Is this one solvable?
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Dr.Peterson
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Are you sure you don't mean x^(x^(x^2018)) = 2018? What you wrote is just x^(2018x^2) = 2018; it's not a lot easier to solve (numerically), but is different.
The other, in fact, is undefined as written, since it's equivalent to x^(x^inf) = 2018.
Steven G
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Yes, that is what I mean. x^(x^(x^2018)) = 2018
Steven G
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No, no, I want x, NOT what x is approximately. Besides, I posted the wrong problem. I hate tex. In any case, the problem is x^(x^(x^2018)) = 2018
Dr.Peterson
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So you do want \(\displaystyle x^{x^{x^{2018}}} = 2018\).
I am reasonably sure that this equation can only be solved by numerical methods (approximately). When I give it to WolframAlpha, it times out on me, which is not a good sign.
The word "solve" does not inherently mean "solve exactly, or in closed form"; "solvable" can be taken in different ways. A numerical solution is a kind of solution.
Steven G
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It actually has a very nice solution, no need for approximation.
Steven G
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I did not go that way, but it may work. I will give you a hint. How would you solve x^2018 = 2018?? Simple enough.Hmmm...unfamiliar grounds for me; is this the 1st step:
x^(x^2018) = log(2018) / log(x)
then:
x^2018 = log(2018) / log(x) / log(x)
Dr.Peterson
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I don't see that that is relevant; your equation is of a very different type.
How would you solve x^(x^2) = 2 exactly?
Steven G
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I can give you the answer but it will be a give away to my problem. To be honest, the hint which I gave Denis will really help a lot.
Steven G
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I would simplify 2^(1/18) * 3^(1/9) to the obvious result.
Steven G
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The answer is right in your face!
Dr.Peterson
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His answer to x^2018 = 2018 is correct, and can't be stated any more simply unless you write it as the 2018th root of 2018. It's approximately 1.003778. Check: 1.003778^2018 = 2017.548.
What are you thinking the answer is?
Steven G
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Please read the post #11 (by Denis) where he says the solution to x^2018 = 2018 is x= 2018^(1/2018). He never answered my original problem.
Dr.Peterson
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Yes, I know. I'm not talking about the original problem at the moment.
I am commenting on your reply to his answer to the little problem you gave him as a "hint". When you say, quoting his answer of "x= 2018^(1/2018)", that "The answer is right in your face", it appears that you are saying his answer to that problem is incomplete. Or are you saying that about the original problem for some reason? I'm trying to understand.
My belief is that your "hint" is not relevant to the original problem. I want to understand what you mean by the hint, in order to see where you are heading, and whether you might be making some sort of mistake.
But your original post contained two problems; when I refer to the "original problem", I am talking about the first, x^(x^(x^2018)) = 2018. I know that there is a solution to your second original problem (the infinite "power tower"), and now that I think about it, I see that that one is related to your hint (though there are some details I have to look into, now that I am thinking about that one). Possibly you think that your first original problem, which is what I have been focusing on, is more closely related to the second than I think it is.
Steven G
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OK, let's clear this all up. My original part a problem, \(\displaystyle x^{x^{x^{2018}}} = 2018\), does have a solution and that solution is x = 2018^(1/2018).
My part b problem, \(\displaystyle x^{x^{x^{...}}}\) =2018, in fact has no solution.
Dr.Peterson
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You're right about the infinite power tower. The well-known trick for solving these is to point out that if we define \(\displaystyle f(x) =x^{x^{x^{...}}}\), then \(\displaystyle f(x) =x^{f(x)}\), so that if \(\displaystyle f(x) =2\), for example, we can conclude that \(\displaystyle 2 =x^2\) and therefore \(\displaystyle x = \sqrt{2}\). But if you do the same thing with \(\displaystyle f(x) =4\), you end up with the same solution: \(\displaystyle 4 =x^4\) implies that \(\displaystyle x = \sqrt{2}\). But \(\displaystyle f(\sqrt{2})\) can't equal both 2 and 4.
The problem is that this work assumes that the equation has a solution. The reality is that f(x) can equal 2, but it can't equal 4. Likewise, it can't equal 2018, so that equation has no solution, as you can see by checking the claimed solution. (These are the extra details I didn't have time to write up last night.)
I had used a spreadsheet to find numerical solutions to both problems, and just discovered an error in a formula, which explains why I continued to think your solution was incorrect. On checking manually, it was obvious why it would work.
But can your method of solution (which I think was based on plugging in the invalid solution of the infinite tower) be called anything more than guess and check? I think I am right that no algebraic method can give an exact solution. (Admittedly, you didn't actually ask for an algebraic method, just an exact answer; but your hints appeared to be about a method, not just a solution: "how would you solve ...".)
This is the problem with presenting a math problem as a puzzle that you can't discuss.