Let $(x_{0}, y_{0})$ be the solution of the following equations \[ \begin{aligned} (2 x)^{\ln 2} & =(3 y)^{\ln 3} \\ 3^{\ln x} & =2^{\ln y} . \end{aligned} \] Then $x_{0}$ is

Basic Mathematics — JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 Question with Solution

Let

(x_{0}, y_{0})

be the solution of the following equations

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

Then

x_{0}

Answer: (C)

From the second equation:

3^{l n x} = 2^{l n y}

. Taking logarithms:

(lo g x) (lo g 3) = (lo g y) (lo g 2)

, so

lo g y = \frac{( l o g x ) ( l o g 3 )}{l o g 2}

.

From the first equation, taking logarithms:

(lo g 2) (lo g 2 + lo g x) = (lo g 3) (lo g 3 + lo g y)

.

Substituting the expression for

lo g y

(lo g 2)^{2} + (lo g 2) (lo g x) = (lo g 3)^{2} + \frac{( l o g 3 ) ^{2} ( l o g x )}{l o g 2}

.

Rearranging:

(lo g 2)^{2} - (lo g 3)^{2} = \frac{( l o g 3 ) ^{2} - ( l o g 2 ) ^{2}}{l o g 2} (lo g x)

.

This gives

- lo g 2 = lo g x

, so

x = \frac{1}{2}

, hence

x_{0} = \frac{1}{2}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option