Let $(x_{0}, y_{0})$ be the solution of the following equations:
${(2 x)}^{ln 2} = {(3 y)}^{ln 3}$
$3^{ln x} = 2^{ln y}$
Then $x_{0}$ is

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Solution

Given equations are
${(2 x)}^{ln 2} = {(3 y)}^{ln 3} (ln 2) ln 2 x = (ln 3) ln 3 y ⟹ {(ln 2)}^{2} + (ln 2) (ln x) = {(ln 3)}^{2} + (ln 3) (ln y)$
and also,
$3^{ln x} = 2^{ln y} ⟹ \frac{ln x}{ln y} = \frac{ln 2}{ln 3} ⟹ ln y = \frac{ln 3}{ln 2} . ln x$ ,
substituting this in the first equation gives,
${(ln 2)}^{2} + (ln 2) (ln x) = {(ln 3)}^{2} + (ln 3) (\frac{ln 3}{ln 2} . ln x) ⟹ {(ln 2)}^{3} + {(ln 2)}^{2} (ln x) = {(ln 3)}^{2} + {(ln 3)}^{2} . ln x ⟹ ln x = \frac{{(ln 2)}^{3} - {(ln 3)}^{2}}{{(ln 3)}^{2} - {(ln 2)}^{2}} ⟹ x_{0} = e^{\frac{{(ln 2)}^{3} - {(ln 3)}^{2}}{{(ln 3)}^{2} - {(ln 2)}^{2}}}$

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