. That wasn't so bad. What if we have x^2+4=3334, though? We can subtract the 4 from both sides and get x^2=333. But now we have to take a square root. You can probably tell that teh square root function only works for very special numbers, like 10101010101010101010 and 55555. But the thing is, unlike the normal number system, we can define the square root as the inverse function of squaring a number. In the normal system, we couldn't do that because of negative numbers. But in this number system, we can't square negatives anyway because of the annoying - sign. Yay!
Our normal number system is called Base-10, or the decimal system. We also have the binary (base-2), hexadecimal (base-16), and other systems. It makes perfect sense for us to call this number system the base-🥔 system. We could also call it the potatary system, since every other base seems to have a cool name like that.
What if we have x^x=333333333? Of course, just like with base-10, there is no extremely simple way of solving this, so we can guess and check. Since there are a lot of 3's, we can try x=3, and sure enough, that works. What if we had a quadratic equation? Does the quadratic formula still work? Let's see. 2x^2 + 3x - 333333333 = 0. Plugging that into the quadratic formula doesn't work very well, so I'll let you do it. Not exactly sure how to solve these things, because completing the square and factoring doesn't work either.
If this is base-🥔, that means that the last digit is worth 1, the second-to-last is worth 🥔, and the third-to-last is worth 🥔^2, or 🥔*🥔. How many is 🥔, you may ask? Great question, no idea. You'll have to write 🥔 a lot though. Exactly 🥔 times, actually.