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I can't solve this question I have tried but I can't find any other websites that help. Thanks

$$\sqrt{x}^{\sqrt{x}^{\sqrt{x}}} = 2^{512}$$

This is not infinite exponent just 2 times

Shaun
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Unknown
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    Can you please put parentheses around the terms which belong together and write it in MathJax. – EuklidAlexandria Sep 15 '18 at 21:27
  • @Unknown Here is a handy tutorial for writing equations in mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Jam Sep 15 '18 at 21:29
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    One solution is to numerically solve $x\log_2\left(x\right)+\log_2\left(\log_2\left(x\right)\right)=9$, although I don't think this is particularly elegant. – Jam Sep 15 '18 at 21:42
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    Compare with $4^{4^4}$ . – user90369 Sep 15 '18 at 21:48
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    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Sep 15 '18 at 21:49
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 15 '18 at 21:49
  • ok i understand I will post better questions – Unknown Sep 16 '18 at 00:18

1 Answers1

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Hint: First define $y=\sqrt x$ to clean things up. Then if you are in the integers note that $y$ must be a power of $2$ so let $y=2^z$. Plug that in and use the laws of exponents. Taking the $\log_2$ of the equation will also clean things up a bit.

If you are in the reals you have $$y^{y^y}=2^{512}\\ y^y\log_2y=512\\y\log_2 y +\log_2(\log_2 y)=9$$ and we can do fixed point iteration with $$y=\frac {9-\log_2(\log_2 y)}{\log_2 y}$$ I find $y=4$ is a solution, so $x=16$

Ross Millikan
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    Why must $y$ be a power of $2$? If we were solving $y^y=2^6$ for example, we'd have $y\approx3.4$, which isn't a power of $2$. – Jam Sep 15 '18 at 21:33
  • Of course, we can see that $\sqrt{x}$ is a power of $2$ but I'm not sure how you've justified it. – Jam Sep 15 '18 at 21:36
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    I was assuming you were working in the integers. If you are in the real you are right. Then just take the $\log_2$ of the $y$ stack twice and do a numeric solution. – Ross Millikan Sep 15 '18 at 21:45
  • That makes sense. It seems like there wouldn't be a closed-form for non-integer solutions. – Jam Sep 15 '18 at 21:46