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Quadratic Equation

Let x0,y0 b...

Question

Let $(x_{0}, y_{0})$ be the solution of the following equations $(2 x)^{l n 2} = (3 y)^{l n 3}$ and $3^{l n x} = 2^{l n y}$
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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6

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Solution

The correct option is C $\frac{1}{2}$
Take, $3^{l n x} = 2^{l n y}$ Taking log on both sides

$(l n x) (l n 3) = (l n y) (l n 2)$

$l n y = \frac{(l n x) (l n 3)}{l n 2}$

take $(2 x)^{l n 2} = (3 y)^{l n 3}$

$(l n 2) (l n 2 x) = (l n 3) (l n 3 y)$

$(l n 2) (l n 2 + l n x) = (l n 3) (l n 3 + l n y)$

$(l n 2)^{2} + l n 2 \times l n x = (l n 3)^{2} + l n 3 \times \frac{(l n x) \times (l n 3)}{(l n 2)}$

$(l n)^{2} [l n 2 + l n x] = (l n 3)^{2} [l n 2 + l n x]$

$[(l n)^{2} - (l n 3)^{2}] [l n 2 + l n x] = 0$

$- l n 2 = l n x \Rightarrow x = \frac{1}{2}$

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${log}_{e} x$ can also be written as ln(x)

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