x^x =0.25 find x


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Question Number 38099 by Cheyboy last updated on 21/Jun/18

$x^{x} = 0.25$ $f i n d x$
Commented by MrW3 last updated on 22/Jun/18

$n o r e a l s o l u t i o n,$ $s i n c e x^{x} a l w a y s ⩾ \frac{1}{^{e} \sqrt{e}} = 0.6922 > 0.25$ $s e e Q 37414$
Commented by MrW3 last updated on 22/Jun/18

Commented by prof Abdo imad last updated on 22/Jun/18

$(e) \Leftrightarrow e^{x l n x} = 4^{- 1} = e^{- l n (4)} = e^{- 2 l n (2)} \Leftrightarrow x l n (x) + 2 l n (2) = 0$ $l e t f (x) = x l n (x) + 2 l n (2) w i t h x > 0$ $f^{^{'}} (x) = l n (x) + 1 ⩾ 0 \Leftrightarrow l n (x) ⩾ l n (\frac{1}{e}) \Leftrightarrow x ⩾ \frac{1}{e}$ $s o f i s i n c r e a s i n g o n [\frac{1}{e}, + \infty [a n d d e c r e a s i n g$ $o n] 0, \frac{1}{e}] f (\frac{1}{e}) = - \frac{1}{e} + 2 l n (2) > 0$ ${l i m}_{x \to 0^{+}} f (x) = 2 l n (2) > 0$ ${l i m}_{x \to + \infty} f (x) = + \infty s o f (x) \neq 0 \forall x > 0$ $s o t h e e q u a t i o n h a v e n t a n y r e a l s o l u t i o n .$
Commented by Cheyboy last updated on 22/Jun/18

$O o h k s i r t h a n k y o u, i t w a s g i v i n$ $t o m e b u t i c o u l d n o t s o l v e i t$
Commented by Cheyboy last updated on 22/Jun/18

$T h a n k y o u M r W 3 a n d p r o f a b d o$
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