Substitute the 2023 on the left side with x^x^2023 and we get x^x^x^x^2023=2023. Repeat infinite and we get x^x^x^x^...=2023. This is the same as x^(x^x^x^...)=2023. Substitute the parentheses with 2023 and we get x^2023=2023. (As we assumed that the limit of x^x^x^... exists, we have to plug this value into the original equation to proof that this is actually a solution.)

I think you do need an extra line of logic at the end to be completely accurate, as y = x^X is not strictly monotonic.

@laurentthais6252

Жыл бұрын

x^x is not monotonic but is bijective with a unique minimum at x=1/e.

@TheMoogleKing93

Жыл бұрын

@@laurentthais6252 x^x is not injective , you can clearly see that from a graph.

@laurentthais6252

Жыл бұрын

@@TheMoogleKing93it is bijective for x above 1 and it's all we need since x cannot be smaller than 1, otherwise the left term in the original equation would be smaller than 1

@TheMoogleKing93

Жыл бұрын

@@laurentthais6252 more specifically its bijective for x > 1 and the maximum value it takes below this point is well below 2023. You'd lose a mark on a university paper for just declaring your one solution was the only one and not considering small x!

@laurentthais6252

Жыл бұрын

@@TheMoogleKing93 Amen great professor. Maybe you can refute my PhD and cancel my 26 referred journal papers ?

Brilliant content as usual please keep up your amazing work

So, in general, if we replace 2023 in this problem with say, “Box”, then any equation in the form of x to the x to the Box = Box will have the solution x = [Box to the 1/Box], since that would result in the LHS always simplifying to [Box to the 1st power], which is Box (as long as Box zero).

@Hyakurin_

Жыл бұрын

No, this is not correct In the last part he uses the fact that f(y)=y^y is injective. But this is true only if x>1. In this case f(y)=f(2023) implies y=2023, because f(y)>1 so y^y>1 and y>1. Now, in ]1,+infty[ f is injective so y=2023. But if you choose 0

Kudos! 👌👌👌

Wow such an outstanding and technical question for welcoming the new year 2023😊😊😊#SUPERMATHS

You come up with some good ones. Thank you so much for what you do. Happy New Year.

An absurd problem we'll never see in the real world. But I'm just drinking beer and relaxing, so here goes. If a^a = b^b then bases a and b are equal. So let's look for the a^a = b^b format. Raise both sides by a freaking absurd power of 2023. KEY: the 2023 on the LHS goes on the x DOWNSTAIRS. So we have the absurdity of x^2023^x^2023 = 2023^2023. There's the format we want. So x^2023 = 2023 and x = a ridiculous 2023^(1/2023). I need another beer.

That's awesome

I LOVE IT! Thank you. ♥

Wow so much nicely explained! Great technique. SO much impressed by the clean writing with different colours!!! Well done. Such a mesmerizing channel. #SUPERMATHS

@jim2376

Жыл бұрын

She's amazing with her writing and different colors.

Good 👍👌

Calligraphy wonderful.

Just raised equation to the 2023 power and answer one step solution

Wonderful channel. Great content! Keep it up! #SUPERMATHS

Thanks ♥♥💕

unfortunately that is only the principal root however there are infinite complex solutions which is why you should use the lambert w function instead. Hope this helps.

I tried it with Lambert-W-Function and succeeded :-) It's WAY more complicated (x=exp{W[2023*ln(2023)]/2023}, but according to wolfram-alpha it's the same value

@pureffm

Жыл бұрын

Can you not just just take the 2023rd root of both sides and get an immediate answer?

@reconquistahinduism346

Жыл бұрын

@@pureffm not at all. Because the L.H S is not 2023rd power of x. It is x^2023th power of x. You can't take 2023rd root on both the sides for that reason.

It is much easier to raise the left expression to the power of 2023.

Good luck.

Solve for both roots of x^(x^(1/2023))=1/2023. That's way more fun.

when my teachers told me to show my work, it never occurred to me to make up so many steps!

Wow😲Its awesome

@RhinocerosMovie9

Жыл бұрын

Yeah definitely

it can be extended infinetly x^x^x^x^x^x^x^x^x^n = n^(1/n)

Happy new year!

wow!!!!

Nice

I solved this in my head intuitively in like five seconds and got 2023 to the 2023rd root. Which is the same as 2023 to the 1/2023 power.

By inspection : "at or greater than 1, but certainly less than 2 - but probably not much greater than 1."

nice

i think you can actually see that just by looking at it

At least, a video from the presenter which isn't lame 👍

I would start by factoring 2023 on the right hand side. "the remainder is left as an exercise for the viewer". Next video!

If I were able to solve this, I wouldn't watch your video on how to.

Or we can also do by Lambert W function

@peamutbubber

Жыл бұрын

Uncreative

It should be equal to euler constant = 2.7281828... . I do'nt know how to find out the value of x. thanks again

Taking both side at power 2023, we see something like x = 2023^(1/2023)

@chiararomano1818

Жыл бұрын

This is the fastest way.

Hi , could you please solve this fomula x^x= (x-1)^(x+1) Thanks

Easy

1

Did you know that this function has two minimums and no maximum?

For what?

It's been a while for me and math, but can't you just immediately raise each side to the 1/2023 power? I can't remember why, but isn't the resulting exponent on the left side then equal to 1? Again, take it easy on me.

@Мопс_001

Жыл бұрын

You'd get x^((x^2023)/2023)

@timower5850

Жыл бұрын

@@Мопс_001 Yes, but can't this exponent be reduced to 1? It's obvious from the solution ( x = 2023^(1/2023) ) that it must be. That is, after raising each side to the 1/2023 power, you get x^((x^2023)/2023) = 2023^(1/2023), and since the solution is x = 2023^(1/2023), x must be equal to x^((x^2023)/2023), meaning ((x^2023)/2023) =1. Not?

@Мопс_001

Жыл бұрын

@@timower5850 try to write it on paper. x^((x^2023)/2023) is not the same as x^((x*2023)/2023) I suppose what you are talking about is the second one where the power be reduced, but even so x^x is not equal to 1

@timower5850

Жыл бұрын

@@Мопс_001That's not what I said, though. For simplicity, let a = x^2023, making the original equation x^a = 2023. Raising each side to the power of 1/2023 yields (x^a)^(1/2023) = 2023^(1/2023), yielding x^(a/2023) = 2023^(1/2023). Since the solution is, per the author, x = 2023^(1/2023), x must be equal to x^(a/2023) since they're equal to the same thing. Given that x can alternately be written as x^1, this means x^1 = x^(a/2023). By inspection, it's obvious that (a/2023) =1, and by reverse substitution, (x^2023/2023) =1, my original point. (as for your "x^((x^2023)/2023) is not the same as x^((x*2023)/2023), I'm not sure where you picked up the (*) sign.)

@Мопс_001

Жыл бұрын

​@@timower5850 Well, I tried tracing it from a = x^2023 x^(a/2023) = 2023^(1/2023) here it's a bit hard for me to understand this since here we would see that x = 2023 but at the same time a = 1 = x^2023 So I'm not sure how it should be solved from here

With logaritma 2 seconds

I solved it with the help of Lambert W, but does she know that she was banned from the Olympics?

just use natural logs

Root at the 2023th of 2023.

The initial substitution is dubious unless you clearly state the problem as x^(x^2023). But normally, as written, it would conventionally be interpreted as (x^x)^2023, which is a different problem.

@Semak2002

Жыл бұрын

Thank you! OMG, it's really frustrating, i was this close to check rules of calculation's order

X=1

Ax is gonna to be lolol

You added too many steps in my opinion. Can we not just take the nth root on both sides *raise to the power of 1/2023*, then it gives the answer immediately?

X=2023^(1/2023).

5h

Next year You will post equation x^x^2024=2024 and so one

뭔가 속고있는 느낌 이거는 뭐지?

Why do you sound like that lol 😂 Where are you from 😂😂😂

so x= the 2023th root of 2023

Biyog

the reason English speaker usually doesn't good at calculating(or mental calculating) is because of the name of number. It is too long. In case of Korean, all the sounds from 1 to 10 are all short one syllables. it is easy to pronounce. but In case of English, 1,2,4,10 are short syllables, but, 3,5,6,8,9 are long syllables because of two reasons. 1) the sound of [ai], [ei] (five, nine, eight) makes pronouncing longer. For example, say fast oneoneoneoneoneone, and say fast eight,eight,eight,eight The speed of the former is faster than the speed of pronouncing eight. 2) the existence of consonants before(or after) one syllable. For example, three is th + ree , -> it sounds two syllable. (bcz the move of tongue is too big) so say fast oneoneoneoneone, and say fast three,three,three,three.. Also the former is faster : that's because three is long syllable. Six is also long syllable : oneoneoneoneone, sixsixsixsixsix -> the former is faster.(the former is easier to say fast) Even the number of seven is two syllables!! it is too long to pronounce. Moreover, from 11, it is a new word. In case of Korean, we call 11 "tenone", 12 "tentwo" , 24 : twotenfour, 41, fourtenone -> it is very intuitive. and... oh my god.. 100 => hundred, = long two syllable.. omz.. why "long" syllable? -> Compare "hundred" to "body" : body is easier to say faster. because body is short two syllable. hundred is long two syllable. in case of korean, 100 is also one short syllable. thousand, = long two syllable. in case of korean, 1000 is also one short syllable. In conclusion, it is hard to say fast in case of English. For example. 358,936 = three hundred fifty eight thousand nine hundred thirty six. => 14 syllables.(10 long syllables!!!) but in case of korean, 11 syllables, and all syllables, are just short syllable, just one sound. That's why the English speaker is too slow to explain the solution or to calculate numbers.

What about x^x^2023=2022 Bullshit

Математические извращенцы

В загальному вигляді: X^(X^A) = A , тоді X = A^(1/A)

Smart girl, I can also solve it like this, no??? (X^2023) ^(X^2023) = 2023^2023 Then, X^2023 = 2023 X = 2023 ^1/2023 Nice or not nice???

@ralkadde

Жыл бұрын

Nice!

@mathwindow

Жыл бұрын

Nice nice!!!

Супер

Exp(1/266) + 1/284926

f(x)=x^{x^[x^(x^n)]}=n>0 . Проверте : x=n^(1/n) . f(x) - монотонно возрастающая функция. Значит решение единственное. С уважением ,lidiy27041943

@peamutbubber

Жыл бұрын

u overcomplicated it buddy

1