Q28 / 69 · JEE Advanced 2011 (Paper 1) · Mathematics

Let $\left(x_0, y_0\right)$ be the solution of the following equations $(2 x)^{\log 2}=(3 y)^{\log 3}$, $3^{\log x}=2^{\log y}$, then $x_0$ is equal to

  • A
    $\frac{1}{6}$
  • B
    $\frac{1}{2}$
  • Correct answer
    C
    $\frac{1}{2}$
  • D
    6

Solution

Taking log on both sides, $$ \log 2 \cdot \log (2 x)=\log 3(\log 3 y) $$ $$ \begin{array}{ll} \Rightarrow & \log 2\{\log 2+\log x\} \\ & =\log 3\{\log 3+\log y\} \\ \text { and } & \log x \cdot \log 3=\log y \log 2 \\ & \log y=\frac{\log x \cdot \log 3}{\log 2} \end{array} $$ From Eqs. (i) and (ii), we get $$ \begin{aligned} & \log 2\{\log 2+\log x\} \\ &=\log 3 \cdot\left\{\log 3+\frac{\log x \cdot \log 3}{\log 2}\right\} \\ & \Rightarrow \quad(\log 2)^2+\log 2 \cdot \log x \\ &=(\log 3)^2+\frac{(\log 3)^2}{(\log 2)} \cdot \log x \\ & \Rightarrow \quad \log x\left\{\frac{(\log 3)^2}{\log 2}-\log 2\right\} \\ &=(\log 2)^2-(\log 3)^2 \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \quad \log x=-\log 2=\log 2^{-1} \\ & \therefore \quad x=\frac{1}{2} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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