Infinite Power!

Algebra Level 1

$\huge{ x^{x^{x^{x^{x^{.^{.^.}}}}}}=2}$

Find the value of $x$ that satisfies the equation above.

Cannot calculate Infinite 2

\sqrt{2}

9 solutions

Satvik Golechha
Oct 10, 2014

By the Tower Rule , we need to evaluate the indices from the right to the left. So $a^{m^n} = a^{(m^n)}$ is correct while $a^{m^n} = (a^m)^n$ is incorrect.

$\huge x^{x^{x^{x^{x^{...}}}}}=2 \quad\Rightarrow\quad x^{\left(x^{x^{x^{x^{...}}}}\right)}=2$

Since the expression inside the bracket is simply equals to $\large x^{x^{x^{x^{...}}}}$ , which equals to 2 as provided in the question, so the entire expression inside the bracket equals to 2.

Thus we have $x^2 = 2\Rightarrow x = \pm \sqrt 2$ . However, because the function $y = \large \left(x^{x^{x^{x^{...}}}}\right)$ is only defined for $x> 0$ , then $x > 0$ only. This means that $x = -\sqrt2$ is not a valid solution. Hence, $x = \sqrt2$ only.

${ x }^{ { x }^{ { x }... } }=4$

${ x }^{ ({ x }^{ { x }..) } }=4$

${ x }^{ 4 }=4$

$x\quad =\sqrt [ 4 ]{ 4 }$

$x\quad =\sqrt { 2 }$ ?

@John Muradeli

Guilherme Naziozeno - 6 years, 8 months ago

@Satvik Golechha

No. This ${x}^{y}=y$ only works when $x<{e}^{\frac{1}{e}}$ Since 4 is more than ${e}^{\frac{1}{e}}$ , it doesn't work. (Maximum value $y$ can take is $e$ )

By doing the above, you have just proved that ( $2=4$ ) See Question

Julian Poon - 6 years, 7 months ago

Yeah, I had eventually worked that mistake out. Thanks @Julian Poon

Satvik Golechha - 6 years, 7 months ago

@Guilherme Naziozeno That was a cool observation! I guess it's because $2$ and $4$ follow the relation $a^b=b^a$ . Also, it means that there are two values of $\huge{ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}}}}$

Satvik Golechha - 6 years, 8 months ago

Now, the last sentence may tick the math community upside down. We've discovered a function that fails the vertical line test!!! WOW??

John M. - 6 years, 8 months ago

@John M. – Actually, that wasn't a function, anymore than x^2 = 2 is.

Whitney Clark - 6 years, 8 months ago

@Whitney Clark – An equation is not a function, but a representation of one variable in terms of another is. And so we can have a function

$x^{x^{x...}}=y$ ,

or,

$y=f(x)$ , $f(x)=x^{x^{x...}}$ .

And so, $f(\sqrt{2})=2,4 \rightarrow$ we got ourselves a function that violates the vertical line test.

John M. - 6 years, 8 months ago

Why there are two values of the above function.

Ronak Agarwal - 6 years, 8 months ago

@Ronak Agarwal – its because both 2 and 4 will give same ans for the given function

kamalakanta jena - 6 years, 6 months ago

So what's the other value?

John M. - 6 years, 8 months ago

I think what this proves is that if there is a solution to this equation, then it is a square root of two. Did somebody prove that there actually exists a solution?

Rene Siles - 6 years, 7 months ago

Well what I did was the following: using the limit when n tends to infinity of x^(x^(x...)) with a number n of x's, it equals when there is a number n-1 of x's, because it makes no difference whether n or n+k with n going to infinity in an expression, k a real number. So by that it can be proved that if (we are not analysing if the expression is true, for it is priorly given) x^(x^(x^...))=b=2, with infinite x's, then x^b=2, and then x^2=2 and x=+-sqrt(2). As the series would give a nule-negative number for x<0, we accept that x>0 and then x=sqrt(2). And that is the only real solution for this equation.

Luiz Felipe - 6 years, 7 months ago

Well, if we find a solution and it fits the original system of equations, in this case one equations, then there is no contradiction in this found solution to be a solution of the system, right? Many a time it is easier and more useful to find the solution and test it than to prove that there is a solution.

Luiz Felipe - 6 years, 7 months ago

I like solving this as a limit of sequence. Here is an if and only if proof. X (n+1)=x^(x n), x 1=x, x>0. So the assumption that the limit of x n exists and is 2. and so letting n to infinity 2=x^2 so x= sqrt{2} ( keeping in mind x>0). So if the limit is 2 it is x=\sqrt{2}. Conversely, One can check that if x (n+1)= (\sqrt{2})^(x n). X 1= \sqrt{2} then x n converges to 2. In fact: Easy to check that x 2>x 1. If x (n+1)>x n then X (n+2)= (\sqrt{2})^(x (n+1))>(\sqrt{2})^(x n) = x (n+1) So by induction x n is increasing. To prove that is bounded. Also proceed by induction.X 1=\sqrt{2}<2 If x n<2 then x (n+1)=(\Sqrt(2))^(x_n) < (\sqrt(2))^2= 2. So by induction xn<2. Bounded and increasing then converges and the limit is \sqrt{2}<L<=2 such that L=\sqrt{2}^L, So L=2.

Ahmad Sabra - 5 years, 2 months ago

For to be sure of this result. First we must to be really sure if the sequence : $a_1=\sqrt{2}$ and $a_{n+1}={\sqrt{2}}^{a_n}$ converges to a finite value in this case 2. Otherwise make this subtitutions is wrong and there is no solution

Pablo Cesar Herrera Ortiz - 6 years, 7 months ago

If x = sqrt(2), the sequence X(1) = x, X(N) = X(N-1)^x should be equal the left size of this equation. But this sequence diverges (X(N) goes to infinite if N goes to infinite). Thus, this question should have none solution. What you have found is a candidate for answer but it does not satisfy the equation. Am I right?

Igor Pontes - 6 years, 8 months ago

While I realize that top-down exponentiation is the generally accepted order of execution -- and so squareroot(2) is the answer -- according to Wikipedia this is not always the order of execution. "For example, Microsoft Office Excel evaluates a^b^c as (a^b)^c" If one uses the Excel order of execution, one gets the solution x^x^(x^x^x^x...) = x^x^(2) = 2. In this case the solution is closer to x=1.3. Since that wasn't one of the answers, I put that it could not be calculated. I realize that this isn't the generally accepted order of execution, but I would have preferred that the top-down order was stated in the problem presented. .

Richard Levine - 6 years, 8 months ago

If there are no brackets given, then we are expected to follow the top-to-down pattern, I guess. @Richard Levine

Satvik Golechha - 6 years, 8 months ago

But you can not do that unless you know the limit of that function exist. It like writing 1-1+1-1+1-1.....=0. but that is also not true because it has no limit.

Mhammd Al Mhammd - 5 years, 4 months ago

The left hand side is plus infinite for any x>1. There should not be any solution.

Sayyed Ahmed Iftekhar - 5 years, 5 months ago

I think $\sqrt{2}$ does not satisfy the equation for x. You can check that out. By the substitution you made, it looks like you added 2 junk roots (± $\sqrt{2}$ ), both of which fail to satisfy the equation for x.

Sayyed Ahmed Iftekhar - 5 years, 5 months ago

John M.
Oct 10, 2014

$x^{x^{x^{x...}}}=2$

$\log {(x^{x^{x^{x...}}})}=\log {2}$

$x^{x^{x^{x...}}} \log {x}= \log {2}$

$2\log {x} = \log {2}$

$\log {x} = \frac{1}{2} \log {2}$

$\log {x} = \log {2^{\frac{1}{2}}}$

$\boxed {x=2^{\frac{1}{2}}}$

GENERAL CASE:

$x^{x^{x^{x...}}}=N$

$x=N^{\frac{1}{N}}$

PROOF:

Replace the 2s with $N$ in each step and you'll get the result.

I do not understand how the infinite powers of x were converted into 2

Ali Haider Noorani - 6 years, 7 months ago

because x^x^x^....... is equal to 2

Syed Gilani - 6 years, 7 months ago

Satvik's solution above mine is more intuitive.

And if you doubt this answer, you can try exponentiating using Wolfram Alpha and see that the limit as the number of exponent increases approaches 2.

Also, this may be of some help.

Cheers,

John M. - 6 years, 7 months ago

Infinity is like a sea, you can add, subtract , multiply or divide by a finite number, it will be same as before. Infinity is amusing.

DreamRunner Moshi - 5 years, 2 months ago

how do u write like this? i cant type like that :(

Aporajita Tume - 6 years, 8 months ago

It's LaTeX! Check out this guide to LaTeX. Or, quick guide:

Type \ (stuffhere \ ) for latex. And don't space the slash and the parenthesis (I did to show you how I did it). Or, if you want your TeX centered, type with brackets instead of parenthesis: \ [... \ ]

So, x^2=4, wrapped, can display as $x^2=4$ .

John M. - 6 years, 8 months ago

Cesar Conterno
Oct 10, 2014

x^x^x^x^x^x... = 2 => x^2 = 2, x= 2^1/2

I'm not too swift in advanced math,but I know the square root of 2 is 1.414. It is not X,x,x...

George Landry - 4 years, 8 months ago

Tanvir Galib
Oct 16, 2014

x^x^x^x^x^x^x.... = 2

ln (x^x^x^x^x^x^x....) = ln 2

x^x^x^x^x^x.... ln(x) = ln 2

2 ln(x) = ln 2

ln x^2 = ln2

x^2 = 2

x = root(2)

Dominic Bryan
Dec 14, 2015

It's simple X^x^x^x... = 2 Assumed ...^x^x^x...=2 X^2=2 X= √2

Gandoff Tan
Apr 7, 2019

$\begin{aligned} { x }^{ { x }^{ { x }^{ { \bullet }^{ { \bullet }^{ \bullet } } } } } & = & 2,\quad { x }^{ { x }^{ { x }^{ { \bullet }^{ { \bullet }^{ \bullet } } } } }=2 \\ { x }^{ ({ x }^{ { x }^{ { x }^{ { \bullet }^{ { \bullet }^{ \bullet } } } } }) } & = & 2 \\ { x }^{ 2 } & = & 2 \\ x & = & \boxed { \sqrt { 2 } } \end{aligned}$

Ashithcalicut .
Jun 11, 2016

X^x^x.....=2 X^2=2 X=√2

Abdullah Al Mamun
Apr 6, 2016

we can use log.

log(x^x^x^x^x......) = log2 (x^x^x^x.....)logx = log2 logx = 1/2 log2 logx = log 2^(1/2) x = 2^(1/2)

Piyush Prasad
Mar 6, 2016

x^x^x.... = 2

Therefore, x^2 = 2

x = + or - √2 Since x < 0 cannot be true as 2 is an even number

Therefore, x = √2

Infinite Power!

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