And then we say that it's not possible

Algebra Level 3

$x$ is a non-negative real number which satisfies the following equation.

$\huge x^{x^{x^{x^{.^{.^{x^{16}}}}}}}=16$

Then find the value of $x$ . Correct your answers to three decimal places.

Enter 666 if you come to the conclusion that no such $x$ exists.

This question is part of the set All-Zebra

The answer is 1.189.

3 solutions

Sabhrant Sachan
May 3, 2016

$\text {we are given : } \large x^{x^{x^{x^{.^{.^{.^{.^{16}}}}}}}}=16 \\ \implies x^{16}=16 \\ \implies \boxed{x=2^{\frac14}=1.189 }$

How do you know that the tower converges to 16?

Otto Bretscher - 5 years, 1 month ago

If it was wrong I wouldn't have posted the answer :P

Sabhrant Sachan - 5 years, 1 month ago

People do post wrong solutions every day ;)

200 people "solved" this one before it was discovered that the given answer was wrong.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – You have a good point Sir ! . I do realize that all functions have domain but i don't really know how to calculate the domain of this infinite tower . can you explain it to me in laymen terms ? i only know that it is related to Lambert W-Function. i researched about it few months ago, i am too young to understand anything related to it .

Sabhrant Sachan - 5 years, 1 month ago

@Sabhrant Sachan – The tower recursively defines a sequence $a_0=16, a_{n+1}=x^{a_n}$ ; we are looking for the limit of this sequence when $x=16^{1/16}$ . Things turn out quite simple: The sequence is constant 16, as we can show by induction: $a_{n+1}=x^{a_n}=16^{a_n/16}=16$ ....done!

Note that if we had an $x$ on top of the tower (or 15 or 17), then there would be no solution.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – I agree with you, that is why this is an exception .

Abhay Tiwari - 5 years, 1 month ago

@Otto Bretscher – The answer is correct because it lies in the interval $\large{ (e^{-e},e^{\frac{1}{e}})}$ but i don't know how to prove this result

Sabhrant Sachan - 5 years, 1 month ago

I think the solution and answer is incorrect. For any number greater than 1, if you tower it like that, it will constantly increase. Also, 16^1.189 = 27.0208284. EDIT: The exponent will need to pass 15. 1.189^15 will DECREASE.

Brian Wang - 5 years, 1 month ago

The power is taken the other way around, $1.189^{16}\approx 16$ . It all works out.

Otto Bretscher - 5 years, 1 month ago

You would need to pass 1.189^15 on the way right(not very good with exponents). That is about 10. It's going the wrong way.

Brian Wang - 5 years, 1 month ago

@Brian Wang – As you evaluate the tower from the top, step by step, you have $x^{16}=16, x^{x^{16}}=16$ etc, all the way down.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Then you would need to prove that it converges to 16. Anyways "at the very top" it would need to by x^x^x........^15 before it hit x^x^x^x^16. But 1.189^15 is around 10. Also, Excel says it converges to 1.239287. You know what this reminds me of? "Turtles all the way down"

Brian Wang - 5 years, 1 month ago

@Brian Wang – Below I'm proving that it converges to 16 since the sequence is constant 16. Don't mind Excel... trust the math! The equlibrium 16 is unstable, so, small round-off errors will push it down to the stable equilibrium.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – sqrt(2)^sqrt(2) to infinity equals 2.

Brian Wang - 5 years, 1 month ago

Sir, how's my solution? ;)

Abhay Tiwari - 5 years, 1 month ago

Brian, I have posted the solution please check it.

Abhay Tiwari - 5 years, 1 month ago

I agree with Brian. First of all there is no need to put 16 in exponent because it doesn't change the answer. Because if answer is 1.189 then 1.189^16 will decrease to 15.955. Now for next term again 1.189^15.955 will decrease, and so on. How-ever the decease in this will also decrease but won't remain same . and also it is for infinite times. so there will be a great change in value of 16. I think if there is no 16 in exponent then it can be easily explained that 1.189^1.189>1.189 and then again we will take exponent then it will again increase but the rate of increase won't increase because it somehow depends on log function and powers of 1.189. So it will give 16 atlast

Akash Shukla - 5 years, 1 month ago

@Akash Shukla – Akash, please check my solution at the last of this comment section.

Abhay Tiwari - 5 years, 1 month ago

@Akash Shukla – No, if the top number were anything but 16, we would not have a solution. Think about it!

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Yes , Sir I got this . I was taking it in a wrong way by considering 1.189 and not $2^1/4$

Akash Shukla - 5 years, 1 month ago

@Otto Bretscher – Because there are infinite x's, does is the 16 really there? For example, 0.0(repeating)1 = 0

Brian Wang - 5 years, 1 month ago

@Brian Wang – We are starting with the 16 ;)

Otto Bretscher - 5 years, 1 month ago

@Brian Wang – Here the answer is $2^1/4$ . so $(2^1/4)^16$ = $2^4)$ = 16 and so on .. At last by passing through infinity when we come to first position then also the power is 16 and base is $2^1/4$ which leads to 16. Now suppose if it is 24 in power then $(2^1/4)^24)$ = $2^6$ and so on and at last we wouldnot get $(2^1/4)^16$ =16. But I think there shouldnot be any number and there must be x only in exponent at infinity to satisfy the answer

Akash Shukla - 5 years, 1 month ago

@Akash Shukla – If you place an $x$ on top instead of 16 then there is no solution.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – initially , I also thought the same but when I calculated by calculator then during each steps the amount very slowly increases . firstly it increases at first position after decimal and then next after that position and so on. So I think that at infinity it can go upto 16. I have noted there that if base is same and power is changing then at less difference it will change very negligibly. If the base is not too big

Akash Shukla - 5 years, 1 month ago

@Akash Shukla – If you make the top entry $x=16^{1/16}\approx 1.19$ , then the tower will approach a limit of about 1.24.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Sir , is there any way to solve this ,as you have discussed above.

Akash Shukla - 5 years, 1 month ago

@Akash Shukla – Yes: we have to study the recursion $x_{n+1}=16^{x_n/16}$ for various initial values $x_0$ ; in this problem we are given the initial value $x_0=16$ , the top entry in the tower. First we need to find the fixed points, where $16^{x/16}=x$ .

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Sir , is it true that for any equation like $x^y=y$ , $1≤x<1.5$

Akash Shukla - 5 years, 1 month ago

@Akash Shukla – As long as $y\geq 1$ , yes

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Thank you, Sir.

Akash Shukla - 5 years, 1 month ago

Abhay Tiwari
May 3, 2016

Since we are not sure of the convergence of the power tower so to make it easy we should not touch the L.H.S. and let's take the R.H.S.

$\large 16={\sqrt[4]{2}}^{16}$

$\large 16={\sqrt[4]{2}}^{{\sqrt[4]{2}}^{16}}$

$\huge 16={\sqrt[4]{2}}^{{\sqrt[4]{2}}^{{\sqrt[4]{2}}^{16}}}$

$\huge16={\sqrt[4]{2}}^{{\sqrt[4]{2}}^{{\sqrt[4]{2}}^{.^{.^{.^{16}}}}}}$

$x=\sqrt[4]{2}=\boxed{1.189}$

This has nothing to do with the Lambert function and with $x\leq e^{1/e}$ ; the 16 on top changes everything. It's much simpler than that: The sequence converges to 16 because it remains constant 16 in every step, as you show.

Otto Bretscher - 5 years, 1 month ago

Ok, sir, actually I meant that the question can be proved by the other way also like you do through Lambert function, which will eventually lead to 1.189 as the answer as it does come under $e^{\frac{1}{e}}$ . This was my point that I wanted to say, and yes the way I have shown has nothing to do with that. Sorry for confusion.

Abhay Tiwari - 5 years, 1 month ago

I'm still confused. I don't see how the Lambert function could help in solving this problem. Can you show us the alternative solution?

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Sir, I am trying to do it, but Sir, when we know that the series is converging as I have shown in my solution, then can't we just write it as $x^{16}=16$ and then solve it. I did it by assuming $x^{16}=16$ only, but I am getting 1.239 as the answer. Please guide me!

$a=-\frac{W(- ln (\sqrt[16]{16}))}{ln (\sqrt[16]{16})}=1.239$

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – Yes, the answer is $x=16^{1/16}\approx 1.189$ . The Lambert function plays no role in this problem... let's not "overthink" it.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Sir, please post the solution on your recent problem on p ower tower, I am getting curious about it. I ticked one of the wrong answers. :P

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – I will wait a little, hoping that somebody else will post a solution.

Otto Bretscher - 5 years, 1 month ago

Arian Tashakkor
May 3, 2016

Let's assume : $A= x^{x^{x^{x^{.^{.^{.^{16}}}}}}}=16$ .Since there is a infinite number of $x$ 's in the power , it is safe to say that the following equation is true : $A = x^A = 16 \Rightarrow x^{16}=16$ which gives $x = 1.189$ up to three decimal places.

And then we say that it's not possible

The answer is 1.189.

3 solutions

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