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Quadratic Equation

Let x 0, y 0 ...

Question

Let $(x_{0}, y_{0})$ be the solution of the following equations $(2 x)^{l n 2} = (3 y)^{l n 3}$ , $3^{l n x} = 2^{l n y}$ then $x_{0}$ is

\frac{1}{6}

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\frac{1}{3}

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\frac{1}{2}

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Solution

The correct option is C $\frac{1}{2}$
We have,
$(2 x)^{l n 2} = (3 y)^{l n 3} \Rightarrow l n 2. l n 2 x = l n 3. l n 3 y \Rightarrow l n 2. l n 2 x = l n 3. (l n 3 + l n y) . . . . (1)$
Also given $3^{l n x} = 2^{l n y}$
$\Rightarrow l n x . l n 3 = l n y . l n 2 \Rightarrow l n y = \frac{l n x . l n 3}{l n 2}$
Substituting this value of $l n y$ in equation (1), we get
$l n 2. l n 2 x = l n 3 [l n 3 + \frac{l n x . l n 3}{l n 2}] \Rightarrow (l n 2)^{2} l n 2 x = (l n 3)^{2} l n 2 + (l n 3)^{2} l n x \Rightarrow (l n 2)^{2} l n 2 x = (l n 3)^{2} (l n 2 + l n x) \Rightarrow (l n 2)^{2} l n 2 x - (l n 3)^{2} l n 2 x = 0 \Rightarrow [(l n 2)^{2} - (l n 3)^{2}] l n 2 x = 0 \Rightarrow l n 2 x = 0 \Rightarrow 2 x = 1 o r x = \frac{1}{2}$

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