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Question Number 200224 by cortano12 last updated on 15/Nov/23

	Answered by Rasheed.Sindhi last updated on 16/Nov/23
	${ ((yx^(log_y x ) =x^(5/2) ...............(i))),((log_4 y.log_y (y−3x)=1...(ii) )) :} (ii)⇒ (1/(log_y 4 ))=(1/(log_y (y−3x) )) y−3x=4 y=x^a (say) log_y y=alog_y x a=(1/(log_y x)) y=x^(1/(log_y x)) (i)⇒x^(1/(log_y x)) ∙x^(log_y x ) =x^(5/2) x^(log_y x +(1/(log_y x))) =x^(5/2) log_y x +(1/(log_y x))=(5/2) 2(log_y x)^2 −5log_y x +2=0 (2log_y x−1)(log_y x−2)=0 log_y x=(1/2),2 ∧ y=3x+4 y^2 =x ∣ y^(1/2) =x (3x+4)^2 =x ∣ (3x+4)^(1/2) =x 9x^2 +23x+16=0_(No real roots) ∣ 3x+4=x^2 x^2 −3x−4=0 (x−4)(x+1)=0 x=4 ,x=−1 y=3(4)+4=16,3(−1)+4=1 y=16,1 (x,y)=(4,16),(−1,1)^(invalid)$
	${\begin{cases} {y x}^{\log_{y} x} = x^{\frac{5}{2}} . . . . . . . . . . . . . . . (i) \\ \log_{4} y . \log_{y} (y - 3 x) = 1 . . . (i i) \end{cases}$ $(i i) \Rightarrow \frac{1}{\log_{y} 4} = \frac{1}{\log_{y} (y - 3 x)}$ $y - 3 x = 4$ $y = x^{a} (s a y)$ $\log_{y} y = a \log_{y} x$ $a = \frac{1}{\log_{y} x}$ $y = x^{\frac{1}{\log_{y} x}}$ $(i) \Rightarrow x^{\frac{1}{\log_{y} x}} \cdot x^{\log_{y} x} = x^{\frac{5}{2}}$ $x^{\log_{y} x + \frac{1}{\log_{y} x}} = x^{\frac{5}{2}}$ $\log_{y} x + \frac{1}{\log_{y} x} = \frac{5}{2}$ $2 {(\log_{y} x)}^{2} - {5 \log}_{y} x + 2 = 0$ $({2 \log}_{y} x - 1) (\log_{y} x - 2) = 0$ $\log_{y} x = \frac{1}{2}, 2 \land y = 3 x + 4$ $y^{2} = x ∣ y^{1 / 2} = x$ ${(3 x + 4)}^{2} = x ∣ {(3 x + 4)}^{1 / 2} = x$ $\underset{N o r e a l r o o t s}{9 x^{2} + 23 x + 16 = 0} ∣ 3 x + 4 = x^{2}$ $x^{2} - 3 x - 4 = 0$ $(x - 4) (x + 1) = 0$ $x = 4, x = - 1$ $y = 3 (4) + 4 = 16, 3 (- 1) + 4 = 1$ $y = 16, 1$ $(x, y) = (4, 16), \overset{i n v a l i d}{(- 1, 1)}$
	Answered by Frix last updated on 16/Nov/23

	$With x, y > 0 . . .$ $. . . the 1^{st} equation is equal to$ $(\ln x - 2 \ln y) (2 \ln x - \ln y) = 0$ $\Rightarrow y = \sqrt{x} \lor y = x^{2}$ $. . . the 2^{nd} equation is equal to$ $3 x - y + 4 = 0 \Rightarrow y = 3 x + 4$ $(1) \sqrt{x} = 3 x + 4 \Rightarrow x \notin R$ $(2) x^{2} = 3 x + 4 \Rightarrow x = 4 \land y = 16$
	Answered by Rasheed.Sindhi last updated on 16/Nov/23
	${ ((y x^(log_y x ) =x^(5/2) .................(i))),((log_4 y.log_y (y−3x)=1....(ii) )) :} (i)⇒y=x^((5/2)−log_y x ) ⇒y^2 =x^(5−2log_y x ) ⇒log_y y^2 =(5−2log_y x)log_y x 2=5log_y x−2(log_y x)^2 2(log_y x)^2 −5log_y x+2=0 log_y x=((5±(√(25−16)))/4)=2,(1/2) y^2 =x ∣ y^(1/2) =x⇒y=x^2 .........(iii) (ii)⇒(1/(log_y 4 ))=(1/(log_y (y−3x) )) ⇒y−3x=4 ⇒y=3x+4...................(iv) (iii) & (iv) : (3x+4)^2 =x ∣ 3x+4=x^2 9x^2 +23x+16=0_(No real roots) ∣ x^2 −3x−4=0 (x−4)(x+1)=0⇒x=4,−1 ⇒y=3(4)+4 , 3(−1)+4=16,1 but y≠1 ∴ (x,y)=(16,4) ✓$
	${\begin{cases} y x^{\log_{y} x} = x^{\frac{5}{2}} . . . . . . . . . . . . . . . . . (i) \\ \log_{4} y . \log_{y} (y - 3 x) = 1 . . . . (i i) \end{cases}$ $(i) \Rightarrow y = x^{\frac{5}{2} - \log_{y} x} \Rightarrow y^{2} = x^{5 - {2 \log}_{y} x}$ $\Rightarrow \log_{y} y^{2} = (5 - {2 \log}_{y} x) \log_{y} x$ $2 = {5 \log}_{y} x - 2 {(\log_{y} x)}^{2}$ $2 {(\log_{y} x)}^{2} - {5 \log}_{y} x + 2 = 0$ $\log_{y} x = \frac{5 \pm \sqrt{25 - 16}}{4} = 2, \frac{1}{2}$ $y^{2} = x ∣ y^{1 / 2} = x \Rightarrow y = x^{2} . . . . . . . . . (i i i)$ $(i i) \Rightarrow \frac{1}{\log_{y} 4} = \frac{1}{\log_{y} (y - 3 x)}$ $\Rightarrow y - 3 x = 4$ $\Rightarrow y = 3 x + 4 . . . . . . . . . . . . . . . . . . . (i v)$ $(i i i) & (i v) : {(3 x + 4)}^{2} = x ∣ 3 x + 4 = x^{2}$ $\underset{N o r e a l r o o t s}{9 x^{2} + 23 x + 16 = 0} ∣ x^{2} - 3 x - 4 = 0$ $(x - 4) (x + 1) = 0 \Rightarrow x = 4, - 1$ $\Rightarrow y = 3 (4) + 4, 3 (- 1) + 4 = 16, 1 b u t y \neq 1$ $∴ (x, y) = (16, 4) ✓$
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