x^x =2^(2048) x=?


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Question Number 161951 by mathlove last updated on 24/Dec/21

$x^{x} = 2^{2048}$ $x = ?$
	Answered by aleks041103 last updated on 24/Dec/21

	$2048 = 2^{11}$ $\Rightarrow x^{x} = 2^{2^{11}}$ $l e t x = 2^{t}$ ${(2^{t})}^{2^{t}} = 2^{t . 2^{t}} = 2^{2^{11}}$ $\Rightarrow t 2^{t} = 2^{11}$ $f (t) = t . 2^{t}$ $f i s m o n o t o n o u s l y r i s i n g \Rightarrow t h e r e i s o n l y 1 s o l u t i o n$ $f (1) = 2 < 2^{11} \Rightarrow t > 1$ $\Rightarrow 2^{11} = t 2^{t} > 2^{t} \Rightarrow t < 11$ $s u p p o s e t \in N$ $t 2^{t} = 2^{11} \Rightarrow t = 2^{11 - t}$ $\Rightarrow t i s a p o w e r o f 2 a n d t \in (1, 11)$ $\Rightarrow t = 2, 4, 8$ $\Rightarrow t = 8$ $\Rightarrow x = 2^{8}, x^{x} = 2^{8 . 2^{8}} = 2^{2^{11}} = 2^{2048}$ $\Rightarrow A n s . x = 2^{8} = 256$
	Answered by mr W last updated on 24/Dec/21

	$2^{2048} = 2^{2^{11}} = 2^{2^{a} 2^{b}} = {(2^{2^{a}})}^{2^{b}} w i t h a + b = 11$ $2^{2^{a}} \overset{!}{=} 2^{b}$ $2^{a} = b = 11 - a \Rightarrow a = 3, b = 11 - 3 = 8$ $2^{2048} = {(2^{8})}^{2^{8}} = x^{x}$ $\Rightarrow x = 2^{8}$
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