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[@damyankorena]

x^x=a
xlnx=lna
lnx * e^(lnx)=lna
lnx=W(lna)
x=e^W(ln a)
Substitute a=5^50
And x=25 in the real branch of the W function

[@vikrantharukiy7160]

Beat me to it

[@bigfgreatsword]

Thanks Oiler!

[@irinaseif9691]

​@@bigfgreatsword*Euler

[@bigfgreatsword]

@@irinaseif9691 Not you-ler, oi-ler

[@irinaseif9691]

@@bigfgreatsword it's pronounced Oiler, but written Euler

[@prithvisinghpanwar6609]

where are rest 24 answers

[@antipneumonic3563]

yeah, this is only e to the positive, principal lambert W of ln of 5^50th power!

[@savitatawade2403]

​@@antipneumonic3563😂😂

[@foodforthought247]

​@@antipneumonic3563Speak of _that_ function is strictly prohibited.

[@weelusmallgi]

Surprisingly, it also works if you flip the 2 and 25. (5^25)^2 is the same exact thing. Idk why I'm acting stupid because I already knew this rule and forgot it lol

[@itzsasha4019]

i↑↑i - can you do that??🤔

[@aplasticplate]

The answer of your question is actually pretty interesting
If my calculations and assumptions are correct then it shall be 1.12458 + 0.06508i

[@ubncgexam]

That's to easy...
It's i^e^(-pi/2)

[@samueljehanno]

​@@ubncgexamProof ?

[@eastonrocket兀]

@@ubncgexam What about i↑↑↑i or i↑↑e?

[@fsponj]

​@@ubncgexamProof

[@theforestita2376]

25 tetration of 2

[@wrsz]

sigma

[@raphaelfrey9061]

But this means that there is an operator O which cancels x^x. It would look like this:
x^x = y | O()
x = O(y)
Question is, what is this operator O?

[@MaloventMouse]

It's inverse of x^x function

[@v8torque932]

half tetration

[@raphaelfrey9061]

@@v8torque932 but is this well defined

[@v8torque932]

x^x can be rewritten as x^^2 which indices recursive powers can have inverses. x^^1/2 will have 2 solutions if e^-1/e < x

[@uggupuggu]

if x^x = y
then x = e^W(ln(y))

[@FederalBereau]

Lucky guess

[@neologicalgamer3437]

How do we prove that that's the only answer though? I know there isn't another positive real solution, but what about negative or otherwise complex?

[@azainkadri9556]

As per the fundamental theorem of algebra, there are 25 complex solutions to this .

[@neilgamer5290]

Isn’t x•x = 5•5?

[@shakila_parveen]

No if x=3 then x^x will be 3^3 which will be 3×3×3. Thus x^x is not equal to x times x

[@India_Pakistan1]

Exponents and multiplication are clearly different things

[@marcusscience23]

x^x is x^^2, it’s a tetration. Addition is repeated succession, multiplication is repeated addition, exponentiation is repeated multiplication, and tetration is repeated exponentiation.

[@neilgamer5290]

Sorry im in sixth standard

[@flatuitous]

like.. 6th grade? bruh ur not meant to even be on yt
minimum age is 13@@neilgamer5290

[@AOOA926]

2.6k likes

[@ChannelTerminatedbyYouTube]

The perfect loop doesn't exis...

[@The_Alvalscompany]

Hello

[@WeirdGoat]

I'm not a mathematician, but your calculation is not right, that's 7th grade math

[@WeirdGoat]

@ihml842 5^50 = 25^49 = 125^48, this is the right way to do it. 5^50 != 25^25, 25^25 = 5^26

[@sonkim6876]

​@@WeirdGoat Bro don't know how do to do math.a^(bc)=(a^b)^c

[@WeirdGoat]

@@sonkim6876 yes, I was wrong

[@mrmimi807]

It's 7th grade math ,and you didn't get it right?

[@WeirdGoat]

@@mrmimi807 Yes, unlike you, I make mistakes, you sir is too smart to make any.

[@bitcoin-trash]

no it is false😂

[@Ostup_Burtik]

It is true.

[@codywohlers2059]

dont so the stupid make it loop thing. its annoying af

[@GillesF31]

Great ... by "chance" 50 = 5² ... Imagine the equation was x^x = 5^47 ... :-))
x^x = 5^47

=> use of W Lambert function (product logarithm)

x^x = 5^47

ln(x^x) = ln(5^47)

x·ln(x) = ln(5^47)

e^ln(x)·ln(x) = ln(5^47)

ln(x)·e^ln(x) = ln(5^47)

W(ln(x)·e^ln(x)) = W(ln(5^47))

ln(x) = W(ln(5^47))

e^ln(x) = e^W(ln(5^47))

x = e^W(ln(5^47))

x = e^productlog(ln(5^47))

----------------------------------------------------
| WolframAlpha: x ≈ 23.849167 |
----------------------------------------------------

[@Emily-fm7pt]

If you need to apply this, it will be taught to you beforehand, considering that the lambert W is just such a hyper-specific thing to know. “Most” of the time when you see problems like this, it’s doable using the method shown in the video

[@mrmimi807]

5² ≠ 50

[@GillesF31]

@@Emily-fm7pt Yes, nevertheless it's great to solve any equation, ... WHATEVER are their input data. 🙂

[@GillesF31]

@@mrmimi807 Oops ... I wanted to write > ...

[@itzsasha4019]

i=x^x | x=? very interesting to knowwww

[@v8torque932]

e^W(i*pi/2)

[@marcusscience23]

Or i^^(1/2).