Put y = x^8. Then you have two equations: x^y = 8 (given), y = x^8 (our substitution). Take the natural log of both sides of each equation, obtaining: y*ln(x) = ln(8), ln(y) = 8*ln(x). Substitute the former into the latter: ln(y) = 8*ln(x) = 8*[ln(8)/y]. Rearrange the latter to get y*ln(y) = 8*ln(8). Thus, y = 8. Now get rid of y (in the original substitution): x^8 = 8. Taking the 8th root yields x = 8^(1/8).
@matiasnicolasmartostano4410
9 ай бұрын
hey yoohoo great solution 🥳
@LightYagamiK
9 ай бұрын
This deserves a pin
@gajsingh1117
9 ай бұрын
Good one buddy
@plygem6927
9 ай бұрын
X^X^8 = 8 (X^X^8)^8=8^8 (X^8)^(X^8) = 8^8 [(a^m)^n=(a^n)^m] Now both side of equation has its power and base as same x^8 = 8 x = 8^(1/8) I tried this without use of complex functions
@matiasnicolasmartostano4410
9 ай бұрын
you have instagram?
@goofy_maxima
9 ай бұрын
Answer for questions at the end 1) x=3^⅓ 2)x=pi^(1/pi) As it can be generalized as x=a^(1/a) Hence it becomes easy to solve! Edit: I put exponents in brackets as suggested by a reply.
@maxgeorge-ruiz5449
8 ай бұрын
can't u just simplify to x=3^1/3=1?? as well as with the other ones
@maha1769
7 ай бұрын
@@maxgeorge-ruiz5449 3^1/3 does not equal 1.
@amrhesham1409
Ай бұрын
Square root 1.53272
@hmmm6200
Ай бұрын
please put everything that belongs in the exponent in parentheses the 1/3 symbol youre using obviously works, but for pi^1/pi one might think thats just pi/pi so 1 same with a^1/a
@goofy_maxima
Ай бұрын
@@hmmm6200 thanks! Did it.
@Mr_Boness
8 ай бұрын
I used the lambert w function. 1) x^x⁸ = 8 ln both sides 2) x⁸lnx = ln8 Multiply both sides by 8 3) 8x⁸lnx = 8ln8 Simplify 4) x⁸lnx⁸ = 8ln8 Rewrite x⁸ as e^lnx⁸ 5) lnx⁸ • e^lnx⁸ = 8ln8 Lambert w function 6) lnx⁸ = w(8ln8) Make x the subject 7) x = ⁸√(e^w(8ln8)) This is the exact answer of what x is equal to.
@ishansh0077
6 ай бұрын
Which branch is positive real
@shouryasharma8186
4 ай бұрын
Also, ln8 = 3ln2 so u can simplify that too
@samirbehera8708
2 ай бұрын
Here I Got 1/e^1/8
@MrZauberwuerfel
9 ай бұрын
I got to the solution with a completely different approach: x^(x^8) = 8 = 8^1 = 8^(a/a) = (8^(1/a))^a Then choose a such that a = x^8. For the above equation to be true we obtain x = 8^(1/a) From a = x^8 we obtain x = a^(1/8) Therefore x = a^(1/8) = 8^(1/a). We can now easily detect, that a=8 is a solution. Therefore x = 8^(1/8)
@bobjoe8182
9 ай бұрын
you're missing x=-2^(3/8) as well as infinitely many imaginary solutions, namely x=e^(1/8 * W(8(2ipin+ln(8)))), where n can be any integer, i is sqrt(-1), W(z) is the product log function, and 2ipin is not equal to -ln(8). not to mention x=e^(1/8 * W_1(8(2ipin+ln(8)))), where the same constraints are true and W_k(z) is the analytic continuation of the product log function. even x=e^(1/8 * W_2(8(2ipin+ln(8)))) works with the same constraints. you're infinitely many solutions short of the answer and did not even provide all the real solutions. so much for "easily detecting" the answer, i suppose :)
@MrZauberwuerfel
9 ай бұрын
@@bobjoe8182impressive comment, but in the beginning of the video it was stated, that x must be a positive real number
@Brid727
9 ай бұрын
If x^x^8 = 8 Then, (x^x^8)^8 = 8^8 We know that, (a^m)^n = a^mn So, x^8x^8 = 8^8 [Here x^8 itself is an exponent so 8 times that is 8x^8] So, (x^8)^(x^8) = 8^8 We know, a^a = n^n given that n>1 So, x^8 = 8 Therefore, x=8th root of 8 If you want to write it another way, it can be written as x=2^(3/8)
@anveshsawarn203
9 ай бұрын
x^y=8... eqn 1 where y=x^8.... eqn 2 Now we can say x=y^(1/8) from eqn 2 substituting value of x in eqn 1 we get y^(y^1/8)=8 which is equal to y^y=8^8 y=8 and x=y^1/8=8^1/8.
@christsdisciple3105
9 ай бұрын
0:36-1:41 this is different from how I was taught all the way through my senior year in 2012. I didn't want to say anything until I looked it up, but everywhere I looked said that, when you take a power to a power, you are supposed to multiply them and then take the base to that power, which will yield 64 rather than 512 in your example. So... not sure if that changes things.
@bosstowndynamics5488
9 ай бұрын
This is partly going to be one of those semantic issues like the deliberately ambiguous divide and multiply questions that go around from time to time, but what I was taught was that you multiply out exponents if the base term and an exponent are inside a bracket and you raise the whole thing to another power - this is a mathematically equivalent shortcut for the second one Presh described anyway. Since you still need some form of notation to describe x^(y^z) it makes sense that option one is used that way, since the stacked exponential terms with parentheses are messier than just adding parentheses to (x^y)^z
@utkarshgautam7896
3 ай бұрын
In the attempt 1 after getting x^x^x^x^.... =8, you could notice that since the number of x are infinite, the tower of all those infinite x is equal to 8 which gives us: x^x^x^x^.... =8 x^8=8 x=8^(1/8)
@flash24g
9 ай бұрын
When you have the infinite power tower, you can substitute the equation into itself: x^x^x^... = 8 => x^(x^x^x^...) = 8 => x^8 = 8.
@flash24g
9 ай бұрын
I've just realised a flaw wtih this argument: I omitted to consider the possibility that the expression has different values for odd and even counts of x in the power tower, in which case the value of the infinite power tower will be undefined.
@Calmerism
9 ай бұрын
I'm sure there is some weird explanation why this technically doesn't work, but why would'nt the equation at 6:27 give you a result? From that infinite tower follows x^8 = 8 and you have the solution for x being the 8th root of 8.
@alexmch-s4651
9 ай бұрын
The infinite sequence for this problem should be the limit of x^(x^(.....(x^8))....)). For x=8^(1/8), this converges to 8. In fact, it just becomes the constant sequence 8,8,8.........
@jackdisbury1047
4 ай бұрын
The infinite x to the power of x can be helpful as you substitute 8 into the exponent x^x^... becomes x^8 and then solve for x since that will equal 8
@mityakiselev
9 ай бұрын
Wow, thanks m8! I'm trying to go for teaching mathematics as a major and I'm usually stuck with typical middle school topics. This feels more like a tricky test or olympiad question, and I have to admit I didn't know how to approach it at first. Like, I know you're supposed to play around with algebraic puzzles to simplify them, but I have little to no experience with "nesting" functions.
@divisix024
9 ай бұрын
In this case the motivation is that most of the times equations of the form “unknown to the power of unknown (may be different from the previous one) equals some constant” is hard to solve, except if it is of the for, x^x=a^a, where x is the unknown and a is a given constant.
@deepfriedlettuce.
9 ай бұрын
i have ateacher who teacges midde schoo l math
@ELLIPSIS797
9 ай бұрын
Hey preshtalwalkar has anyone has yet proved that 3x+1 maths problem
@vgtcross
9 ай бұрын
No
@BeefinOut
9 ай бұрын
It's unsolved, and there's a good chance it's simply unsolvable. Veratasium has a good video on it
@glitchquitch
9 ай бұрын
Is that his name? Which country is it from?
@finnwilde
9 ай бұрын
@@glitchquitch*he
@glitchquitch
9 ай бұрын
@@finnwilde "it" was referring to the name lol
@dskinner6263
9 ай бұрын
I appreciate the clarity and variety of your videos - thank you! I watched a video today on another channel about algebraic identities and it completely confused me.
@LimeLogan
9 ай бұрын
5:42 You can actually proceed from here though with a solution. What you can do is, since removing the first "x" is equal to this equation, you can do x^8=8, then you can clearly see that x=8th root of 8.
@usrnewxnew5227
9 ай бұрын
Hey presh, are you of indian descent by any chance? Your surname sounds marathi (its a spoken language in india here, in the state of maharashtra)
@philipframpton9428
9 ай бұрын
Attempt 1 could have worked. After trending to an infinite series, you can then substitute the inifinity of x^x.... terms that equal 8 back into the same equation as the initial index, thus giving x^8 = 8 and x=8^⅛ .
@manaiakilani3837
9 ай бұрын
I consider x as a real number. x^(x^8) = 8 => (x^8)*ln(x) = ln(8) => ln(x) * exp(8*ln(x)) = ln(8) We multiply by 8 from both sides : 8*ln(x) * exp(8*ln(x)) = 8*ln(8) On the right side, we notice 8 = exp(ln(8)) to get : 8*ln(x) * exp(8*ln(x)) = ln(8)*exp(ln(8)) Then we use the Lambert Function ( W( X * exp(X) ) = X ): - On the left side(with X = 8*ln(x)) : W(8*ln(x) * exp(8*ln(x))) = 8*ln(x), - On the right side (with X = ln(8)): W(ln(8)*exp(ln(8))) = ln(8) So back to the equation, we get : 8*ln(x) = ln(8) => ln(x^8) = ln(8) => x^8 = 8 Finally the solution : x = 8^(1/8) = sqrt(sqrt(sqrt(8)))
@archimidis
9 ай бұрын
Blackpenredpen, yay!
@mohannad_139
9 ай бұрын
i got x = e^(W(8×ln8)/8) which is right but could not figure out it's basically ⁸√8
@user-xr1xw4oo9i
9 ай бұрын
Tell me how 8^1^8 isn't a better solution?
@user-xr1xw4oo9i
9 ай бұрын
I,m confused a little 😢
@plygem6927
9 ай бұрын
X^X^8 = 8 (X^X^8)^8=8^8 (X^8)^(X^8) = 8^8 [(a^m)^n=(a^n)^m] Now both side of equation has its power and base as same x^8 = 8 x = 8^(1/8) I tried this without use of complex functions
@abbuthurab9697
9 ай бұрын
I think another method may be 8^(1/8)^8^(1/8)^8=8 by using (a^m)^n=a^mn if x=8^(1/8) it make both sides equal. 8^(1/8)=1.2968395547(approximately).
@parvaraval4339
9 ай бұрын
x=2^y, by visualisation y=3/8, hence x=8^1/8
@Musterkartoffel
9 ай бұрын
I'm still waiting for the day when you finally do something with the lambert w function
@pierrecurie
9 ай бұрын
Somebody else got the answer as x = e^(W(8ln8)/8), where W is any of the branches.
@redroach401
9 ай бұрын
Here's how i solved it: take ln of both sides, multiply boths sides by 8. Take w of both sides and you get ln(x^8) = ln(8). Make evrything base e to cancel out ln and you get x^8=8. Take 8th root of both sides. X = 8^1/8
@davidbrisbane7206
9 ай бұрын
It turns out that if 1/e < a < 1, then there are two real solutions to x^(x^a) = a. Example. Let a = 1/2 then the two solutions for a are ... x = 1/4, which is x = a^(1/a), but x = 0.0625 is also a solution. Also, if a is an even integer then x = -1 * a^(1/a) is also solution. Of course, there are also an infinite number of complex solutions.
@marcasrealaccount
9 ай бұрын
The way I decided to solve it was by taking x^(x^8) = 8 and turning that into e^(8 ln x) ln x = ln 8, from there I chose a = ln x by which I could now use the lambert omega function or power logarithm which for the form axe^(bx) = y has the solution W(by/a)/b = x, filling it in gives us a = W(8ln8)/8, and because W(x ln x) = ln x, which gives us the solution ln x = (ln 8) / 8, which is x = 8^(1/8) or 2^(3/8).
@phalanxutsav6300
9 ай бұрын
in these types of question to get a general soln we can also use lambert W function i used it to get x=(e^W(8*ln8))^1/8 although it is very complex it gives us a graph for analysis
@mikebrau5354
9 ай бұрын
The infinite power attempt and graph is wrong because Presh forgot the ^8 at the end. If you include that, then x^8=8 is on the graph as expected and proven later.
@77elite9
9 ай бұрын
What happens when you leave the 8 at the end? You substitute it again! So it is infinite.
@mousemaps9168
9 ай бұрын
@@77elite9 It's infinite but it still needs an 8 at the... er... end.
@77elite9
9 ай бұрын
@@mousemaps9168 the problem is that there is no end because it is infinite. Case closed.
@mousemaps9168
9 ай бұрын
@@77elite9 what about the sequence that is the reverse of the digits of pi? That's also an infinite sequence, but it ends at 3. Case reopened?
@77elite9
9 ай бұрын
@@mousemaps9168 case closed, it’s disputed what’s correct.
@justpaulo
9 ай бұрын
Thanks for the Math explanation and for the Excel tutorial too.
@_Dearex_
9 ай бұрын
I've tried sqr(2) and got quite close, playing around with it I landed on sqr(2)^0.75 as solution 😄
@VitaliyCD
9 ай бұрын
That IS really close!! Almost like they could be equal... 🤔
@wearron
9 ай бұрын
thats because sqrt(2) = 2^(1/2), and 0.75= 3/4. then when you raise sqrt(2) to the 3/4, you get 2^(3/8) which is (2^3)^(1/8) = 8^(1/8) which is exactly what the answer is!
@heco.
9 ай бұрын
I used a different method to solve this problem. x^(x^8) = 8 ln(x^(x^8)) = ln(8) (x^8)ln(x) = ln(8) ln(x)*e^(ln(x^8)) = ln(8) [e^ln(x) == x] ln(x)*e^(8*ln(x)) = ln(8) 8*ln(x)*e^(8*ln(x)) = 8*ln(8) [multiply both side by 8] W(8*ln(x)*e^(8*ln(x))) = W(8*ln(8)) [use the LambertW function] 8*ln(x) = W(ln(8)*e^(ln(8))) [W(x*e^x) == x] 8*ln(x) = ln(8) ln(x) = (1/8)ln(8) ln(x) = ln(8^(1/8)) x = 8^(1/8) or x = 2^(3/8)
@mikebrau5354
9 ай бұрын
That's a complicated way of raising to the 8th power and adjusting order of exponentiation, as Presh did. It's hidden under the logarithms.
@CheckmateSurvivor
9 ай бұрын
Just published the most difficult puzzle in the world. No computer in the world should be able to solve it. Why don't you give it a try?
@necrolord1920
9 ай бұрын
There are also infinitely many complex solutions which can be defined in terms of the Lambert W function .
@timothyodonnell8591
9 ай бұрын
I love the bonus Excel tip!
@Mathematical-Mind
9 ай бұрын
Another method which is also worth considering: x^x^8=8 and (x^8)^(x^8)=8^8. From this we can get that x^8=8. Now we can see x=8^(1/8).
@flash24g
9 ай бұрын
I'm guessing you raised both sides to the 8th power, giving (x^x^8)^8 = x^(8 x^8) = (x^8)^(x^8)? This shows that x^8 = 8 is a possible way to satisfy the equation, but it doesn't show that it's the only way.
@Fouriersuir
9 ай бұрын
So if you look back in the first few moments of the video, When there are no brackets, you cannot evaluate a^b^c as a^c^b Because you evaluate the powers from top to bottom, What you did is raised both sides to the 8 but you cannot move them around as there are no brackets and can only evaluate them top to bottom. I believe it’s just a coincidence that it happened to be the answer but it’s just not good math unfortunately. Good idea tho.
@flash24g
9 ай бұрын
@Fouriersuir I haven't tried to evaluate a^b^c as a^c^b, and I'm not sure that Victor has either. My interpretation of his comment is based on (a^b)^c = a^(bc). Rereading my comment, I see that it's a bit confusing. My three expressions separated by '=' are successive rearrangements of the LHS after raising to the 8th power, then equated with the RHS becoming 8^8.
@sadnanjuhib
9 ай бұрын
Nice pun at 1:58
@shashwatgupta4613
9 ай бұрын
X^x^8 = 8 Simply rais3 both sides to power 8 (X^8)^(x^8) = 8^8 X^8 = 8 Also by graphs; we can find that there is only 1 solution
@plygem6927
9 ай бұрын
X^X^8 = 8 (X^X^8)^8=8^8 (X^8)^(X^8) = 8^8 [(a^m)^n=(a^n)^m] Now both side of equation has its power and base as same x^8 = 8 x = 8^(1/8)
@roddywishart4613
9 ай бұрын
Thanks for the excel tip, that will come in very handy one day and I'll think of you.
@JAirra-z1d
Ай бұрын
Yes, the function of 8 is 8 with the notion that you see that a parentheses should be around x and it still = 8 it's just the logic of dividing 8 about 3 times.
@shaurryabaheti
9 ай бұрын
what if we raise both sides to 8... (x^(x^8))^8 = 8^8 we rearrange exponents, and replace 8 with 2^3 on the right hand side we get (x^8)^(x^8) = (2^3)^(2^3) we get x^8 = 2^3 so x = 2^(3/8) or x = 8th root of 8
@ToyKeeper
9 ай бұрын
I just worked this out from the thumbnail, and what I found is wild. No matter how many "x raised to the" are placed at the beginning, the answer never changes. And it works for any number, not just 8. X is the Nth root of N, and X raised to the X raised to the X raised to the X infinite times, raised to the N... produces N.
@chaosredefined3834
9 ай бұрын
Let u = x^8. So, x^u = 8. Take ln of both statements ln u = 8 ln x u ln x = ln 8 Multiply the first of those equations by u. Multiply the second one by 8 u ln u = 8 u ln x 8 u ln x = 8 ln 8 Notice that the RHS of the first is the same as the LHS of the second. So, u ln u = 8 ln 8. As f(u) = u is positive and monotonically increasing over the domain (0, inf), and f(u) = ln u is monotonically increasing over the same domain, the product (u ln u) is monotonically increasing over the domain. Furthermore, because the product isn't defined outside the domain, that means that u ln u is injective. That is, if a ln a = b ln b, then a = b. With that in mind, given that u ln u = 8 ln 8, we have u = 8, and therefore 8 = x^8, or x = 8^(1/8).
@brickie9816
9 ай бұрын
just raise both sides to the 8th power and we will have (x^8)^(x^8) = 8^8 so x^8 = 8 idk how to prove there are no other solutions tho
@pwmiles56
9 ай бұрын
Yes good approach. The function x^x is less than 1 for 0=1 there is exactly one solution for positive x. In this case a=8^8 which is greater than 1 so there is only one solution for x^8, namely 8.
@brickie9816
9 ай бұрын
@@pwmiles56 thank you for clarification!
@huzefa6421
9 ай бұрын
That fun thing was actually my solution thinking that there is no solution 😂😅
@drawforge3920
9 ай бұрын
The spreadsheet one 😂😂😂
@gdmathguy
9 ай бұрын
INTERESTING trick. Gonna be using that more often
@shooshx
9 ай бұрын
But what was the mistake with the first try?
@ZevVeli
9 ай бұрын
So the other trick I realized. if x^x^a=a and therefore x^a=a then if you take the natural log of both sides you get ln(x^a)=ln(a) using the power rule of logs you get a*ln(x)=ln(a) rearrange and you get ln(x)=ln(a)/a. Solve for x and you get that x=e^(ln(a)/a) So for the bonus solutions when a=3 then x is about 1.442 and when a=PI then x is about 1.440
@t3h_m0nk3y
9 ай бұрын
4:35 but why stop there? I love it.
@kristofferkejser7594
9 ай бұрын
Congrats on 3 million subs!
@mr.d8747
9 ай бұрын
*This equation can also be solved using the Lambert W function, the inverse function of x•e^x. [W(x•e^x)=x]* *x^x^8 = 8 /ln* *x^8 • ln x = ln 8* *(e^ln x)^8 • ln x = ln 8 /•8* *8 • (e^8 • ln x) • ln x = 8 • ln 8* *8•ln(x) • e^(8•ln(x)) = ln(8) • e^ln(8) / W* *8•ln(x) = ln(8) /exp* *(e^ln x)^8 = e^ln(8)* *x^8 = 8 /( )^⅛* *x = 8^⅛ = ⁸√8*
@adityakumarstuffdoer9591
9 ай бұрын
Answer is approximately [sqrt(2) - 0.13] by intuition
@_sahildahat_
9 ай бұрын
for the practice questions, please check my answer: 3^(1/3) & pi(1/pi)
@kevinshenshiqi8210
8 ай бұрын
Answer to the two questions: Question 1: if x^(x^3) = 3, find the value of x. Solution: x = 3^(1/3), which is approximately equal to 1.44224957031 Question 2: if x^(x^(x^π)), find the value of x. Solution: x = π^(1/π), which is approximately equal to 1.43961949585
@user-yg97f5hfvh
9 ай бұрын
Plug in x=2^(3/8) and that's the answer. I put 3 in the numerator cuz 8=2³
@anandkishore3898
9 ай бұрын
Easy one this. I firstly thought of polynomial root way, but then switched to logarithms, and that's where I sensibly figured out to raise both sides to 8th power. Real root by comparison : 8^(1/8) = 2^(3/8) ☺️
@captainpolar2343
9 ай бұрын
put x^(x^y)=y in a graphing calculator it looks cool
@zzstoner
9 ай бұрын
You are stranded on a deserted island. You're allowed to bring only ONE item with you. What do you choose? .... Excel, of course!!!
@GalaxyJayden2266
9 ай бұрын
As a second grader, what are you on
@moixemi
17 күн бұрын
you’re too far from home💀
@GalaxyJayden2266
17 күн бұрын
@@moixemitime to fail life🎉
@Keldor314
8 ай бұрын
"Let's just say you're stuck on a desert island, and you only have access to a spreadsheet..." What is this? StandUpMaths??
@jayktomaszewski8738
4 ай бұрын
cubert(3) & pithrt(pi)
@jaimintankariya6955
9 ай бұрын
What about the natural log when we expans the( a^b)^c
@shajiabbas5342
9 ай бұрын
General formula is nth root of the power where n is the power
@naveenramkumar6123
9 ай бұрын
Shouldn't -(8)^(1/8) also be a solution or am i just mistaken?
@davidhowe6905
9 ай бұрын
I initially thought the same, but early in the video (but not the thumbnail) 0:15 he specifies positive real values.
@naveenramkumar6123
9 ай бұрын
@@davidhowe6905 aaah good catch! i missed this
@ZombieKillerThe
8 ай бұрын
there are 8 solutions since x^8=8 but only 2 real numbers, 1 positive and 1 negative. The other 6 are imaginary/complex numbers
@adellutri
9 ай бұрын
Wait a minute, at 7:46 how did you jump from x^y = 8 --> y^1/8=x ? I don't get that step.
@ahojg
9 ай бұрын
x^x^a = b x = exp(W(a*ln(b))/a) In that special case of a = b: very nice piece of work :)
@mathmachine4266
9 ай бұрын
x^x⁸ = 8 x⁸ln(x) = ln(8) x⁸ln(x⁸)/8 = ln(8) x⁸ln(x⁸) = 8ln(8) The Lambert W function is defined as the inverse of the function xe^x. W(xe^x) = x. Note, for all values of x, there are countably infinitely many solutions to this equation (almost all of which are complex), therefore this function is sometimes also specified with an integer to represent the branch number. The principle branch is branch 0. ln(x⁸)e^ln(x⁸) = 8ln(8) ln(x⁸) = W(8ln(8)) x⁸ = e^W(8ln(8)) x = e^(W(8ln(8))/8 + 2πNi/8), where N is an integer According to Wolfram Alpha, when evaluating the 0th branch and N=0, we get 2^(3/8). Indeed, (2^(3/8))^(2^(3/8))^8 = (2^(3/8))^2³ = (2^(3/8))^8 = 2^3 = 8. However, that's just one solution.
@ELLIPSIS797
9 ай бұрын
Pi)^1/pi and same way 3 one .
@creounity
9 ай бұрын
Interesting approach with excel's goal seek. Thank you!
@Minefreak022
9 ай бұрын
the stacking X's are crooked..... i CANT unsee it. it hurts.
@attackhelicopteriscool
4 ай бұрын
i realized that x^8 = 8 is the same to x^(x^8) because if you substitute 8 in x^8 with x^8 you get the same equation as x^(x^8) = 8. so the positive real answer is 8^(1/8) edit: i found the answer before the video even solved it haha
@adityajha2889
9 ай бұрын
Raise both sides to the oower 8 And let x^8=y ( did this so that i could present my method in this comment list) => y^y = 8⁸ =>y=8 =>x⁸=8 =>x=8^(1/8) =1.294
@dennisd7
9 ай бұрын
Finally some excel gang representation! I feel seen. :D Lovely.
@jamiebonsall430
9 ай бұрын
Petition for Presh Talwalkar to solve all future problems in Microsoft Excel
@Greed24
9 ай бұрын
Man I hate when I am stuck on a desert island with only a spreadsheet (I’m forced to numerically solve x^x^8=8)
@tobyfitzpatrick3914
9 ай бұрын
It's true what they say about pi appearing everywhere..!
@takshraval.
9 ай бұрын
Then what was the mistake when we substituted it infinite times?
@renatorossi6237
4 ай бұрын
Ottima esposizione
@user-gd2ir1gg6v
9 ай бұрын
We need 10 hours version To the power of x to the power of x to the power of x
@Doeniz1
9 ай бұрын
I didn't get where the mistake was, when he concluded that 8^(1/8) also solves the infinite power tower equation.
@Neodynium.the_permanent_magnet
9 ай бұрын
And why x^x^x^x^.... = 8 has no solution as we just follow the same replacement trick (infinitely)?
@justafanofalphabetlore
9 ай бұрын
I know the rule of infinite tetration . So z^8=8 for obtaining eighth root of 8.
@c-6969
9 ай бұрын
Congrats on 3m!
@pranjalpathak4498
8 ай бұрын
Presh your decisions I'm mind your walker
@MUJAHID96414
9 ай бұрын
Hey, belive me when I saw the thumbnail, I guessed correct.
@Geoff_G
9 ай бұрын
I solved it the easiest way. I just watched the video.
@GottfriedLeibnizYT
9 ай бұрын
So why is the infinite exponents false?
@cannot-handle-handles
9 ай бұрын
My guess: It's wrong because x^x^x^x^… is interpreted as the limit of the sequence x, x^x, x^x^x, and so on, but not the limit of the original sequence x^8, x^x^8, x^x^x^8, and so on. So, if you plug in 8^(1/8) for x, then x, x^x, x^x^x, … will converge, but not to 8. There's also a part of the video showing al values that x^x^x^x^… can take, and 8 is not one of them.
@kaizoisevil
9 ай бұрын
Let x = 2^m. Then we have 2^(m*2^(8m)) = 8. Only real solution for the exponent is 3, so we have m*2^(8m) = 3. Multiply that by 8 and we get (8m)*2^(8m) = 24. Let n = 8m. Then n*2^n = 24. For n < 0, the function n*2^n is negative. For n >0, the function is positive and strictly increasing, since both n and 2^n are strictly increasing. So there is only one real solution where n*2^n = 24. Looking at factors of 24, we can plug in 3, and now we know that is the only solution. So n = 3, which means m = 3/8, which means x = 2^(3/8).
@justafanofalphabetlore
9 ай бұрын
This can also works for decimal iterate for exponential function.
@Ben_Ja_Man
9 ай бұрын
The answer at the end is pi=3
@ELLIPSIS797
9 ай бұрын
It was a good ques i answered easily
@Rishabh_Joshi_
9 ай бұрын
Interesting!
@krispyking2450
9 ай бұрын
how do u do the excel trick on google sheets or excel mobile
@nothingtoseehere2189
9 ай бұрын
What’s wrong with attempt 1 that’s what I did and then you realise if you put x to the power of that infinite power it’s just the orignal power which is the same as x^8
@TanmaY_TalK
9 ай бұрын
Bro just give me reason to use Excel
@DarkRaiderDoesNotExist
9 ай бұрын
cool video!
@sahildas.
9 ай бұрын
I graphed this on desmos and x is approximately 0.652
@welcomb
9 ай бұрын
Looks like a fixed point combinator
@AayushLath
9 ай бұрын
If x^x^x till infinity = y then y=x^y. So 8 = x^8, so x = 8^1/8. The attempt 1 demonstration of the video is therefore wrong.
@mikebrau5354
9 ай бұрын
That's "18th Century non rigorous mathematics". If x=8^(1/8), then the tower of exponentials doesn't converge to 8. The ^8 is critical. That's why Presh's graph is wrong (which he should have known because he already computed the answer in Excel!) If you include that, then you have x^(x^... ^8)=8, or x^8=8, and the numerical approximation succeeds.
@AayushLath
9 ай бұрын
@@mikebrau5354 for a maximum of e^1/e the tower will always converge and not diverge and e^1/e > 8^1/8. Please check before making a fool of yourself next time!
@mike1024.
9 ай бұрын
Hey that was pretty clever! It reminds me of Banach's Fixed Point theorem.
@mittarimato8994
9 ай бұрын
Good video. These pillar equations are not that common in the school. Also, the visuals were nice.
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