Question-200224

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

Answered by Rasheed.Sindhi last updated on 16/Nov/23

{\begin{cases} {y x}^{\log_{y} x} = x^{\frac{5}{2}} \dots \dots \dots \dots \dots (i) \\ \log_{4} y . \log_{y} (y - 3 x) = 1 \dots (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 \dots

\dots 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}

\dots 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

{\begin{cases} y x^{\log_{y} x} = x^{\frac{5}{2}} \dots \dots \dots \dots \dots . . (i) \\ \log_{4} y . \log_{y} (y - 3 x) = 1 \dots . (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} \dots \dots \dots (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 \dots \dots \dots \dots \dots \dots . (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) ✓