Basics & LogarithmsmediumJEE Advanced 2011Free

Basics & Logarithms — JEE Maths practice question (JEE Advanced 2011)

JEE Maths question with a full step-by-step solution.

Question

Let

(x_{0},y_{0})

be the solution of the following equations:

(2x)^{\ln 2}=(3y)^{\ln 3}

and

3^{\ln x}=2^{\ln y}

. Then

x_{0}

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

correct

6

Solution

Step 1: Take

\ln

of both given equations. From

(2x)^{\ln 2}=(3y)^{\ln 3}

\ln 2\cdot(\ln 2+\ln x)=\ln 3\cdot(\ln 3+\ln y)

(\ln 2)^{2}+\ln 2\cdot\ln x=(\ln 3)^{2}+\ln 3\cdot\ln y\quad\ldots(1)

Step 2: From

3^{\ln x}=2^{\ln y}

\ln x\cdot\ln 3=\ln y\cdot\ln 2\;\Rightarrow\;\ln y=\dfrac{\ln x\cdot\ln 3}{\ln 2}\quad\ldots(2)

Step 3: Substitute (2) into (1):

(\ln 2)^{2}+\ln 2\cdot\ln x=(\ln 3)^{2}+\ln 3\cdot\dfrac{\ln x\cdot\ln 3}{\ln 2}

(\ln 2)^{2}+\ln 2\cdot\ln x=(\ln 3)^{2}+\dfrac{(\ln 3)^{2}\cdot\ln x}{\ln 2}

Step 4: Rearrange:

(\ln 2)^{2}-(\ln 3)^{2}=\dfrac{(\ln 3)^{2}\cdot\ln x}{\ln 2}-\ln 2\cdot\ln x

(\ln 2)^{2}-(\ln 3)^{2}=\ln x\left(\dfrac{(\ln 3)^{2}}{\ln 2}-\ln 2\right)

(\ln 2)^{2}-(\ln 3)^{2}=\ln x\cdot\dfrac{(\ln 3)^{2}-(\ln 2)^{2}}{\ln 2}

Step 5: Simplify:

-((\ln 3)^{2}-(\ln 2)^{2})=\ln x\cdot\dfrac{(\ln 3)^{2}-(\ln 2)^{2}}{\ln 2}

-1=\dfrac{\ln x}{\ln 2}\;\Rightarrow\;\ln x=-\ln 2=\ln\dfrac{1}{2}

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

Answer: (3)

\dfrac{1}{2}

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