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Answer 1

AnswEr :

Given System of Equations,

$\sf \: {4}^{ \frac{x}{y} + \frac{y}{x} } = 32 \\ \\ \longrightarrow \: \sf \: 2 {}^{2( \frac{ {x}^{2} + {y}^{2} }{xy}) } = {2}^{5}$

• The bases are equal

$\longrightarrow \: \sf \: 2( {x}^{2} + {y}^{2} ) = 5xy - - - - - - - - (1)$

Also,

$\sf \: log_{3}(x - y) + log_{3}(x + y) = 1$

We know that,

• log_b(a) = log(a)/log(b)

Thus,

$\sf \: \dfrac{ log(x - y) }{ log(3) } + \dfrac{ log(x + y) }{ log(3) } = 1 \\ \\ \longrightarrow \: \sf \: log(x + y) + log(x - y) = log(3)$

We know that,

• log a + log b = log(ab)

$\longrightarrow \: \sf \: log( {x}^{2} - {y}^{2} ) = log(3)$

The ratios of the logarithm functions are equal

Thus,

$\sf \: \ {x}^{2} - {y}^{2} = 3 - - - - - - - - - - - (2)$

Consider Equation (2),

$\implies \: \sf \: {y}^{2} = {x}^{2} - 3 - - - - - - - - (3)$

Putting value of y² in Equation (1),

$\implies \: \sf \: 2( {x}^{2} + {x}^{2} - 3) = 5xy \\ \\ \implies \: \sf \: 4 {x}^{2} - 6 = 5xy \\ \\ \implies \: \sf \: y = \dfrac{4x {}^{2} - 6}{5x}$

Squaring on both sides,

$\implies \: \sf \: {y}^{2} = \dfrac{16 {x}^{4} - 48 { x }^{2} + 36 }{25 {x}^{2} } - - - - - - - - - (4)$

Equating equations (3) and (4),

$\dashrightarrow \: \sf \: {16x}^{4} - 48 {x}^{2} +36 = 25 {x}^{4} - {75x}^{2} \\ \\ \dashrightarrow \: \sf \: 9 {x}^{4} - 27 {x}^{2} - 36 = 0$

Applying the Quadratic Formula,

$\sf \: {x}^{2} = \dfrac{ - ( - 27) \pm \sqrt{ {27}^{2} - 4(9)(-36) } }{2(9)} \\ \\ \implies \: \: \sf \: {x}^{2} = \dfrac{27 \pm \sqrt{729 + 1296} }{18} \\ \\ \implies \: \sf \: {x}^{2} = \frac{27 \pm \sqrt{2025} }{18} \\ \\ \implies \: \sf \: {x}^{2} = \dfrac{27 + 45}{18} \: or \: \dfrac{27 - 45}{18} \\ \\ \implies \boxed{ \boxed{ \sf \: {x}^{2} = 4 \: or \: - 1}}$

Consider Equation (3),

When x² = 4,

$\sf \: y {}^{2} = 4 - 3 \\ \\ \longrightarrow \: \sf \: {y}^{2} = 1$

When x² = - 1,

$\sf \: y {}^{2} = - 1 - 3 \\ \\ \longrightarrow \: \sf \: {y}^{2} = - 4$

Thus,

$\implies \boxed{ \boxed{ \sf \: {y}^{2} = 1 \: or \: - 4}}$

We had to solve the above equations for value of x :

If values of are 'x' is real and equal,then x² = 4

$\leadsto \sf x = \pm \sqrt{4} \\ \\ \leadsto \sf x = \pm \ 2$

Thus,the value of 'x' is 2