the sum of all x values (x^2 −6x+9)^((x−3)/(x−1)) = (x−3)^((x−4))


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Question Number 110471 by abdullahquwatan last updated on 29/Aug/20

$the sum of all x values$ ${(x^{2} - 6 x + 9)}^{\frac{x - 3}{x - 1}} = {(x - 3)}^{(x - 4)}$
	Answered by Rasheed.Sindhi last updated on 29/Aug/20
	$(x^2 −6x+9)^((x−3)/(x−1)) = (x−3)^((x−4)) ( (x−3)^2 )^((x−3)/(x−1)) = (x−3)^((x−4)) ^(• ) ((2(x−3))/(x−1))=x−4 2x−6=x^2 −5x+4 x^2 −7x+10=0 (x−2)(x−5)=0 x=2 ∨ x=5 ^• x−3=0,1⇒x=3 ∨x=4 But x=3 produces indeterminate form 0^0 So Solution Set is {2,4,5}$
	${(x^{2} - 6 x + 9)}^{\frac{x - 3}{x - 1}} = {(x - 3)}^{(x - 4)}$ ${({(x - 3)}^{2})}^{\frac{x - 3}{x - 1}} = {(x - 3)}^{(x - 4)}$ $^{∙} \frac{2 (x - 3)}{x - 1} = x - 4$ $2 x - 6 = x^{2} - 5 x + 4$ $x^{2} - 7 x + 10 = 0$ $(x - 2) (x - 5) = 0$ $x = 2 \lor x = 5$ $^{∙} x - 3 = 0, 1 \Rightarrow x = 3 \lor x = 4$ $B u t x = 3 p r o d u c e s i n d e t e r m i n a t e$ $f o r m 0^{0}$ $S o S o l u t i o n S e t i s$ ${2, 4, 5}$
	Commented by abdullahquwatan last updated on 29/Aug/20

	$3 included ?$
	Commented by Rasheed.Sindhi last updated on 29/Aug/20

	$S o r r y {i t}^{'} s e x c l u d e d b e c a u s e x = 3$ $g i v e s i n d e t e r m i n a t e f o r m 0^{0}$
	Commented by Rasheed.Sindhi last updated on 29/Aug/20

	$T h a n k s a b d u l l a h f o r$ $m e n t k o n i n g e r r o r .$
	Commented by abdullahquwatan last updated on 29/Aug/20

	$thank you sir$
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