JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 5)
Let $$\left( {{x_0},{y_0}} \right)$$ be the solution of the following equations
$$\matrix{ {{{\left( {2x} \right)}^{\ell n2}}\, = {{\left( {3y} \right)}^{\ell n3}}} \cr {{3^{\ell nx}}\, = {2^{\ell ny}}} \cr } $$
Then $${x_0}$$ is
$$\matrix{ {{{\left( {2x} \right)}^{\ell n2}}\, = {{\left( {3y} \right)}^{\ell n3}}} \cr {{3^{\ell nx}}\, = {2^{\ell ny}}} \cr } $$
Then $${x_0}$$ is
$${1 \over 6}$$
$${1 \over 3}$$
$${1 \over 2}$$
$$6$$
Explanation
We have,
$${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$$ ...... (1)
$${3^{\ln x}} = {2^{\ln y}}$$ ....... (2)
$$ \Rightarrow (\log x)(\log 3) = (\log y)\log 2$$
$$ \Rightarrow \log y = {{(\log x)(\log 3)} \over {\log 2}}$$ ........ (3)
Taking log both sides of Eq. (1), we get
$$(\log 2)\{ \log 2 + \log x\} = \log 3\{ \log 3 + \log y\} $$
$${(\log 2)^2} + (\log 2)(\log x) = {(\log 3)^2} + {{{{(\log 3)}^2}(\log x)} \over {\log 2}}$$ from Eq. (3)
$$ \Rightarrow {(\log 2)^2} - {(\log 3)^2} = {{{{(\log 3)}^2} - {{(\log 2)}^2}} \over {\log 2}}(\log x)$$
$$ \Rightarrow - \log 2 = \log x$$
$$ \Rightarrow x = {1 \over 2} \Rightarrow {x_0} = {1 \over 2}$$.
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