MHB Challenge: Create 64 with two 4's

[soroban]

Using any of \{+,\;-,\;\times,\;\div,\;x^y,\;\sqrt{x},\;x!\}

. . create 64 with two 4's.

 

[MarkFL]

Re: Challenge

Spoiler

In base 15:

44

 

[eddybob123]

Re: Challenge

(Clapping)

 

[mathbalarka]

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 :

$\sqrt{\sqrt{\sqrt{4^{4!}}}}$

 

[soroban]

Re: Challenge

Hello, mathbalarka!

That's it! . . . Nice reasoning!

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

 

[mathbalarka]

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)