$x$ and $y$ are two literals and logarithm of the quotient of sum of $x$ and $y$ by number $3$ is equal to half of the summation of the logarithms of $x$ and $y$.

$\log \Bigg[\dfrac{x+y}{3}\Bigg]$ $=$ $\dfrac{1}{2} (\log x + \log y)$

This logarithmic equation has to be solved to find the value of $\dfrac{x}{y} + \dfrac{y}{x}$.

Move the quantity half to left hand side of the equation.

$\implies 2 \log \Bigg[\dfrac{x+y}{3}\Bigg]$ $=$ $\log x + \log y$

The multiplying number $2$ can be moved to exponent position by the power law of logarithm.

$\implies \log {\Bigg[\dfrac{x+y}{3}\Bigg]}^2 $ $=$ $\log x + \log y$

$\implies \log {\Bigg[\dfrac{x+y}{3}\Bigg]}^2 $ $=$ $\log (xy)$

Both sides are having common logarithm functions. So, functions should be equal and then continue the simplification to obtain required result.

$\implies {\Bigg[\dfrac{x+y}{3}\Bigg]}^2 $ $=$ $xy$

$\implies \dfrac{{(x+y)}^2}{3^2}$ $=$ $xy$

$\implies \dfrac{{(x+y)}^2}{9}$ $=$ $xy$

$\implies {(x+y)}^2$ $=$ $9xy$

$\implies x^2 + y^2 +2xy$ $=$ $9xy$

$\implies x^2 + y^2$ $=$ $9xy-2xy$

$\implies x^2 + y^2 = 7xy$

Move the literal coefficient of $7$ to left hand side of the equation.

$\implies \dfrac{x^2 + y^2}{xy} = 7$

$\implies \dfrac{x^2}{xy} + \dfrac{y^2}{xy} = 7$

$\implies \require{cancel} \dfrac{\cancel{x^2}}{\cancel{x}y} + \dfrac{\cancel{y^2}}{x\cancel{y}} = 7$

$\therefore \,\,\,\,\,\, \dfrac{x}{y} + \dfrac{y}{x} = 7$

Therefore, it is the required solution for this logarithmic problem and the value of $\dfrac{x}{y} + \dfrac{y}{x}$ is $7$.

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