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AnswEr :

Given System of Equations,

• The bases are equal

Also,

We know that,

• log_b(a) = log(a)/log(b)

Thus,

We know that,

• log a + log b = log(ab)

The ratios of the logarithm functions are equal

Thus,

Consider Equation (2),

Putting value of y² in Equation (1),

Squaring on both sides,

Equating equations (3) and (4),

Applying the Quadratic Formula,

Consider Equation (3),

When x² = 4,

When x² = - 1,

Thus,

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

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

Thus,the value of 'x' is 2

Answer:

AnswEr :

Given System of Equations,

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

4

y

x

+

x

y

=32

⟶2

2(

xy

x

2

+y

2

)

=2

5

The bases are equal

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

2

+y

2

)=5xy−−−−−−−−(1)

Also,

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

3

(x−y)+log

3

(x+y)=1

We know that,

log_b(a) = log(a)/log(b)

Thus,

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

log(3)

log(x−y)

+

log(3)

log(x+y)

=1

⟶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)⟶log(x

2

−y

2

)=log(3)

The ratios of the logarithm functions are equal

Thus,

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

2

−y

2

=3−−−−−−−−−−−(2)

Consider Equation (2),

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

2

=x

2

−3−−−−−−−−(3)

Putting value of y² in Equation (1),

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

⟹2(x

2

+x

2

−3)=5xy

⟹4x

2

−6=5xy

⟹y=

5x

4x

2

−6

Squaring on both sides,

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

2

=

25x

2

16x

4

−48x

2

+36

−−−−−−−−−(4)

Equating equations (3) and (4),

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

⇢16x

4

−48x

2

+36=25x

4

−75x

2

⇢9x

4

−27x

2

−36=0

Applying the Quadratic Formula,

\begin{lgathered}\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}}\end{lgathered}

x

2

=

2(9)

−(−27)±

27

2

−4(9)(−36)

⟹x

2

=

18

27±

729+1296

⟹x

2

=

18

27±

2025

⟹x

2

=

18

27+45

or

18

27−45

⟹

x

2

=4or−1

Consider Equation (3),

When x² = 4,

\begin{lgathered}\sf \: y {}^{2} = 4 - 3 \\ \\ \longrightarrow \: \sf \: {y}^{2} = 1\end{lgathered}

y

2

=4−3

⟶y

2

=1

When x² = - 1,

\begin{lgathered}\sf \: y {}^{2} = - 1 - 3 \\ \\ \longrightarrow \: \sf \: {y}^{2} = - 4\end{lgathered}

y

2

=−1−3

⟶y

2

=−4

Thus,

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

y

2

=1or−4

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

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

\begin{lgathered}\leadsto \sf x = \pm \sqrt{4} \\ \\ \leadsto \sf x = \pm \ 2\end{lgathered}

⇝x=±

4

⇝x=± 2

Thus,the value of 'x' is 2.

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