Let be the solution of the following equations Then is

JEE MainMathematics/ ModerateLogarithm

Let $\left( x_{0},y_{0} \right)$ be the solution of the following
equations
$(2x)^{\mathcal{l}n2} = (3y)^{\mathcal{l}n3}$
$3^{\mathcal{l}nx} = 2^{\mathcal{l}ny}$
Then $x_{0}$ is

Show answer and solution

(2x)^{\mathcal{l}n2} = (3y)^{\mathcal{l}n3}

\Rightarrow \mathcal{l}n2\ \mathcal{l}n(2x)\mathcal{= l}n3\ \mathcal{l}n(3y)\mathcal{= l}n3\ (\mathcal{l}n3 + \mathcal{l}ny)

....(1)
Also

3^{\mathcal{l}nx} = 2^{\mathcal{l}ny}

\Rightarrow \mathcal{l}n{x\mathcal{l}n3}\mathcal{= l}ny\ \mathcal{l}n2

....(2)
By (1)

\Rightarrow \mathcal{l}n2\ \mathcal{l}n(2x) = \mathcal{l}n3(\mathcal{l}n3 + \mathcal{l}ny)

\Rightarrow \mathcal{l}n2 \cdot \mathcal{l}n(2x) = \mathcal{l}n3\{\mathcal{l}n3 + \frac{\mathcal{l}n \times \mathcal{l}n3}{\mathcal{l}n2}\}

\Rightarrow \mathcal{l}n^{2}2\ \mathcal{l}n2x = {\mathcal{l}n}^{2}3(\mathcal{l}n2 + \mathcal{l}nx)

\Rightarrow ({\mathcal{l}n}^{2}2 - {\mathcal{l}n}^{2}3)(\mathcal{l}n2x) = 0

\Rightarrow \mathcal{l}n2x = 0 \Rightarrow x = \frac{1}{2}

Source: Grass-G / M-16 Mathematics Paper 25 Nov First.docx

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