Can 0^x=2? (I don’t really get this solution, but it’s extremely cool!)

Added: 2026-05-26

[00:00:00]The goal for today is to solve the equation 0 to the x power is equal to two. How is this possible, right? I'm going to demonstrate solution by a viewer and I'll show you guys the comment at the end of the video.

[00:00:13]This is really really cool. It's kind of similar to the equation I did before.

[00:00:18]One to the x power is equal to two. But here we are not going to be using complex solutions. Instead we'll be using what we call the dual numbers.

[00:00:31]What is it? Well, let me show you. We have some epsilon and that's actually not equal to zero. But if you square epsilon, you get zero. H I can give you a quick example. Consider Z4. You can just think about this as like the world with just four numbers. zero one two three not four though because four will be considered zero. This is like the remainders when you have a whole number divided by four. So in C4 two is not equal to zero because 2 is two but 2^ 2 is equal to 4 which is considered zero in this world. It's kind of like that.

[00:01:19]Anyways here we have this to work with.

[00:01:22]zero is the same as epsilon squared.

[00:01:25]We'll put that here and then raise that to the x power. So we can just multiply the exponents and that's equal to two. Then let's take natural log both sides.

[00:01:40]However, do not cancel this and that because this is not e unlike that e over there. This is epsilon.

[00:01:48]So I'm actually just going to bring the power to the front. And we have 2x * ln epsilon equals ln2.

[00:01:58]Then we'll divide both sides by 2 n ln epsilon.

[00:02:02]So x is equal to ln2 over 2 ln epsilon.

[00:02:10]And that's pretty much it, huh? However, we can actually simplify ln epsilon a little bit. Check this out.

[00:02:20]When we have a dual number, we also have the standard form very similar to a complex number which is a plus b i. So right here, let's first figure out a form for a + b * epsilon.

[00:02:37]And to do so, we actually have to use the power series for e to the some power. And I'm talking about that e over there. So let's say e to the u expanded you get 1 + u + u ^2 over 2 factorial plus u to the third power over 3 factorial and so on so on so on.

[00:02:59]This is good for all use. Right now I'm going to plug in not b epsilon though I will actually use c epsilon. You will see why later. If I plug in c epsilon into all the u's we get 1 + c epsilon plus c epsilon squared over 2 factorial plus c epsilon third power over 3 factorial and so on so so on now check this out epsilon square is zero so this is out epsilon cube is epsilon square * epsilon which is 0 * epsilon this is out anything higher than epsilon square of course they will all be zero. So these are all out.

[00:03:47]So e to the c epsilon is just 1 + c epsilon.

[00:03:57]Okay.

[00:03:58]Well I kind of want to have a right here. So let's take this equation and multiply everybody by a. So we get a e c epsilon equals a + a c epsilon. But I want to have b right here though, right? So let's go ahead and call that b.

[00:04:23]And of course here we have c. So we can divide both sides by a. So c is equal to b over a. And of course we have to say something like a is negle syable blah blah blah. Well keep that in mind that's all. So here a is just a and then we have e c is what is b over a and then that's equal to oh epsilon and then a here plus that it's the b and then epsilon.

[00:04:57]So have a look a + b epsilon is equal to a * e to the b over a * epsilon. This is very similar to what the polar form of a complex number a + b i is equal to r e i theta. How cool is that? Right now how are we going to utilize this though?

[00:05:23]Check this out. Here I'm going to make a note starting with epsilon I'm going to look at it as 0 + 1* epsilon and then of course zero is epsilon square. So this means epsilon is equal to epsilon squared plus 1 * epsilon.

[00:05:47]Now here we have what? This is our a and this is our b.

[00:05:54]So we can actually put this into this form which is going to give us a which is epsilon squared time e raised to the b / a power which we have 1 / epsilon squared and then times epsilon.

[00:06:16]So as you can see I can say this epsilon equals epsilon squared and then here we have just e to the reduce that we have one over epsilon and I know this is insanity because there are a lot of things going on first isn't epsilon square just equal to zero and when you multiply by e raised to 21 over epsilon isn't that whole thing just equal to zero.

[00:06:48]Moreover, we actually divided by zero right here already. So, what exactly is going on?

[00:06:55]I I don't know too much to be honest.

[00:06:58]Let's just go with it and see what you guys think about this at the end. But I do have a lot of like questions myself as well. Like, is this literally legitimate? Well, let's just continue.

[00:07:13]How though? Well, it's almost I want to figure out an expression for ln epsilon.

[00:07:19]And to do so, let's just take ln here and ln here. So, we are going to get ln of epsilon equals ln of the first.

[00:07:31]That's a multiplication, right? So, I can say that's adding ln of the second.

[00:07:38]And I know this is bad because we have ln of zero. But as I said just put the pole to the front all that and this and that cancel. So ln epsilon equals 2 ln epsilon + 1 / epsilon.

[00:07:54]Now we can subtract well depends how you want to do it. Subtract 2 ln epsilon both sides so they cancel and that'll give us negative ln epsilon which will give us 1 / epsilon. So of course ln epsilon is equal to -1 / epsilon.

[00:08:19]Okay.

[00:08:23]Yeah. Somehow l and epsilon we end up with a pretty nice expression -1 / epsilon. So if you come back here and replace that right here we get x = ln2 / 2 * -1 / epsilon put that to the top and puts a negative.

[00:08:44]So ladies and gentlemen, we have a solution. X is equal to negative epsilon, let's put the negative in blue as well, * L12 over 2.

[00:09:00]So what do you guys think about that?

[00:09:04]Well, I do have so many questions, but this is the solution I saw.

[00:09:11]It's really cool. However, as I said, my concern is epsilon square is zero already. So why cycle is going on, right? And secondly, when we take the natural log both sides, do we have to add something to it? Just like the complex numbers, when you have the polar form, when you take the natural log, it actually is multivalue, right? You have to consider different branches and all that stuff. Do we have to put some more things here? I don't know.

[00:09:42]But that's a solution. Here's a comment.

[00:09:45]Thank you so much. So cool. I enjoy a lot.