x^x^x^x = 4^x^3

[@FlashAm]

I don't know how you can get angry comments! You're one of the most wholesome, understandable math teacher on this site! Keep doing what you do. Much support

[@PrimeNewtons]

Wow! Glad you think so.

[@mrjerry9918]

@@PrimeNewtonsyou making tough maths easier to understand with basics😮

[@HelloIAmAnExist]

Yeah but people get mad when he makes mistakes or overcomplicates things, which I can understand

[@yuyuvybz]

Wish i could send a million 😭🙏🏻 I wish i can pass my final exams …also what could be the mathematical explanation for the "guess" that you made

[@VideoGames_Wizard]

Intuition

[@Pablo-v4n3z]

How can someone be aggressive towards someone as peaceful as this guy?

[@james789c]

Not towards him but towards other people in comment section ;)

[@nandoaires]

@@james789c "Welcome to the Internet, have a look around" (BURNHAM, Bo)

[@ProactiveYellow]

We define 0!=1 because of the pattern that A!=A(A-1)! Which means 1!=1•0!. This combines with our use case of the factorial as the total permutations of A objects because if you have 0 objects, you only have 1 meaningful way to "permute" it: identity. I think it makes sense to define ⁿ0=1 according to its limit, just like how we define 0!, or even how we define multiplication on irrational numbers (the notion of "repeated addition" kind of falls apart for e•π, because what does it mean to "add e to itself π times?") We commonly see these kinds of extensions on functions to fill in the gaps.there really are only three options: define tetration only over the positive reals, define tetrations of 0 as 0, or as 1. Including 0 is nice for closure over the natural numbers, and between defining it as 0 or 1, the particular function seems to "want" it to be 1. It's more convenient to not have a jump discontinuity there, and since this seems a matter of definition for us, we are free to choose the nicest option. So long as we are consistent with our choice, it seems perfectly reasonable within the rules of mathematics (recognizing that this acts to extend the original definition of tetration, rather than strictly defining it as power towers).

[@wbell539]

First time I've ever seen an approach like this, thank you. I'm sure you're aware that wikipedia offers a recursive definition of tetration that has the delightful property of making any problem with the zeroth tetration simply disappear!

[@T.Furtig]

Thank you for the interesting problem and your wonderful explanation! We can also take log with base 2 (log_2) to solve this equation. x^x^x^x = 4^x^3 x^x^x^x = (2^2)^x^3 = (2)^2x^3 x^x^x log_2 (x) = 2x^3 log_2 (2) log_2 (x) = log_2 (2) =1 therefore x^x^x = 2x^3 x^x^x : x^3 = 2 x^(x^x -3) = 2^1 (x^x - 3) = 1 x^x = 4 =2^2 x = 2

[@juanpabloarroyo2313]

so you basiclly induced the same same than he had, you got a equation which structure is similar in both sides and then assume and check if the exponents are the same. Thats what you did when stating that x^x-3 = 1. Am I right? or am I missing something, if not does this procedure have a name and is it rigurous or just a heuristic?

[@coreymonsta7505]

It could be defined or not defined motivated by limits. But there’s two to consider. As x -> 0, x^x -> 1 with no problem. But as (x, y) -> (0, 0), x^y can approach different values. Maybe there’s algebra issues with defining 0^0 = 1

[@winstonridgewayhardy]

I know you will not let any nasty viewers get you down. God is shining through you. If God is with you, who can be against you? I will continue watching and learning from your videos. Be well my friend.

[@carltompkins3282]

Re: some difficulties you’re having with flaming comments lately. I can venture an answer. You’re demonstrably a very nice guy and professional mathematician and enthusiastic teacher- overall of academic methodology and mind. Due that you’ve taught subject material like calculus and trig etc on your channel you mainly appealed to students and scientists previously. Since you have broadened to googology after one year you are encountering more non-professional and argumentative viewers. These topics are nowhere near as mathematically involved as your actual field is- I concede as I’m not a mathematician like you are and found your channel due to the googology content as most new viewers obviously have done so. This is why the responses are more juvenile/asinine now as not all casual math fans, who like “big numbers” but not serious applied mathematics are reasonable or mild mannered as say myself or an academic who is just more likely to be so. A further factor is because your understanding of googology and things like zero to the power of zero equals 1 are common concepts to lay video watchers but not to a pro mathematician like yourself your mistakes and struggles here are amplified out of all proportion. But your videos on anything else you do utterly perplexes them and me mostly as well and that is why you don’t get asinine comments on things like your calculus videos because they’re out of their depth there. All I can say is you touched on a field that’s very popular but definitely rougher/less educated overall in demographic. The main body of work you do I can see is utterly brilliant, the best example on KZbin and that this latest stuff is intended as more “fun” and I hope that of you can get this popular foray under control so that it can work for you to help popularise your serious stuff more. Kind regards 😊

[@enxel493]

This gentleman has expressed exactly what I wanted to say and in a much better way that I possibly could, so I'm just gonna support this comment, congratulate you on how well you explain these subjects and motivate people to learn, and thank you for your videos!

[@Hanible]

It's ok to be enthusiastic and argumentative, It reminds me of my university days, arguing was the norm back then, mid-sentence you'd say "ohhh.. never mind I'm an idiot!" and move on! I used to argue with my teachers too. I was right sometimes, but it's not about who's right and who's wrong, it's about the exchange! Elitism is not the answer, it harms the field by alienating the youth ,the juvenile who are the future of math! Let them argue, let them get excited and let them be intrigued! We've all been there before. Isn't that right?

[@fredschneider7475]

PN is absolutely correct. 0^0 is undefined

[@fossilofmed5421]

This guy speak my mind but in a much much better way!! Totally agree dude 😊

[@DidarHaq-bg4qn]

A shiny and frictionless black board i ever seen .

[@glorrin]

if x^x^x^x is not defined on 0 then 0 cannot be an answer

[@domahidipeter6092]

I think you are right

[@satunnainenkatselija4478]

I don't even know what the left side means.

[@glorrin]

@@satunnainenkatselija4478 I wanted to reply to explain this but then I saw prime newton video today, he explain what the domain is. 0 is not part of the domain of x*x*x*x.

[@bunty2]

​@@satunnainenkatselija4478 L.H.S at x=0 becomes 0^0^0^0. now question arise 0^0^0^0 is equal to 1 or not? because according to desmos it is equal to 1. how?

[@freedomisaverb6750]

0⁰ is 1

[@bjrnlsriedelriedel7500]

Indeterminate forms: In certain scenarios, especially in calculus, 0^0 might arise as part of an indeterminate form. For example, if you evaluate the limit of expressions like x^x or 0^0, the outcome might depend on how both the base and the exponent approach zero. Some paths could yield 1, while others might result in different outcomes, like 0 or infinity. In these cases, 0^0 could be considered undefined or indeterminate, meaning no single value consistently applies. So basically If we assume 0^0=1 mathematics is consistent, but if it indeed is wrong then every formula with 0^0 would be undefined. These two conflicting rules create ambiguity, so if we decide that neither rule applies uniformly in all contexts, we might declare x^0=1 ( if x not equal to 0) Suggests that 0^0=1 While 0^x=0 (for any x>0) suggests that 0^0=0 Therefore we might even declare 0^0 as undefined @primenewtons

[@ibrahimmassy2753]

I enjoy so much see your videos. I would like to explain my point of view:The case x=0 is most the limit of the tretration of x^^4 but at x=0 is an inteterminate form. On the other hand, taking log base x on both sides twice you get x^x=3+(ln(ln(4))-ln(ln(x))/ln(x). This equation is transcendental so you can use Newton method and is easier than the previous one. Exploring the growing of x^x, 1/ln(x) and ln(ln(x))/ln(x) you can prove that x=2 is the only solution for x>0

[@estudarelegal8473]

Great job dude! Congrats from Brazil!

[@elephantdinosaur2284]

If you use power series then you already assume 0^0 = 1. If f(x) = sum a_n x^n, then most people don't have a problem saying that f(0) = a_0 which assumes 0^0 = 1. Strictly speaking 0^0 isn't defined but when it comes to notation as a shorthand it's 1.

[@proximitygaming8253]

You could argue that the exponent expression on the left of the logs cannot have any logarithm solutions, so therefore the ln parts must be equal and so ln(x) = ln(2)

[@hesaamsadeghi8954]

Hey man just a little note for desmos, it can also ‘solve’ the equations given above, if you input into desmos 4^x = 4^x^3 it will show you the intersections on a graph, when you do put it into desmos it shows that x=2 is the only solution!

[@Sigma.Infinity]

I like the use of logs for this. I got to the same final expression by repeatedly raising both sides by 1/x, but that was far more long-winded.

[@Kishorpatil-ir9sq]

I am from india and i love the way you explain sir❤❤

[@sunitagarg9560]

Let x tetration 4 be y, Therefore y=4 raise to power x raiseto power 3. Take log on both side with base 2 Solving you will get Log 2 with base 2=2(x) ^3 Now plot a graph with these points and you will get your answer. OR Lg y=0.602(x)^3 OR Ln y = 1.38(x) ^3

[@jamesharmon4994]

n! equals n * (n-1)... but if n = 0, this starts out as 0 * (anything). And we know that 0 times anything equals 0, so why does 0! = 1?? Answer, because it's defined to be 1. I propose the same for zero to the zero power equals one because any number to the zero power equals 1.

[@PrimeNewtons]

You make sense!

[@jamesharmon4994]

@@PrimeNewtons I have been impressed with your videos, and I take this praise with great honor.

[@isaiaholaru5013]

It would be neat, and only break 1 rule; 0^x = 0 Besides, Desmos would finally be correct.

[@platypi_otbs]

I think there may be applications where 0⁰ must be 1 for it to work but it's an impossible value similar to a negative distance or "sanity check" when a denominator would be 0.

[@jensdittrich8178]

In calculus there are two justifications to define 0^0. For that consider the limit x->0 for x>0 of two funtions f(x)=0^x and g(x)=x^0. One easily reaches f->0 and g->1, and indeed 0^0=0 and 0^0=1 are used as potential definitions. Consequently here, one would have to look at the limit process for the left and right handed side, and can define the continuation for x=0. Secondly, with your implication of ln x = ln 2 you already assumed x=2 thus not adding additional value. Maybe it is possible to show that there is no solution x > 2, then you'd have to check only 0\leq x\leq 2

[@holyshit922]

How good you are in calculating sums ? I can give you finite sum which is calculated by Wolfram Alpha incorrectly

[@PrimeNewtons]

Please email them. If wolfram is wrong, I bet I'd be sweating through them..

[@holyshit922]

@@PrimeNewtons yes i sent this sum

[@PrimeNewtons]

Thanks. I'll look into it. I'm sure I'll be learning many new things 😂

[@holyshit922]

@@PrimeNewtons have fun

[@holyshit922]

@@PrimeNewtons you wrote me that i can send you this sum and you even didn't read it In this sum n and m are fixed, k is running index and inside sum there is product of two binomial coefficients

[@Yourex19]

As soon as 4^x^3 is equal to 2^2x^3 and is an integer since x is searched inside integer set. Then the left side has only 2 as prime diviser. Therfore x^x^x^x has to be a power of 2. And then x should be equal to 2, elswhere if x had a divider p odd, p have to divide the left side which is a power of 2 : impossible ! Hence, it remain just to verify that match the equality.

[@surendrakverma555]

Very good. Thanks 🙏

[@mustafaseyitt]

Its already impossible to find negative solution. But you can try to find a solution in an interval by approaching. To look for integer, left side already going to the moon after 3 (literally). So 2 is okay

[@infofer1586]

Hi, sorry if i am wasting your time, but i had a problem with solving a math olympiad question from croatia (for the 1st grade of high school or 9th grade). I find it really interesting and when I saw the online solution I didn't understand it. If you would reply to this comment so I can send you the task that would be awsome!! Thank you in advance!!

[@infofer1586]

The question states: Find all natural numbers for n>2 for which there is a dividor d of the number n for which there is an equasion n=a³+d³ for which a is the smallest dividor of n larger than 1.

[@CosmoShinobi-h5w]

If you interpret the tetration as representing some combinatorial configuration then it would make sense that there is only 1 way to choose something even 0^0 (I.e how many ways to choose nothing given no options). However, I do agree with you that by itself it would not generalize well with some limit being taken.

[@daviesabraham177]

The Greatest ❤🔥

[@relshi]

0^0 being 1 makes much more sense in set theory. Since there is one function from the empty set to any set, the answer to 0^0 is 1. The only other options I can give are that multiple different limits lead to 0^0 being 1, and that an empty product defaults to the multiplicative identity just as the empty sum defaults to the additive identity, the multiplicative and additive identities being 1 and 0 respectively.

[@mantadele3291]

Great job and keep the hard work

[@IRanOutOfPhrases]

My take on 0^0: I'm....fine....with it being an undefined value. It competes with two contradictory rules: x^0 is always 1 0^x is always 0 So which rule takes precedence here? Its 'impossible' to say. But I feel like in other areas of math, we're comfortable with just "making up" values when the time calls for it. We do it when we say 0! = 1. That's a value we as people chose to assign to that function. It's not really dictated by hard logic, but rather a soft argument that reverses fractorials and observes a pattern of working from say 4! backwards to 1! and then 0!. In the case of these two rules regarding 0^0, I consider which one is more 'powerful' than the other. In the case of 0^x, it is only true for positive numbers. It does NOT work with negative numbers. As soon as you throw a negative at that rule, it blows up in your face. In the case of x^0, it works for ALL numbers. Except when x=0. Because we don't want to hurt the other rule's feelings, I guess. So the math degree in me agrees with you, that 0^0 has no value. But I can't blame someone for thinking 0^0 should abide by the x^0 rule. It works for literally every other single number.

[@Sigma.Infinity]

Wolfram Alpha does not include x=0 in the domain of y = x^(x^(x^x)) - 4^(x^3).

[@aryamanghosh692]

0 raised to 0 in inderterminate and that is why we use limits as its is in the neighbourhood of 0

[@diogochadudmilagres4533]

Making 4 = 2², in right-side we've 2^(2*x^3), and in left we've x^(x^x^x). By Comparison to (even 'cause are integer what we are looking for), x = 2, and x^x^x = 2*x^3, but x = 2, so: 2^(x^x^x) = 2^(2*x^3)--> exponents are equal: 2^(2^2) = 2*(2^3) --> 2^4 = 2*8 --> 16 = 16 (TRUE)

[@Wjuw-dq3rg]

BEST MATH TEACHER EVER

[@baidonchandipo2804]

Am from Zambia, I love your explanations

[@pdwag198]

Does the program knows to but a circle when a point is non-inclusive? It could be that the curves doen't cross at x=0, the line thickness make it look like it does.

[@stonedigit5030]

Pls Solve 1-cos4x divided by 1-cos6x

[@jamesalewis]

x ↑↑ 4 = 4 ^ x ^ 3 x^x^x^x = 4^x^3 Rule: a^b = u^v → a = u^(v/b) 4 = x^(x^x^x / x^3) = x^y for y = x^x^x * x^-3 y = x^x^x * x^-3 = x^(x^x-3) 4 = x^x^(x^x-3) = z^(z-3) for z = x^x 4 = z^(z-3) → z = 4 (4-3=1 is an obvious only solution for "z") z = x^x = 4 → x = 2 This doesn't symbolically prove it, but it gets much closer

[@misterj.a91]

You can do this identification since we want integers solutions.

[@dirklutz2818]

Look! 🎯 0.001^0.001=0.993116 and 0.000001^0.000001=0.9999862, so 0^0=1

[@belcavendishny]

the limit argument disagrees with itself, however. look! 0^0.001=0 0^0.000001=0 0^0.00000000000001=0 your conclusion is correct but it needs better support

[@dirklutz2818]

@@belcavendishny The function I used is x^x. When x goes to 0, the result is 1. The function 0^x is always zero.

[@belcavendishny]

@@dirklutz2818 "the function 0^x is always 0" well you seem to think 0^x is 1 when x=0

[@dirklutz2818]

@@belcavendishny The claim (not the proof) : 0^0 = (x-x)^(x-x) = (x-x)^x / (x-x)^x = 1

[@gerardvanwilgen9917]

As I see it, if we say, that 0 is outside of the domain of the function f(x) = ⁴x, which seems entirely reasonable, there is no intersection of both graphs at x = 0 because the graph of the first function does not exist there.

[@Yourex19]

@primenewtons What do you think about the above? Thanks for your job !

[@GOLDman4856]

0 works too

[@Reaper69377]

👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻

[@Dash14479]

Love from india ❤

[@PriestlyEdits]

where are u from man... wow pls where u from who are u

[@namo2299]

❤❤

[@MazenMohamed-zh1ze]

I have know new things such as tetration and penetration and more.. I don't know how can I even thank you for you explaination and all what you say😁☺

[@Hanible]

x=2 works! x^x^x^x > 4^x^3 for x >2 because x^x>4 and x>=3 and 0,1 doesn't work, so the only solution is 2.

[@abhishekrajamanthri3635]

Isn't there a way to solve this math problem without comparing both sides? sir

[@tamilselvanrascal5956]

🎉🎉🎉

[@MichaelOsei-pi3xm]

Please how do we solve 4^x=4x³-16

[@kaw57_]

0^0 = 1 because a^b = 1 * a * a * a ... } b times. by making b equal 0, you are multiplying 1 by a 0 times, which will only ever give 1

[@prabhatsingh3305]

Wow x = 0 was easy but x= 2 a Brain teaser . thanks 😊

[@CC--qn4gf]

Not sure how correct this is but I compared both sides of the equation and realized that on the right x was raised to a power and that on the left 2 was raised to the power 1, so then I let the powers equal one another and got x = 2.

[@spinothenoooob6050]

Due to e^0 Taylor series i guess then 0^0=1

[@wubbybubby1276]

wait how can you divide by x^3 on both sides if a solution to this problem is 0

[@LuisAlejandroVegaMerchan]

hand wavy argument. 0^0=1 under set theory

[@loai8854]

How can it be solved? x^x^x=9^9 thank you 🌹🌹

[@hanyahamba-AlKhaliq]

Sorry sir.. But if we solve x=2 .. The answer is not same.. is it? 2^4 = 16 4^2^3 is not 16 or am i missing something

[@rudrapatel6890]

0^0 is indeterminant form

[@PriestlyEdits]

no i think square of 0 would be one by comparison... given there are no negative squares in nature

[@PriestlyEdits]

cubic of -1 would be a positive integer i guess

[@WhiteGandalfs]

(3:57) "i'm going to write my ln(4) as..." is an auguration of the result - which i meanwhile call a "math jugglers trick", since such steps only make sense in the case that the author already knows the solution he is "pretending to develop" in advance. IF we admit that our reformation steps only make sense on pre-auguration of the result, the result could have been drastically easier tested in the original expression (which i did in this case before watching the video). Thus, finally: I'm not really impressed by this jugglers trick :D

[@ezxd5192]

Wolframalpha says 0^0 is undefined so that's enough evidence

[@jumpman8282]

Without having analyzed it properly, it appears that 𝑥 = −1 is the only negative integer that makes ⁴𝑥 real. However, 𝑥 = −1 is not a solution to the equation. So, if we assume that 𝑥 is a positive integer, then 𝑥^(𝑥^𝑥 − 3) is a positive integer. Thereby ln(4) ∕ ln(𝑥) must also be a positive integer. Let 𝑛 = ln(4) ∕ ln(𝑥) ⇒ 4 = 𝑥^𝑛 𝑥, 𝑛 ∈ ℕ ⇒ (𝑥, 𝑛) ∈ {(2, 2), (4, 1)} 𝑥 = 2 is a solution to the original equation, but 𝑥 = 4 is not. So, it appears that 𝑥 = 2 is the only integer solution (unless we accept 0⁰ = 1, or there exists some negative integer 𝑥 that also solves the equation).

[@Forei-m7p]

wow!

[@GOLDman4856]

0^0=1 doesn't need an argument it's definitionally true

[@nonnibaxtyar94]

can u come with me to my exam?

[@adarshgupta8215]

x = 2, simple hit and trial

[@yvanricardoecarrigomez]

2 Helen ex, ex Helen 4

[@sudedere3487]

legend

[@lateefkareem]

2

[@kuku-nunu3681]

Show me da graph

[@redroach401]

a new problem for you: x = x^((x^x)-3) solve for x

[@bowlteajuicesandlemon]

(x^(x^x))/(x^3) = x x^(x^x) = x^4 If x != 1, x^x = 4 e^(ln(x)*x) = ln(4) ln(x)*x = ln(4) ln(x)e^(ln(x)) = ln(4) ln(x) = W(ln(4)) x = e^(W(ln(4))) X can also be 1.

[@redroach401]

@@bowlteajuicesandlemon good job but can you derive a solution for 1

[@dinabandhusaha5520]

But what is the value of x

[@shilopkala9178]

Um I kinda assumed but the answer came out to be 2

[@Sigma.Infinity]

I'm wondering if there could be other solutions when x0. The LHS would be >0 if x^x^x was even. I am sure there aren't, but I don't know whether I could prove it. The Desmos graph doesn't show negative x for the LHS expression, presumably because the domain when x is negative consists of isolated values of x. Wolfram Alpha gives the domain of y = x^(x^(x^x)) - 4^(x^3) as: {x ∈ ℝ : (x≤-1 and x^(x^x), x^x ∈ ℤ and x ∈ ℤ) or x>0}, which indicates that there are negative values of x which give real values of y. The question is, are any of these values zero? If so, we have found another solution.

[@astellagaming]

:0