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Question Number 210371 by Abdullahrussell last updated on 08/Aug/24

	Answered by Spillover last updated on 08/Aug/24

	$u s e g r a p h i c a l m e t h o d t h e n f i n d p o i n t o f$ $i n t e r s e c t i o n$ $l e t$ $y =^{x} \sqrt{\frac{2}{x - 1}} d r a w i t g r a p h$ $y = {(x - 2)}^{(x - 2)} d r a w i t g r a p h$
	Answered by Spillover last updated on 08/Aug/24

	Commented by Spillover last updated on 08/Aug/24

	The point where the two curves intersect represents the solution to the equation. From the graph, you can see that the intersection occurs at approximately x=2.414
	Answered by Frix last updated on 08/Aug/24

	$Let x > 1$ $2^{\frac{1}{x}} {(x - 1)}^{- \frac{1}{x}} = {(x - 1)}^{x - 2}$ $2^{\frac{1}{x}} = {(x - 1)}^{\frac{{(x - 1)}^{2}}{x}}$ $\frac{\ln 2}{x} = \frac{{(x - 1)}^{2} \ln (x - 1)}{x}$ $\frac{\ln 2}{\ln (x - 1)} = {(x - 1)}^{2}$ $x > 1 \Rightarrow let x = 1 + \sqrt{t}$ $\frac{2 \ln 2}{\ln t} = t$ $2 n 2 = t \ln t$ $t = 2$ $x = 1 + \sqrt{2}$
	Commented by Spillover last updated on 08/Aug/24

	$p e r f e c t$
	Answered by mr W last updated on 08/Aug/24

	$x > 0$ $\frac{2}{x - 1} = {(x - 1)}^{x (x - 2)}$ $2 = {(x - 1)}^{x (x - 2) + 1}$ $2 = {(x - 1)}^{{(x - 1)}^{2}}$ $2^{2} = {[{(x - 1)}^{2}]}^{{(x - 1)}^{2}}$ $\Rightarrow {(x - 1)}^{2} = 2$ $\Rightarrow x = 1 + \sqrt{2}$
	Commented by ajfour last updated on 08/Aug/24
	Absolutely. Hello sir mrw
	Commented by mr W last updated on 08/Aug/24

	$h i! n i c e t o s e e y o u a g a i n! h o p e y o u$ $a r e d o i n g w e l l!$
	Commented by ajfour last updated on 08/Aug/24

	$y e s, t h a n k s!$
	Commented by Spillover last updated on 08/Aug/24

	$g r e a t$
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