x^x^x^x^x=5?

[skeptical scientist]

I was told the equation x^x^x^x^x=5 has a solution expressible with elementary functions. (Associating the exponential function in the interesting way, of course.) I can't vouch that this is the case, as nobody I know has been able to solve it. Any ideas?

[Wnderer]

for x^x^x^x^x = 5; x is almost the ln(5); x~= ln(4.912)

but x^x^x^x^x = 6; is actually closer; x~= ln(5.001)

[rigwarl]

Associating the exponential function in the interesting way, of course.

Can you clarify what this part is asking?

Also, Wnderer's numbers are off I believe.

[Malle]

Associating the exponential function in the interesting way, of course.

Can you clarify what this part is asking?

I'm assuming that he means that
x^x^x^x^x = x^(x^(x^(x^x)))
and not
x^x^x^x^x = (((x^x)^x)^x)^x

Example: x = 2 then
2^2^2^2^2 = 2^2^2^4 = 2^2^16 = 2^65536
which is roughly
2.00*10^19728

which is clearly several orders of magniture more interesting than
2^2^2^2^2 = 4^2^2^2 = 16^2^2 = 256^2 = 65536

[Wnderer]

Wolfram Alpha gives the answer as x = 1.5934088....

http://www.wolframalpha.com/input/?i=x^x^x^x^x+%3D+5

and the answer to ln(x) = 1.5934088 is x= 4.92049...

http://www.wolframalpha.com/input/?i=ln ... +1.5934088

and x^x^x^x^x = 6 ; x=1.60962774663638...

http://www.wolframalpha.com/input/?i=x^x^x^x^x+%3D+6

and ln(x) = 1.60962774663638; x = 5.000949 ...

http://www.wolframalpha.com/input/?i=ln ... 6277466363

If I try different parenthesis I get

Spoiler:

xxxxx5.gif (1.29 KiB) Viewed 8724 times

(log is the natural logarithm)
(W is the product log function)

[mfb]

Is that possible for other numbers than 5, too?
Or is there anything special about 5?

[jestingrabbit]

Is that possible for other numbers than 5, too?
Or is there anything special about 5?

Could you be a little less specific?

@Wnderer: If you bracket stuff the wrong way, as ((((x^x)^x)^x)^x) = 5, you may as well just have x^(x^4) = 5. From there, you can get that

(x^(x^4))^4 = x^(4*(x^4)) = (x^4)^(x^4) = 5^4.

And that gives you a form that wp talks about, y^y = z, and solves for you as y = ln(z)/W(ln(z)), which in this instance gives

x^4 = ln(5^4)/W(ln(5^4)).

Not much more work gets the form that alpha gives you.

@OP: the only way to even begin trying to get an answer, at least as far as I can see, is some sort of wacky application of Lambert W. But even from there, I'm looking and I'm not seeing. Can you source the claim?

[skeptical scientist]

@OP: the only way to even begin trying to get an answer, at least as far as I can see, is some sort of wacky application of Lambert W. But even from there, I'm looking and I'm not seeing. Can you source the claim?

Friend-of-a-friend type of deal, so as I said in the OP, I can't vouch that a solution exists. My office-mate's ex-boyfriend gave her the problem (without the solution), and she shared it with the rest of the office in the hope that someone would be able to solve it. None of us have.

[mfb]

Is that possible for other numbers than 5, too?
Or is there anything special about 5?

Could you be a little less specific?

Under the assumption that there is a nice expression for the solution of x^x^x^x^x=5:
Is there a similar expression for the solution of x^x^x^x^x=6 or other real numbers instead of 5?

But as we don't know if it exists at all...

[jestingrabbit]

Is that possible for other numbers than 5, too?
Or is there anything special about 5?

Could you be a little less specific?

Under the assumption that there is a nice expression for the solution of x^x^x^x^x=5:
Is there a similar expression for the solution of x^x^x^x^x=6 or other real numbers instead of 5?

But as we don't know if it exists at all...

One would hope that a resolution of the problem would be able to answer this, but before there's any resolution, no one's going to be able to give you any answers here.

Anyway, was thinking maybe the binomial series (applied to something like (1 + (x-1))^(x^x^x^x), and then to something similar for the exponent there), but that seems like just asking for trouble, but its at least one other line of inquiry.

[t1mm01994]

If I know my log-rules right, this should be equivalent to x^4ln(x)=ln(5). Not that that really gives you any help in any way, but yeah.. Just throwing my highschool math out there hoping one of you could do something usefull with it

[skeptical scientist]

If I know my log-rules right, this should be equivalent to x^4ln(x)=ln(5). Not that that really gives you any help in any way, but yeah.. Just throwing my highschool math out there hoping one of you could do something usefull with it

It's not. I'm not quite sure what you did to get this, but I suspect the error was forgetting that exponentiation is not associative.

[t1mm01994]

never mind that, indeed. Failed in associating. Going down the ln-route you actually get x^x^x^xln(x)=ln(5), or x^x^xln(xln(x))=ln(ln(5)), and so on. not too much helpfulness, just ugliness :$

[jaap]

never mind that, indeed. Failed in associating. Going down the ln-route you actually get x^x^x^xln(x)=ln(5), or x^x^xln(xln(x))=ln(ln(5)), and so on. not too much helpfulness, just ugliness :$

No, from
x^(x^(x^(x^x))) = 5
we get
(x^(x^(x^x))) * ln(x) = ln(5)

Then taking logs again we get
ln( (x^(x^(x^x))) * ln(x) ) = ln(ln(5))

which 'simplifies' to
ln((x^(x^(x^x)))) + ln(ln(x)) = ln(ln(5))
(x^(x^x))ln(x) + ln(ln(x)) = ln(ln(5))

I very much doubt that it really has an elementary solution.

[t1mm01994]

I'll head back to my cave of lurking :X

[Wnderer]

It reminds me of the e^pi - pi = 20 joke. It's almost true.

If I solve

x^x^x^x^x = e^x; x~= 5/pi; e^(5/pi) ~= 4.9113...

(x/pi)^(x/pi)^(x/pi)^(x/pi)^(x/pi)= e^(x/pi); x~= 5.00040422...

e^phi the golden ration is almost 5, too 5.04316...

[rigwarl]

Maybe it could be possible that the solution whoever came up with the problem was looking for is something like:

x = 0+1
0+1^0+1^0+1^0+1^0+1 = 5

[jestingrabbit]

Well, if that's what they wanted, they suck worse than a really sucky thing.

I'm noodling around with Lambert W, getting nowhere. Might try it from an infinite binomial series angle for a while now.

[notzeb]

Inverse Symbolic Calculator says no dice:

http://oldweb.cecm.sfu.ca/cgi-bin/isc/l ... ype=simple

[tomtom2357]

How about we try proving whether it is rational or irrational. I can prove that x^x=n (n an integer of course) implies that if x is not an integer, its transcendental, and that x^{(x^x)}=n (n is still an integer) implies that if x is not an integer, its irrational. I shall spoiler my proofs below.

Spoiler:

I shall first prove that x must be irrational (this is a necessary part of my proof) If x^x=n is rational and not an integer, then x=a/b (b>1 and a/b in lowest terms), and (a/b)^(a/b)=n. So now we have n^b=(a/b)^a, but since a/b is in lowest terms, then a^a/b^a is in lowest terms too, but n^b is an integer, therefore integer=non-integer, which is impossible. Therefore x is not rational. Next to prove (by contradiction, again) it is transcendental: now that we know that it is not rational, we have that x^x=n where x is algebraic. But then n= an algebraic number to the power of an irrational algebraic number, which, by the gelfond-schneider theorem, is transcendental. Therefore x cannot be algebraic, and is therefore transcendental.

Next is my proof that x^{x^x}=n implies that x is either an integer, or irrational.

Spoiler:

If x is rational, then it is of the form a/b. However, then (a/b)^{(a/b)^(a/b)}[/sup]=n, but by the above proof, (a/b)^a/b is algebraic-irrational, and therefore we have an algebraic number to the power of an irrational algebraic number, which is transcendental (by the gelfond-schneider theorem, again). But an integer is not transcendental, therefore x cannot be rational, and is therefore irrational. QED

Can anyone prove any stronger results?

[t1mm01994]

f) If x^x=n is rational and not an integer, then x=a/b (b>1 and a/b in lowest terms), and (a/b)^(a/b)=n. So now we have n^b=(a/b)^a, but since a/b is in lowest terms, then a^a/b^b is in lowest terms too, but n^b is an integer, therefore integer=non-integer, which is impossible. Therefore x is not rational.

What happens there? Because right now, I dont understand.. a^a/b^b has nothing to do with and (a/b)^(a/b) afaik.

[mfb]

What happens there? Because right now, I dont understand.. a^a/b^b has nothing to do with and (a/b)^(a/b) afaik.

I think it is a typo, as the logic is correct with a^a/b^a instead of a^a/b^b.

The interesting part is to extend this proof up to the stack of 5 x.

[tomtom2357]

Oops, sorry, it is a^a/b^a, but my proof still holds doesn't it? Also, the results I used (the gelfond-schneider theorem) has no stronger reults that I know of, and to prove more you have to take a rational number to the power of a transcendental number, which no one (that I know of) has proven any (useful) results on.
Edit: Woops, didn't look far enough (again), Baker's Theorem and the lindenmann-weierstrass theorem might be interesting tools to use, however there are no theorems (that I know of, so I might be wrong, again) about how power towers of height >2 relate to transcendentality.
Can anyone use these theorems to help? And is this line of reasoning helpful to the problem?

[tomtom2357]

Maybe a double Lambert W?

[jestingrabbit]

Maybe a double Lambert W?

Unless you describe what that means, it means precisely nothing.

Anyway, I'm going to write up the best bits of what I had. I never got anywhere but the most productive path that I tried for was as follows.

I started with the most basic example of this sort of question, z = x^x. You can check that x = exp(W(ln(z))) solves this. So, we've made it to the second of five rungs on our little ladder. From there, life gets hard.

The formula, x = exp(W(ln(z))), suggests that we might try something similar to get to the next rung. If we let x = exp(W(W(ln(z)))), then you can work out that z = x^(x^(x+1)). That might not seem like a huge problem, but I've found it pretty tough to work out what needs changing.

[tomtom2357]

Sorry, I meant what you did, repeated application of the lambert w function.

[mfb]

I started with the most basic example of this sort of question, z = x^x. You can check that x = exp(W(ln(z))) solves this. So, we've made it to the second of five rungs on our little ladder. From there, life gets hard.

The formula, x = exp(W(ln(z))), suggests that we might try something similar to get to the next rung. If we let x = exp(W(W(ln(z)))), then you can work out that z = x^(x^(x+1)). That might not seem like a huge problem, but I've found it pretty tough to work out what needs changing.

Well, but what did you gain by stating that x = exp(W(ln(z)))? You used a function which quite ugly in its definition. You cannot express it with elementary functions or express it with a single power series.
In that way, it is easy to find a solution: Define W_i(z) as as the largest real number which satisfies the equation x^x^...(i times)...^x=W_i(z). Now, the solution to the initial problem is trivial: x=M₅(5).

[jestingrabbit]

I started with the most basic example of this sort of question, z = x^x. You can check that x = exp(W(ln(z))) solves this. So, we've made it to the second of five rungs on our little ladder. From there, life gets hard.

The formula, x = exp(W(ln(z))), suggests that we might try something similar to get to the next rung. If we let x = exp(W(W(ln(z)))), then you can work out that z = x^(x^(x+1)). That might not seem like a huge problem, but I've found it pretty tough to work out what needs changing.

Well, but what did you gain by stating that x = exp(W(ln(z)))? You used a function which quite ugly in its definition. You cannot express it with elementary functions or express it with a single power series.
In that way, it is easy to find a solution: Define W_i(z) as as the largest real number which satisfies the equation x^x^...(i times)...^x=W_i(z). Now, the solution to the initial problem is trivial: x=M₅(5).

Firstly, what's elementary is entirely a thing of convention. We weren't given the elementary functions on gold tablets by an angel, its purely a convention. To my mind, it makes sense to include the Lambert W function in that collection, its certainly no harder to work with than ln, and it appears in a great many computer language's maths libraries. I also don't really think its ugly. I've taken W and conjugated it by the exponential function. Put another way, if the inverse of the function g(x) = xe^x is W, the inverse of the function f(x) = x^x is f^(-1) = exp W exp^(-1). I think that's pretty nice.

By way of contrast, your W_i's are really arbitrary, and your solutions don't really say anything. They pretty much just say that the solution is the solution.

I've also reduced the number of times that x appears, and we're trying to find values for x. This is a step forward.

[Vaniver]

Maybe it could be possible that the solution whoever came up with the problem was looking for is something like:

x = 0+1
0+1^0+1^0+1^0+1^0+1 = 5

I would wager money this was the intent of the problem creator.

[BlueSoxSWJ]

I'm betting, instead, that the x^s are supposed to be infinite. In other words, solve x^(x^(x^(x^(x^(.....))) = 5.

Spoiler:

The simple trick is realizing that the infinite exponentiation also equals x raised to the same infinite exponentiation. Thus, the general solution:
x^(x^(x^(x^(...)))=n
x^n = n
x = nth root of n

[battlesnake]

x^(x^(x^(x^x))) = 5

Using X^Y = e^(Y ln(X) )

x^(x^(x^x)) = ln(5)/ln(x)

x^(x^x) = ln(ln(5)/ln(x)) / ln(x)

x^x = ln(ln(ln(5)/ln(x)) / ln(x)) / ln(x)

x = ln(ln(ln(ln(5) / ln(x)) / ln(x)) / ln(x)) / ln(x)

ln(ln(ln(ln(5) / ln(x)) / ln(x)) / ln(x)) / ln(x) - x = 0

Then a non-linear solver (aka trial and error) gives 1.5934... which is approx ln(5).