Good luck

You're absolutely right! We are learning so many complex algebra equations but not the LambertW function. And because of that we can't solve stuff like x^x = 2. I am in my last year of high school but we don't have LambertW function in college. So, I learned it entirely from internet.

It's the inverse function of xe^x. Mathematician found there is no way to get x from xe^x, so they figured out what the inverse function would be and how it worked and called it lambert w function or product log. It became a useful function for solving certain equations.

I’m 13 years old too!if you’re interesting in maths,I suggest you to learn calculus it’s really fun!

@@manyifung5411 bruh why you 13 yr olds are getting into this complicated math thing ?you have a beautiful life to enjoy . also , you have to learn calculus anyways after 3 or 4 year later. why not enjoy now

@@prateek1.9 some people enjoy math yk.

i solved it in my head too. took me like a minute since i am not really used to tetrations.

Because math is great!

Here comes algebra...

Because most sciences are actually fun when you know what the hell is going on

Same but 15 year here

@@samthedjpro cooll! we can b friends!

Welcome aboard!

Thanks, looking forward to what's ahead!@@PrimeNewtons

Working on it

You are already a pro

@@PrimeNewtons I'm an engineer so yeah in a way I'm pro lmao nice videos you're a great teacher!

hahahha ur like mee

Mathematician obviously since they deal with all maths related and probably math that involves studying space since the universe is infinite and involve big numbers like from billion up to sextillion and more light years

Well since the function isn't continuous, only defined for positive integers, you'd have to do something _real_ creative to differentiate/Integrate something like 2↑↑x, but it would probably be easier to find a way to extend the function to the reals.

The derivative of the tetration function is as of 2023 not found, and an unsolved problem in mathematics!

@@aguyontheinternet8436 I would imagine continuation to at least real number line is not an issue: logarithms are known, no matter what "degree" of exponentiation, so something like 1.234^3.678^6.845 is technically defined. It's a mess to write it in tetration form, but something like x↑↑ x etc is defined (Idk how to write tetration on keyboard, but I am abusing notation from Graham's number to get the point across). So technically, a tetration of variable real numbers is defined and continuous.

​@@maxborn7400there's no elementary functions that express the integral of x^x. It simply does not exist.

It must be a positive integer

That's an interesting question.

I might be misunderstanding, but 1.5^x≠x tetrated to 1.5. It would be 1.5 to the power of x. For tetration, use ^^. For example, the equation is 3^^x^^2=27

@@sportsloverbaseball I see my error. Thanks.

@@kimutaiboit8516 np :)

i think that has yet to be defined

@@trifonmag4205 I thought so, asked my professor he said you don't need to know this 😑

I imagined such a problem. To keep it simple let's start with ½^X That will be 1/(²X) Now for 1.5^X it will be ³X/²X. Just my thoughts.

@@kimutaiboit8516 I also thought that but I was wrong as x time x is x² or x^2/1 and x^1/2 will be opposite which is how many times it should be multiple so it makes x which is √x but when we talk superpowers (supeman 😂) ²x will means x power x but x superpower 0.5 must means how many times x should be exponentiate to make x which neither logs or any function can define, if you have any answer please comment 👍

@samthedjpro let's say x=3 ²3 is 3³ ¹3 is 3 My claim was ½3 is 1/3³ But now that you mention it, I think I have to relook it. The expression I made is for -²3 I think ½3 is undefined as you said. 🤔 PS: What could ⁰3 be? I guess undefined too.

I remember reading somewhere that we haven’t defined tetration for non-natural numbers yet but I can’t remember where I read that so that could be a complete lie

okh bro still helpful tho :)@@freddier47

@@freddier47Real numbers are defined We can understand better with super Logarithm (inverse of Tetration) By definition sLog2 (2^^3) = 3 NOTE: "sLog" is a notation for super Logarithm. Like how Logarithm cancels the base leaving the exponent ex. Log2 (2^3) = 3 super Logarithm does the same with Tetration leaving the super power. We can use super Logarithm to solve non integer super powers since super Logarithm is repeated Logarithm by definition. Let's let sLog2 (16) = 3+x Where 0 ≤ x < 1 (represents a 0 or decimal) sLog2 (2^^3) = sLog2 (2^2^2) => Log2(2^2^2) = 2^2 => Log2(2^2) = 2 =>Log2(2) = 1 At this point we've taken three logs representing our integer part of the solution (given by the fact that the answer is equal to 1). We just take log again for the decimal x (the remainder of 2's that we need.) Log2 (1) = 0 Thus sLog2 (16) = 3+0 = 3 Well let's look at what happens when we go backwards through the same process to see what happens to the remainder. Log2 (Log2 (Log2 (Log2 (16)))) = 0 Log2 (Log2 (Log2 (16))) = 2^0 Log2 (Log2 (16)) = 2^2^0 Log2 (16) = 2^2^2^0 16 = 2^2^2^2^0 = 2^2^2 = 2^^(3+0) The remainder adds an extra '2' to the top of the power tower and the additional 2 is raised to the power of the remainder For 0 ≤ x ≤ 1 By definition sLog a(a^^3+x) => a^a^a^a^x By definition of Tetration a^^3+x = a^a^^2+x = a^a^a^^1+x = a^a^a^a^^x a^a^a^a^^x = a^a^a^a^x a^a^a^^x = a^a^a^x a^a^^x = a^a^x a^^x = a^x by definition for 0 ≤ x ≤ 1 We can take for example 3^^e ≈ 3^3^3^(e-2) ≈ 227,914.4 It's complex numbers (i.e. 2^^i) that it's not defined

For real numbers a^^x = a^x for 0 ≤ x ≤ 1 for negative super powers we use the above definition to extend to negatives -1 ≤ x ≤ 0 a^^x = Log a (a^^x+1) NOTE: because -1 ≤ x ≤ 0 then 0 ≤ (x+1) ≤ 1 thus the above rule applies a^^x = Log a (a^^x+1) = Log a (a^x+1) = (x+1)*Log a(a) = x+1: For -1 ≤ x ≤ 0 a^^-2 is undefined because a^^-2 = Log a(a^^-1) and a^^-1 = 0 by definition so a^^-2 = Log a(0) and you can't have 0 in the Log

@@ryanman0083 actually, complex numbers have been defined too.

x must be an integer

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