x^(√x) =(√x^x ) find x


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Question Number 175554 by Linton last updated on 02/Sep/22

$x^{\sqrt{x}} = \sqrt{x^{x}}$ $f i n d x$
	Answered by mr W last updated on 02/Sep/22

	$x = 0$ $x^{x^{\frac{1}{2}} - \frac{x}{2}} = 1$ $\Rightarrow x = 1$ $\Rightarrow x^{\frac{1}{2}} - \frac{x}{2} = 0 \Rightarrow x = 4$
	Commented by BaliramKumar last updated on 02/Sep/22

	$x = 0 undefiend$
	Commented by mr W last updated on 02/Sep/22

	$u n d e f i n e d f o r y o u, b u t n o t f o r a l l!$
	Commented by mr W last updated on 02/Sep/22

	Commented by BaliramKumar last updated on 02/Sep/22

	$y o u a r e r i g h t s i r$ $3 s o l u t i o n v a l i d$
	Commented by BaliramKumar last updated on 02/Sep/22
	$x^(√x) = (√x^x ) x^x^(1/2) = x^(x/2) x^(1/2) log(x) = (x/2)log(x) log(x){(√x) −(x/2)} = 0 log(x){2(√x)−x} = 0 log(x){2−(√x)}(√x) = 0 log_e (x) = 0 2−(√x) = 0 (√x) = 0 x = e^0 (√x) = 2 x = 0 x = 1 x = 4 x = 0$
	$x^{\sqrt{x}} = \sqrt{x^{x}}$ $x^{x^{\frac{1}{2}}} = x^{\frac{x}{2}}$ $x^{\frac{1}{2}} l o g (x) = \frac{x}{2} l o g (x)$ $l o g (x) {\sqrt{x} - \frac{x}{2}} = 0$ $l o g (x) {2 \sqrt{x} - x} = 0$ $l o g (x) {2 - \sqrt{x}} \sqrt{x} = 0$ $\log_{e} (x) = 0 2 - \sqrt{x} = 0 \sqrt{x} = 0$ $x = e^{0} \sqrt{x} = 2 x = 0$ $x = 1 x = 4 x = 0$
	Commented by BaliramKumar last updated on 02/Sep/22

	Answered by Rasheed.Sindhi last updated on 02/Sep/22

	$x^{\sqrt{x}} = \sqrt{x^{x}}$ $x^{x^{1 / 2}} = x^{x / 2}$ $x^{1 / 2} = x / 2$ $x^{1 / 2 - 1} = 1 / 2$ $\frac{1}{x^{1 / 2}} = \frac{1}{2}$ $x^{1 / 2} = 2$ $x = 2^{2} = 4$
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