MHB Challenge: Create 64 with two 4's

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The challenge is to create the number 64 using two 4's with various mathematical operations. Solutions presented include complex expressions involving logarithms and factorials, such as $-\lg \, \log_{\sqrt{4}} \, \underbrace{\sqrt{\sqrt{...\sqrt{4}}}}_{\text{65 times}}$ and $4^{\left(\log \sqrt{\sqrt{\sqrt{e^{4!}}}}\right)}$. An alternate solution is $4^{\left(\sqrt{\sqrt{\sqrt{4!}}}\right)}$, which simplifies the approach without using logarithms. The term "Perelman-like" is used to describe the logarithmic method, referencing a related problem from Yakov Perelman's work. The discussion highlights the creativity and complexity involved in solving mathematical challenges.
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Using any of \{+,\;-,\;\times,\;\div,\;x^y,\;\sqrt{x},\;x!\}

. . create 64 with two 4's.
 
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Re: Challenge

In base 15:

44
 
Re: Challenge

(Clapping)
 
Re: Challenge

A Perelman-like solution suffices :

$-\lg \, \log_{\sqrt{4}} \, \underbrace{\sqrt{\sqrt{...\sqrt{4}}}}_{\text{65 times}}$

Or even,

$4^\left ({\log \sqrt{\sqrt{\sqrt{e^{4!}}}}} \right )$

But as $\log$ is not desired, a twist around the base of logarithm of the latter should do :

$\sqrt{\sqrt{\sqrt{4^{4!}}}}$
 
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Re: Challenge

Hello, mathbalarka!

That's it! . . . Nice reasoning!

Alternate form: .4^{\left(\sqrt{\sqrt{\sqrt{4!}}}\right)}
 
Re: Challenge

Okay, thanks! The other form didn't click to me, really nice!

Now, in my post, the first form using logarithm is referred to as "Perelman-like" as another related, but twisted problem was given in Yakov Perelman's Mathematics Is Fun. He refers a challenger in a congress of physicist in Odessa. I couldn't find further reference, so just named this after him. (a year ago when I found out this kind of approach is in general very doable for these problems)
 

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